Chemical Calculations SALISBURY-LONG TABLE OF ATOMIC WEIGHTS Element Symbol Atomic Weight Element Symbol Atomic \\ eight Aluminum Al 27.1 Neodmium Nd 1 14.3 Antimony Sb 120.2 Neon Ne 20.2 Argon A 39.88 Nickel Ni 58.68 Arsenic As 74.96 N iton Barium Ba 137.37 (radium emanation) Nt 222.4 Bismuth Bi 208.0 Nitrogen N 1 1.01 Boron B 11.0 Osmium Os 1 90.9 Bromine Bi- 79.92 Oxygen O 16.00 Cadmium Cd 112.40 Palladium Pd 106.7 Caesium Cs 132.81 Phosphorus P 31.01 Calcium Ca 40.07 Platinum Pt 195.2 Carbon C 12.005 Potassium K 39.10 Cerium Ce 140.25 Praseodymium Pr 140.9 Chlorine Cl 35.46 Radium Ra 226.0 Chromium Cr 52.0 Rhodium Rh 102.9 Cobalt Co 58.97 Rubidium Rb 85.15 Columbium Cb 93.5 Ruthenium Ru 101.7 Copper Cu 63.57 Samarium Sa 150.1 Dysprosium »y 162.5 Scandium Sc 14.1 Erbium Er 167.7 Selenium Se 79.2 Europium Eu 152.0 Silicon Si 28.3 Fluorine F 19.0 Silver Ag 107.88 Gadolinium Gd 157.3 Sodium Na 23.00 Gallium Ga 69.9 Strontium Sr 87.63 Germanium Ge 72.5 Sulfur s 82.06 Glucinum G1 9.1 Tantalum Ta 181.5 Gold Au 197.2 Tellurium Te 127.5 Helium He 4.00 Terbium Th 159.2 Holmium Ho 163.5 Thallium T1 201.0 Hydrogen II 1.008 Thorium Th 232.1 Indium In 11 1.8 Thulium Tin 168.5 Iodine I 126.92 Tin Sn 1 18.7 Iridium Ir 193.1 Titanium Ti 48.1 Iron Fe 55.84 Tungsten W 181.0 Krypton Kr 82.92 Uranium U 238.2 Lanthanum La 139.0 Vanadium V 51.0 Lead Ph 207.20 Xenon Xe 130.2 Lithium Li 6.94 Ytterbium Lutecium Lu 175.0 (Neoytterbium) Yb 1 73.5 Magnesium Mg 21.32 Yttrium Yt 88.7 Manganese Mn 54.93 Zine Zu 65.37 Mercury Hg 200.6 Zirconium Zr 90.6 Molybdenum Mo 96.0 Chemical Calculations BY S? H. SALISBURY, Jr., B. S., M. S. 444 - FORMERLY ASSISTANT PROFESSOR OF INDUSTRIAL CHEMISTRY LEHIGH UNIVERSITY AND J. S. LONG, Ch. E., M. S. ASSISTANT PROFESSOR OF INORGANIC CHEMISTRY LEHIGH UNIVERSITY COPYRIGHT, 1918, BY S. H. SALISBURY, Jr., AND J. S. LONG INTRODUCTION XPERIENCE indicates that most students do not get a firm grasp on the fundamental principles of In- organic Chemistry until the work on the non metals is largely completed. For this reason we, here at Le- high, review the uniquely chemical principles during the second half of the year in connection with the course in Qualitative analysis. The majority of men taking the course are engineer- ing students. We therefore approach pure science, not forget- ting the viewpoint of the engineer. This little book is written in an attempt to do this. We use the calculation work to help the student understand the principles involved and to indicate the dependence of these principles upon experiment, and to connect the principles to useful applications. Each chapter has been made of approximate length for I recitation (based on standards here at Lehigh). The presentation of principles is therefore condensed. No attempt is made to supercede the theoretical and descriptive work of inorganic chemistry. Wherever possible, however, descriptive matter is worked in (cf 7 Methods of Salt Formation, Chapter III, Tabulation of Oxyacids in Problems, Chapter IV, Prob- lems on the Rarer Elements, etc.) ' It is assumed that the student has had at least a little trial at equation writing before he takes up the work in Chapter III arid following ones. We start this calculation work at Lehigh near the end of the first term,-say after the student has had 1 0 weeks of work on Elementary Inorganic Chemistry,-and continue the work until the end of the second term. In some respects our presentation is radical. We dis- courage the use of proportion as ordinarily used:-This is to this as this is to this, and ask the student to go back to the fundamental "1." This idea is further carried out in the calculation of gas volumes for temperature and pressure cor- III IV CHEMICAL CALCULATIONS PV rections. We do not use the = K equation, but aim to T have the student visualize the operation by deciding in his own mind whether increase or decrease, say of volume, will be caused by the temperature and pressure changes, taking each separately. Atomic weights and answers are given to the second place of decimals in formal presentation to encourage the habit of accuracy and thorough-going-ness and to give him an early flight-to show him that careful work has been done by others. Where the time allotted to calculation work is short and followed by a rigorous course in Quantitative Analysis, the individual instructor can often permit the use of the Slide Rule in calculations. This is true if the student recognizes that the slide rule is simply an expedient to allow him more time for consideration of principles. This book is naturally affected by methods of presenta- tion used in books on elementary chemistry which have been used as texts in our experience-Mellor, Kahlenberg, Smith, Senter and others. It would be difficult to give credit for individual points. Our deep acknowledgement is due and joyfully given to Dr. H. M. Ullmann for many kind helps in this work. S. H. SALISBURY, JR. J. S. LONG Lehigh University, Bethlehem, Pa. TABLE OF CONTENTS I. Conversion of Temperature. Absolute Zero. Absolute Density. Relative Density. Specific Gravity. II. Interrelation of Temperature, Pressure and Volume of Gases. Partial Pressures. III. The Law of Definite Proportions. IV. The Law of Multiple Proportions. The Atomic Theory. V. Avogadro's Hypothesis. Proof that the molecules of Hydrogen, Oxygen and Chlorine each contain 2 atoms. Determination of Molecular Weights. VI. Combining Weights. Chemical Equivalents. VII. Chemical Equations. VIII. Derivation of Empirical and Molecular Formulas. IX. Volumetric Analysis. Normal Solutions. X. Gas Analysis. XI. Calorific Power. Calorific Intensity. Answers to Problems. V CHAPTER I OUTLINE Measurement of Temperature. 1. Comparison of Centigrade and Fahrenheit Scales. 2. Absolute Temperature -273° 3. Solution of Problem. Density-Specific Gravity. GENERAL. 1. Absolute Density. Mass per unit volume. 2. Standard Conditions of Temperature and Pressure O°C & 760 mm. 3. Relative Density. Effect of Temperature and Pressure GASES. 1. Vapor Density or Relative Density. 2. Solution of a Problem. SOLIDS AND LIQUIDS. 1. Specific Gravity. SOLIDS. 1. Archimedes Principle. 2. Body heavier than water. Insoluble in water. 3. Body heavier than water. Soluble in water. 4. Solution of a Problem. 5. Powder-Solution of a Problem. LIQUIDS. I. Pyknometer Method. 2- Hydrometer. Problems. 7 8 CHEMICAL CALCULATIONS THE MEASUREMENT OF TEMPERATURE. The temperature of a body is usually expressed in DE- GREES, on one of the three scales. (a) Centigrade (b) Fahrenheit (c) Absolute. A fourth scale, the Reamur is used in a few industries and in Russia. Comparison of the three scales may be made by means of water. This is shown graphically in Figure 1. Scales Freezing point of pure water Boiling point of pure water Centigrade 0 100 Fahrenheit 32 212 Absolute 273 373 Between the freezing point and boiling point of pure water there are 100 Centigrade degrees or 180 Fahrenheit de- grees, hence 180 9 1 °C = -cF 100 5 100 5 1 °F = = - °C 180 9 Boiling Point of Pure Water 0cC corresponds to 32 °F 1 00 °C corresponds to 2 1 2 °F Fahrenheit' Cen<<<jrade Freezing Point of Pure Water i CHEMICAL CALCULATIONS 9 ABSOLUTE TEMPERATURE. 1 All gases contract of their volume at 0°C for each 273 1 C they are cooled (pressure remaining constant at 760 mm). Thus if they start with 2 73 cc of a gas at 0cC and cool it to -100£C, the volume will contract to 173 cc. If the 2 73 cc at 0£C were cooled to -200 cC the volume would contract to 73 1 cc and so on, the volume diminishing of its volume at 273 zero for each degree C it is cooled. When a temperature of -2 73 is reached the volume will theoretically have become zero. This temperature (-273), at which the volume of gases would become zero, is called the absolute zero. The lowest tempera- ture reached to date, -2 71 C, was reached in liquefying the rare gas helium. Centigrade temperatures can be converted to Absolute temperatures by adding 2 73cC. Thus 68 cC is 68 Centigrade degrees above the freezing point of pure water. The Absolute zero is 2 73 centigrade degrees below the freezing point of pure water, hence 68 cC is 68° plus 2 73° = 34 1 0 Absolute. Absolute temperature is generally expressed in Centigrade degrees. Problem 1. What temperature on the Centigrade scale corresponds to 120° on the Fahrenheit scale? 120°F is 120 - 32 = 88 Fahrenheit degrees above the freezing point of pure water. 5 88 Fahrenheit degrees = 86 X - = 48.8 Centigrade 9 degrees. The freezing point of water on the Centigrade scale is 0°C. 48.8 Centigrade degrees above 0°C is 48.8 °C. 48.8 °C then corresponds to 120 °F. 10 CHEMICAL CALCULATIONS DENSITY-SPECIFIC GRAVITY. The Absolute Density of a substance is the number of units of mass of the substance contained in a unit of volume. In scientific work it is usually expressed in grams per cubic centimeter or grams per liter. One liter of air at (0cC and under 760 mm pressure) weighs 1.293 grams. The absolute density of air is then 1.293 grams per liter. It is customary to adopt a temperature of 0 cC and a pressure of 760 mm as standard conditions for the comparison of the density of gases. Any gas is said to be measured at standard conditions when measured at a temperature of 0 cC and a pressure of 760 mm of mercury. (1 atmosphere). When the weights of equal volumes of two substances are compared, the RELATIVE DENSITY is obtained, i.e. weight of a given volume of substance Relative Density = weight of an equal volume of another substance taken as a standard The volume (and consequently the density) of both solids, liquids and gases is changed by a change of tempera- ture, consequently whenever the density of a substance is ex- pressed the temperature should be given. Variation of pressure has practically no effect on the vol- ume of solids or liquids, hence the pressure need not be stated in giving the density of solids or liquids. In the case of gases, however, a variation of pressure will cause a proportionate change in the volume. The value of the pressure should hence be stated when giving the density of gases. The Relative Density of a gas is often called the VAPOR DENSITY. Hydrogen is generally used as the standard and other gases compared to it. i.e. weight of a given volume of a gas Vapor Density weight of an equal volume of hydrogen at 0°C & 760 mm Since hydrogen is the lightest known gas this offers the advantages that the densities of all other gases will be greater GASES. CHEMICAL CALCULATIONS 11 than 1. Air is, however, sometimes used as the standard, and occasionally oxygen or other gases. Unless otherwise stated in this work, we will use hydrogen as the standard. I liter of hydrogen at standard conditions weighs 0.08973 g. Problem. 2600 cc of Carbon Dioxide at standard conditions were found by experiment to weigh 5.148 grs. Calculate- (a) The Absolute Density. (b) The Relative Density of Carbon Dioxide. mass per unit volume 5.148 g. Absolute Density = = = in grams per liter 2.6 liter 1.98 g. per liter, liter of hydrogen at standard conditions weighs 0.08973 g. weight of 1 liter of Carbon Dioxide Relative Density weight of 1 liter of Hydrogen at 0°C & 760 mm 1.98 Relative Density = = 22.06 0.08973 SOLIDS AND LIQUIDS. • In determining the Relative Density of Solids and Liquids water at its maximum density (at 4cC) is used as the standard for comparison. The relative density of solids or liquids, com- pared to that of water is generally known as SPECIFIC GRAVITY. weight of a certain volume of the substance Specific Gravity = weight of an equal volume of water at 4 cC The Specific Gravity of a substance may be expressed as determined at a certain temperature and compared to that of water at 4 cC. Thus the Specific Gravity of Mercury might be expressed as- 15° D = 13.559, which means that 1 cc of mercury at 1 5 °C 4° weighs 1 3.559 times as much as 1 cc of pure water at 4 C. 12 CHEMICAL CALCULATIONS (A) Solids. As stated above the Specific Gravity of a Solid is equal to weight of a definite volume of the solid weight of an equal volume of water at 4 °C. The weight of the volume of water equal to the volume of the solid is obtained by first weighing the solid in air and then weighing it in water; the difference of the two weights be- ing the loss in weight of the solid in water which is the same as theweight of avolume of water equal to the volume of the solid. This is according to Archimedes' principle which states that a body wholly submerged in a fluid is bouyed up by a force which is equal to the weight of ITS volume of the fluid. This can be illustrated as follows: Let A be a cube 1 cm square immersed in water and supported by a cord so that it exerts no down- ward pressure and so that its upper surface is just 1 cm below the surface of the water. The pressure of a horizontal surface under water is equal to the weight of the column of water having an area equal to that of the surface and a height equal to the distance the surface is below the surface of the water. The pressure down upon ab is, then, one gram, and the pressure up on cd is equal to two grams. Consequently the pressure up being greater than the pres- sure down, there is a bouyant force ex- erted upon the body, and this force is equal to the difference between the up- ward and downward forces, or, in this case, one gram. The volume of water displaced is one cubic centimeter which, of course, weighs one gram. Therefore the body is bouyed up by a force equal to the weight of the volume of water displaced. Hence the difference in weight of a solid when weighed in air and in water is equal to the weight of its volume of water. In determining the specific gravity of solids we have the following cases: Z CHEMICAL CALCULATIONS 13 1. When the body is heavier than water, the specific grav- ity is equal to the weight of the solid in the air, divided by its loss of weight in water. As shown above the loss of weight in water is equal to the weight of the volume of water displaced by the solid, this volume of water being of course equal to the volume of the solid. 2. The body is heavier than water but is soluble in it. In this case a liquid of known specific gravity, in which the sub- stance is insoluble, must be used. In order to determine the specific gravity of a substance soluble in water, it is first weighed in air and then in some liquid in which it is insoluble, in petroleum ether, for example. The difference between these two weights is the weight of the volume of petroleum ether which is equal to the volume of the solid. From this the specific gravity (compared of course to water) can be calculated. For example: Problem. A lump of sugar weighs 4.00 g. in air and 2.375 g. when immersed in petroleum ether. Specific gravity of petroleum ether is 0.65. Calculate the specific gravity of sugar. Weight of sugar in air 4.000 g. Weight of sugar immersed in water 2.375 g. 1.625 g. = the weight of volume of petroleum ether equal to the volume of the lump of sugar. But the specific gravity of petroleum ether is only 0.65 as great as that of water. The volume of water equal 1.625 to the volume of petroleum ether would weigh - = 2.5 g. 0.65 4.00 g. The specific gravity of the sugar is therefore - 1.6 2.5 g. 14 CHEMICAL CALCULATIONS 3. A Powder. Insoluble in water. Weigh a small flask or pyknometer (a con- venient kind is shown in the sketch). Sup- pose the weight is 8.754 g. Fill the flask with water and again weigh say this weight is 20.004 g. The difference in these weights is, of course, the weight of water which just fills the flask. This difference is 1 1.250 g. Weight of the powder in air 3.556 g. Take out the stopper "S." Introduce the powder into the flask. Replace the stopper. The powder introduced will displace its volume of water and this will over- flow through the capillary when the stopper is replaced. Now weigh again. Weight of flask, powder and enough water to fill = 21.782 g. Difference between (A) weight of flask filled with water plus weight of powder in air and (B) flask with powder and enough water to fill it, gives the weight of water displaced by the powder. 20.004 g. 3.556 Weight A is of course 23.560 g. Weight B was found to be 2 1.782 g. A. flask filled with water plus weight of powder in air = 23.560 g. B, flask with powder and enough water to fill it = 21.782 g. Difference = weight of water displaced = 1.778 g. i.e. Weight of powder in air = 3.556 g. Weight of its volume in water = 1.778 g. 3.556 Specify gravity of Powder ==2.0 1.778 Ct/nllary F'A 3 ( B ) Liquids. The Specific Gravity of liquids may be determined by means of (a) a Pyknometer (b) Hydrometer (c) Westphal balance. CHEMICAL CALCULATIONS 15 In using the pyknometer (A) the weight of the liquid which just fills the pyknometer is de- termined, (B) that weight of pure water at 4 cC which just fills the pyknometer is deter- A mined. The ratio - B is of course the specific gravity of the liquid. Hydrometers are weighted bulbs which float in the liquid, the depth to which they sink is determined by the density of the liquid and is observed on a scale in the stem of the bulb. They are prepared by observa- tion with liquids of known spe- cific gravity. A convenient type is shown in the sketch. F|<J H. PROBLEMS. 1. Convert 60 cC to the corresponding temperature on the Fahrenheit scale. 2. 200cF corresponds to what temperature on the Centi- grade scale? 3. Convert the following temperatures to their corre- sponding values on the Absolute scale; 30 F; -20cC; O F; -125 °F. 4. What points on the Centigrade and Fahrenheit scales correspond to 260 CA? 5. What temperature F corresponds to 40 cC? 6. A Fahrenheit thermometer placed in a cooling solu- tion shows a drop of 187°. What is the corresponding drop in Centigrade degrees? 7. A specific gravity flask weighed 10.2157 g. Filled with pure water it weighed 20.462 7 g., and when filled with 16 CHEMICAL CALCULATIONS chloroform 25.6972 g. Calculate the specific gravity of the chloroform. 8. What is the relative density of oxygen if 2000 cc weigh 2 858 g. at standard conditions. 9. 2600 cc of sulphur dioxide were found to weigh 7.460 g. what is the vapor density of the gas? I 0. What is the absolute density of carbon dioxide if 300 cc of it weigh 0.591 g. ? 1 1 What is the relative density of nitric oxide referred to air when 500 cc of it weigh 0.672 g. ? 12. A piece of iron, the specific gravity of which is 7.86, weighs 500 g. Calculate its volume in cc. 1 3. The weight of a liter of water vapor at standard con- dition of temperature and pressure is 0.8100 g. What is its relative density? 14. Determine the specific gravity of a piece of metal which weighs 44.890 g. in air and 39.3900 g. in water. 15. A piece of metal weighed 8.92 g. in air, 5.91 g. in water and 6.40 g. in alcohol. Calculate the specific gravity of the alcohol. 16. A sample of sodium hydroxide weighs 1 1.070 g. in air and 7.187 g. in an oil. The specific gravity of the hydroxide referred to water is 2.015. What is the specific gravity of the oil? 1 7. Calculate the specific gravity of a sample of ground rock from the following data: Weight of sample in air 3.50 g. Weight of specific gravity flask filled with water 2 7.50 g. Weight of flask with sample and filled up with water 29.50 g. 18. The density of iron is 7.8 g. per cc. What is the density in pounds per cu. inch? What is the weight of an iron bar 30 ft. long and 5 /i square inches in cross section? 19. A bottle weighs 50.62 g. and 288.93 g. when filled with water. What is the cubic contents of the bottle? When the bottle is filled with oil it weighs 239.2 g. What is the spe- cific gravity of the oil? 20 A piece of glass weighs 260.7 g. in air and 153.8 g. CHEMICAL CALCULATIONS 17 in water at 20cC. The same piece of glass weighs 92.2 g. in sulphuric acid. What is the specific gravity of the acid? 21. A piece of lead weighs 233.6 g. in air and 212.9 g. in water. What is the specific gravity of lead? 22. An acid carboy holds 198.5 pounds of sulphuric acid or 107.9 pounds of water. What is the specific gravity of the acid? What is the capacity of the carboy in gallons? 23. If a body weighs 9.25 g. in air, 8.2 g. in water and 8.36 g. in gasoline. What is the specific gravity of gasoline? CHAPTER II OUTLINE Kinetic explanation of the behavior of a gas. Boyle's law-Volume inversely proportional to pressure, if temperature is kept constant. Density of a gas directly proportional to the pressure. Gay Lussac's law-Volume directly proportional to Absolute temperature-if pressure is kept constant. Solution of a problem. Increase in temperature decreases the density-if pressure is constant. Effect of temperature and pressure together. Solution of a problem. Partial Pressure-Dalton's Law. Vapor tension of water-see last page. Problems. 18 CHEMICAL CALCULATIONS 19 EFFECT OF TEMPERATURE AND PRESSURE ON THE VOLUME OF GASES. Very careful measurements have indicated that the small particles (molecules) of which substances are composed, are very nearly incompressible, i.e. when subjected to tremendous pressure, each molecule contracts only a very small amount in size. This indicates that in solids and liquids, the molecules are closely packed together; for solids and liquids decrease only slightly in volume when subjected to tremendous pressure. In the case of gases we think that the molecules are separated from one another. This belief is confirmed by the fact that the volume of a gas is decreased by increasing the pressure on the gas. We know that the molecules of gases are moving- that they are traveling in sensibly straight lines at high speeds, colliding, rebounding and continuing their travel in other directions. It is assumed that the molecules are perfectly elastic, in accord with the fact that no energy is lost by the collisions. A great many of the molecules are, of course, continually colliding with the walls of the vessel. These re- peated collisions exert a pressure on the walls and balance the pressure acting on the gas. Thus consider a gas contained in a cylinder subject to a pressure P as shown in the Fgure. The pressure caused by the collision of the particles against the walls balances the pressure P and prevents the piston from mov- ing down. The speed of each moving particle is dependent on the temperature. The higher the temperature, the greater the speed of the molecules and consequently the greater the energy it possesses. If a gas be heated the speed of the molecules is increased, the energy increased, the number of collisions r 20 CHEMICAL CALCULATIONS per minute Increased and the pressure on the walls, due to these collissions, is increared. If the temperature is decreased, the speed of the molecules is decreased ; their energy decreased ; the number of collisions per minute on the walls decreased ; and the pressure on the walls decreased. If the temperature of the gas is kept constant and the pressure increased, the molecules will be crowded together- i.e. the volume of the gas will be decreased. Due to this crowding the collisions will become more frequent as the molecules are crowded together, until a point is reached where the increased number of collisions will cause increased pressure on the walls, sufficient to balance the increased external pres- sure. The exact numerical effect of increase of pressure upon the volume of gases was first found by Boyle in 1 662. Boyle compressed air in the short closed leg of a U tube, using varying amounts of mercury as a compressing agent. He concluded from his experiments that, temperature remaining constant, THE VOLUME OF A GAS IS INVERSELY PRO- PORTIONAL TO THE PRESSURE. This is known as Boyle's law of gases. It may be illustrated as follows: 300 C.C. Hfo cc 300 c e. L. Consider a cylinder "S," containing 900 cc of a gas, under a pressure P. If the pressure is doubled the gas will be compressed until it occupies one-half its original volume, i.e. its volume, under a pressure 2P, will be 450 cc. If the pressure is increased to 3P, the volume will be reduced to 300 cc and so on. CHEMICAL CALCULATIONS 21 We have found that the volume of a gas is inversely proportional to the pressure. From this it can be seen that the Absolute Density of a gas is directly proportional to the pressure. When a gas is compressed the volume decreases proportionately. The weight remains the same, i.e. the same weight of gas is caused to occupy a smaller volume. The density is increased. Thus, suppose that the 900 cc of gas 4.5 g in S weighs 4.5 g. The absolute density is = 0.005 900 cc g. per cc or 5 g. per liter. When the volume is decreased to 450 cc by doubling the pressure, the absolute density becomes 4.5 = 0.01 g. per cc or 10 g. per liter, for the total weight 450 of the gas remains the same = 4.5 g., i.e. by doubling the pressure the absolute density is doubled. Similarly trebling the pressure would treble the absolute density, etc. EFFECT OF TEMPERATURE. It has been found by experiment that a gas will expand 1 -- of the volume it occupies at 0cC (273° abs) for each 273 one degree centrigrade it is heated. Similarly it will contract 1 of its volume at 0 C for each one degree centigrade 273 it is cooled. It is assumed of course that the pressure is kept constant, thus:- 2 73 cc of any gas at 0cC (273° Absolute) 274 cc " lcC (274° " ) 275 cc " 2°C (275° " ) 373 cc " 100cC (373° " ) becomes 272 cc " " " " - I °C (272° " ) 271 cc " " " " -2°C (271° " ) 173 cc " '-100cC (173° " ) etc. This relation between increase in volume and increase in temperature, was first stated in the form of a law by Gay 22 CHEMICAL CALCULATIONS Lussac, in 1801-PRESSURE REMAINING CONSTANT, THE VOLUME OF A GAS IS DIRECTLY PROPOR- TIONAL TO THE ABSOLUTE TEMPERATURE. Let us illustrate this by an example:- 10 liters of a gas measured at 0°C will occupy what volume at 100cC? 0°C = 2 73° Absolute. 