Laws of stoichiometry

Stoichiometry is the branch of chemistry that deals with the calculation involving mass and volume relationship among reactant and product in a chemical reaction.


Some laws of stoichiometry
  • Law of conservation of mass
  • Law of constant proportion
  • Law of multiple proportions
  • Law of reciprocal proportions
  • Gay Lussac’s law of gaseous volume

Law of conservation of mass

It states that “Mass is neither created nor destroyed in a chemical reaction or The total mass of reactant is equal to the total mass of product in a chemical reaction”.

For a general reaction,

A + B = C + D

let w1 gm of A reacts with w2 gm of B to give w3 gm of C and w4 gm of D, then according to this law,

w1 + w2 = w3 + w4

Illustration,

H_{2} + Cl_{2} \rightarrow 2HCl\\ \begin{align*} Mass\ of\ reactant &= 2 \times mass\ of\ H + 2 \times mass\ of\ Cl\\ &= 2 \times 1 + 2 \times 35.5\\ &=73\\ Mass\ of\ product &= 2(mass\ of\ H + mass\ of\ Cl)\\ &= 2(1+35.5)\\ &= 73 \end{align*}

Solved examples

What weight of sodium chloride reacts with 6.8 gm of silver nitrate, if 5.74 gm of silver chloride and 3.4 gm of sodium nitrate are formed?
Solution:

\begin{align*}
\underset{w_{1}(?)}{Nacl} + \underset{w_{2}(6.8gm)}{AgNO_{3}}\ \rightarrow \underset{w_{3}(5.74gm)}{AgCl} + \underset{w_{4}(3.4gm)}{NaNO_{3}}\\
\end{align*}\\

According\ to\ law\ of\ conservation\ of\ mass,\\
w_{1} + w_{2} = w_{3} + w_{4}\\
or,\ w_{1} + 6.8 = 5.74 + 3.4\\
\therefore w_{1} = 2.34\ gm


Law of constant or definite proportion

It states that “The same compound always contains the same elements combined together in the same proportion by weight whatever the origin or mode of formation of compound”.

Illustration:

1. Let us take an example of water. Water can be obtained from a tap, river, ocean, lake, etc. It also can be synthesized in the lab. When the pure sample of water is analyzed, it is found to be composed of hydrogen and oxygen which combine together in a 1:8 ratio by mass irrespective of the source of water.

ii. Carbon dioxide may be obtained from various sources like

C + O2 → CO2
CaCO3 → CaO + CO2

Regardless of the source of CO2, the mass ratio of C and O = 12:32 = 3:8.


Solved Example

1.81 gm of copper was obtained by heating 2.26 gm of copper oxide. In a separate experiment, 2.13 gm of copper was dissolved on nitric acid, copper nitrate so formed was ignited to give 2.67 gm of copper oxide. Show that the above experimental data explain the law of definite proportion.
solution:

Case I:
Weight of copper oxide = 2.26 gm
Weight of copper = 1.81 gm
Weight of oxygen = 2.26 – 1.81 = 0.45 gm
Ratio of weight of copper to oxygen = 4:1

Case II:
Weight of copper oxide = 2.67 gm
Weight of copper = 2.13 gm
Weight of oxygen = 2.67 – 2.13 = 0.54 gm
Ratio of weight of copper to oxygen = 4:1

In both the case ratio of the weight of copper to oxygen is the same. So the given data explain the law of definite proportion.


Law of multiple proportions

It states that “When one element combines with another element to form two or more different compounds, then the weight of one of the element which combines with the constant weight of other bears the simple whole-number ratio to one another”.

Illustration I:
Let us consider two elements hydrogen and oxygen that combine with each other to form water and hydrogen peroxide.

In H2O, 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen.
In H2O2, 2 parts by weight of hydrogen combine with 32 parts by weight of oxygen.
Thus, the weight of oxygen (16 parts and 32 parts) combined with the same weight of hydrogen (2 parts) bears to each other in a simple whole-number ratio = 16:32 = 1:2.

