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