100cC = 373° Absolute. We have increased the temperature from 273° to 373° 373 Absolute; i.e. the new temperature = of the old. 273 Volume is directly proportional to the temperature. 373 The new volume will be of the old. 273 373 i.e. the new volume will be X 1 0 liters = 1 3.66 liters. 273 Since increase in temperature causes the volume to in- crease, it causes the absolute density to be decreased. The weight remains the same, while the volume it occupies is in- creased. Thus suppose that 1 liter of a gas weighs 5 g. at 0cC (2 73° Absolute). The absolute density is 5 g. per liter or 0.005 g. per cc. If the temperature is raised to 2 73 cC (546c Absolute) the volume will be doubled, i.e. the volume will become 2 liters. The weight remains the same, 5 g. The 5 g. absolute density is, therefore, = 2.5 g. per liter or 2 liters 0.0025 g. per cc; i.e. by doubling the temperature the abso- lute density is halved. EFFECT OF TEMPERATURE AND PRESSURE. When both temperature and pressure change, the effect on the volume is determined by considering each separately, thus keeping pressure constant calculate the effect of the change of temperature. Then keeping the temperature con- stant, calculate the effect of change of pressure. These are illustrated by the following problem: CHEMICAL CALCULATIONS 23 A certain weight of gas occupies 250 cc at 20°C and 700 mm pressure. Calculate the volume this will occupy at 5 cC and 740 mm pressure. If the pressure had remained constant and only the tem- perature had changed the effect would be as follows1 20cC = 293° Absolute. 5°C = 2 78°C. We are lowering the temperature from 293° to 278°, 278 i.e. the new temperature is of the old. 293 Now the volume is directly proportional to the absolute tem- perature. Lowering the temperature lowers the volume, i.e. the 278 278 new volume = of the old, i.e. X 250 cc = 237.2 cc. 293 293 We have now found the volume the amount of gas will occupy at the lower temperature if the pressure had remained the same, 700 mm. But the pressure also changed. 740 The new pressure is of the old. We know that the volume 700 is inversely proportional to the prssure. Increase of pressure causes decrease of volume. Therefore the volume under the 700 new conditions will be X 740 '2 78 X 250 cc 293 = 224.3 cc. Example: 2. A certain quantity of a gas measured 500 cc at a temperature of 1 5 °C and 750 mm pressure. What pres- sure is required to compress this quantity of gas into a 400 cc vessel at a temperature of 50cC? 1 5 °C = 288° Absolute. 50 °C = 323° Absolute. In solving these problems first put down what is to be found. In this case it is the pressure. New pressure == ? The old temperature was 288°. The new temperature is 323°. If we keep the volume constant, raising the tem- perature necessitates a raise of pressure; i.e. as far as the tem- perature change is concerned. 323 323 New Pressure = old pressure X = 750 mm X 288 288 24 CHEMICAL CALCULATIONS 288 To multiply 750 mm by would lower the pressure. 323 323 To multiply 750 mm by will raise the pressure. 288 We know that when the temperature is raised the pressure must be raised if volume is to stay constant, i.e. we know that 323 288 the New pressure = 750 X and NOT 750 X 288 323 Now at the higher temperature we want to compress 500 cc to 400 cc. To do this the pressure must be raised (if temperature is kept the same), i.e. the pressure must be 500 made greater to do this, i.e. 400 323 750 X 288 500 X = 1 05 1.3 mm. 400 New pressure = 500 400 We know that it is and not because the pressure 400 500 400 is to be increased. If we used the fraction the pressure 500 would be decreased. PARTIAL PRESSURES. Consider two vessels as shown in Figure 7, each con- taining 1000 cc of a different gas. If communication be es- CHEMICAL CALCULATIONS 25 tablished between the vessels by means of the cock C, in a short time it will be found that the gas from A has diffused into B and that in B has diffused into A until the compo- sition of the mixture of the two gases is the same at EVERY POINT IN EITHER VESSEL. The I 000 cc of either A and B has expanded until it occupies 2000 cc. Since the volume of each gas has thus been doubled, the pressure acting upon it must have been halved. (Bolye's law) i. e. the gas A is P now under a pressure -. Likewise the gas B is under a 2 P pressure of -. The sum of the pressures on the two gases 2 P P is then - plus - which equals and balances the pressure P 2 2 on the piston. If this were not the case the piston would move down, i.e. THE TOTAL PRESSURE OF A MIXTURE OF GASES MAY BE REGARDED AS THE SUM OF THE PRESSURES WHICH EACH WOULD EXERT IF IT ALONE OCCUPIED THE WHOLE SPACE. THIS IS KNOWN AS DALTON'S LAW OF PARTIAL PRESSURES. Water gives off vapor, independent of any other gaseous substances that may be present in the space about it, but DEPENDENT ON THE TEMPERATURE. At any tem- perature water will vaporize and fill the space about it until the partial pressure exerted by the water vapor is the maxi- mum which it can exert at that temperature. If then some gas is collected and measured over water, at say atmospheric pressure, some water will vaporize and this water vapor will diffuse uniformly throughout the other gas and will exert a certain vapor pressure. The pressure actually acting on the gas measured will not then be equal to the atmospheric, but will be the atmospheric minus the vapor pressure of water at that temperature, for, the vapor pressure of water will balance a small part of the pressure acting on the moist gas. A table of the values of Vapor Pressure of Water at temperatures from 0cC to 100cC is given on the last page. 26 CHEMICAL CALCULATIONS PROBLEMS. 1. A given weight of oxygen measures 500 cc at 2 73 °C. What is its volume at 0°C? What is its volume at 546cC? 2. A given volume of gas was cooled to 40°C, at which temperature it was found to occupy one-half of its original volume. What was the original temperature? 3. A given weight of carbon dioxide occupied 250 cc at a pressure of 200 mm. What will be its volume at a pressure of 740 mm? 4. The absolute density of nitrogen is 1.255. What would be the weight of one liter of nitrogen collected at 25 cC and 720 mm pressure? 5. Reduce 1 000 cc of hydrogen at a pressure of 760 mm to the following conditions; the temperature remaining con- stant: (a) 750 mm; (b) 700 mm; (c) 600 mm; (d) 500 mm; (e) 350 mm; (f) 250 mm; (g) 100 mm. Plot your results using the pressures as ordinates and the volumes as abscissae. What sort of curve is the graph? What is its equation? Explain. 6. The temperature remaining at 0°C, at what pressure will one liter of oxygen weigh one gram? One liter of oxygen weighs 1.429 g. at standard conditions. 7. A certain volume of nitrogen measured 500 cc at a temperature of 1 8 cC and a pressure of 742 mm. On the fol- lowing day its volume was 490 cc, when the temperature was 7 °C. What was the barometric reading? 8. A given volume of oxygen measured 1 000 cc at 25 °C and 700 mm pressure. If the temperature is raised to 100°C what pressure must be applied to the gas in order that it may occupy its original volume? 9. A certain weight of gas at a temperature of 200 °C and a pressure of 750 mm was cooled to 22cC, the volume being kept constant. What is the new pressure? 1 0. At a temperature of zero C what pressure must be applied in order that the density of dry oxygen shall be equal to that of dry hydrogen under standard conditions ? CHEMICAL CALCULATIONS 27 11. A flask can stand an internal pressure equivalent to 3088 mm of mercury. It is filled with air at 20°C and 760 mm pressure. Above what temperature will the flask burst? 12. What increase in temperature will be necessary to expand 200 cc of a gas at 0°C and 700 mm pressure to a volume of 250 cc at 725 mm pressure? 13. 200 cc of a gas with an absolute density of 5 is heat- ing to 100 °C and the pressure lowered to 700 mm. What weight of the gas under the new conditions can be contained in a vessel of 100 cc capacity? 1 4. What is the absolute density of oxygen at 2 I °C and 700 mm pressure? One liter of it weighs 1.429 g. at standard conditions. 15. Reduce 50 cu. ft. of hydrogen measured at 29.4 inches pressure to a pressure of 28 inches. 1 6. A liter of dry oxygen measured at standard condi- tions was subjected to a pressure of 40 atmospheres and a temperature of 100cC. Calculate its new volume. 1 7. Under constant pressure the weight of a liter of car- bon dioxide was found to have changed from 1.97 g. to 1.50 g. What was the change in temperature? 18. One liter of mercury vapor measured at 20 °C and 740 mm pressure weighed 9 g. If the temperature is changed to 10 °C and the pressure to 760 mm, what weight of the gas can be contained in a vessel of 250 cc capacity? 19. 20 g. of a gas measured at 500 mm and -48 °C were heated to 177 °C and the pressure decreased to 400 mm; 500 cc of this rarified gas weighed 0.8 g. Calculate (a) the original volume of the gas, (b) the original density? 20. 250 cc of a gas are measured over water which is at a temperature of 20°C, the barometer reading 763.2 mm. What will be the volume when the temperature rises to 2 5 °C and the barometer falls to 761 mm? 21. A volume of hydrogen at 1 7 °C measures 3.5 liters. The gas is heated until the volume increases to 5 liters. If the pressure remains constant, what is the temperature of the gas at the increased volume? 22. A liter of Chlorine under standard conditions weighs 28 CHEMICAL CALCULATIONS 3.167 g. The pressure is increased to 770 mm and the tem- perature is changed under these conditions a liter of gas was found to weigh 1.5 g. At what temperature is the gas? 23. A volume of a certain gas measured 750 cc over mercury at a temperature of 75 cC. The barometric pressure is 750 mm and the level of the mercury inside the measuring tube is 45 mm above that on the outside. What volume will the gas occupy at 60°C and 760 mm? 24. Each of two vessels contains 3 liters of gas, the one being under a pressure of 740 mm and the other under a pressure of 725 mm and both being at a temperature of 85 °C. If the vessels are now placed in communication with each other and all the gas compressed into one of the vessels and during the operation the temperature is increased to 90 cC, what will be the pressure acting of the gas? 25. 3 liters of carbon dioxide and 4.5 liters of nitrogen both at the same temperature are contained in separate vessels and are under pressure of 2.5 kg and 1.4 kg per square cm. respectively. Both these gases are now mixed together in a vessel having a capacity of 5 liters. If there is no change in temperature what pressure will obtain in the last vessel? CHAPTER III OUTLINE Jean Rey observed that there is a limiting proportion of air which a metal will absorb. A given weight of Mg. will combine with a definite weight of Oxygen. Not more. In general a definite weight of one element will combine with a definite weight of another. When a mixture of 32.06 g. of S. and 55.84 g. of Fe is heated, all of the S. and all of the Fe combine. None of the S. or of the Fe remains uncombined. If a mixture of 32.06 g. of S. and more than 55.84 g. of Fe is heated, some Fe remains uncombined. If a mixture of more than 32.06 g. of S. and 55.84 g. of Fe is heated, some S. remains uncombined. S. and Fe combine in a definite proportion 32.06 to 55.84 parts by weight. The Law of Definite Proportions:-When elements unite to form a compound, they do so in a definite proportion by weight. A compound can often be prepared by several different methods. The compound always contains the same elements combined in the same proportions by weight, no matter which method was used to prepare it. Dalton's explanation of the law of definite proportions- Chemical combination consists in the union of atoms in simple combinations. Examples of this. Solution of 2 problems. 29 30 CHEMICAL CALCULATIONS THE LAW OF DEFINITE PROPORTIONS. It was noted by Jean Rey in 1630 that when a metal was heated for some time in the air "the weight of the metal increased from the beginning to the end, but when the metal is saturated it can take up no more air." He says further, "Do not continue the calcination (heating) in this hope; you would lose your labor." Experiments like the following illustrate this:-24.32 grams of magnesium were heated in a porcelain crucible ex- posed to the air. The magnesium united with oxygen from the air: Its lustre disappeared and a white powder was formed. The weight increased and became greater and greater as the lustre of the magnesium disappeared, until all of the magnesium had been changed to the white powder. The white powder was composed of magnesium and oxygen. Its chemical name is magnesium oxide. The weight of the white powder was found to be 40.32 grams. After this first weighing, the crucible and its contents were heated for ten minutes longer, exposed to the air as before and then again weighed. The weight had not changed. It was then heated 20 minutes more and again weighed. There was no gain in weight during the twenty minutes additional heating. The weight of the white powder was 40.32 grams. We decide then from this experiment that 24.32 grams of magnesium, when all of it is converted to the white powder, will unite with 16.00 grams of oxygen-NOT MORE-no matter how much oxygen surrounds it, nor how long the heating is con- tinued. Magnesium and oxygen then combine in the propor- tion of 24.32 parts by weight of magnesium with 16.00 parts by weight of oxygen. Further-it has been observed a great many times, by other chemists, that when magnesium unites with sulphur, or when iron unites with surphur, or in fact when any element unites with any other, that-a definite weight of the one element always unites with a definite weight of the other and forms a definite weight of the new substance. The elements unite in DEFINITE proportions by weight to form the new substance. CHEMICAL CALCULATIONS 31 To illustrate this further-Iron and Sulphur can be mixed together in any proportions. From this mixture the iron could be separated by means of a magnet (or) the sulphur could be dissolved from the mixture by carbon disulphide. If, however, the finely ground mixture of Iron and Sulphur be heated, say over a Bunsen flame, iron and sulphur com- bine and form a new substance, iron sulphide. This substance has new properties, different from those of Iron or of Sulphur. We find that the Iron and the Sulphur combine to form this new substance, in the proportion of 32.06 parts by weight of Sulphur to 55.84 parts by weight of Iron. Thus if a finely ground mixture of 32.06 grams of Sulphur and 55.84 grams of Iron is carefully heated, they will combine completely. If we powder the mass after heating, and examine it, we will find that none of the iron remains uncombined. A magnet will not remove any of it from the mass. ALL of the iron has united with the sulphur. Similarly we would find that ALL of the sulphur has combined with the iron. None remains. Carbon disulphide will not remove any of it. We know then that Iron and Sulphur unite in the proportion of 55.84 parts by weight of Iron to 32.06 parts by weight of Sulphur. If we heat a mixture of 32.06 grams of Sulphur and say 165.00 grams of Iron (more than 55.84 grams) we find that not all of the Iron combines with the sulphur to form ferrous sulphide. Part of the iron remains uncombined, and if the mass is powdered after heating, this iron can be removed by means of a magnet. We would find that 1 09.1 6 grams of Iron would be left uncombined and retaining its original properties. The rest 165.00 -109.16 = 55.84 grams of Iron had united with the 32.06 grams of Sulphur. This is the same proportion in which they united when ALL of the Sulphur and all of the Iron combined. We see then that Iron and Sulphur combine in the proportion of 32.06 parts by weight of Sulphur with 55.84 parts by weight of Iron, no matter whether more Iron is present or not. Excess Iron is left uncombined. In the same way if we heated carefully a mixture of 55.84 grams of Iron with say 250.00 grams of Sulphur, instead of 32.06 grams, we would find that only 32.06 grams of the Sul- phur united with the 55.84 grams of Iron. The remainder 250.00 32 CHEMICAL CALCULATIONS - 32.06 = 217.94 grams of Sulphur would be left uncom- bined. Again we see that Iron and Sulphur unite in the pro- portion of 55.84 parts by weight of Iron to 32.06 parts by weight of Sulphur, despite the fact that more Sulphur was present. Sulphur and Iron then combine in a definite propor- tion 32.06 to 55.84 parts by weight. From a great many observations like these which showed, in every case, that the elements combine in a definite proportion by weight, the underlying law has been deduced. It is known as the Law of Definite Proportions. WHEN ELEMENTS UNITE TO FORM A COMPOUND THEY DO SO IN A DEF- INITE PROPORTION BY WEIGHT. Another Way of Stating This Law. Very often a compound can be made by several different methods, but, no matter what method was used to make the compound, we find that the elements united to form it in the same proportions by weight, in each method. We usually find this by breaking up the compound into the elements by some chemical means. The process is known as Analysis. Let us consider an example. Sodium Sulphate may be prepared in 7 ways. 1. By the oxidation of Sodium Sulphide. Sodium Sulphide -fi Oxygen = Sodium Sulphate as represented by the chemical equation Na2S + 202 = Na2SO4 2. By treating some sodium salts with H2SO4 Sodium Chloride + Sulphuric Acid Sodium Sulphate * Hydro- chloride Acid 2NaCl + H-SOr = . Na-.'SOi + 2HC1 3. By treating Sodium Hydroxide with Sulphuric Acid. Sodium hydroxide-(-Sulphuric acid=Sodium sulphate-]-water 2NaOH + H,SO4 = Na2SO4 + 2H,O 4. By treating Sodium with Sulphuric Acid. Sodium -|- Sulphuric acid = sodium sulpha e -f- Hydrogen 2Na + H2SO4 = Na,SO4 + Ha 5. By treating Sodium Oxide with Sulphuric Acid. CHEMICAL CALCULATIONS 33 Sodium oxide -f- Sulphuric acid = Sodium Sulphate -(- water Na2O + H2SO4 = Na2SO4 + H2O 6. By treating sodium oxide with SO3, the anhydride of H2SO4 Sodium Oxide-(-Sulphuric acid anhydride = Sodium Sulphate Na2O + SO3 = Na2SO4 7. By union of the elements, sodium, sulphur, and oxygen Sodium A Sulphur 4- Oxygen = Sodium Sulphate 2Na + S + 202 = Na2SO4 If we examine the sodium sulphate made by any one of these 7 methods we find that it is identical in composition and properties with that made by any of the other methods. That it is always composed of three elements, sodium, sulphur and oxygen and that THE RELATIVE WEIGHTS OF EACH PRESENT ARE ALWAYS THE SAME. 46.00 parts by weight of Sodium 32.06 " " " " Sulphur 64.00 " " " " Oxygen in 142.06 " " " " Sodium Sulphate We can then express the Law of Definite Proportions in another way: A DEFINITE CHEMICAL COMPOUND AL- WAYS CONTAINS THE SAME ELEMENTS COMBINED IN THE SAME PROPORTIONS BY WEIGHT. H2SO4 Atomic Theory Explanation of This Law. The question now comes up, WHY is it that the ele- ments always combine in EXACTLY the same proportions by weight in a given compound? A very satisfactory ex- planation for this is found in the Atomic Theory offered by Dalton in 1802. Dalton conceived 'he idea that chemical combination consists simply in the union of the atoms in simple combinations, and in whole numbers, for atoms are indivisible-Two, 3, 4, 5, 6, or more atoms unite to form larger particles called molecules. Let us consider our former and now familiar examples on this basis. (a) One atom of iron (weight of an atom of iron - 34 CHEMICAL CALCULATIONS 55.84) unites with one atom of Sulphur (weight of atom of sulphur = 32.06). Fe 4" S = FeS 1 atom 4- 1 atom = 1 molecule 55.84 32.06 87.90 parts by weight, such as grams, ounces, pounds, etc. (b) As a second example: Two atoms of Hydrogen (weight of one Hydrogen atom = 1.008), unite with one atom of oxygen (weight - 16 00) ard form a molecule of water H, 4- O = H2O 2 atoms 1 atom = 1 molecule 2 X 1008 1 X 16.00 = 2.016 -4- 16.00 parts by weight 2.016 16.00 = 18.016 (c) As a third example: Two atoms of sodium (weight = 23.00) unite wi'h one atom of Sulphur (weight == 32.06) and four atoms of Oxygen (weight == 16 00) and form a molecule of sodium sulphate. 2 Na 4- S + 4 0= Na2SO4 2 atoms 1 atom 4 atoms = 1 molecule 2 X 23.00 + 1 X 32.06 + 4 X 16.00 = 46 4- 32.06 4- 64 46 32.06 64 = 142.06 parts by weight. In writing formulas, the number of atoms of each element in the compound, is designated by a small subscript number. Thus when sodium sulphate is formed 2 atoms of Na unite with 1 atom of S and 4 atoms of Oxygen and form 1 molecule of Sodium Sulphate. The formula is written Na.' Si Oi, or simply Na2SO4. The molecule of sodium sulphate contains 2 4- 1 T 4 = 7 atoms. The weight of the molecule is nat- urally the sum of the weights of all the atoms in it. Two atoms of Na (weight 23.00) unite with one atom of S (weight 32.06) and 4 atoms of Oxygen (weight 16.00) and form 1 molecule of Na2SO4 (weight 142.06). We see then why the relative weights of Na, S and O in 142.06 parts by weight of Na2SO4 are always exactly as 46.00 to 32.06 to 64.00. If the atoms united in some other way, say 2 atoms of Na (weight 23) with 2 atoms of S (weight 32.06) and 3 CHEMICAL CALCULATIONS 35 atoms of oxygen (weight 16), the relative weights of the elements present would be 46.00 of Na 64.12 of S 48.00 of O in 158.12 of the compound. This substance would not be sodium sulphate and it would not have the properties of sodium sulphate. We will find (next chapter) that some elements have the power to combine with one another in several different ways, different substances being formed in each case. The proportions in which they combine are, however, DEFINITE in each case. Let us now apply the Law of Definite Proportions to the solution of some problems. I. (a) What weight of Sulphur is there in 10 grams of Sodium Sulphate? The Law of Definite Proportions tells us that the relative weights of Na, S and O in Na2SO4 must always be the same, no matter how the Na2SO4 was formed. Weight of 2 atoms of Na - 46.00 Weight of 1 atom of S = 32.06 Weight of 4 atoms of O = 64.00 Weight of 1 molecule of Na2SO4 = 142.06 There are always 32.06 parts by wt of S in 142.06 parts by wt of NaQSO4 " " " 32.06 grams of S in 142.06 grams " " " " 32.06 " " " 142.06 " " " " 1 gram 32.06 In 1 0 g. of Na.,SO4 there are I 0 X = 2.25 g. of S. 142.06 (b) What weight of Sodium is there in 10 grams of Sodium Sulphate? There are always 46.00 parts by wt of Na in 142.06 parts by wt of Na2SO4 " " " 46.00 grams " Na " 142.06 grams of Na SO 46.00 • 142.06 grams " Na " 1 gram of Na2SO4 36 CHEMICAL CALCULATIONS 46.00 In 1 0 g. of Na.,SO4 there are 10 X 3.2 3 g. of Na. 142.06 (c) What weight of SO3 is there in 10 grams of Sodium Sulphate. Groups of elements can be considered just the same as elements on the above principle. In fact we have seen that sodium sulphate can be made by the direct union of SO, and Na2O (Method 6-page 33). Weight of 1 atom of S = 32.06 Weight of 3 atoms of O = 48.00 Weight of 1 molecule of SO, = 80.06 There are 80.06 grams of SO. in 1 42.06 grams of Na2SO4 80.06 There are grams of SO., in 1 gram of Na.,SO4 142.06 80.06 In 10 grams of Na,SO4 there are 10 X 5.63 grams of SO3 ' 142.06 II. Calculate the percentage purity of a sample of Silver Chloride Ore which contains 65.00% of Ag. The ore contains other things besides Silver Chloride. The chemical substance silver chloride has the formula AgCl. Weight of 1 atom of Ag = 107.88 Weight of 1 atom of Cl = 35.46 Weight of 1 molecule of AgCl = 143.34 There are 1 07.88 grams of Ag in 1 43.34 grams of AgCl 107.88 There are grams of Ag in 1 gram of AgCl 143.34 107.88 In I 00 grams of AgCl there are 1 00 X 75.26 grams of Ag 143.34 PERCENTAGE MEANS PARTS IN ONE HUNDRED PARTS: grams in 100 grams; pounds in 100 pounds. When we say that candy contains 60 per cent, of sugar, we mean that out of I 00 parts by weight of candy, 60 parts by weight CHEMICAL CALCULATIONS 37 are sugar, no matter what the parts by weight may be ounces, pounds, grams, etc. In 1 00 parts of AgCl there are 75.26 parts of Ag. The percentage of Ag in AgCl is then 75.26. Now the Ore contains only 65.00 % of Ag. If it were wholly AgCl it would contain 75.26% of Ag. 65.00 In contains then only as much Ag as it would if it 75.26 were composed wholly of AgCl. 65.00 It is then X 100% - 86.38% pure. 75.26 100.00 86.38 1 3.62 % of this ore is something else, other than Silver Chloride. In solving the following problems always try to keep before you the principle involved. The Law of Definite Pro- portions: 1. Hew many parts by weight of Sulphur are there in 8 7.91 parts by weight of ferrous sulphide? How many parts by weight of S in 100 parts by weight of ferrous sulphide? What, then, is the percentage of S in ferrous sulphide? 2. How many parts by weight of mercury are there in 232-66 parts by weight of cinnabar, HgS? How many parts by weight of mercury are there in 1 00 parts by weight of cin- nabar? What, then, is the percentage of Hg in cinnabar? 3. How many parts by weight of water are there in 239.492 parts by weight of NgSO4. 7H2O? How many grams of water are there in 1 00 grams of MgSO4. 7H2O? What, then, is the percentage of water in MgSO4. 7H2O? 4. How many parts by weight of Sulphur are there in 233.43 parts by weight of BaSO4? How many grams of S are there in 100 grams of BaSO4? What, then, is the percentage of Sulphur in BaSO4? 5. How many parts by weight of Sulphur Trioxide, SO3, are there in 233.43 parts by weight of BaSO4? How many 38 CHEMICAL CALCULATIONS grams of SO3 are there in 100 grams of BaSO4? What is the percentage of SO3 in BaSO4 ? 6. How many grams of Mg are there in 222.72 grams of Mg2P2O7? (b) How many grams of Mg are there in 1 00 grams of Mg2P2O7? (c) How many grams of MgO are there in 1 00 grams of Mg2P2O7? What, then is (d) the percentage of Mg in Mg2P2O7 (e) the percentage of MgO in Mg2P2O7? 7. How many grams of Iron are there in 159.68 grams of Fe2O3? How many grams of iron are there in 100 grams of Fe2O3? How many grams of iron in 100 grams o f Fe3O4? What then is the percentage of iron in Fe2O3, in Fe3O4? 8. What weight of potassium is present in 50 grams of potassium sulphate, K2SO4? 9. What weight of phosphorus is present in 1 00 grams of di-sodium hydrogen phosphate, Na2HPO4? 10. What weight of hydrogen is present in 200 grams of hydrogen peroxide, H2O2? 11. What weight of manganese dioxide contains 50 grams of oxygen? 12. What weight of potassium perchlorate contains 40 grams of chlorine? 1 3. What weight of sulphuric acid can be made from 100 pounds of sulphur? 14. Calculate the weight of potassium in a sample of pure sylvite KC1, which analyzed for 2.230 grams of chlorine. 1 5. What weight of zinc is present in that weight of pure zinc sulphate, ZnSO4.5H2O, which contained 29 grams of sulphuric anhydride? 1 6. Calculate the percentage purity of a sample of mag- nesite, MgCO3, which contains by analysis 20.2 per cent, mag- nesium. 1 7. Calculate the percentage of potassium chloride in a sample of carnallite, KCl.MgCl2.6H2O, which contains 34.75 per cent, of chlorine. 1 8. Calculate the percentage of magnesium oxide, MgO, present in a sample of dolomite, which contains 42.7 per cent, carbon dioxide. CHEMICAL CALCULATIONS 39 19 0.3207 gram of common salt yielded 0.7842 gram of silver chloride. Determine the percentage of chlorine pres- ent in the salt 20. How much sodium chloride can be made from (a) 50 grams of scdium, (b) 50 grams of sodium hydroxide? 21. How much oxygen can be obtained from (a) '100 gms. of KClO.j, (b) 200 gms. of HgO? 22. A sample of alum, K2A12(SO4)4. 24 H2O gave 7.92 % of potassium on analysis. What is the percentage purity of the samples 23. What is the percentage purity of a sample of silver, 20 gms. of which when dissolved in HNO3 and treated with hydrochloric acid, yielded 21.1 gms. of AgCl? 24. What is the percentage of KC1 in carnellite, KCl.MgCl2.6 H2O? 25. How much magnesium sulphate can be formed from 20 grms. of MgO? 26. What is the percentage purity of a sample of mar- ble 20 gms. of which yielded 3.0 liters of CO2 at 30 °C and 740 mm pressure? One liter of CO2 at standard conditions weighs 1.974. 2 7. How many pounds of aluminum can be obtained from a ton of bauxite which contains 60% A12O3? 28. What is the percentage of (a) A12O3, (b) moisture in common alum, K2SO4.A12(SO4)3.24H2O? 29. How much Paris Green [Cu(C2H3O2) 2. Cu3As2O6] can be made from the copper obtained from one ton of a copper ore which contains 40% of CuS? CHAPTER IV OUTLINE Dalton observed while studying Methane and Ethylene that the different weights of C which are united with the same weight of H, are in the proportion of 2 to 1. In the oxides of C the different weights of C which are combined with the same weight of O, are in the proportion of 2 to 1. In the 6 oxides of Nitrogen the weights of O which are combined with the same weight of N, are in the proportion 1 :2:3:4:5:6. Dalton thus showed that when 2 elements combine to form several different compounds, the different weights of the one, which are united with a fixed weight of the other, are simple whole number multiples. This is Dalton's law of Mul- tiple Proportions. Dalton's explanation of this law: The Atomic Theory. The Weights of Atoms. Relative Weights. Actual Weights. Examples. Solutions of two Problems. 40 CHEMICAL CALCULATIONS 41 THE LAW OF MULTIPLE PROPORTIONS. We have observed that if elements unite to form a com- pound they always do so in a definite proportion by weight. It has been mentioned, however, (page 35) that in some cases the same elements can unite in several different proportions to form several different compounds. In these cases, when the proportions, in which the elements unite, are different, the compounds are different in their properties. For any one of these compounds the proportions in which the elements unite are always the same. We further find that the different weights of the one element, which unite with a fixed weight of the second are whole-number multiples of each other. Dalton observed this about 1803 while he was analyzing the 2 gases Ethylene and Methane. These are different sub- stances with different properties. Dalton found, however, that each of them contained the elements carbon and hydrogen and NO others. He further found that the proportions of carbon and hydrogen in the two compounds were different. His re- sults were of course not very accurate for the methods of analysis, and the apparatus he used at that time were rather crude. The actual proportions of C and H in these com- pounds are ■ c parts by weight H parts by weight Ethylene contains. . . . 24.00 for each 4.032 Methane contains. . . . 12.00 4.032 The proportion, by weight, of C or of H in either com- pound, is always constant, but it will be observed in the two compounds, the weights of C, combined with the same weight of H, are as 2 to 1. Having discovered this astounding result for 2 compounds of carbon and hydrogen, Dalton was led to analyze the 2 com- pounds of carbon and oxygen. He found that the different weights of carbon, which in the 2 compounds were combined with the same weight of oxygen, were as 1 to 2. 42 CHEMICAL CALCULATIONS c o parts by weight parts by weight Carbon Monoxide contains Carbon Dioxide contains. . 12.00 for each 16.00 6.00 16.00 The student should be able to reckon the proportion which exists between the different weights of Oxygen, which are combined with the same weight of Carbon in these 2 com- pounds. Dalton's suspicion that there must be a law underlying these simple figures grew into conviction when he again found simple proportions in the oxides of Nitrogen. We know 6 oxides of Nitrogen. N parts by weight O parts by weight Nitrous Oxide contains. . . 28.02 16.00 1 Nitric Oxide contains. . . . 28.02 for each 32.00 2 Nitrous Acid Anhydride contains 28.02 48.00 3 Nitrogen Peroxide contains Nitric Acid Anhydride 28.02 64.00 4 contains 28.02 80.00 5 Nitrogen Hexoxide con- tains 28.02 96.00 6 The proportion of N or of O in any one of the compounds is always constant, but in the 6 compounds, the different weight of O, combined with the same weight of N, are in the proportion of 1 :2:3:4:5:6. Dalton thus showed that WHENEVER THE SAME ELE- MENTS UNITE WITH EACH OTHER TO FORM SEVERAL DIFFERENT COMPOUNDS, THE DIFFERENT WEIGHTS OF THE ONE, WHICH UNITE WITH A FIXED WEIGHT OF THE OTHER ARE WHOLE NUMBER MULTIPLES. Dalton could find no exceptions to this idea, so he advanced it as a law. It is known as Dalton's LAW OF MULTIPLE PRO- PORTIONS. This law has been more amply confirmed by an enormous number of experiments like those originally made by the pioneer Dalton. Dalton's mind was not satisfied with the deduction of this CHEMICAL CALCULATIONS 43 remarkable law. He sought an explanation for it. This was afforded in a brilliant and elegant way by his Atomic Theory of Matter, of which the following statements give a brief out- line: I. Every element is made up of small particles called Atoms. The atoms are real particles of matter, which cannot be subdivided by any known chemical process. II. Chemical compounds are formed by the union of the atoms of different elements, I atom of one element with 1 atom of another element; I atom of an element with 2 atoms of another element; or 1 with 3; 2 with 2; 2 with 3; 3 with 2; 2 with 4, etc. III. All atoms of the same element have the same weight. Each atom of Hydrogen has the same weight as any other atom of Hydrogen. Each atom of Iron has the same weight as any other atom of Iron. IV. Atoms of different elements have different weights. The weight of an atom of Iron is different from that of an atom of Hydrogen or from that of an atom of Chlorine, etc. We do not know the ACTUAL weight of an atom of any element. No one has succeeded in isolating and actually weighing an atom of any element. The actual weights of atoms have been calculated from reliable observations and are known to be extremely small numbers; thus the weight of an atom of Hydrogen is approximately 0.000000000000000,- 0000000029 grams. However, we do know, and know very exactly, the RELATIVE weights of the atoms. We know that the RELATIVE weights of an atom of Hydrogen and an atom of Oxygen are 1.008 and 16.00. If we take the weight of an atom of Oxygen as 16.00 then the weight of an atom of Hydrogen is 1.008 (slightly greater than 1.) The oxygen atom is very nearly 1 6.00 times as heavy as the Hy- drogen atom (15.88 times.) If we take the weight of an atom of Oxygen as 16.00 then the weight of an atom of Hy- drogen is 1.008. If we take the weight of an atom of Hydro- gen as 1. then the weight of an atom of Oxygen is 15.88. It is more convenient to take the weight of an atom of Oxygen as 16.00, thus the weight of an atom of Hydrogen becomes 1.008 (a little greater than 1.) Now since Hydrogen is the 44 CHEMICAL CALCULATIONS lightest element, no elements will have atomic weights less than 1. This gives convenient numbers. Oxygen is chosen as the standard or base because it has been found to combine with all known elementsexcept fluorine and the rare gases of the atmosphere and thus forms compounds in which we can determine directly the Relative weights in which nearly all the elements unite with Oxygen. From these figures we can get the atomic weights. When we speak of the weight of an atom (atomic weight), we do not mean the actual weight of 1 atom. We mean the RELA11VE weight of one atom of the element, taking the weight of an atom of Oxygen as 1 6 00. When we say that the atomic weight of iron is 55.84 we mean that, taking the weight of an atom of Oxygen as 16.00 the weight 55.84 of an atom of Iron is 55.84. The Iron atom is times 16.00 as heavy as the Oxj gen atom. A table giving the atomic weights cf all (he known elements is given on the inside of the front cover. We must remember that atoms have not yet been seen. In fact there seems to be good evidence that we may never be able to actually see the individual atoms. The Atomic Theory is, however, fully in accord with the great mass of evidence, with the facts cf chemistry, and has been extremely useful to scientists. It is therefore universally accepted and used. Let us now see how we can use the Atomic Theory to explain some of the examples of elements that combine in several proportions to form several different compounds. (a) Carbon Monoxide. From the evidence of repeated analysis we have the fact that 1 gram of Carbon Monoxide c o contains 0.4286 gram 0.5 7 1 4 gram 28 grams of Carbon Mon- oxide contain - 1 2.00 grams 1 6 00 grams In Carbon Monoxide 12.00 parts by weight of C are combined with 16.00 parts by weight of O. CHEMICAL CALCULATIONS 45 The weight of 1 atom of C - 12 The weight of 1 atom of O = 16 We decide then that in Carbon Monoxide 1 atom of C is combined with 1 atom of O. (b) Carbon Dioxide. C o 1 gram of Carbon Dioxide contains - - - 0.2 72 7 gram 0.7273 gram 44 grams of Carbon Dioxide contain - - - - 1 2.00 grams 32.00 grams In Carbon Dioxide 12.00 parts by weight of C are com- bined with 32.00 parts by weight of O. The weight of an atom of C is 12.00. The weight of an a'om of O is 16 00. We decide then that in Carbon Dioxide 1 atom of C is combined with 2 atoms of Oxvgen. In Carbon Monoxide there are 1 atom of C and 1 atom of O. In Carbon Dioxide there are 1 atom of C and 2 atoms of O. Carbon Dioxide contains twice as many Oxygen atoms, for each Carbon atom, as Carbon Monoxide does. We can see then why the weights of O, combined with the same weight of C, are as 2 to 1. (b) The Oxides of Nitrogen. We have learned that from the evidence of repeated analyses, the composition of the Oxides of Nitrogen is N O Parts Parts by Weight by Weight Nitrous Oxide contains . . 28.02 16.00 Nitric Oxide contains . . 28.02 32.00 Nitrous Acid Anhydride contains. . . 28.02 48.00 Nitrogen Peroxide contains . . 28.02 64.00 Nitric Acid Anhydride contains. . . . . 28.02 80.00 Nitrogen Hexoxide contains . . 28.02 96.00 We know that the weight of an atom of N (atomic weight of N) is 1 4.0 1. We know that the weight of an atom of O (atomic weight of O) is 16.00. 46 CHEMICAL CALCULATIONS We can reckon then that in the 6 compounds, the relative numbers of atoms of each element in the molecules are N O Number of Atoms Number of Atoms 28.02 16.00 Nitrous Oxide 2 14.01 16.00 28.02 32.00 2 - _ 2 14.01 16.00 28.02 48.00 Nitrous Acid Anhydride. . = 2 = 3 14.01 16.00 28.02 64.00 Nitrogen Peroxide - 2 . -4 14.01 16.00 28.02 80.00 Nitric Acid Anhydride. . . 2 = 5 14.01 16.00 28.02 96.00 Nitrogen Hexoxide 2 6 14.01 16.00 In each of the 6 compounds there are 2 atoms of Nitro- gen, while in the 6 compounds the number of Oxygen atoms present are 1, 2, 3, 4, 5 and 6. We can see then why the weights of Oxygen, combined with the same weight of Nitro- gen, are as in the proportion of 1 :2:3:4:5:6. The Atomic Theory thus offers a beautiful explanation of the laws of Definite and Multiple proportions. The atoms of elements unite in simple proportions. The weights of the atoms are constant, therefore the elements must unite in simple proportions, corresponding to the weights of the atoms, or else in simple multiples of these, if the elements unite in several different proportions. PROBLEMS. 1. The 2 chlorides of Iron have the following composi- tion : Ferrous Chloride Fe = 44.05% Cl = 55.95% Ferric Chloride Fe = 34.42% Cl = 65.58% CHEMICAL CALCULATIONS 47 Show how these illustrate the law of Multiple Proportions. By expressing the composition of the substances in per cent , the evidence of Multiple Proportions is hidden but can easily be brought out by calculating the amount of Cl, in each case, combined with 1 part by weight of Iron. Thus in ferrous chloride 44.05 grams of Fe are combined with 55.95 grams of Cl. 1 1 gram of Fe is combined with X 55.95 = 1.2 70 grams of Cl. 44.05 In ferric chloride 34.42 grams of Fe are combined with 65.58 grams of Cl. 1 1. gram of Fe is combined with ----- X65.58 =1.905 grams of Cl. 34.42 The weights of Cl which are united with the same weight of Fe ( 1 gram of Fe) are in the proportion of 2 to 3 for 1-270 1.905 = 0.635 and 0.635 2 3 2. 3 acids of Sulphur have the following composition: Sulphurous Acid H = 2.456% S = 39.061% O = 58.482%. Sulphuric Acid H = 2.055% S = 32.688% O = 65.255% Pyro Sulphuric Acid H = 1.131% S = 35.994% O = 62.874% Show that these acids illustrate the law of Multiple Pro- portions: Sulphurous Acid. 2.456 g. of H are combined with 39.061 g. of S and 58.482 g. of O. 1 1 g. of H is combined with X 39.061 g. of S = I 5.904 g. ofS. 2.456 1 I g. of H is combined with X 58.482 = 23.81 1 g. of Oxygen. 2.456 2.055 g. of H are combined with 32.61 1 g. of S and 65.255 g. of O. 1 1 g. of H is combined with X 32.688 g. of S = 1 5.906 g. ofS. 2.055 Sulphuric Acid. 48 CHEMICAL CALCULATIONS 1 1 g. of H is combined with X 65.255 = 31.754 g. of Oxygen. 2.055 Pyro Sulphuric Acid. 1.131 g. of H are combined with 35.994 g. of S and 62.874 g. of O. 1 1 g. of H is combined with X 35.994 g. of S = 31.824 g. ofS. 1.131 1 1 g. of H is combined with X 62.874 - 55.591 g. of Oxygen. 1.131 The weights of S which are combined wi'h 'he same weight of H ( 1 gram) in the 3 compounds, are 1 5.90, 1 5.90, 31.824, or are in the proportion 1:1:2. The weights of O, which are combined with 1 gram of H. are, in the 3 compounds, 23.81 g., 31.754 g., and 55.591 g., or are in the proportion of 3:4:7 for 23.81 31.754 55.591 = 7.941 = 7.94 = 7.94 3 4 7 If we know the molecular formula of any ore of the three, we can write the molecular formulas of the others. Thus if we know that the molecular formula of Sulphurous Acid is H2SO3, then that of the Sulphuric Acid must be H SO4, and of Pyro Sulphuric Acid H2S2O7, for the O in the three com- pounds is in the ratio of 3:4:7 for the same amount of H. The Sulphur is the same in the first two and twice as great in the third. (3) The 2 chlorides of copper have the following com- position : Cuprous Chloride Cu = 64.192% Cl = 35.807% Cupric Chloride Cu = 47.267% Cl = 52.732% Show that these illustrate the law of Multiple Proportions. (4) The 3 oxides of phosphorus have the following com- position : 1 P = 56.395 % 0 = 43.604% 2 P = 49.238% 0 = 50.761% 3 P = 43.693% 0 = 56.306% CHEMICAL CALCULATIONS 49 The molecular weight of No. 1 has been determined as 110.08. Show that these compounds illustrate the law of Multiple Proportions and write their molecular formulas. (5) Tin exhibits valences of 2 and 4. It forms 2 com- pounds with Oxygen, which have the following composition: 1 Sn = 88.121 % 0 =1 1.878% 2 Sn = 78.765 % 0 = 21.234% Show that these compounds illustrate the law of Multiple Proportions and write the molecular formulas and chemical names of these compounds. (6) Potassium chlorine and oxygen form 4 compounds which have the following composition: 1 K = 43.176% 01 = 39.156% 0 =17.667% 2 K= 36.692% 01 = 33.277% 0 = 30.030% 3 K = 25.853% 01 = 31.503% 0 = 42.643% 4 K = 28.220% 01 = 25.591% 0 = 46.188% Show that these illustrate the law of Multiple Proportions. No. 3 is potassium chlorate, KC1O3; write the names and the molecular formulas of the others. (7) Hydrogen and Oxygen form 2 compounds of the following composition: 1 H = 1 1.190% O = 88.819% 2 H= 5.926% 0 = 94.073% Show that these illustrate the law of Multiple Proportions. No. 1 is water. Give name and formula of No. 2. (8) Two atoms of Oxygen unite to form a molecule of ordinary Oxygen. Three atoms of Oxygen will unite under (he influence of a silent electric discharge and form a molecule of Ozone. Write the formulas which represent a molecule of Ozone and of Oxygen respectively and state the molecular weight of each. (9) Sodium, Sulphur and Oxygen from 3 compounds of the following composition: 1 Na = 36.490% S = 25.432% 0 = 38.077% 2 Na = 32.380% S = 22.567% 0 = 45.051% 3 Na = 29.091% S = 40.551% 0 = 30.356% Show that these illustrate the law of Multiple Proportions. No. 1 is Sodium Sulphite, Na2SO3. Write the names and for- mulas of the others. 50 CHEMICAL CALCULATIONS (10) Carbon and Hydrogen form a great many com- pounds with each other. Show that the 5 of the following composition illustrate the law of Multiple Proportions: 1 H = 25.149% C= 74.850% 2 H = 20.127% C= 79.872% 3 H= 14.383% C = 85.616% 4 H= 18.300% C = 81.699% 5 H= 17.355% C = 82.644% (11) Three acids of Phosphorus have the following com- position : 1 P = 38.776% H= 1.259% 0 = 59.964% 2 P= 37.824% H = 3.684% 0 = 58.490% 3 P = 31.652% H = 3.083% 0 = 65.265% Show that these illustrate the law of Multiple Propor- tions. No. 1 is Metaphosphoric Acid, HPO3. Write the names and formulas of the others. (12) Three Potassium Salts have the following composi- tion : 1 K = 49.412% S = 20.257% 0 = 30.331% 2 K = 44.876% S= 18.398% 0 = 36.726% 3 K = 30.748% S = 25.212% 0 = 44.040% Show that these illustrate the law of Multiple Proportions. No. 2 is Potassium Sulphate. Write the names and formulas of the others. (13) Two Acids of Nitrogen have the following com- position : 1 H = 2.143% N = 29.797% 0 = 68.059% 2 H= 1.599% N = 22.231% 0 = 76.168% Show that these illustrate the law of Multiple Proportions. No. 2 is Nitric Acid. Write the name and formula of No. 1. ( 14) Two Oxides of Antimony have the following com- position : 1 Sb = 83.356% 0 =16.643% 2 Sb = 75.031 % 0 = 24.968% Show that they illustrate the law of Multiple Proportions. The molecular weight of No. 1 was found to be 168.2. Write the names and molecular formulas of these oxides. (15) Two Oxides of Arsenic have the following com- position: CHEMICAL CALCULATIONS 51 1 As = 75.747% 0 = 24.252% 2 As = 65.205% 0 = 34.794% Arsenic exhibits valences of 3 and 5. Show that these compounds illustrate the law of Multiple Proportions and give their names and empirical formula. ( 1 6) Two Sulphates of Sodium have the following com- position : 1 Na2O = 43.643% SO, = 56.356% 2 Na2O = 27.912% SO3 = 72.087% Show that these illustrate the law of Multiple Proportions. The molecular weight of No. 1 is I 42.06. Write the molecular formulas and name of these compounds. ( 1 7) Na, H, P and O form 2 compounds of the follow- ing composition: 1 Na = 32.383*% H = 0.7069% P = 21.851% O = 45.056% 2 Na = 32.155% H = 1.409 % P = 21.697% O = 44.737% Show that these illustrate the law of Multiple Proportions. (18) K, Cr and O form 2 compounds of the following composition: 1 K = 40.267% Cr = 26.776% 0 = 32.955% 2 K = 26.580% Cr = 35.350% 0 = 38.069% Show that these illustrate the law of Multiple Proportions. (19) FLO and P..O- form compounds of the following composition: 1 FLO = 10.628% P2O5 = 89.372% 2 FLO = 2 7.55 7% P2O5 = 72.443% Show that these illustrate the law of Multiple Proportions. (20) Two Sulphates of Iron have the following com- position : 1 Fe = 36.761% S = 21.106 % 0 = 42.132% 2 Fe = 27.929% S = 24.053% 0 = 48.016% Shew that these illustrate the law of Multiple Proportions and write their formulas assuming Iron to exhibit valences of 2 and 3. (21) Calculate the Percentage of each element in each of the two Sulphides of Arsenic As2S3 and As2S-. (22) Calculate the Percentage of each element in each of the two Nitrates of Mercury HgNO3 and Hg(NO3)2. CHAPTER V OUTLINE Gay Lussac found by experiment that when gases unite the volumes which unite are in the proportion of simple whole numbers: the volume of any product which is a gas is related to the volumes of the gases which reacted, in the proportion of simple whole numbers. This is Gay Lussacs Law of Vol- umes. Two examples which indicate that equal volumes of gases contain the same number of molecules, (when measured under the same conditions of temperature and pressure). Avogadro's Hypothesis. Avogadro's Hypothesis explains Gay Lussac s Law of Volumes. Distinction between an atom and a molecule. Proof that the molecules of each of the gases, Hydrogen, Oxygen and Chlorine, contain 2 atoms, and should hence be written respectively H2, O2 and Cl2. Significance of the figure 22.3. Problems. 52 CHEMICAL CALCULATIONS 53 GAY LUSSAC'S LAW OF GASES. AVOGADRO'S HYPOTHESIS. RELATION BETWEEN VOLUME AND MOLECULAR WEIGHT. We have observed that when elements unite, they do so in sinple proportions by weight and, in some cases, we have seen, that the same elements unite in several proportions. The weights of the one united with the same weight of the other are whole number multiples. We have seen how from an enormous number of careful observations, the laws of Definite and Multiple Proportions were deduced. We have further seen that the Atomic Theory affords a simple and beautiful explanation of these two laws, (and that the laws in turn serve to strengthen the statements of the atomic theory). When the elements which unite, are gases, we find that the action is governed by another simple law. The VOLUMES of the gases which unite, are whole number multiples. If the product formed is a gas, its volume will be related to the volumes of the gases which united, by a simple proportion. This was first observed by Gay Lussac and Humboldt in 1805. They found when Hydrogen and Oxygen unite to form water, that 2 parts by volume of Hydrogen united with exactly I part by volume of Oxygen. If more Hydrogen or more Oxygen was used, than required by the proportion 2 to 1, the excess of Hydrogen or Oxygen remained uncom- bined. Gay Lussac further proved that when Hydrogen and Chlo- rine unite, I part by volume of Hydrogen unites with exactly I part by volume of Chlorine and forms exactly 2 parts by volume of Hydrochlpric Acid. When Ammonia and Hydro- chloric Acid unite, 1 part by volume of Ammonia unites with 1 part by volume of Hydrochloric Acid and forms 1 part by volume of Ammonium Chloride Vapor. He further showed that 2 parts by volume of Ammonia were formed from the union of 1 part by volume of Nitrogen and 3 parts by volume of Hydrogen. In all the cases he observed the volumes of gases which united were in simple proportion. One part by 54 CHEMICAL CALCULATIONS volume of a gas united with 1 part by volume of another or 1 part Vy volume of a gas with 2 parts by volume of another, or 1 with 3, or 2 with 1, or 2 with 2, or 2 with 3, etc. He did not find any case where, for example, 1 part by volume of a gas united with a complex proportion such as 2.1 77 parts by volume of another. Gay Lussac, therefore, concluded that "gases always unite in the simplest proportions by volume." This conclusion of Gay Lussac's has been amply confirmed by a large number of experiments since then and we now state it as a law. Gay Lussac's Law of Volumes:-WHEN GASES UNITE WITH ONE ANOTHER, THE VOLUMES WHICH UNITE ARE IN THE PROPORTION OF SIMPLE WHOLE NUMBERS: IF THE PRODUCT IS A GAS, ITS VOLUME IS RELATED TO THE VOLUMES OF THE GASES WHICH REACTED, IN THE PROPORTION OF SIMPLE WHOLE NUMBERS. Gay Lussac suspected that the explanation of the law he had discovered, would be afforded by Dalton's Atomic Theory. He was, however, unable to work out this explana- tion, although he came close to it. We will see that the key- note of the explanation was given by Avogadro in 1811. First, however, let us consider several examples of combina- tion of gases and see what conclusion we can make from them. (1) Consider the action of NHb on HC1. It is represented by the equation NH, -p HC1 = NH4C1 (The equation is based on facts.) It is also an experimental fact that each 1 liter of NHb unites with exactly 1 liter of HC1. We know from the action which has taken place (as represented by the equation) that each one molecule of NHy unites with 1 molecule of HC1. Now assume that in 1 liter of NH, under these standard conditions there are 30,700,000,000,000,000,000,000 mole- cules of NH3. Then since each molecule of NH., unites with 1 molecule of HC1, it follows that 1 liter of HC1 under same conditions must contain exactly the same number of molecules as 1 liter of NH, namely, 30,700,000,000,000,000,000,000. (2) Consider the action of Hydrogen on Chlorine, as represented by the equation H2 -f- Cl2 = 2HC1. (The equa- tion is based on facts). It is an experimental fact that each 1 liter of Hydrogen CHEMICAL CALCULATIONS 55 unites with exactly I liter of Chlorine. We know from the action which takes place and the product formed, that each 1 molecule of Hydrogen united with 1 molecule of chlorine. This is of course represented in the chemical equation. Assume that 1 liter of Hydrogen under standard condi- tions contains 30,700,000,000,000,000,000,000 molecules, then since we know that each 1 molecule of Hydrogen unites with 1 molecule of chlorine, it follows that 1 liter of chlorine under the same conditions must contain the same number of molecules as 1 liter of Hydrogen, 30,700,000,000,000,000,- 000,000. Since the days of Avogadro we have been able to to talk in terms like this. Avogadro stated that EQUAL VOLUMES OF GASES CONTAIN THE SAME NUMBER OF MOLECULES WHEN BOTH GASES ARE AT THE SAME TEMPERATURE AND BOTH SUBJECTED TO THE SAME PRESSURE. This as- sumption was made by Avogadro in 181 1. It is known asAvo- gadro's Hypothesis. If we accept it we can see, as was brought out in 2 examples taken, why gases must unite in simple propor- tions by Volume-1 liter of one gas with 1 liter of another; or 1 liter of one gas with 2 liters of another; or 1 liter with 3 liters; or 1 with 4; 2 with 1 ; 2 with 2; 2 with 3, etc. Why 1 liter of a gas cannot unite with say 0.442 liters or 2.717 liters of another gas. Assume that 1 liter of every gas contains the same number of molecules-30,700,000,000,000,000,- 000,000-when at the same temperature and subjected to the same pressure. Then if 1 liter of a gas united with 0.442 liter of an- other, 30,700,000,000,000,000,000,000 molecules of the one gas would unite with 0.442 X 30,700,000,000,000,000,- 000,000 molecules of the other, i. e. each 1 molecule of the one gas would unite with 0.442 molecule of the other. But this cannot be the case. The MOLECULES can only unite in simple proportions, 1 molecule with 1 molecule or 1 mole- cule with 2 molecules, or 1 with 3; 2 with 1 ; 2 with 2 ; 2 with 3; not one molecule with 0.442 molecule. 0.442 molecule 442 cannot exist. We cannot have of a molecule. The 1000 molecule can be split up into the atoms which compose 56 CHEMICAL CALCULATIONS it, but no further. We see then that the simple assump- tion that equal volumes of gases contain equal numbers of molecules helps us to explain why the volumes that unite are in simple proportion. The molecules unite in simple pro- portions and the volumes that react must also be in this same simple proportion. The Hypothesis is entirely in accord with the enormous mass of experimental data on gases. It is there- fore universally accepted by chemists. The number of mole- cules in I liter of any gas at standard conditions has been esti- mated in several reliable ways. This number is approximately 30,700,000,000,000,000,000,000 -i. e. 307 X 1020. It is well here to try to distinguish clearly between an atom and a molecule. The atom as we know is the smallest particle of an element which retains the properties of the ele- ment. It cannot be further subdivided by any known chem- ical means. When 2 or more atoms combine (chemically) they form a larger particle which we call a MOLECULE. The atoms which combine may be different,-thus 1 atom of Carbon unites with 2 atoms of Oxygen and forms a Molecule of Car- bon Dioxide. This is represented by the equation C O, == CO2. The atoms which unite may be atoms of the same ele- ment, thus 1 atom of Oxygen unites with another atom of Oxygen and forms one molecule of Oxygen-this is repre- sented by the equation O - O = 0-2. There are 2 atoms of Oxygen in 1 molecule of Oxygen. The molecule of each of the elementary gases Hydrogen, Nitrogen and Chlorine, con- tains 2 atoms. Let us see how we know this in the cases of H, Cl and O. In the case of some elements the atom and the molecule appear to be identical. The molecule of the elements contain 1 atom. This is true for the elements Argon and Helium, which are gaseous at ordinary temp at which we live, and for the vapor of Hg, an element which is liquid at the temperature at which we live. We do not know how many atoms there are in 1 molecule of most of the elements which are solids at the temperature at which we live. Proof that the molecules of Hydrogen, Chlorine and Oxygen contain 2 atoms. We know from experiment that 1 part by volume of Hydrogen unites with 1 part by volume CHEMICAL CALCULATIONS 57 of Chlorine, and forms 2 parts by volume of Hydrochloric Acid Gas. Let us take 1 liter as the part by volume, then assume that 1 liter of any gas contains 307 X 1 O20 molecules at standard conditions. (Avogadro's Hypothesis). H + Cl i=2 HC1 2 2 Hydrogen + Chlorine = Hydrochloric Acid 1 liter + 1 liter = 2 liters (307 X 1020) molecules + (307 X 1020) molecules = 2(307 X 1020) molecules The liter of Hydrogen contains 307 X 1 02° molecules. The liter of Chlorine contains 307 X 1 O20 molecules. Each of the 2 liters of HC1 contains 307 X 1 O20 molecules. We notice that there are 2 (307 X 1 O20) molecules of HC1. Now each of these must contain at least one ATOM of Hydro- gen. We cannot have a half or other fraction of an atom, therefore in the 2(307 X 1 O20) molecules of HC1 there are at least 2(307 X IO20) ATOMS of Hydrogen. But we had 307 X 1020 MOLECULES of Hydrogen before combination with Chlorine took place, i.e. 307 X1020 molecules of Hydrogen furnished 2(307 X 1 O20) ATOMS of Hydrogen. 1 molecule of Hydro- gen contains 2 atoms of Hydrogen. We have proved that 1 molecule of Hydrogen contains at least 2 atoms. Since in no known case does 1 molecule of Hydrogen produce more than 2 atoms of H, we conclude that 1 molecule of Hydrogen con- tains 2 atoms. This conclusion is confirmed by other chemi- cal data. The student should follow the same line of reason- ing for Chlorine instead of Hydrogen, and thus show that the Chlorine molecule contains 2 atoms. We know from experiment that 2 parts by volume of Hydrogen unite with 1 part by volume of oxygen and form 2 parts by volume of water vapor. Let us take 1 liter as the part by volume. Then, by Avogadro's Hypothesis, 1 liter of any gas contains 30,700,000,000,000,000,000,000 molecules un- der standard conditions of temperature and pressure. 2H2 + O2 = 2H2O Hydrogen Oxygen Water Vapor 2 liters -|- 1 liter = 2 liters 2(307 x IO20) + 307 X 1 O20 = 2 (307 X IO20) molecules 58 CHEMICAL CALCULATIONS Each of the 2 liters of Hydrogen contains 307 X 1020 molecules. The liter of Oxygen contains 307 X IO20 molecules. Each of the 2 liters of Water Vapor contains 307 X IO20 molecules. We notice that there are 2 (307 X 1020) molecules of Water Vapor. Now each of these must contain at least 1 atom of Oxygen, i. e. there are 2(307 X 1 02 ) atoms of Oxygen. 30 7 X 10 molcules of Oxygen furnished 2(307 X 1 020) atoms of Oxygen. I molecule of Oxygen contains 2 atoms of Oxygen. This really proves that 1 molecule of Oxygen contains at least 2 atoms. Since, however, we know of no cases in which the molecule of Oxygen is divided into more than 2 parts, we conclude that the molecule of Oxygen contains 2 atoms of Oxygen. This conclusion is confirmed by other data. The weight of one atom (atomic weight) of Oxygen is I 6.00. There are 2 atoms of Oxygen in 1 molecule of Oxygen. 1 he weight of 1 molecule (molecular weight) of Oxygen is then 2 X 1 6.00 = 32.00. Similarly since the molecules of Hydrogen, Chlorine and Nitrogen each contain 2 atoms, the weight of 1 molecule of Hydrogen is 2 X 1-008 = 2.016 weight of 1 molecule of Chlorine is 2 X 35.46 = 70.92 weight of 1 molecule of Nitrogen is 2 X 14.00 = 28.02 Now we find by experiment that 1 liter of Oxygen weighs 1.429 grams 1 liter of Hydrogen weighs 0.08987 grams 1 liter of Chlorine weighs 3.22 grams 1 liter of Nitrogen weighs 1.2505 grams at standard conditions- O°C and 7 60 mm pressure If we take THAT number of grams of each of these gases, which is numerically equal to the molecular weight, we find that (under like conditions of temperature and pressure) the vol- ume this weight occupies is the same in each case. Thus-at standard conditions, 1 liter of Oxygen weighs 1.429 grams, or, stating this backwards, 1.429 grams of Oxygen occupy 1 liter. 1 1 gram of Oxygen occupies X 1 =0.7 liter. 1.429 CHEMICAL CALCULATIONS 59 32.00 grams of Oxygen occupy 32.00 X 0.7 ~ 22.40 liters. Similarly: 0.0897 grams of Hydrogen occupy 1 liter. 1.0 1.0 gram of Hydrogen occupies = 11.14 liter 0.0897 2.016 grams of Hydrogen occupy 2.016 = 22.43 liters. 70.92 Chlorine, 70.92 grams occupy X = 22.02 liters. 3.22 28.02 Nitrogen, 28.02 grams occupy = 22.41 liters. 1.250 44.00 Carbon Dioxide, = 22.25 liters. 1.977 The average of these is approximately 22.3 liters. We have deduced a remarkable principle:-That number of grams of a gas, which is numerically equal to the molecular weight, occupies 22.3 liters at standard conditions, (0°C and 760 mm pressure). At other temperatures and pressures the volume occupied would of course not be 22.3 liters. We will now apply these principles (Gay Lussac's Law, Avogadro's Hypothesis) to the solution of some problems. 1. 2 750 cc of Sulphur Dioxide gas at 0 X 760 weigh 7.87 grams. Calculate the molecular weight of Sulphur Dioxide. 2 750 cc of Sulphur Dioxide Gas weigh 7.87 grams. 1 1 cc of Sulphur Dioxide Gas weighs X grams = 0.002863 grams. 2750 1000 1 000 cc. ( 1 liter) of Sulphur Dioxide Gas weigh X 0.002863 - 2.863 grams. 2750 22.3 liters of Sulphur Dioxide Gas weigh 22.3 X 2.863 = 64.06 grams. But we know that that number of grams of any gas which is numerically equal to the molecular weight, occupies 22.3 liters at standard conditions. 60 CHEMICAL CALCULATIONS 64.04 grams Sulphur Dioxide occupies 22.3 liters at standard conditions. .'. 64.06 is the Molecular Weight of Sulphur Dioxide. PROBLEMS. 1. One liter of a gas (standard conditions) weighed 1.98 grams, 1 liter of hydrogen (standard conditions) weighs 0.0897 grams. Calculate the molecular weight of the gas. 2. The molecular weight of carbon dioxide is 44. Cal- culate the weight of 1 liter of this gas. 3. Calculate the weight of 1 liter of hydrogen sulphide at standard conditions. What, then, is the absolute density of hydrogen sulphide? 4. Calculate the absolute density of nitrous oxide gas. 5. Calculate the absolute density of ammonia gas. Cal- culate the relative density of ammonia gas, compared to hydrogen. 6. It is an experimental fact that 1 liter of nitrogen will unite with 3 liters of hydrogen and produce 2 liters of am- monia gas. Starting with the premise that the nitrogen mole- cule contains 2 atoms of nitrogen, prove that the formula of ammonia is NH... 7. Eight grams of oxygen occupy what volume at 20 lC and 700 mm pressure. 8. 8000 cc of a gas, measured at 20 °C and 740 mm pressure weighed 14 grams. Calculate the molecular weight of the gas. 9. 0.062 gram of a gas occupies 25 cc at 100cC and 760 mm. Calculate the molecular weight of this substance. 10. What is the molecular weight of mercuric chloride, the vapor density of which is 1 34.68. 11. 1 20 cc of water vapor weigh 0.056 g. at 1 80cC and 740 mm. Show from this data that the molecular weight of water is 1 8.0 1 6. CHEMICAL CALCULATIONS 61 I 2. The weight of 3840 cc of a certain vapor at standard conditions is 24 grams. Calculate the molecular weight of the gas. 13 Eight grams of oxygen are mixed with 10.08 grams of hydrogen. Both gases were measured at standard condi- tions. Calculate (1 ) the volume occupied by the mixture (2) the absolute density of the mixture (3) the relative density of the mixture. 14. 22 grams of CO_, were mixed with 20.16 grams of hydrogen. Both gases were measured at standard conditions. Calculate (1) the volume occupied by the mixture (2) the absolute density of the mixture (3) the relative density of the mixture. 15 What is the relative density of hydrogen chloride (molecular weight 36.46) referred to each of the following: (a) Hydrogen. (b) Oxygen. (c) Chlorine. 16. 14.01 g. of nitrogen were mixed with 35.46 g. of chlorine. Both gases were measured at standard conditions. Calculate (1) the volume occupied by the mixture (2) the absolute density of the mixture (3) the relative density of the mixture. 1 7. Calculate the weight of oxygen which will occupy the same volume as 1.008 grams of hydrogen. Both at standard conditions. 18. 2500 cc of a gas measured over water at 30cC and under barometric pressure of 750 mm weigh 5 grams. Calcu- late the molecular weight of the gas. 19. The absolute density of a certain gas is 2.5. Calcu- late the molecular weight of the gas. 20. The relative density of a certain gas is 20. Calculate (1) the absolute density of the gas (2) the molecular weight of the gas. 21. It is an experimental fact that 1 liter of nitrogen unites with 1 liter of oxygen and produces 2 liters of nitric oxide gas. Starting with the premise that the nitrogen mole- cule contains 2 atoms show that the formula of nitric oxide is NO. CHAPTER VI OUTLINE 32.685 g. of Zn = chemically 9.033 g. of Al = 23.00 g. of Na = 2 7.92 g. of Fe, for each of these quantities will lib- erate the same weight of hydrogen (1.008 g.) from acids. These weights are chemically equivalent. In water each 8 g. of oxygen are united with 1.008 g. of hydrogen. .'. 8 g of oxygen are equal chemically to 1.008 g of hydrogen. 8.00 parts by weight of oxygen is used as the basis of combining weights. That weight of an element which com- bines with 8.00 parts by weight of oxygen is the combining weight of the element. Chemical combination takes place of course between the atoms. Examples: Combining weights of ( 1 ) Mg from MgO (2) Al from ALO.. (3) Cl from MgCL. Combining weights of compounds-8.00 parts by weight of oxygen is also the basis for the combining weights of com- pounds. Examples: Combining weights of ( 1 ) an Acid (2) a Base (3) a Salt (4) an Oxidizing Agent. 62 CHEMICAL CALCULATIONS 63 COMBINING WEIGHTS-CHEMICAL EQUIVALENTS. Most of us have observed in the laboratory that many of the metals dissolve in acids and that Hydrogen is liberated when they dissolve. By dissolving weighed amounts of the various metals in acids, and weighing the Hydrogen evolved, it has been found that for example: 32.685 grams of Zinc liberate 1.008 grams of Hydrogen 9.033 grams of Aluminium liberate 1.008 grams of Hydrogen 23.00 grams of Sodium liberate 1.008 grams of Hydrogen 2 7.92 grams of Iron liberate 1.008 grams of Hydrogen To liberate 1.008 grams of Hydrogen from acids, we see that 32.68 grams of zinc are necessary, or if we used sodium instead of zinc 23.00 grams of sodium would be necessary. W e would decide from this that 32.68 grams of zinc is equiva- lent in its chemical action to 23.00 grams of sodium. Like- wise we would need 9.033 grams of aluminium or 2 7.92 grams of iron to liberate 1.008 grams of Hydrogen. 9.033 is the number of grams of aluminium which is equivalent in its chem- ical action to 2 7.92 grams of iron or to 32.685 grams of zinc or to 23.00 grams of sodium or to 1.008 grams of Hydrogen. These weights of these elements are equivalent to each other chemically. If we pass an electric current through water, the water is split up into hydrogen and cxygen, the elements which compose it. The hydrogen collects at the cathode; the oxygen at the anode. If we weigh the quantity of gas evolved at each elec- trode, we find that for each 1.008 grams of hydrogen which collect at the cathode, 8.00 grams of oxygen collect at the anode. We decide at once that in water each 1.008 of hydro- gen are united with 8.00 grams of oxygen. This is the pro- portion in which they combined. 8.00 grams of oxygen is, therefore, chemically equivalent to 1.008 grams of hydrogen. 8.00 grams of oxygen must therefore be chemically equivalent to 32.685 grams of zinc or to 9.033 grams of aluminum or to 23.00 grams of sodium or to 2 7.92 grams of iron, i.e. 32.685 grams of Zn = 1.008 g. Hydrogen = 8.00 g. Oxygen 9.033 grams of Al = 1.008 g. Hydrogen = 8.00 g. Oxygen 64 CHEMICAL CALCULATIONS 23.00 grams of Na = 1.008 g. Hydrogen = 8.00 g. Oxygen 2 7.92 grams of Fe = 1.008 g. Hydrogen = 8.00 g. Oxygen We would expect then if Zn or A 1 or Na or Fe combined with Oxygen that they would combine in the proportions 32.685 grams of Zn with 8.00 grams of Oxygen 9.033 grams of Al with 8.00 grams of Oxygen 23.00 grams of Na with 8.00 grams of Oxygen 2 7.92 grams of Fe with 8.00 grams of Oxygen The results of repeated analyses confirm our expectations and show them to be true. 8.00 parts by weight of Oxygen is chosen as the basis for comparison of combining weights or chemical equivalents. THAT WEIGHT OF AN ELEMENT WHICH WILL COM- BINE WITH 8.00 PARTS BY WEIGHT OF OXYGEN IS KNOWN AS THE COMBINING WEIGHT OF THE ELE- MENT. The parts by weight may be grams, ounces, pounds, tons or any other units of weight. The combining weight is frequently called the Equivalent weight or the Chemical Equivalent of the element. Some other element might be used as the basis for com- bining weights, but Oxygen is more suitable. Hydrogen was formerly used as the basis of combining weights and that weight of an element which either combined with or replaced 1 part by weight of Hydrogen, was taken as the combining weight. Hydrogen, however, combines with only a few ele- ments, while Oxygen combines with all the elements except Fluorine and the rare gases of the atmosphere. Oxygen is therefore a much more suitable basis for combining weights. When the same elements unite in more than one propor- tion to form different compounds the weights of each element in the different compounds will either be the combining weights or simple multiples of the combining weight; for ex- ample, the combining weight of Chlorine is 35.46. Mercurous Chloride contains 35.46 parts by weight of Chlorine for each 200.6 parts by weight of Hg. Mercuric Chloride contains 2 X 35.46 parts by weight of Chlorine for each 200.6 parts by weight of Hg. The weights of Cl, combined with the same weight of Hg, are in the proportion of 1 to 2 in the two compounds. CHEMICAL CALCULATIONS 65 We must remember that the combination of elements is through the atoms. The atoms combine in simple proportions. The weight of each atom is constant, so the weight of the elements which combine must be proportional to the atomic weights, i. e. the combining weights must be proportional to the atomic weights. This can be seen from examples like the following. I. When Magnesium and Oxygen unite they form Mag- nesium Oxide, MgO; each 1 atom of Mg unites with 1 atom of Oxygen, to form 1 molecule of MgO. Each million atoms of Mg unite with 1 million atoms of Oxygen and form 1 mil- lion molecules of MgO. The weight of 1 atom of Mg is 24.32 The weight of 1 atom of O is 1 6.00 Therefore, no matter how many million atoms of Mg unite with exactly the same number of atoms of O, to form molecules of MgO, the proportions in which they unite will always be 24.32 parts by weight of Mg with 1 6.00 parts by weight of O, or 12.16 parts by weight of Mg with 8.00 parts by weight of O. 12.16 is therefore the combinirg weight of Mg. It is worth notice in passing that the combining weight of Mg is '/2 of the atomic weight of Mg. II. If we know the number of atoms of each element which combine to form the molecule, we can readily calculate or derive the combining weight. For example, we know that each 1 molecule of Aluminium Oxide contains 2 atoms of Al and 3 atoms of O. This is represented by the formula A12O3. The weight of 2 atoms of Al is 2 X 2 7.1 - 54.2 The weight of 3 atoms of O is 3 X 1 6,00 = 48.00 Therefore when aluminium and oxygen unite to form aluminium oxide they will do so in the proportion. 48.00 parts by weight of O with 54.2 parts by weight of Al. 54.2 1.00 part by weight of O with parts by weight of Al. 48.00 54.2 8.00 parts by weight of O with 8 X - 9.033 parts by weight of Al. 48.00 9.033 is therefore the combining weight of Aluminium. 66 CHEMICAL CALCULATIONS It is worth notice in passing, that the combining weight of Al is '/3 of the atomic weight of Al. III. Determine the combining weight of Chlorine frcm the following data: Magnesium Chloride has the composition Mg = 25.535 % Cl = 74.464%. From example No. 1 we know that the combining weight of Mg is 12.16 because 12.16 parts by weight of it unite with 8.00 parts by weight of O. That weight of chlorine which unites with 12.16 parts by weight of Mg will be the combining weight of chlorine. In magnesium chloride Mg and Cl are found by analysis to be combined in the proportions of 25.5 35 parts by weight of Mg with 74.464 parts by weight of Cl 74.464 1 part by weight of Mg with = 2.916 parts by weight of Cl. 25.535 12.16 parts by weight of Mg with 12.16 X 2.916 = 35.46 parts by weight of Cl. 35.46 is therefore the combining weight of Cl. It is worth notice in passing that the combining weight of Cl is the same as the atomic weight of Cl. The combining weight of an element can thus be deter- mined from a compound of the element with some other ele- ment whose combining weight we know. It is not necessary to always End the weight of the element which is chemically equiv- alent to 8.00 grams of Oxygen. COMBINING WEIGHTS CHEMICAL EQUIVALENTS OF COMPOUNDS. The principle of combining weights or chemical equiva- lents may be extended from elements to compounds thus Experiment shows that 23.00 grams of Sodium will react on water and form 40.008 grams of NaOH, as represented by the equation Na + HOH = NaOH + H 23.00 18.016 = 40.008 1.008 grams further these 40.008 grams of NaOH will exactly neutralize 36.468 grams of HC1 40.008 grams of NaOH will exactly neutralize 49.038 grams of H SO 2 4 40.008 grams of NaOH will exactly neutralize 22.00 grams of CO 2 CHEMICAL CALCULATIONS 67 We would say at once that 40.008 grams of NaOH are chemically equal to 36.468 grams of HC1 or to 49.038 grams of H2SO4 or to 22.00 grams of CO2 or to 23.00 grams of Sodium. As the basis we again adopt 8.00 parts by weight of O or 1.008 parts by weight of H. That weight of a com- pound which is chemically equal to 8.00 parts by weight of O or 1.008 parts by weight of H is the combining weight or chemical equivalent of the compound. (a) Combining weight or chemical equivalent of acids. That weight of an acid which contains 1.008 parts by weight of (Ionizable) H is of course the combining weight. Since 1.008 parts by weight of H is chemically equal to 8.00 parts by weight of O this weight of the acid will be equal to 8.00 parts by weight of O. Example: Sulphuric Acid H2SO4. I molecule contains 2 atoms of H = 2 X 1.008 parts by weight I molecule contains 1 atom of S = I X 32.06 parts by weight 1 molecule contains 4 atoms of O 4X1 6.00 parts by weight 1 molecule contains 98.076 parts by weight There are 2 X 1-008 parts by weight of H in 98.076 parts by weight of H.,SO4 98.076 There are 1.008 parts by weight of H in = 49.038 parts by weight of H2SO4 2 Therefore 49.038 is the combining weight of H2SO4. (b) Combining weight or Chemical Equivalent of Bases. That weight of a base which contains 1 7.008 parts by weight of (OH) is the combining weight or chemical equiva- lent. The reason for this can be seen by considering water H2O or HOH. We consider that 1 molecule of water contains 1 atom of H and 1 (OH) group. 1.008 parts by weight of H. 16.00 4- 1.008 = 17.008 parts by weight of OH. Therefore 1 7.008 parts by weight of OH is chemically equal to 8 parts by weight of O. Example:-Find the combining weight of Fe(OH)3. 1 molecule of Fe(OH)3 contains 68 CHEMICAL CALCULATIONS 1 atom of Fe, weight 1 X 55.84 = 55.84 3 (OH) groups, weight 3 X 17.008 = 51.024 Weight of 1 molecule - - - - 106.864 There are 2 X (17.008) parts by weight of OH in 106.864 parts by weight of Fe(OH)3. 106.864 There are 1 7.008 parts by weight of OH in - 3 35.621 parts by weight of Fe(OH)3. . : 35.621 is the combining weight of Fe(OH)3. (c) Combining weight of Salts. The combining weights of Salts can be determined from the weight of metal present. Find the combining weight of the metal of the Salt (that weight of the metal which is chem- ically equal to 8.00 g. of O). That weight of the Salt which contains this determined weight of the metal is the combining weight or chemical equivalent. Example:-Find the combining weight of Cr2(SO4)3. In Cr2 (SOj)3 the Cr exhibits a valence of 3. The com- pound can be considered as Cr2O;!.33SO3, i. e. made up of 2 oxides:-one basic, the other acidic, considering Cr2O3. 3X16= 48 parts by weight of O are combined with 2x52 = 1 04 parts by weight of Cr. 1 1 part by weight of O is combined with- X 104 = 2.166 parts by weight of Cr. 48 8 8 parts by weight of O are combined with - X 1 04 = 1 7.33 parts by weight of Cr. 48 1 7.33 is then the combining weight of Cr. That weight of Cr2(SO4)3 which contains 1 7.33 g. of Cr will be the combining weight of Cr2(SO4)3. Cr2(SO4)3. (2 X 52) + (3 X 32.06) + (12 X 16) = 392.18. There are 1 04 parts by weight of Cr in 392.1 8 parts by weight of Cr2(SO4)3. 392.18 There is 1 part by weight of Cr in parts by weight of Cr2(SO4)3. 104 392.18 There are 1 7.33 parts by weight of Cr in 1 7.33 X - 65.35 parts by weight of Cr2(SO4)3. 104 CHEMICAL CALCULATIONS 69 65.35 is then the combining weight of Cr2(SO4):i. (d) Combining weights of Oxidizing Agents. That weight of the oxidizing agent which will furnish 8.00 grams of Oxygen. Example:-Find the combining weight of K2Cr2O7. When K2Cr2O. acts as an oxidizing agent, 1 molecule of it gives 3 atoms of nascent O as indicated by. the equation K2Cr2O7 = K2O ~F Cr2O3 % 30. Weight of 2 atoms of K = 2 X 39.1 0 - 78.20 Weight of 2 atoms of Cr = 2 X 52.00 = 1 04.00 Weight of 7 atoms of O = 7 X 1 6.00 = 1 1 2.00 Weight of 1 molecule of K.2Cr2O7 = 294.20 294.2 parts ty weight of K-'Cr-'O; furnish 3 X 16.00 - 48.00 parts by weight of O. 294.2 parts by weight of K2Cr2O- furnish 1 part by weight of 48 [Oxygen. 294.2 X 8 = 49.033 parts by weight of K2Cr2O7 furnish 8 48 [parts by weight of Oxygen. 49.033 is then the combining weight of K2Cr2O7. PROBLEMS. 1. Silver Oxide contains by analysis Silver 93.097%, Oxygen 6.903%. Calculate the combining weight of Silver. 2. Lead Chloride is found by analysis to contain Lead 74.51 %, Chlorine 25.49%. The combining weight of Cl is 35.46. Calculate the combining weight of Lead. 3. By analysis 2 grams of Ferric Oxide were found to contain 1 39.87 grams of Iron. Calculate the combining weight of Iron. 4. Cupric Bromide was found to contain Cu 28.45%, Br 71.55%. Calculate the chemical equivalent of Cu. The chemical equivalent of Br is 79.92. 5. Five grams of Ag2O were treated with excess HC1. The AgCl formed weighed 6.184 grams. Calculate the chemical equivalent of Chlorine. 70 CHEMICAL CALCULATIONS 6. Two grams of K2O were treated with excess HBr. The KBr formed weighed 5.054 grams. Calculate the chem- ical equivalent of Br. 7. It was found that 63.6 parts by weight of an unknown metal combined with 36.4 parts by weight of S. Find the combining weight of the metal. The combining weight of Sulphur is 1-6.03. 8. 1.560 g. of liquid Chloride of Phosphorus were de- composed by excess water. The resulting liquid, after adding AgNO:i, filtering, washing and drying the precipitated Silver Chloride, yielded 4.881 g. of AgCl. From this data calculate the combining weight of Phosphorus. The combining weights of Silver and Chlorine are, respectively, 107.88 and 35.46. 9. The equivalent of Mercury is 100.3 Calculate the combining weights of Oxygen, Hydrogen, Copper, Sulphur and Chlorine from the following data: Mercuric Oxide con- tains 92.614% of Hg. Cupric Oxide contains 79.892 % of Cu. Cupric Chloride contains 47.267 % of Cu. Hydrogen Sulphide contains 94.08 % of S. Hydrogen Chloride contains 9 7.2 35 % of Cl. 10. 5.00 grams of Zinc, when dissolved in HC1, evolved 1705 cc of Hydrogen at standard conditions. Calculate the chemical equivalent of Zinc. 1 1. 2.0 grams of Aluminium, when dissolved in H2SO4, evolved 2468 of Hydrogen at standard conditions. Calculate the chemical equivalent of Aluminium. 12. Calculate the chemical equivalent of KMnO4 when this substance is used as an oxidizing agent. 1 3. Calculate the chemical equivalent of each of the fol- lowing: H(NO3), Ba(OH)2, K3PO4, KC1O3. 1 4. Calculate the chemical equivalent of each of the fol- lowing: HBr, MnSO4, Na2Cr2O7, Rb(OH). 15. Calculate the chemical equivalent of each of the fol- lowing: HF, A1(OH)3, Ca3(PO4)2, La2O3. 1 6. Calculate the chemical equivalent of each of the fol- lowing: K4[Fe(CN)G], Ba3(PO3)2, MnO2, Cs(OH). CHEMICAL CALCULATIONS 71 1 7. Calculate the chemical equivalent of each of the fol- lowing: H2'O2, O3, PH3, V,O.. 1 8. Calculate the chemical equivalent of each of the fol- lowing: AsH3, Cu2S, K(CN), Sc2O,. 19. Calculate (he chemical equivalent of each of the fol- lowing: Cerium Oxide, Thorium Oxide, Selenic Acid, Beryl- lium (Glucinum) Hydroxide. 20. Calculate the chemical equivalent of each of the fol- lowing: Germanium Dioxide, Titanic Oxide, Auric Hy- droxide, Tellurous Acid. 2 1. Write the formula of each of the following oxides and from these calculate the combining weight of the element in each case: Uranium Oxide, Tungstic Oxide, Cobalt Oxide, Cadmium Oxide, Zirconium Oxide. 22. The molecular weight of Radium Bromide is 385.84. The chemical equivalent of Br is 79.92. Calculate the atomic weight of Radium. 23. The molecular weight of Platinic Chloride is 336.04. The combining weight of Cl is 35.46. Calculate the atomic weight of Pt. 24. Two g. of Zinc were added to a solution of HgCl,. The Zn dissolved. 6.137 g. of Hg was liberated by this action. The chemical equivalent of Zinc is 32.685. Calcu- late the chemical equivalent of Hg11. 25. A piece of Iron which weighed 20 grams was im- mersed in a solution of CuSO4 for some time. After remov- ing it was found to have lost 2.5 grams. 2.846 g. of Copper was liberated by the action which took place. Calculate the chemical equivalent of Cu. The combining weight of Fe is 22.92. CHAPTER VII OUTLINE The chemical equation Equality. Chemical action takes place between molecules in simple com- binations. The equation indicates the number of molecules which react and the number of molecules of products formed. The equation represents the relative weights .of the substances which react and of the products formed. The equation represents the volume of any gas which is one of the reacting substances or of the products. 22.3. Summing up-What the equation represents. The equation represents facts. These facts can be derived from (A) Actual experimental tests. (B) Some general principles or laws. Examples illustrating equation writing using A and B. Solution of a problem. Problems. 72 CHEMICAL CALCULATIONS 73 CHEMICAL EQUATIONS. We are of course familiar with the fact that when two or more substances react chemically to form new ones, there is no gain or loss in weight. The sum of the weights of the prod- ucts is equal to the sum of the weights of the original substances. Thus when carbon burns the carbon is not lost. The carbon has simply combined with a definite weight of oxygen from the air and formed a new substance, carbon dioxide. The weight of carbon dioxide formed is equal to the sum of the weights of the carbon and of the oxygen with which it com- bined. Since this is the case, the chemical action can be rep- resented in the form of EQUALITY-A CHEMICAL EQUA- TION. In this case the action is represented by the equation C + Oj = CO.> symbols being used *to make the impression more vivid. Chemical actions take place by molecules. The mole- cules react with one another in simple combinations-1 mole- cule of one compound with one molecule of another compound or, one molecule of one compound with two molecules of an- other, or one with three, or two with three, or three with two, etc. .'. The EQUATION which represents the action must, of course, indicate the numbers of molecules which react. Thus when sulphuric acid is neutralized with sodium hydroxide, each one molecule of sulphuric acid reacts with two molecules of sodium hydroxide. This is indicated by the equation: HJSOJ + 2 NaOH = Na,(SO4) + 2H2O 1 molecule + 2 molecules = 1 molecule + 2 molecules 98.076 + 2(40.008) = 142.06 + 2(19.016) Now we know the relative weights of the molecules. The weight of the molecules is in each case the sum of the weights of the atoms in it. Thus the weight of the molecule of H.,SO4 is 2 X 1.008 + 32.06 + 4 X 16.00 = 98.076. The weight of the molecule of NaOH is 23.00 + 1 6.00 + 1.008 = 40.008. Knowing then the numbers of molecules which take part in the action and the relative weight of each we can see then that the equation will also indicate the RELATIVE WEIGHTS 74 CHEMICAL CALCULATIONS of the compounds or elements which take part in it, either as reacting substances, or as products formed. It must be remembered that the equation represents only RELATIVE weights, thus: The molecular weight of H2SO4 is not 98.076 grams nor is the molecular weight of NaOH 40.008 grams nor ounces nor pounds nor any other unit. The equation indi- cates that the RELATIVE weights of H2SO4 and NaOH which react in this case are 98.075 and 40.008 i.e. H (SO ) + 2 NaOH = Na (SO ) + 2 H O 2 < 2 4 2 98.076 + 2(40.008) = 142.06 + 2 (18.016) parts by weight-parts by weight = parts by weight-parts by weight where the parts by weight can be any convenient unit, such as grains, grams, ounces, pounds, tons, etc. In laboratory calculations grams and ounces are most frequently used, while operations pounds and tons are most generally used. IF GASES TAKE PART IN THE REACTION. We have learned that in the case of gases, that weight of the gas in grams, which is numerically equal to the molecular weight, occupies approximately 22.3 liters at standard condi- tions. Thus the molecular weight of carbon dioxide, CO2, is 44. 44 grams of carbon dioxide occupy 22.3 liters at standard conditions. If then one of the reacting substances or the prod- ucts formed is a gas, not only can the weight of it be calculated with the aid of the equation but the volume occupied by the gas can also be calculated. Each molecular weight of it will occupy 22.3 liters at standard conditions. SUMMING UP. The Chemical equation represents: I. The substances which react, and the products formed. II. The relative numbers of molecules of the reacting substances and of products formed. III. The relative weights of reacting substances and of products formed. IV. The volume of any gas which is one of the reacting substances or products. CHEMICAL CALCULATIONS 75 A chemical equation represents experimental facts. It is evident that the facts must be known before the equation is written. It cannot be too strongly emphasized that the student should not attempt to write the equation representing a chem- ical action unless he knows that the substances do react, and knows the nature of the products formed. In some cases it is necessary to know the actual products formed. In other cases the general nature of the products is sufficient (the composition of the products can be deduced from some generalization). To illustrate these two cases: When potassium chlorate is heated oxygen is given off. This can be shown by means of a glowing splinter. We can indicate the action to this extent as an equation KCIOb = o + ? The equation cannot be completed to represent the ex- perimental facts unless we show what the other product is KC1O, KC1O or KC1. Thus three different equations are pos- sible. KC1O3 = O + KC1O, KC1O, - 20 + KC1O KCIO3 = 30 + KC1 If, when the equation has been completed, we test the residue and find that it is KC1, we are then in a position to complete and balance the equation. It is only necessary to keep in mind the fact that we are representing an equality. There must be the same number of atoms of each element before and after the action. They will be united in new com- binations, but none are lost. The equation is simply a graphic illustration of the principle of conservation of matter. To consider the other case-where the action can be inferred from general principles-consider the neutralization of an acid by a base. The equation can be completed from a general principle that when an acid neutralizes a base, the prod- ucts formed are water and a salt. In general the composition of the salt is inferred from the nature of the acid and the base. Thus if HC1 acts on NaOH, we are sure that the action is rep- resented by the equation HC1 + NaOH = H2O + NaCl. 76 CHEMICAL CALCULATIONS It is not necessary to show by tests that the salt is NaCl. Further, if any hypothetical acid H2R neutralizes a base M(OH) we know that the action will be correctly represented by the equation 2MOH + H2R = 2H2O + M2R. Similarly there are other types of chemical action, such as the action of an acid on a metal, or of an oxidizing agent on a reducing agent, etc., which, in many cases, allow us to write the equation representing the facts from general prin- ciples rather than from information on the actual example at hand. It is profitable to take advantage of a generalization wherever possible, and thus systematize the process of writing equations, but it must be clearly understood that unless the facts (either general or specific) are known, no attempt should be made to express the action in the form of an equation. Let us now consider a problem and see how information can be obtained by calculation, from an equation. Problem. Twenty grams of Aluminium were treated with sulphuric acid. Calculate (a) the weight of H2SO4 necessary to react with the 20 grams of Al; (b) the weight of Aluminium Sul- phate formed; (c) the volume of Hydrogen liberated (stand- ard conditions). The equation representing the action can be written from the general principle that when an acid acts on a metal Hy- drogen is evolved, and a salt is formed. The salt in the case will be the Aluminium salt of sulphuric acid, aluminium sul- phate. 2A1 + 3H2(SO4) = A12(SO4)3 + 3H2 2(27.1) + 3(2.016 + 32.06 + 64) = (54.2 + 96.18 + 192) + 6.048 54.2 + 294,228 = 342.38 + 6.048 parts by weight, such as grams, ounces, tons, etc. The equation shows that 54.2 g. of Al require 294.228 g. of H.,SO4 1 1 g. of Al requires X 294.228 g. of H.,SO4 54.2 CHEMICAL CALCULATIONS 77 1 20 g. of Al require 20 X X 294.228 g. of H,SO. = 108.5 7 g. of H2SO4 54.2 (b) 54.2 g. of Al form 342.38 g. of A12(SO4)3 1 1 g. of Al forms X 342.38 g. of Al.,(SOj).t 54.2 1 20 g. of Al form 20 X X 342.38 = 126.33 g. of A12(SO4)3 54.2 (c) One molecular weight of Hydrogen, H2, in grams, will occupy 22.3 liters at standard conditions. If taken in grams, 3 molecular weights 3H2 will occupy 3 X 22.3 liters = 66.9 liters. The equation then indicates that 54.2 g. of Al form 66.9 liters of Hydrogen (standard condi- tions) . 1 1 g. of Al forms X 66.9 liters of Hydrogen. 54.2 1 20 g. of Al forms 20 X X 66.9 = 24.68 liters of Hy- 54.2 drogen at standard conditions. PROBLEMS. 1. A solution contains 20 grams of Cu SO4. H2S is passed into this until precipitation is completed. Calculate the weight of CuS precipitated. 2. What weight of BaSOi is precipitated by the action of excess NasSO+ on a solution containing 10 grams of BaCls ? What weight of sodium chloride remains in solution? 3. 20 grams of Na2SO4 are added to a solution containing 25 grams of BaCh. What weight of BaSOi is precipitated ? 78 CHEMICAL CALCULATIONS 4. A dime (weight 2.485 g., contains Ag = 92.5%, Cu - 7.5%) is dissolved in HNO3. Excess NaCl is added. Cal- culate the weight of AgCl precipitated. 5. How much iron ore containing 50% Fe2O3 is neces- sary to produce 1 ton of iron? 6. Prussian blue may be made by adding IGFe (CN)« solution to FeCl3. Calculate the weight of FeCl3 necessary for the production of 10 Tbs. of Prussian Blue. 7. A solution of CaCl, is added to a solution which con- tains 25 grams of K2CO3. Calculate ( 1 ) the weight of CaCO, precipitated. (2) the weight of KC1 left in the solution. 8. 20 grams of Aluminium are treated with excess H2SO4. Calculate (1 ) the weight. (2) the volume at standard condi- tions of the Hydrogen liberated. 9. A balloon has a capacity of 5000 liters. Calculate the weight of HC1 necessary to produce enough Hydrogen to fill it by action of the HC1 on some metal. 10. What weight of FeO is made by roasting 1 ton of Copperas, FeSO4.5H2O? 11. A red pigment consisting of oxides of iron and of CaSO4 is made by roasting a mixture of FeSO4.5H2O and Ca(OH)2. The finished pigment must not contain free lime. It is desirable that no SO, should be given off. Calculate the proper proportions of Ca(OH)2 and FeSO4.5H2O to use. 12. The use of Thermite in welding depends on the action of Al on Fe,O3, the Al being oxidized to A12O3. The Fe2O., reduced to Fe. Calculate the theoretical proportion of powdered Al and Fe2O3 in the mixture. 13. 50 grams of NaCl are treated with excess H.'SOi. Cal- culate the volume of HC1 evolved (standard conditions). 14. 2000 cc of Hydrogen measured at 30°C and 700 mm pressure was evolved by the action of an acid on a quantity Fe. Calculate the weight of the Fe. 1 5. What weight of NH4NO3 must be heated in order to furnish 10 liters of laughing gas, N2O? (standard conditions). CHEMICAL CALCULATIONS 79 1 6. What weight of HNO., is necessary to prepare 5 liters of Nitric oxide (standard conditions) by action on Cu? 1 7. What weight of Ca(OH)2 is necessary to neutralize 1 0 grams of H2SO4? 18. What weight of H2SO4 is necessary to neutralize 20 grams of KOH? What would be formed if half of this quan- tity of H2SO4 were used? 1 9. It is desired to precipitate the lime from a hard water by means of Na2CO3. How much Na2CO3 is necessary to purify 100 gallons of water which contain 0.5 grs. of CaSO4 per gallon? 20. HgO costs 20c. per oz. KC1 costs 8c. per oz. Cal- culate the relative costs to prepare a definite amount of oxygen from each of these materials. 21. A sample of hard water contains 0.2 g. of CaSO4 and 0.25 g. of MgSO4 per gallon. Calculate the weight of Na2CO3 necessary to precipitate the Ca and Mg from 100 gallons of water. CHAPTER VIII OUTLINE Chemical Formulas are derived from the results cf Analysis. Chemical Formulas represent weights. The weight of an element present, divided by the weight of one atom of that element, gives the number of atoms of that element present. The Empirical formula represents the RELATIVE num- ber of atoms of each element present. The Molecular formula represents the ACTUAL number of atoms of each element present. 1 o determine the Molecular formula it is necessary to know the Molecular weight of the compound or to have other data of similar nature. Formulas of minerals. 80 CHEMICAL CALCULATIONS 81 DERIVATION OF CHEMICAL FORMULAS. The composition of a compound, is, of course, derived from experiment, from analysis, from breaking up the com- pound into its parts. The chemical formula of the compound stands for or represents this composition. The chemical for- mula is, of course, derived from the analysis. Suppose, for example, that an unknown substance was given to you and you were asked to determine what it was. A qualitative analy- sis of the substance would first be necessary to determine which elements were present. Suppose that you found by a quali- tative analysis that lead and chlorine were present. No others. You would then determine how much of each was present (determine the proportions of each present). You would make a quantitative analysis of the compound. We will as- sume some figures. Suppose you found as the result of your quantitative analysis of the compound that each 1 gram of it contained lead-0.7450 gram chlorine-0.2550 gram This would give a much more complete knowledge of the compound than the qualitative analysis alone. It would not be final. You would probably express the result of your quantitative analysis in terms of per- centage and say that the compound contains Pb 74.50% Cl 25.50% Most chemists express the results of their analysis in this way. They state the percentage of each element present. Now from this how can we derive or deduce the formula which represents the composition? It must be clearly recognized that the chemical name expresses the composition just as com- pletely as does the formula. The formula, however, visual- izes the composition more clearly. The name Sodium Sul- phate. for example, means a definite substance whose composi- tion has been established by experiment. The formula Na2SO4 does not do any more, but we can recognize at once the rela- tive weights present by inspection of the formula. While if we relied on the name alone, it would be neces- sary to refer constantly to the results obtained by the analysis 82 CHEMICAL CALCULATIONS or else to carry them in memory. 1 he formula is derived from the analysis. It is a convenient way to visualize the results of the analysis in terms of atoms whose relative weights we know. A formula is valuable because the symbols in it stand not only for a certain element but also for a certain weight of that element. Thus S stands not only for S but for 32.06 parts of weight of S. O not only for Oxygen but for 16.00 parts by weight of O, the parts by weight being relative. Now how can we deduce the formula from the analysis, as expressed in percentage. Let us consider 100 parts by weight of the compound, say 1 00 grams. By analysis we found that this contained Pb-74.50 g. Cl-25.50 g. Now the weight of I atom of Pb is 207.2 Now the weight of 1 atom of Cl is 35.46 We have 74.50 parts by weight of Pb. The weight of 1 atom of Pb is 207.2. How many atoms of lead are there- logically we say 74.50 - 0.3595 atoms of Pb. 207.2 Likewise; there aref25.5O grs. of Cl. The weight of 1 atom of Cl is 35.46. How many atoms of Cl are present? Logically it follows: 25.50 0. 7 1 9 1 atoms of Cl. 35.46 Now these are only relative numbers. The weight of 1 atom of Pb is not 207.20 grams nor ounces nor pounds; neither is the weight of an atom of Cl 35.46 g, etc. We have learne 1 that the actual weight of an atom of either of these elements is extremely small. Too small to be determined by actually weighing on a balance. The atomic weights are relative num- bers. When we say that the atomic weight of Pb is 207.2 and that of Cl 35.46 we mean relatively. We mean that the lead 207.2 atom is as heavy as the chlorine atom. The numbers we 35.46 CHEMICAL CALCULA 101 83 get then, 0.3595 and 0.7191, do not represent the actual num- ber of atoms of each element present. They represent the RELATIVE numbers. They mean that for each 0.3595 atoms of Pb in this compound there are 0. 7 1 9 1 atoms of Cl. Further -we know that fractions of atoms do not exist. Atoms can- not be divided into parts, so it will be well to at once calculate from these, the relative number of atoms present, as small whole numbers. For each 0.3595 atoms of Pb there are 0.7191 atoms of Cl. For each 1 atom of Pb there are 0.7191 - 2 atoms of Cl. 0.3595 We can then say that the formula is Pb! Cl2 or simply PbCL. This is the EMPIRICAL formula. It represents the RELATIVE number of atoms present. The figures we have derived do not tell us that the formula is Pb Cl2 and not Pb2Cl4 or Pb.,Cl(! or Pb4Cl8, etc. If we wish to know the actual number of atoms of each eiement present in 1 molecule of the compound, it is necessary to have more (experimental) information about the compound. It is necessary to know its molecular weight or to have some other data which will help us to decide the actual number of atoms of each element present. Suppose for example that by experiment we know the molecular weight of lead chloride to be 2 78. Now we can decide the actual number of atoms pres- ent. The molecular weight is of course the sum of the weights of the atoms which compose it. The weight of 1 atom of lead is 1 X 207.2 = 207.2 The weight of 2 atoms of Cl is 2 X 35.46 = 70.92 If the formula is PbCL themolecular weight will be 278.1 2 If there are 2 atoms of Pb and 4 of Cl, i. e. the formula is Pb2Cl4 then weight of 2 atoms Pb is 2 X 207.2 = 414.4 weight of 4 atoms Cl is 4 X 35.46 = 141.84 Weight of the molecule is 556.24 If there are 3 atoms of Pb, 6 atoms of Cl, i. e. the formula is Pb3Cl0 then 84 CHEMICAL CALCULATIONS weight of 3 atoms of Pb = 3 X 267.2 = 621.6 weight of 6 atoms of Cl = 6 X 35.46 = 212.72 Weight of the molecule is ' 834.36 We see that in this case the formula PbCL represents correctly the experimental facts. This formula, which repre- sents not only the Relative numbers of the atoms in 1 molecule but also the ACTUAL number of atoms of each element, is known as the Molecular Formula. Formula of Minerals. A great many minerals are composed of oxides, i.e. the various elements are combined with oxygen forming oxides and these oxides combined with one another. The result of an analysis of such a mineral is stated in terms of the per- centages of the various oxides present: thus the mineral olivine may be considered as composed of 2 oxides - MgO and SiO2. Analysis would determine the percentage of each of these oxides present rather than the percentage of each of the ele- ments Mg, Si and O. We would find MgO 5 7.215% SiO2 42.784% The formula of the mineral will be most clearly represented by the relative num- bers of molecules of these oxides considered as units. Thus- the molecular weight of MgO is 24.32 -j- 16.00 = 40.32. The molecular weight of SiO2 is 28.3 -f- 2 X 16.00 = 60.3. . - .The relative numbers of molecules of MgO and SiO2 are 57.215 = 1.4190 molecules of MgO 40.32 42.784 = 0.7095 molecules of SiO., 60.3 or, since 1.4190 : 0.7095 as 2 : 1, the relative numbers of molecules of MgO and SiO2 present are 2 molecules of MgO to each 1 molecule of SiO2. The formula of the mineral is hence (MgO) 2.SiO2. When a mineral contains small amounts of other oxides which are similar in chemical nature to the several oxides CHEMICAL CALCULATIONS 85 present in major amounts, these small amounts of oxides are regarded as impurities. It is assumed that these oxides have replaced chemically equivalent amounts of the principal oxides. The relative numbers of molecules of these are added in with the numbers of molecules of the oxides they probably replaced, thus giving the relative numbers of molecules of the major oxides before replacement took place. This procedure re- moves unnecessary complication of the formula of the min- eral and at the same time compensates the effect of the small amounts of oxides which are regarded as impurities. PROBLEMS. 1. What is the formula of the substance which gave by analysis 5.88 per cent. Hydrogen and 94.1 2 per cent. Oxygen? 2. What is the formula of the substance which gave on analysis 20.00 per cent. Carbon, 26.67 per cent. Oxygen, and 5 3.33 per cent. Sulphur? 3. What is the formula of the substance which gave on analysis 28.73 per cent. Potassium, 0.73 per cent. Hydrogen, 23.52 per cent. Sulphur and 47.02 per cent. Oxygen? 4. What is the formula of the substance which gave on analysis 9.76 per cent. Magnesium, 13.01 per cent. Sulphur, 26.01 per cent. Oxygen and 51.22 per cent. Water? 5. What is the formula of the substance which gave on e.nalysis 22.70 per cent. Zinc, 11.15 per cent. Sulphur, 22.28 per cent. Oxygen and 43.87 per cent. Water? 6. Derive the formula from the following composition: Calcium, 38.72; Phosphorus, 20.0; Oxygen, 41.28 per cent. 7. Derive the formula of the substance which has a rela- tive density of 4.0(0 = I) and the composition: 93.75 per cent. Carbon and 6.25 per cent. Hydrogen. 8. Derive the formula of the subtance having the follow- ing percentage composition: Sodium 32.79, Aluminium 1 3.02, and Fluorine 54.19. 9. Derive the formula of the substance with an absolute 86 CHEMICAL CALCULATIONS density of 1.1 89 and the following composition: Carbon 92.1, Hydrogen 7.85 per cent. 1 0. What is the formula of the substance which gave on analysis 45.95 per cent, of Potassium, 16.45 per cent, of Nitrogen and 37.60 per cent, of Oxygen? 1 1. Derive the formula of the compound which gave on analysis 26.1 per cent. Carbon, 4.35 per cent. Hydrogen, and the rest Oxygen. The molecular weight was determined ap- proximately as 46. 12. Derive the formula of the compound produced by the combustion of 43.45 grams of Lead with 4.48 grams of Oxygen. 1 3. The percentage composition of a certain Salt is: 2 7.51 percent. Calcium, 22.15 of Sulphur, 1.02 of Hydrogen, 49.32 of Oxygen. Ten grams of this crystallized salt lost 0.6 grams of water upon dehydration. What is the formula? 1 4. Assuming that Silicon Chloride contains only one atom of Silicon, calculate the atomic weight of Silicon from the following data: Analysis shows the percentage to be 16.47 per cent. Silicon, 83.53 per cent. Chlorine. Its vapor density is 85. (H - 1). Atomic weight of Chlorine 35.5. 1 5. The mineral kerolite gave the following percentages on analysis. Calculate its formula: SiO2 46.96 MgO 31.26 H2O 21.22 1 6. A specimen of cobalt-bloom was found to have the following composition. Determine its formula: As2O5 38.43, CoO 36.52, FeO 1.01, H2O 24.10. 1 7. Calculate the formula of soda feldspar from the following analysis: SiO2 68.45, Al.O., 18.71, Fe2O3 0.2 7, CaO 0.50, MgO 0.18, K2O 0.65, Na2O 11.24 18. Derive the formula of Eudidymite from the follow ing analysis: Silica 73.11, Beryllium Oxide 10 62, Sod? 12.24, Water 3.79. CHEMICAL CALCULATIONS 87 1 9. Derive the formula of the mineral Albite from the following analysis: Silica 69.00, Alumina 19.43, Lime 0.20, Soda 11.47. 20. Derive the formula of the mineral Paragonite from ths following analysis: Silica 62.24, Magnesia 30.22, Ferrous Oxide 2.66, Water 4.9 7. 21. A compound was found to contain C-82.65 %, H-17.35%. Its vapor density is 29. Derive the molecular formula. 22. An Oxide of Nitrogen has a vapor density of 46. It contains N-30.44%, 0-69.56%. Derive its molecular for- mula. 23. The molecular weight of a Chloride of Iron is 324.44. It contains Iron 34.32%, Chlorine 65.57%. Derive its molecu- lar formula. 24. Calculate the molecular formula of a hydrated Sul- phate of Iron from the following data: 1 gram of it yielded 0.3296 g. of Fe2O3 upon analysis. A second 1 gram sample of it, when treated with BaCL solution, yielded 0.9640 g. of Ba(SOJ. 25. Derive the formula of that substance which contains by analysis, Nitrogen 7.145%, Hydrogen 2.056%, Sulphur 16.350%, Oxygen 32.641 %, Iron 14.239%, Water of Crys- tallization 7.569%. CHAPTER IX OUTLINE Estimation of the proportion of acid in sugar juice. An example of Titration. Further examples of Titration. Determination of the proportion of NaCl in sea water by Titration. Normal Solution-one which contains in each 1 000 cc of solution THAT number of grams of the compound which is numerically equal to the combining weight. All Normal solutions are equal. 1 cc of a N solution of some substance = 1 cc of a N solution of any other substance. Normal solutions of (A) Acids HNO3 (B) Bases Ca(OH)2 (C) Salts NaCl (D) Oxidizing Agents K2Cr2OT Problems. 88 CHEMICAL CALCULATIONS 89 VOLUMETRIC ANALYSIS. Granulated sugar is obtained from the juice pressed from sugar cane. This juice contains water and a great many soluble substances from the cane in addition to the sugars. As it stands in the storage tanks of the sugar factory it fer- ments and becomes sour, due to the formation of acids in the juice. These acids, of course, have to be removed in some way, the common practice consisting of the addition of sufficient lime to neutralize them. The quantity of lime added must be enough to neutralize all of the acid in the liquor, but should not be much greater than this or else we would have to remove it later. We would certainly not want lime in the sugar as we eat it. How then can we tell how much lime to add? The works Superintendent would formerly have an- swered this by pouring in a couple of bucketsful of lime and stirring it in, and would then taste the liquor and observe whether all the acid was neutralized. The fact that possibly he had added too much lime would not cause him much con- cern. The practice of a modern works is quite different from this. To some young technically trained man is assigned the duty of determining exactly how much lime is necessary to neutralize the acids. He stirs the juice well, then dips out say a quart of the juice and takes it to his laboratory. How can he determine exactly how much lime to add? Most of us would say at once, weigh on an accurate scale or balance a small quantity of lime and dissolve this in the juice. After it is dissolved test the juice with litmus paper. If the juice is still acid (turns the litmus red), weigh a second small quantity of lime and dissolve this in the juice and again test with litmus; repeat this procedure of adding very small weighed quantities of lime until sufficient lime has been added to just turn the paper blue. We now know what weight of lime is necessary to just neutralize the acid in one quart of juice. Four times this weight would be the quantity necessary to add to each gallon of juice in the storage tank. The method of adding repeated small weighed quantities of lime is rather tedious, so that a more rapid and more exact method is used. 90 CHEMICAL CALCULATIONS A solution is made by dissolving cay 50 grams of lime in sufficient water to make just 1 liter (1000 cc) of solution. 1 Each cubic centimeter (cc) of this solution will contain 1000 of 50 - 0.05 gram of lime. This solution is then poured into a graduated glass tube fitted with a stopcock at the bot- tom. Look at the illustration. This graduated tube is called a burette. Burettes are usually marked off in cubic centimeters (cc). These marks are permanent ones made by etching the glass. The usual size of burette holds 50 cc. Each cc is di- vided into tenths. Now we start with the burette filled with the lime solution up to the top mark 0.0 cc. The sugar juice is held in a beaker below the burette. A piece of litmus paper, or a few drops of a solution of litmus in water, is added to the sugar juice and the lime solution is then slowly run into the sugar juice until the litmus just changes to blue. Since the lime solution can be added a few drops or even one drop at a time, we can tell with great exactness just how much of the lime solution was necessary to neutralize the acid in the quart of sugar juice. Assume for example that when enough lime solu- tion has been added to just change the litmus to blue that we find that the level of lime solution in the burette is at 27.0 then 27 cc of lime solution has been delivered from the burette into the sugar juice. Since each cubic .centimeter of the lime solution contains 0.05 gram of lime, then 27 X 0.05 = 1.35 grams of lime are necessary to neutralize the acids in 1 quart of sugar juice. This operation of determining the amount of acid in the juice is an example of what is known as TITRA- TION. The technically trained man would say he had titrated the acid in the sugar juice. In thousands of laboratories all over the world it is one of the duties technical men to titrate liquids of all kinds. Titration is used to determine the amount of acid in sugar Fi'uj 8. CHEMICAL CALCULATIONS 91 juice, to determine the amount of acid in vinegar and in a great variety of other liquors and juices. Further it is not limited to finding the amount of acid. We could reverse the process and could determine the amount of alkali, in a soap liquor for example, by putting in the burette a solution which contained a definite weight of some acid, say HC1, in 1 000 cc. We would in this case add the acid solution until the litmus added to the soap solution just changed from blue to red and would measure how much acid had been run out from the burette to do it. Further, titration is not even limited to acids and bases. Suppose we wanted to know how much common salt (NaCl) there is in a gallon of sea water. We would put in the burette a solution of silver nitrate containing say 240 grams of silver nitrate in 1 000 cc of solution. When silver nitrate is added to common salt a reaction takes place-sodium nitrate and silver chloride are formed. Silver chloride formed is not soluble in the solution and is therefore precipitated. Experiments have showed that parts by weight of silver nitrate are necessary to react with 58.46 parts by weight of sodium chloride and form sodium nitrate and silver chloride. This is represented by the equation: AgNO:i + NaCl = NaNO8 + AgCl 107.88 + 14 + (3 X 16) 23 + 35.46 169.88 58.46 We would add the silver nitrate solution to the sea water a few drops at a time as long as a precipitate of silver chloride continued to form. When sufficient silver nitrate has been added to react with all the sodium chloride then the addition of a one further drop of silver nitrate will not cause any further precipitate to form. We then observe the level of solution in the burette. Suppose 190 cc of silver nitrate solution have been required then since there is 0.240 gram of silver nitrate in 1 cc 190 X 0.240 = 45.60 grams of silver nitrate was necessary to react with the sodium chloride in the pint of sea water. The experiments on which the above equation is based showed us that 1 69.88 grams of silver nitrate were necessary for 58.46 NaCl: 92 CHEMICAL CALCULATIONS 1 I g. of silver nitrate is . : necessary for of 58.46 NaCl. 169.88 1 45.60 g. of silver nitrate is . : necessary for 45.6 X 169.88 X 58.46 = 15.69 NaCl. In general Titration is a very convenient method of or the analysis of substances. Solutions are used which contain weighed amounts of the titrating substance in each liter of solution. These solutions whose strength is known are usually spoken of as STANDARD SOLUTIONS. Normal Solutions. It has been found especially convenient and useful in analysis by this method, to make up standard solutions in which the number of grams of the substance dissolved in each liter of solution is the same numerically as the CHEMICAL EQUIVALENT of that compound. It will be recalled that in developing our knowledge of chemical equivalents we de- cided to consider the Chemical Equivalent of a compound as that weight of it which is equal chemically to 8.00 grams of Oxygen or to 1.008 grams of Hydrogen. Therefore, a normal solution of any substance is a solution which contains in each 1 000 cc of solution that weight of the substance which is equivalent to 8.00 grams of Oxygen or to 1.008 grams of Hydrogen. One liter ( 1 000 cc) of a Normal Solution of some sub- stance must be equivalent chemically to one liter of a Normal Solution of any other substance. This is true in each case, because the weight of the substance dissolved in one liter of the solution is chemically equivalent to 1.008 grams of Hy- drogen. The following examples illustrate this: (A) Normal Solution of Acids. Calculate the weight of HNO3 necessary to make 1 liter of a Normal Solution of Nitric Acid. We found it convenient (page 67) to consider as the CHEMICAL CALCULATIONS 93 Chemical Equivalent of an acid, that weight of it which con- tains 1.008 grams of replacable Hydrogen. The molecular weight of HNO., is 1.008 -|- 14.01 • (3 + 16.00) = 63.018. 63.018 grams of HNO.. contain 1.008 grams of replacable Hydrogen. 63.018 grams of HNO. is then the chemical equivalent cf HNO3. If then we dissolve 63.018 grams of HNO.( in sufficient water to make 1 liter of solution, we will have a normal so- lution of HNO... (B) Normal Solutions of Bases. Calculate the weight of Ca(OH)2 necessary to make 1 liter of N Calcium Hydroxide Solution. We have found (page 67) that that weight of a base which contained 1 7.008 grams of OH is the chemical equiva- lent of the base and we found it very convenient to calculate the chemical equivalent of a base in this way. The molecular weight of Ca(OH)2 is 40.07 -f- 2(16.00 4- 1.008) = 74.086. 74.086 grams of Ca(OH)2 contain 2 ( 1 7.008) grams of (OH). 74.086 = 37.043 grams of Ca(OH)2 contain 17.008 grams 2 ■ of (OH) 37.043 grams of Ca(OH)2 is then the Chemical Equivalent of Ca(OH)2. If we dissolve 37.043 grams of Ca(OH)2 in sufficient water to make 1 liter of solution, we will have a normal solu- tion of Ca(OH)2. One liter of this N Ca(OH)2 solution contains the chem- ical equivalent of Ca(OH)2. This is just enough Ca(OH)2 to exactly neutralize all the HNO.. in 1 liter of N HNO., solu- tion, i.e. 1 liter ( 1 000 cc) of N Ca(OH)2 solution = I liter ( 1 000 cc) N HNO3 solution. 1 cc of N Ca(OH)2 solution = 1 cc of N HNO, solution. (C) Normal Solution of Salts. We found (page 68) that the chemical equivalent of a 5alt could be very conveniently found from the weight of 94 CHEMICAL CALCULATIONS the metal present. That weight of the metal which will com- bine with 8.00 grams of Oxygen is the chemical equivalent of the metal. That weight of the compound which contains this weight of the metal is, of course, the chemical equivalent of the compound. Example: Calculate the weight of NaCl necessary to make 1 liter of N NaCl solution. Na and O combine in the proportion of 2(23.00) parts by weight of Na to 16.00 parts by weight of O and form sodium oxide, Na. O 2(23.00) + 16.00 2(23.00) parts by weight of Na combine with 16.00 parts by weight of O. 23.00 parts by weight of Na combine with 8.00 parts by weight of O. Therefore 23.00 is the combining weight of Na. The molecular weight of NaCl is 23.00 -j- 35.46 = 58.46 58.46 grams of NaCl contain 23.00 grams of Na. 58.46 is then the chemical equivalent of NaCl. Therefore to make 1 liter of N NaCl solution we would dissolve 58.46 grams of NaCl in sufficient water to make 1 liter. Example 2: Calculate the weight of AgNO.. necessary to make 1 liter of N AgNO., solution. The combining weight of Ag is 1 07.88, as can be seen from inspection of the compound Silver Oxide Ag.O 2(107.88) + 16.00 The molecular weight of AgNO.. is 107.88 -|- 14.01 -|- 3 X 16.00 = 169.89. 169.89 grams of AgNO.. contain 107.88 g Ag (the chemical equivalent of Ag). Therefore to make I liter of N AgNO, solution we would dissolve 1 69.89 grams of AgNO3 in sufficient water to make 1 liter of the solu- tion. Silver Nitrate and NaCl react as indicated by the follow- ing equation: AgCl is precipitated. AgNO3 + NaCl = AgCl +NaNO, I liter of N AgNO., solution contains just enough AgNO. to exactly neutralize the NaCl in 1 liter of N NaCl solution. The amounts of AgNOa and NaCl are chemically evuivalent. CHEMICAL CALCULATIONS 95 (D) Normal Solution of Oxidizing Agents. We found that (page 69) the chemical equivalent of an oxidizing agent was that weight of it which liberates 8.00 parts by weight of oxygen. Problem: Calculate the weight of K.2Cr2O7 necessary to make I liter of N K2Cr,O7. When K2Cr2O7 acts as oxidizing agent 1 molecule of it liberates 3 atoms of Oxygen. This may be represented by the equation K2Cr2O7 = K..O + Cr,O3 + 30 2(39.1) + 2(52.0) + 7(16.00) = 3(16.00) 294.2 = 48 294.2 grams of K,Cr.,O7 liberate 48 grams of Oxygen. 294.2 48 grams of K.,Cr,O7 liberate 1 gram of Oxygen. 294.2 8 X 49.033 g. of K,Cr,O7 liberate 8.00 g. of Oxygen. 48 49.033 g. of K,Cr2O7 is the weight of K,Cr,O7 neces- sary to make 1 liter of N K,Cr,O7 solution. 1 A solution which contains in each liter as much of 10 some compound as is necessary to make a Normal Solution of that compound would, of course, be a Tenth Normal Solution N ( solution). One which contains, in each liter, as much 10 of the compound as is necessary for a normal solution, a Half N Normal Solution (- solution), etc. These are sometimes used. 2 Example: Calculate how many cubic centimeters of N HC1 solution are necessary to neutralize a solution which con- tains 4 grams of Ba (OH),. This neutralization is represented by the chemical equation. Ba(OH), +2HC1 = BaCh + 2H.'O 137.37 + 2(16.00 + 1.008) + 2(1.008 + 35.46) = 171.386 72.936 The molecular weight of Ba(OH)2 is 1 71.386. 96 CHEMICAL CALCULATIONS The chemical equivalent of Ba(OH)2 is one-half of this or 171.386 85.693. (Why is this? It is well to refer to page 2 68 for a moment). A normal solution of Ba(OH)2 would then contain 85.693 g. of Ba(OH)2 in each liter (1000 cc) or 85.693 0.085693 g. Ba(OH)., in 1 cc. Now 1 cc of any N 1000 solution = 1 cc of any other N solution. 1 cc of N HC1 = 1 cc of N Ba(OH)2 = 0.085693 g. of Ba(OH)2, or stating this backwards: 0.085693 g. of Ba(OH)2 = 1 cc of N HCL 1 1.0 g. of Ba (OH) X 1 cc = 1 1.66 cc of N HC1 0.085693 4.0 g. of Ba(OH)2 = 4 X 1 1.66 = 46.64 cc of N HC1 i.e. it will take 46.64 cc of N HC1 to neutralize 4 grams of Ba(OH)2. PROBLEMS. 1. Calculate the number of grams of each of the follow- ing necessary to make 1 liter of a N solution of that compound: (a) HNO2 (b) A1(OH)3 (c) CS2(SO4) (d) KC1. 2. Calculate the number of grams of each of the following necessary to make 1 liter of a N solution of that compound: (a) NH,OH (b) H2SOt (c) Rb(COJ (d) MgO. 3. Calculate the number of grams of each of the follow- ing necessary to make 1 liter of a N solution of that compound: (a) K.O (b) Hl (c) Fe(OH)3 (d) RaBr2. 4 How many cc of N KOH solution would be necessary N to exactly neutralize 120 cc of H.SOj solution? 2 5. Calculate the number of cc of N NaOH solution nec- essary to neutralize 1 gram of H2SO4. CHEMICAL CALCULATIONS 97 N 6. Calculate the number of cc of - Ba (OH) ., solution 2 necessary to exactly neutralize 3.0 grams of H2(C2O4). 7. A solution contains 1.5 grams of K.,SOt. How many cc of N BaCl2 solution would be necessary to react with all of this to form BaSO( and KC1? 8. An excess of Na2SO, solution is added to 25 cc of N BaCI. solution. Calculate the weight of BaSOt which will be precipitated. 9. A sample of potassium carbonate required 14 cc of N H2(SO4) to neutralize I gram of the salt. Calculate the per- centage purity of the salt. 10. 2.25 grams of an ammonium salt required 40 cc of N NaOH solution to drive off the ammonia. Calculate the per cent, of ammonia in the salt. 11. 3 grams of a sample of copper sulphate required 30 cc of N NaOH for complete precipitation, leaving no excess of NaOH in the solution. Calculate the percentage of copper in the solution. N 12. What is the amount of HCl in 2 liters of - HC1? 10 N How many cc of this - solution must be taken to prepare I 10 liter of 0.004 N solution? N 13. What is the amount of H.,(SO,) in 2 liters of ■- 10 N H..SO,? How many cc of this - solution would be required 10 to neutralize 5 grams of MgCO.. ? 14. How much KMnO4 should be dissolved in enough water to make 1 liter of solution so that 1 cc of the solution will yield 1 milligram of Oxygen when acting as an oxidizing agent? 15. Calculate the percentage of CaCO., in a sample of 98 CHEMICAL CALCULATIONS chalk which on analysis gave the following data: 38 cc of N NaOH were required to neutralize 1 0 cc of a solution of HC1. 1 gram of the chalk was dissolved in 1 0 cc of the HC1 solution and it was found that 24.5 cc of N NaOH solution were re- quired to neutralize the excess of acid. I 6. By analysis, a solution of KOH was found to have a strength of 0.252 Normal. Calculate how many cc of this solution should be taken and diluted with water in order to N make I liter of - KOH solution. 10 1 7. How many cc of a 0.465 N HC1 solution should be taken and diluted with water in order to prepare 500 cc of N - HC1? 5 18. Two grams of KOH were dissolved in enough water to make 100 cc of solution. 50 cc of this solution required 10 cc of N HNO., for exact neutralization. Calculate the percentage purity of the KOH. 19. How many cc of N HBr will be required to dissolve 1 0 grams of Al? 20. Calculate the number of grams of each of the fol- lowing required to make 1 liter of a N solution of that sub- stance: K,Cr2O7, KC1O,, H .fSeOj), Cs ,O. 2 1. Calculate the number of grams of each of the fol- N lowing required to make 500 cc of a - solution of that sub- 2 stance: Na2Cr2O7, H2O2, H2(TeO1), CeO2. N 22. How many cc of - KMnO4 would be required to 2 oxidize a solution which contains 0.5 gram of H..SO? 23. How many cc of N K2Cr2O7 solution will be required to oxidize an acid solution which contains 1.0 g. of H..S? 24. Calculate the weight of AgCl precipitated when 1 0 cc of 2 N HC1 solution are added to excess AgNO.;. 25. Calculate the weight of PbS precipitated when excess H2S is passed into 100 cc of N Pb(NO3)2 solution. CHAPTER X OUTLINE The volumes of gases which unite are in simple proportion. The Relative volumes of gases which unite can be seen by inspection of the chemical equation. Problem-To calculate the volume of oxygen or of air necessary for the combustion of a given volume of a com- bustible gas. Gas Analysis-The quantitative composition of a mixture of gases can be calculated from a few simple measurements together with inspection of the chemical equations representing the combustion of the gas mixture. Solution of a Gas Analysis problem. 99 100 CHEMICAL CALCULATIONS COMBINATION OF GASES. Gas Analysis. We have seen that an enormous number of experiments on the combination of gases led us to a general law, Gay Lussac's law of volumes-When gases unite, the volumes which unite are in the proportion of simple whole numbers, 1 to 1 ; 1 to 2 ; 1 to. 3; 1 to 4; 2 to 3; etc.; and if the product is a gas, its volume will be in simple proportion with the vol- umes of the gases which combined. We have further noticed that that number of grams of any gas, which is numerically equal to the molecular weight, oc- cupies 22.3 liters at O"C and 760 mm pressure (standard conditions). Thus O.< represents 1 molecule of oxygen, which is, of course, 32.00 parts by weight of oxygen. 32.00 grams of oxygen then occupy 22.3 liters at O"C and 760 mm. Similarly CO2 stands for 1 molecule, 44.00 parts by weight of carbon dioxide. 44.00 grams of carbon dioxide will then occupy 22.3 liters at O"C and 760 mm. This relation between molecular weight and volume is extremely useful. It allows us to calculate the volumes of gases from their weights. 4 hus consider an action such as the burning of CO represented as follows: 2CO + O2 = 2CO 2(12 -f- 16) 4- 32 - 2(12 4 32)grams 2(28) 4~ 32 - 2(44) grams 2(22.3) 4- 22.3 = 2 (22.3) liters 2 -j- 1 = 2 liters 28 grams CO occupy approximately 22.3 liters at O'C and 760 mm. 32 grams O occupy approximately 22.3 liters at O"C and 760 mm. 44 grams CO., occupy approximately 22.3 liters at O"C and 760 mm. The actual volumes which combine, when the weights represented by the equation are in grams, are: 2(22.3) liters of CO + 22.3 liters of O._, = 2(22.3) liters of CO2. CHE Ml CA L CA ECU LA TI0 NS 101 The RELATIVE volumes represented are: 2 parts by volume of CO ( 1 part by volume of O„ = 2 parts by volume CO2. The parts by volume can be cc, liters, gallons, cubic feet or any other unit of volume. We notice that the volumes which unite are in the same proportion as the number of molecules. This we, of course, expect, knowing that equal volumes of gases contain equal numbers of molecules at the same conditions of temperature and pressure (Avogadro's Hypothesis. Each molecule represents 1 part by volume. Let us now consider a problem. Calculate (a) the volume of Oxygen, (b) the volume of Air, necessary for the complete combustion of 25 liters of c2h. The action is represented by the equation 2C2H2 + 5O> 4CO2 + 2H2O 2 (22.3) + 5 (22.3) = 4(22.3) % 2.(22.3) liters. 2 5 = 4 -f- 2 liters. i.e. the equation indicates that each 2 liters of C2H2 unite with 5 liters of O2 and form 4 liters of CO2 and 2 liters of water vapor. 2 liters of C.,H., require for combustion 5 liters of O.,. 1 1 liter of CoH., requires for combustion - X 5 liters of O,>. 2 5 (a) 25 liters of C2H2 require for combustion 25 X - = 62.5 liters of O2. 2 Neglecting the gases present only in small amounts- argon, CO2, moisture, helium, etc., air contains Oxygen, 20.8% by volume Nitrogen, 79.2% by volume i.e. in 100 liters of air there are 20.8 79.2 liters of O2 liters of N2 (b) To burn the 25 liters of C2H2 required 62.5 liters of oxygen. How much air would be necessary? There are 20.8 liters of O2 in 100 liters of air. 100 There is 1 liter of O., in - liters of air. 20.8 CHEMICAL CALCULATIONS 102 100 There are 62.5 liters of O2 in 62.5 X = 313.4 liters of air. 20.8 GAS ANALYSIS. We find then that from the chemical equation we can cal- culate what volume of oxygen is necessary to unite with a given volume of any combustible gas, and we can also calculate the volume of each of the products formed. Conversely if we know the volume of the products formed by the combustion of a gas, we can of course calculate the volume of the gas which was burned. The equation indicates the RELATIVE volumes of gases which react and of gaseous products formed. This is of use in gas analysis. We can calculate the volume of each gas in a mixture of gases from a few simple measurements of the volume of the products formed. An example will illustrate this. A mixture of gases consists of CO, CEh, and C2H2. To 50 cc of this mixture, 1 00 cc of Oxygen was added. (This is more than enough for complete combustion of the 50 cc of gas). This mixture was forced into a strong glass bulb in which 2 platinum wires are sealed, and the mixture made to explode by passing a spark betwen the points of the 2 platinum wires. After explosion the volume was found to be 137.5 cc. The water vapor formed by the combustion was condensed to liquid water by cooling the gas. This caused a decrease in the volume. The volume after cooling was 72.5 cc. Calculate the number of cc of each of the gases H, CHi and C2H2 in the 50 cc of original gas from this data. Let X = number of cc of CO in the 50 cc. Y = number of cc of CH4 in the 50 cc. Z = number of cc of C2H2 in the 50 cc. Then X -|- Y -f-Z = 50cc. The reactions which took place when the spark started the combustion are represented by the following equations, which, of course, also indicate the relative volumes of the gases which combined during the explosion, and also the relative volumes of the products formed: CHEMICAL CALCULATIONS 103 2 CO + O2 = 2CO2 2(22.3) liters A 22.3 liters == 2(22.3) liters 2 liters A 1 liter =, 2 liters 2 cc A 1 cc = 2 cc X X CC A cc - X cc 2 The equation shows that for the complete combustion of 2 cc of CO, 1 cc of oxygen is necessary and 2 cc of CO2 are formed. Therefore X cc of CO will require | X cc of O2 for complete combustion. X cc of CO2 will be formed. CH4 + 2 0, = CO2 + 2 H2O 22.3 liters A 2(22.3) liters = 22.3 liters A 2(22.3) liters 1 liter A 2 liters - 1 liter A 2 liters 1 cc -f- 2 cc = 1 cc A 2 cc Y cc A 2Y cc = Y cc -|- 2Y cc 2CH 50 = 4 CO + 2HO 2 2 2 2 2 2(22.3) liters 4- 5(22.3) liters = 4(22.3) liters + 2(22.3) liters 2 liters + 5 liters = 4 liters + 2 liters 2 cc 5 cc = 4 cc 4- 2 cc 5 Z cc 4- - Z cc = 2 Z cc 4- Z cc X We notice that X cc of CO unite with - cc of O„ and form 2 X X cc of CO.,. A mixture of X cc of CO and - cc of O., 2 3X X ( cc total) form X cc of CO,. A contraction of - cc 2 ' 2 takes place. In the case of CH4 a mixture of Y cc of CH4 and 2 Y cc of O2 form Y cc of CO2 and 2Y cc of water vapor. After the explosion the sum of the volume of the products is the same as the sum of the volume of the gases which combined. No contraction took place. In the case of the C.,H„-a mixture of Z cc of 5Z 7Z C,H., and cc of O ( cc total) form 2Z cc of 2 ' 2 CO2 and Z cc of water vapor. A total of 3Z cc. A contraction CHEMICAL CALCULATIONS 104 Z [7Z 1 of cc I - 3Z I takes place. Therefore the result of the explosion of the three gases mixed with excess of O will be a contraction of X z CC d cc. 2 2 By actual measurement the volume before the explosion was 150 cc and after explosion 137.5 cc. A contraction of 12.5 cc, i. e. X z - cc -| cc - 1 2.5 cc II. 2 2 We observe from the equations that Y cc of CH4 form 2Y cc of water vapor and Z cc of C,H2 form Z cc of water vapor The total volume of water vapor formed by the com- bustion of the mixture is therefore - 2Y cc -f- Z cc. • Now by cooling the gas and condensing the water vapor to liquid water it was noticed that the volume of water vapor in the mixture after combustion was 65 cc, i.e. Volume before cooling to condense water vapor. . . 137.5 cc Volume after cooling 72.5 cc Volume of water vapor 65.00 cc then 2Y cc + Z cc = 65 cc III. We now have three simultaneous equations in X, Y and Z We can, of course, solve these and get X, Y and Z, which we recall are the number of cc of CO, CH4, C.,H2, respectively, in the 50 cc of original gas. I. X + Y + Z = 50 cc „ 5 + z-,is„ 2 2 HI. 2Y + Z = 65 CC X + Z = 25 Z = 15 Y =10 X + Y + Z = 50 cc X + Z = 25 Y 25 cc 2Y + Z = 65 2Y = 50 Z =15 CH EMICA L CA ECU LA TIONS 105 i.e. X = number of cc of CO = 1 0 cc Y = number of cc of CH4 =25 cc Z = number of cc of C2H2 = 1 5 cc 50 cc We notice from the equation that X cc of CO2 was formed by the combustion of the X cc of CO. Y cc of CO_> by the combustion of the Y cc of CHj, and 2Z cc of CO2 from the combustion of the Zee of C2H2; i. e. the total CO2 formed is X + Y + 2Z cc. If we had bubbled the gas after ex- plosion, through KOH solution, this CO2 would have combined with the KOH and been absorbed according to the equation CO2 -J- 2 KOH = K2CO3 -f- H2O. By measuring the volume after bubbling through the KOH solution the decrease in volume would be noticed. This decrease would be the volume of CO formed by the combustion. In the above case it would have been 65 cc, i.e. X 4- Y + 2Z = 65. This would have given another simultaneous equation (relation between X, Y and Z) to use in solving for X, Y and Z. Very often the volume of CO., is determined in this way and this data used to obtain another equation for calculating X, Y and Z. PROBLEMS. 1. How many liters of air at standard conditions of tem- perature and pressure are necessary for the complete combus- tion of 25 liters of (a) marsh gas, (b) acetylene, (c) hy- drogen sulphide? 2. If 700 cc of carbon monoxide are burned, what vol- ume of oxygen will be necessary to complete the reaction? 3. 100 Is. of gas containing 30% of hydrogen, 10% of ethane, 40% of marsh gas and 20% of hydrogen sulphide were mixed with 1 60 liters of oxygen and exploded by means of an electric spark. What is the volume and composition of the resulting gases at standard conditions? 4. If a mixture consisting of 70 volumes of air and 56 106 CHEMICAL CALCULATIONS volumes of hydrogen is exploded by means of an electric spark, is there any gas remaining? If so, what is its com- position and volume at standard conditions? 5. After adding 70 cc of hydrogen to 70 cc of oxygen and nitrogen contained in a tube closed with a water seal, the mixture was exploded by an electric spark. After cooling and measuring the residual gases (nitrogen and hydrogen) they were found to occupy 64.4 cc. Calculate the compo- sition of the original mixture. 6. 140 cc of oxygen were added to 70 cc of gas con- sisting of carbon monoxide and methane in an explosion pipette. After the explosion the cooled residual gas meas- ured 164.5 cc. What was the composition of the mixture? 7. 1 82 cc of a mixture of carbon monoxide, marsh gas and nitrogen were mixed with 210 cc of oxygen in a eudio- meter tube and exploded by a spark. The volume of product after explosion measured 364 cc and this product was passed through a tube containing calcium chloride, after which the volume was observed to be 224 cc. Assuming a constant temperature of 100 °C, calculate the original volumes of each of the constituents in the mixture. 8. A mixture of 2 1 0 cc of carbon monoxide and 1 40 cc of marsh gas were burned together with 420 cc of pure oxygen. Assuming that the temperature was 1 00 C, calculate the volume of the gaseous product. How much Aqueous vapor was formed ? 9. 42.5 cc of a mixture of nitrous and nitric oxides, to- gether with nitrogen, gave on analysis the following data: After mixing with hydrogen the total volume was 92.45 cc, and after the explosion the cooled volume of gas was 42.5 cc. The hydrogen used was in excess and this excess was deter- mined by mixing the residual gas with oxygen and exploding it. The volume, after the addition of the oxygen, was 63.75 cc and the volume after the explosion was 48.0 cc. Calculate the percentage composition of the mixture, assuming that all readings were taken under the same conditions and over water. 10. 245 cc of a mixture of acetylene and oxygen were CHEMICAL CALCULATIONS 107 exploded in a eudiometer tube and the volume of the dry gaseous product was 192.5 cc. How much acetylene was there in the original mixture? 11. 250 cc of a gaseous mixture consisting of carbon monoxide and acetylene, with 625 cc of oxygen, were burned in an explosion pipette. The volume of the residual gases, after cooling, was 650 cc, and upon passing this gas through a solution of potassium hydroxide, all but 300 cc were ab- sorbed. What was the composition of the original mixture? 12. 28 cc of a mixture of marsh gas, nitrogen and hy- drogen were burned with 7 cc of oxygen. After cooling the volume of the product was found to be 25.55 cc. This vol- ume was decreased to 23.45 cc after passing the gas through a solution of potassium hydroxide and further diminished by treatment with an alkaline solution of pyrogallic acid, which absorbs the excess of oxygen, to 22.4 cc. What was the percentage composition of the original mixture? 13. 2 1 0 cc of a mixture of marsh gas, air and hydrogen sulphide were added to an excess of oxygen measuring 350 cc and the whole exploded. The volume of the gaseous product measured 504 cc, all the water being in the form of vapor. After cooling the gas to remove the water vapor and figuring the volume at the original temperature, i.e. 100 C, this was found to be 224 cc. What was the composition of the orig- inal mixture? 1 4. Calculate the percentage composition of a gas which consisted of carbon monoxide, hydrogen, nitrogen and ethylene and gave on analysis the following data: Original volume of gas was 82.0 cc; volume after treatment with bromide to absorb the ethylene, 71.64; after treatment with ammoniacal cuprous chloride to absorb the carbon monoxide, 37.28; after mixing with oxygen and air, 328.88; and after explosion, 2 75.2. 15. After treatment with potassium hydroxide solution, 1 1 4.1 cc of coal gas were reduced to 1 13.26 cc, and upon subsequent treatment with fuming sulphuric acid to absorb the ethylene, the volume was further reduced to 108.5 cc. Of 108 CHEMICAL CALCULATIONS this residual gas 66.5 cc were burned in an explosion pipette with 420 cc of oxygen. After the gaseous product had cooled it measured 377.65 cc, and after treating it with potash solu- tion the volume measurd 344.75 cc. Assuming that this gas contains no nitrogen and that the only saturated hydrocarbon it contains is methane, calculate the percentage composition of the original mixture. 16. Into a stoppered tube (90 cc capacity) filled with chlorine, a small quantity of concentrated ammonia water (excess) was admitted. After shaking the mouth of the tube was opened under a dilute acid solution contained in a tall cylinder. The excess of ammonia, together with the hydro- chloric acid formed in the reaction were thus absorbed. What is the residual volume of gas and what is its volume; tempera- ture and pressure constant? 1 7. 7 cc of a gaseous mixture consisting of carbon monoxide, methane and ethane were mixed with 28 cc of oxygen and burned in an explosion pipette. After cooling the residual gases were fcund to consist of 8.4 cc of carbon dioxide and 16.1 cc of unconsumed oxygen. What was the composi- tion of the mixture? 1 8. What volume of gas will be produced by passing a cubic foot of steam over red hot coal and what will be its com- position? If the steam be passed over heated copper what gas will result and how much will be produced by a liter of steam? 19. A liter of hydrogen containing small amounts of methane and nitrogen was treated with palladium which re- moved all but 38.8 cc. 99.6 volumes of this were mixed with air and oxygen to a total volume of 566.5 and exploded. Its volume then was 413.20, and after treatment with potash 402.1. What were the percentages of nitrogen and methane in the original gas? 20. 560 cc (an excess) of oxygen were added to a mix- ture of ammonia, ethylene and hydrogen, measuring 2 1 7 cc, and the total mixture was exploded. I he resulting gaseous product measured 444.5 cc. After absorption of the carbon dioxide by caustic potash the volume of the dry gas measured 304.5 cc. What was the composition of the original mixture? CHEMICAL CALCULATIONS 109 2 1. 60.2 cc of a gas mixture consisting of carbon dioxide, carbon monoxide, hydrogen, methane, ethylene and nitrogen were treated with caustic potash solution which absorbed all but 59.43 cc. The remaining gas was treated with fuming sul- phuric acid to remove ethylene, further reducing the volume to 5 7.82 cc which were mixed with 39.04 cc of oxygen and 228.35 cc of air, and exploded in a eudiometer tube.. The resulting volume was 268.72 cc which, on treatment with caustic potash solution, decreased to 238.43 cc. The original gas contained 1.61 % of nitrogen. Calculate the percentage composition of the mixture. 22. A tube contains 40.50 cc of hydrogen and 20.25 cc of oxygen at a temperature of 125 C. In what respects do the contents differ from steam at the same temperature? What effects will be observed upon passing a spark through the mix- ture? 23. Given 70 cc of a mixture of CH4 and C2H2. To this was added 175 cc of oxygen and the mixture exploded. Vol- ume after explosion 22 7.5 cc. Calculate the number of cc of each gas in the original mixture. 24. Given 100 cc of a mixture of H2, CO and CHt. To this was added 1 00 cc of oxygen and the mixture exploded. Volume after explosion 1 65 cc. When this gas after explosion was passed through KOH its volume was reduced to 85 cc. Calculate the volume of each of the three gases present in the original mixture. 25. Given 500 cc of a mixture of H2, C,H2 and CO. To this was added 800 cc of oxygen and the mixture exploded. 7he volume of the cooled gas after explosion was 650 cc. By bubbling through KOH this was further reduced to 150 cc. Calculate the volume of each of the three gases in the original mixture. 26. Given 1 liter of a mixture of CH(, CO, N2 and C2H2. 7o this was added 500 cc of oxygen and the mixture exploded. Volume after explosion was 1400 cc. The water vapor was removed by condensing it. 7 his caused the volume to de- crease to 1 1 00 cc. By bubbling through KOH to remove CO2 the volume was decreased further to 700 cc. Calculate the volume of each of the four gases in the original mixture. CHAPTER XI OUTLINE Heat is evolved or absorbed when a chemical reaction takes place. The calorific power of a substance is the heat evolved (calories) during the combustion of a unit weight of a sub- stance. In the case of gases it is often considered as the heat evolved by the combustion of a unit volume of the gas. Calorific power of some familiar substances. Solution of an Example-Calorific power of a sample of coal. Calorific Intensity-The highest theoretical temperature to which the products of a combustion are raised by the heat evolved. Specific Heat of a Gas-The quantity of heat (calories) required to raise unit volume (usually 1 liter) of the gas through 1 °C. Specific heats of N2, H2, O2, CO, CO2, H..O vapor. Solution of a Problem-The temperature of the flame of Hydrogen when burning (a) In oxygen (b) In air. Problems. 110 CHEMICAL CALCULATIONS 111 CALORIFIC POWER AND CALORIFIC INTENSITY. When a chemical reaction takes place, all or part of the original substance disappears and new substances are formed. Coincident with this change in the state of matter there occurs a transformation of energy which is usually, but not always, accompanied by the evolution or absorption of heat; in other words, the temperature of the product of the reaction either rises or falls. A reaction during which heat is evolved is known as an EXOTHERMIC REACTION; and one in which heat is ab- sorbed an ENDOTHERMIC REACTION. In considering the calorific power of substances, we will deal exclusively with exothermic reactions of a type included under the term COMBUSTION. By the combustion of a substance we commonly under- stand that it burns, and in so doing emits light and heat. Such a phenomenon implies a chemical reaction and can be expressed by an equation. When a piece of charcoal burns in the air the carbon unites with oxygen to form carbon monoxide or carbon dioxide, depending on whether there is a deficiency or an excess of air. At the same time heat is evolved. C -f- O2 = CO2 heat. When a piece of magnesium ribbon burns in the air the reaction is represented by the equation. Mg T O = MgO heat. When iron and sulphur are heated together under the proper conditions they combine with the evolution of light and heat as represented by the equation Fe-|-S = FeS-|- heat. The heat evolved in these various reactions can be meas- ured by carrying out the chemical change in an instrument known as a calorimeter. The heat evolved is expressed in Calories or British thermal units. 112 CHEMICAL CALCULATIONS The calorie (cal.) is the quantity of heat necessary to raise the temperature of one gram of water one degree Cen- tigrade. The large or kilogram calorie (Cal.) is equal to I 000 calories. The British thermal unit is the quantity of heat necessary to raise the temperature of one pound of water one degree Fahrenheit. In a great many cases it is more convenient, though not as accurate, to arrive at the heat evolved in a combustion reaction by calculation instead of making an actual calorimetric test, and in order to do this we have recourse to what are known as the calorific powers or the heats of combustion of substances. These have previously been determined by care- ful experiment. The CALORIFIC POWER of a substance is the heat evolved per unit mass when it burns and is usually expressed as calories per gram molecular weight; Calories per kilogram molecular weight or B. T. U. per pound. This number of units is a constant for any given substance irrespective of the manner in which the combustion is carried on so long as it is complete; that is, the calorific power of a substance is the same whether the substance is burned in air or in pure oxygen. The following table gives the calorific power in calories per gram molecular weight of various substances. For in- stance, (C,O2) 12 + 32 = 44 97,200 simply means that when 12 grams of carbon unite with 32 grams of oxygen, 44 grams of carbon dioxide are formed and at the same time 9 7,200 calories of heat are evolved. Hence when carbon burns to carbon dioxide the heat evolved amounts to 97,200 calories per gram molecular weight of carbon or 97,200 Calories per kilogram molecular weight. The calorific power per gram of carbon would be 97200 = 8, 1 00 cal. 12 CHEMICAL CALCULATIONS 113 Molecular Weights (Approx.) Heat Evolved (cal.-mol. wt.) *(C,O2) 12 + 32 = 44 97,200 (H2O) 2+16= 18 **68,9 7 7 liquid **58,060 gas (S,O2) 32 4- 32 = 64 69,260 (PA) 62 4- 80 = 142 365,300 (CO.O) 28 4- 16 = 44 68,040 (Zn,O) 65 4 16 = 81 84,800 (Mn,O) 55 4-16 = 71 ' 90,900 (Si,O2) 28 4- 32 = 60 180,000 (Sn,O,) 1 18 4- 32 = 150 141,300 (Fe2.O3) 1 12 + 48 = 160 195,600 The calorific power of gases is often expressed as calories per liter or calories per cubic meter, and it is advisable in gas calculations to use the calorific power per unit volume rather than the calorific power per unit weight, since in most cases it is more desirable to ascertain the volumes of gases than their weights. If we use the calorific powers per unit volume we can calculate the heat loss directly instead of first converting the volumes into weights and with less chance of error. Since in engineering work the only kind of combustion considered for the production of heat is the combination of various kinds of fuel with oxygen, let us calculate the heat evolved by the combustion of a gram of coal, the ultimate analysis of which is shown to be In general the calorific power of a compound will NOT be the sum of the calorific powers of the elements of which it is composed. This is due to the fact that some of the heat is utilized in undoing the work of combination of the com- pound. * See "Metallurgical Calculations" Richards, Vol. I, pp 1 6. ** In ordinary work the moisture formed by the com- bustion of hydrogen, passes off as vapor, but if it be condensed, as is the case in calorimetric measurements, then taking the products cold, the heat of condensation of water (606.5 cal- ories per gram of water) is given off and the total heat evolved = 58,060 + (18 X 606.5) = 68.977. 114 CH EM ICA L C A LCULATIO NS Carbon 93.00% Hydrogen 2.00% Oxygen 4.00% Nitrogen 1.00% 100.00% Since the combustion is a chemical reaction it can be indi- cated by the following equations: (1) c + o2 = co2 (2) " 2H, 4- O, = 2H,O (vapor) I he nitrogen does not enter into consideration since it does not combine with oxygen under these conditions but passes off as such with the carbon dioxide and water vapor. As the carbon is 93.00 per cent, of the whole we have 0.93 gram of carbon. Turning to the table on page 1 I 3 we find that when 12 grams of carbon burn, 97,200 cal. of heat are evolved. This is, of course, 8 1 00 cal. for one gram. Heat evolved by the carbon in the 1 g. of coal = 0.93 X 8100 = 7533 cal. The hydrogen unites with oxygen to form water vapor. In the case of fuel we assume that all the oxygen shown to be present by analysis is already combined with part of the hydro- gen, thus leaving only a portion of the hydrogen to unite with the oxygen of the air. And it is only this portion of the hydro- gen which evolves heat when the fuel is burned. The first step then is to calculate the portion of hydrogen already combined with the oxygen in the coal. From the reaction we know that 16 grams of oxygen require 2.016 grams of hydrogen or 2.016 1 gram of oxygen requires grams of hydrogen 16 Analysis shows the coal to contain 0.04 gram of oxygen. Therefore the hydrogen necessary to combine with this is 0.04 X 2.016 = 0.005 gram of hydrogen I he hydrogen left over or available hydrogen, as it is called, is O.O2-O.OO5 = 0.015 gram. From the table we see that when 2 grams of hydrogen unite with 1 6 grams of oxygen to form 1 8 grams of water vapor CH EMICA L CALCULA TIONS 115 58060 58.060 cal. of heat are evolved, or = 29,030 cal per 2 gram. (When fuel burns the temperature of the products of combustion is high enough to keep the water in the form of vapor, hence the use of the figure 58,060 instead of 68,977). Heat evolved by the available hydrogen is . 0.015 X 29,030 = 435.45 cal. 7 o sum up: bleat from carbon 75 33.00 cal. Heat from available hydrogen 435.45 I otal heat evolved per gram of coal 7968.45 cal. I he following are the calorific powers of some familiar substances: Heat Evolved Name Formula cal. gram Marsh Gas ch4 13,060 Benzine c6hg 9,910 Petroleum 1 1,400 Alcohol C.H-OH 7,180 Ether C2H5OC2H5 9,020 Anthracite Coal 8,000 Bituminous Coal 7,500 Ethylene c'h' 1 1,090 CALORIFIC INTENSITY. CALORIFIC INTENSITY is the maximum theoretical temperature to which the products of combustion of a sub- stance can be raised starting from zero Centigrade. When a substance burns, definite weights and volumes of combustion products are formed and a definite amount of heat is liberated. This heat raises the temperature of the The composition of gases is usually expressed as per cent, by volume and it must be remembered that a gram molecular weight of any gas at standard conditions of temperature and pressure occupies 22.3 liters. This enables us to calculate the weight of a liter of any gas. 116 CHEMICAL CALCULATIONS products of combustion. In order to calculate what this tem- perature will be in any given case we must first calculate the volumes or weights of the products, together with the heat evolved; knowing the amount of heat necessary to raise a unit weight or volume of the products of combustion one de- gree Centigrade, the final temperature to which they are raised is not difficult to ascertain. The number of heat units expressed in calories, required to raise the temperature of a certain amount of any substance one degree Centigrade, is called the Specific Heat-Capacity or SPECIFIC HEAT of the substance. In the case of solids the specific heat is expressed as calories required to raise the temperature of one gram of the substance one degree Centi- grade. In the case of gases, however, it is more convenient to express it as calories required to raise one liter one degree Centigrade. The following are the mean specific heats of gases be- tween 0 and tcC. They are given in cal. per liter per 1 C. z Nitrogen Hydrogen Oxygen Carbon Monoxide (0.303 + 0.000027t) Carbon Dioxide (0.370 -|- 0.00022t) Water Vapor (0.340 -fi 0.000 1 5t) In order to show the application of the foregoing remarks let us calculate the calorific intensity of hydrogen (a) when burned in oxygen: (b) when burned in air. When hydrogen burns we have the equation 2H, H- O2 - 2H..O (vapor) 2 volumes I volume = 2 volumes One liter of hydrogen requires /i liter of oxygen for complete combustion and produces one liter of water vapor. The amount of heat produced by one molecular weight of hydrogen when burned to water vapor is 58,060 cal. The volume of one molecular weight in grams is 22.3 liters, conse- quently the heat of combustion of one liter of hydrogen is 58,060 = 2603 cal. 22.3 CHEMICAL CALCULATIONS 117 Disregarding the heat utilized in raising the temperature of surrounding bodies we can consider that, theoretically, all the heat produced by the burning of a substance is utilized in raising the temperature of the products of combustion. In the case under consideration the heat raises the temperature of the one liter of water vapor. The mean specific heat of water vapor per liter is (0.340 + 0.0001 5t) cal. Therefore the total heat required to raise one liter of water vapor from O C to t° Centigrade is t X (0.340 + 0.0001 5t) cal. The total heat available in this case is 2603 cal., and there- fore we have the equation t(0.340 + 0.00015t) - 2603 Solving this equation* we find that the theoretical tem- perature is 3183cC. When hydrogen is burned in the air the products of com- bustion will consist of water vapor and nitrogen. The heat produced per liter of hydrogen is the same as in case (a) 2603 Cal.; but the heat is not only absorbed by the water vapor but also by the nitrogen. As in case (a) the volume of water vapor produced is one liter. The volume of the nitrogen is obtained from the oxygen required as follows: Volume of oxygen required by one liter hydrogen 0.5 liter. Composition of air by volume is 79.1 per cent, nitrogen and 20.9 per cent, oxygen. Hence the volume of nitrogen 79.1 accompanying 0.5 liter of oxygen = 0.5 X - 1-892 liter. ' 20.9 * The above equation being an affected quadratic may be reduced to the form ax2 -f- bx -f- c - O. The solution of this equation is b I b2 - 4ac x - -- 2a where x - t a = coefficient of the term t2 b = coefficient of the term t c = absolute term, including the sign of the term. 118 CHEMICAL CALCULATIONS Heat necessary to raise one liter of water vapor to t C is 1 X (0.340 + 0.0001 5t)t cal. Heat necessary to raise 1.892 liter of nitrogen to t C is 1.892 X (0.305 + O.OOOO27t)t cal. = 0.577 + 0.00005 lit2 Heat in water vapor = 0.340t -{- 0.000 1 5t-' Heat in nitrogen = 0.5 77t -|- 0.000051 It2 Total heat = 0.9 1 7t + 0.000201 It2 = 2603 Solving this equation we find that t = 1982 C. PROBLEMS. 1. Calculate the calorific power of water gas which con- tains hydrogen 52.00%, carbon monoxide 44.00% ; nitrogen 4.00%. 2. Calculate the calorific intensity of carbon monoxide* when burned in (a) oxygen, (b) in air. 3. Calculate the calorific intensity of benzene when burned in oxygen. 4. Calculate the calorific power of anthracite coal of the following composition: Carbon 95.00%, moisture 3.00%, ash 2.00%. 5. A sample of natural gas contained by analysis marsh gas 94.00%, ethylene 0.5%, nitrogen 3.00%, carbon dioxide 0.5%, hydrogen 2.00%. Calculate its calorific power, ex- pressing it as calories per liter. 6. Calculate the calorific power of a bituminous coal which contained by analysis, carbon 75.00% , sulphur 0.52% , hydrogen 5.46%, moisture 19.02%. 7. Calculate the calorific intensity of pure carbon when burned in air. 8. Calculate the temperature of the oxyhydrogen flame. 9. Calculate the temperature of the flame of an air blast lamp which burns carbon monoxide and uses 25% excess air over that needed for combustion. CHEMICAL CALCULATIONS 119 10. Calculate the maximum theoretical temperature which can be reached in a Bunsen burner using gas of the composition; Carbon monoxide 31.00%, ethylene 3.00%, nitrogen 66.00%, no excess of air being used. 1 1. How much heat will be lost in the stack gases from a furnace per pound of coal burned on the grate, the coal analyzing as follows: Carbon 75.44%, hydrogen 5.37%, nitrogen 1.22%, sulphur 1.00%, oxygen 8.87%, ash 8.42%, moisture 0.68%. Assume that the air used for combustion is dry and that no carbon monoxide appears in the gas. 1 he gases enter the stack at a temperature of 800 cC. 12. Calculate the theoretical temperature of combustion of a gas of the following composition: Carbon monoxide 80.00%, marsh gas 20.00%. 1 3. In the above problem assume both gas and air to be preheated to 350 C. 14. Using same data as in Problem 13, assume 10.00% excess air is used. ANSWERS CHAPTER I. 1. 140cF 2. 93.33°C 3. 30°F = 271.89° Abs -20cC = 253° Abs 0cF = 255.23° Abs -125°F = 185.78° Ab 3 4. 8.6CF and -13°C 5. 104cF 6. 103.88 centi- grade degrees 7. 1.51 8. 15.92 9. 31.97 10. 1.97 11. 1.039 12. 63.61 cc 13. 9.027 14. 8.161 15. 0.8372 16. 0.7069 .17. 2.333 18. 556.38 pounds 19. 238.31 cc 0.7073 20. 1.632 21. 11.28 22. 1.839 23. 0.847 CHAPTER II. 1. 250 cc, 750 cc 2. 353cC 3. 67.56 cc 4. 1.089 g. 5. pv = K 6. 532.0 mm 7. 728.52 mm 8. 876.1 7 mm 9. 468.1 mm 10. 47.85 mm 11. 9 1 7 °C 12. 80.44 C 1 3. 1.684 g. 14. 1.222 g. per liter 15. 52.5 cu. ft. 16. 34.15 cc 17. 85.54cC 18. 2.39 g. 19. 5.000 cc 4 g. per liter 20. 257.2 cc 21. 141.28cC 22. 310.67cC 23. 665.7 cc 24. 742.7 mm 25. 2.76 Kg per sq cm. CHAPTER HI. 1. (a) 32.06 (b) 36.46 parts (c) 36.46% 2. (a) 200.6 (b) 86.22 parts (c) 86.22% 3. (a) 119.112 (b) 49.73 g. (c) 49.73% 4. (a) 32.06 (b) 13.73g. (c) 13.73% 120 CHEMICAL ( ALCULALIONS 121 5. (a) 80.06 (b) 34.29 g. (c) 34.29% 6. (a) 48.64 g. (b) 2 1.83 g. (c) 36.20 g. (d) 21.83% (e) 36.20% 7. (a) 11 1.68 g. (b) 69.94 g. (c) 72.35 g. (d) 69.94% (e) 72.35 % 8. 22.43 g. 9. 21.85 g. 10. 1 1.85 g. 11. 135.82 g. 12. 156.28 g. 1 3. 305.9 1 pounds 14. 2.458 g. 15. 23.67 g. 16. 70.04% 17. 24.34% 18. 39.12% 19. 60.46% 20. 127.08 g. 73.07 g. 39.16 g. 21. 14.77 g. 22. 96.23% 23. 79.40% 24. 26.83% 25. 59.86 g. 26. 59.04% 2 7. 636.39 pounds 28. 10.74% 45.55 % 29. 1292.78 CHAPTER IV. 3. CuCl 4- CuCl, 4. P,O3, P2O4, P.O, 5. SnO-2, SnOi 6. KC1O, KC1O„ KC1O3, KC1O4 7. H,O, H..O, 8. O,,O; 9. Na,SO3, Na„SO4, Na,S,O3 10. ch4, c,hb, c2h4, c3h„ c4h10 1 1. HPO,, H3PO3, H;iPO4 12. K,SO3, K,SO4, K2S2Ot 1 3. HNO„ HNO3 1 4. Sb2O3, Sb.O, 15. As A, As,O5 16. Na,O.SO3> Na,O, 2SO, 1 7. Na.,HPO4, NaH..PO4 18. K,CrO4, K2Cr.O7 19. HPO3, H;PO4 20. FeSO4, Fe,(SO,)3 2 I. As,S.; - As = 60.9 19 % S = 39.08 I % As.,S, - As == 48.32 7% S = 51.673% 22. HgNO; - Hg= 76.389 N = 5.334 0 =18.276 Hg(NO3)2 Hg = 61.795 N = 8.631 0 = 29.573 122 CHEMICAL CALCULATIONS CHAPTER V. 1. 44.49 2. 1.973 g. 3. 1.528 g. 1.528 g. per liter 4. 1.9 74 g. per liter 5. 0. 7638 g. per liter 8.515 7. 6496 cc 8. 42.92 9. 75.59 10. 271.52 12. 139.375 14. 234.1 5 liters 0.1 8005 g. per 2.006 [liter 15. 18.08 1.139 0.514 16. 22.3 liters 2.218 g. per liter 24.71 17. 16.00 g. 18. 52.36 19. 55.75 20. 1.7946 g. per liter 40.01 1 3. 1 1 7,075 cc 0.1 544 g. per 1.72 [liter CHAPTER VI. 1. 107.88 2. 103.6 3. 18.61 4. 31.785 5. 35.46 6. 79.92 7. 28.0C 8. 10.35 9. 8.00 16.03 1.008 35.46 32.685 10. 32.685 I 1. 9.033 12. 31.606 13. 63.018, 85.693, 70.78, 20.427 14. 80.928, 75.49, 43.66, 102.458 15. 20.008, 26.01 1, 51.715, 54.33 16. 92.08, 95.03, 43.46, 149.818 17. 17.008, 24.00, 11.355, 18.2 18. 25.995, 79.6, 65.1 15, 22.7 19. 43.06, 66.1, 72.608, 21.558 20. 26.12, 20.025, 82.708, 88.758 21. UO3 WO, CoO CdO ZrO. 47.7, 38.66, 37.48, 64.2 30.65 22. 226, 23. 195.2, 24, 100.3 25, 31.785 CHEMICAL CALCULA'LIONS 123 CHAPTER VII. 1. 12.21 g. 2. 1 1.20 g. 3. 32.86 g. 4. 3.051 g. 5. 5716 pounds 6. 7.55 pounds 7. 18.10g. 8. 2.231 g. 24,686 cc 9. 1635 g. 1 0. 296.9 pounds 11. 1 pound of Ca (OH) 2 to each 3.265 pounds of FeSO4.5H.,O 12. 1 part of Al to each 2.947 parts of Fe2O3 13. 19.07 liters 14. 4.155 g. 15. 35.89 g. 16. 56.51 g. 17. 7.553 g. 18. 1 7.47 g. 19. 38.19 g. 20. HgO -$2.70 K.CIO, - for 1 ounce of oxygen 21. 37.59 g. CHAPTER VIII. 1. H2O2 2. COS 3. KHSO4 4. MgSO4.7H,O 5. ZnSO4.7H.,O 6. Ca,(PO4)J 7. C1OHS 8. Na.,AlFt. 9. COH, 10. KNO, 1 1. HCOOH 12. Pb3O4 13. (CaSO4)2H2O 14. 28 15. 2MgO.2SiO,.3H.,O 16. 3CoO.As,O,.8H,O 17. Na.,O.Al.,O.,.6SiO? 18. No„O.2BeO.6SiO.,.H.,O 19. Na.,O.Al.,O.,.6SiO., 20. 3 (Mg,Fe)O.4SiO.,.H..O 2 1. C4H1() 22. N2O4 2 3. Fe2Cl(i 24. FeSO4.5H.,O 25. (NH.l).,SO4.FeSO4.6H.O CHAPTER IX. 1. 47.018 g., 26.041 g., 180.84 g„ 74.56 g. 2. 35.05 g„ 49.038 g., 115.45 g„ 20.16g. 3. 47.1 g., 127.928 g., 35.62 g„ 192.92 g. 124 CHEMICAL CALCULATIONS 4. 60 cc 5. 20.39 cc 6. 133.2cc 7. 17.21 cc 8. 2.91 g. 9. 96.74% 10. 30.28% 11. 31.78% 12. 7.2936 g. 40 cc 13. 9.8076 g. 1 18.5 cc 14. 3.95 g. 15. 67.54 cc 1 6. 396.8 cc 17. 2 1 5 cc 18. 56.10% 19. 1107cc 20. 49.033 g„ 61.28 g., 72.608 g., 140.81 g. 21. 10.91 g., 4.259 g.. 24.189 g„ 10.765 g. 22. 243.6 cc 2 3. 58 69 cc 24. 2.866 g. 25. 1 1.963 g. CHAPTER X. 1. (a) 50 liters (b) 62.5 liters (c) 37.5 liters 2. 350 cc 3. CO, - 60 liters SO, - 20 liters 4. 26.6 volumes of H, remain. 55.3 volumes of N, 5. 25.2 cc O2, 44.8 N2 ' 6. CO - 63 cc ; Cbb - 7 cc 7. CO - 56 cc; CH4 - 70 cc; N.> - 56 cc 8. 665 cc; 280 cc H,O vapor 9. N,O - 43.5 per cent.; NO - 49.4 per cent; N, - 7. 1 per cent. 10. 35 cc 11. 150 cc CO; 100 cc C2H2 1 2. CH4 - 7.5 per cent.; N, - 80 per cent.; H, - 12.5 per cent. 1 3. CH, - 84 cc; air - 1 4 cc; H,S - 1 1 2 cc 14. CO - 41.9%; H, -43.6%; N, - 1.9%; C,H4 - 12.6%. 15. CO, - 0.7%; C,H4 etc - 4.2%; H, - 48 % ; CH4- 40 % ; CO - 7 % 1 6. 30 cc 17. CO - 2.8 cc; CH4 - 2.8 cc; GHu - 1.5 18. H, and CO - 2 cu. ft.; 1 liter of H, 19. CH,- 0.43%; N, - 0.043% 20. NH, - 1 12 cc; C,H4 - 70 cc; H2 - 35 cc 21. CO,- 1.28%; C2H4 -2.68%; CO -48.7%; H, - 44.1%; CH4- 1.66%; N2 - 1.61% CHEMICAL CALCULATIONS 125 22. A contraction of 20.25 cc 23. CH( - 35 cc; CM., - 35 cc 24. H2 - 20 cc; CO - 50 cc; CH4 - 30 cc 25. M - 200 cc; CM, - 200 cc; CO - 100 cc. 26. CHt - 100 cc; CO - 1 00 cc; N, - 700 cc; CM,- 100 cc. CHAPTER XI. 1. 2. 2975=C 2O33cC 3. 3592°C 4. 7695 cal. 5. 9397 cal. 6. 7671 cal. 7. 1939 cal. 8. 3183°C 9. I252°C 10. 1592cC 11. 12. 3124cC 13. 321O°C 14. TENSION OF AQUEOUS VAPOR IN MILLIMETERS Temp. °C. Pressure Temp. °C. Pressure 0 4.6 28 28.1 5 6.5 29 29.8 8 8.0 30 31.5 9 8.6 31 33.4 10 9.2 32 35.7 II 9.8 33 37.4 12 10.5 34 39.6 13 11.2 35 41.8 14 11.9 40 54.9 15 12.7 45 71.4 16 13.5 50 92.0 17 14.4 55 117.5 18 15.4 60 148.9 19 16.3 65 187.1 20 17.4 70 233.3 21 18.5 75 288.8 22 19.7 80 354.9 23 20.9 85 433.2 24 *22.2 90 525.5 25 23.6 95 633.7 26 25.1 100 760.0 27 26.5 126 DATA Length. 1 0 millimeters (mm) = 1 centimeter (cm) 1 0 centimeters (cm) = 1 decimeter (dm) 1 0 decimeters (dm) = 1 meter 1 inch - 2.54 centimeters (cm). 1 cm - 0.393 inch Volume. 1 cubic decimeter = 1000 cubic centimeters (cc) = 1 liter 1 cubic inch = (2.54 cm)3 = 16.39 cubic centimeters (cc) 1 gallon = 231 cubic inches = 3785.4 cubic centimeters (cc) 1 quart = 946.35 cc = 0.94635 liter 1 gallon of water weighs 8.34 pounds avordupois 1 cubic centimeter of water at 4 cC weighs 1 gram (g.) 1000 grams (g.) = 1 kilogram (Kg) 1 pound avordupois = 453.6 grams 435.6 1 ounce avordupois = = 28.35 grams 16 1 calorie is the amount of heat required to raise the tempera- ture of 1 gram of water 1 cC Latent heat of vaporization of water = 537 calories Latent heat of fusion of water = 80 calories