Illustration II:
Nitrogen and oxygen combine to form five different types of oxides:

Oxides of nitrogenMass of nitrogenMass of oxygenMass of oxygen to 14 parts of mass of nitrogen
N2O28168
NO141616
N2O3284824
NO2143232
N2O5288040

Here, the mass of oxygen which combines with the same parts by mass of nitrogen is in the ratio 8:16:24:32:40 which is equal to simple ratio 1:2:3:4:5.


Solved example

Phosphorous forms two compounds with chlorine. In one compound, 1.94 gm of P combines with 6.64 gm of Cl and in another, 0.66 gm of P combines with 4.51 gm of Cl. Are these data in accordance with the law of multiple proportion?
solution:

First case:
Weight of phosphorous = 1.94 gm
Weight of chlorine = 6.64 gm
1.94 gm of phosphorous combines with 6.64 gm of chlorine.
1 gm of phosphorous combines with 3.42 gm of chlorine.

Second case:
Weight of phosphorous = 0.66 gm
Weight of chlorine = 4.51 gm
0.66 gm of phosphorous combines with 4.51 gm of chlorine
1 gm of phosphorous combines with 6.83 gm of chlorine

The mass of chlorine combining with a fixed mass of phosphorous is in the ratio of 3.42:6.83 equal to 2:1 which is a simple whole-number ratio. So these data are in accordance with the law of multiple proportion.


Law of reciprocal or equivalent proportion

It states that “When two different elements combine separately with the same weight of third element, the ratio in which they do so will be same or some simple multiple of the ratio in which they unite with each other”.

Illustration:
The combination of three different elements carbon, hydrogen, and oxygen forms carbon dioxide, water, and methane.

In CO2,
12 gm of carbon combines with 32 gm of oxygen
1 gm of carbon combines with 32/12 gm of oxygen

In CH4,
12 gm of carbon combines with 4 gm of hydrogen
1 gm of carbon combines with 4/12 gm of hydrogen
The ratio of the weight of hydrogen and oxygen which combine separately with the same weight of carbon is 4:32 or 1:8.

In H2O,
2 gm of hydrogen reacts with 16 gm of oxygen
The ratio of the weight of hydrogen to oxygen = 2:16 or 1:8
So the ratio of the weight of hydrogen to oxygen is 1:8 which is the same as in the case of CO2 and CH4. Therefore CO2, CH4, and H2O follow reciprocal proportion.


Solved example

Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.97% of carbon, and sulphur dioxide contains 50% of sulphur. Show that these data are in agreement with the law of reciprocal proportion.
solution:

In carbon dioxide:
% of carbon = 27.27
% of oxygen = 72.73
27.27 gm of carbon combine with 72.73 gm of oxygen.
1 gm of carbon combine with 2.667 gm of oxygen.

In carbon disulphide:
% of carbon = 15.97
% of sulphur = 84.03
15.97 gm of carbon combine with 84.03 gm of sulphur
1 gm of carbon combine with 5.262 gm of sulphur

The weight of sulphur and oxygen which combine with the fixed weight of carbon is in the ratio 5.262:2.667 or 2:1.

In sulphur dioxide:
% of sulphur = 50
% of oxygen = 50
50 gm of sulphur combine with 50 gm of oxygen
Ratio of sulphur to oxygen = 50:50 = 1:1

The ratio of 2:1 is a simple multiple of the ratio 1:2. So these data are in agreement with the law of reciprocal proportion.


Gay Lussac’s law of gaseous volume

It states that “Under the similar conditions of temperature and pressure, the volume of gas reactant and gas product bears a simple whole-number ratio”.

Illustration:
Hydrogen combines with nitrogen to give ammonia gas.

 \underset{3\ vol}{Hydrogen} + \underset{1\ vol}{Nitrogen}\ \rightarrow \underset{2\ vol}{Ammonia}

The ratio of hydrogen, nitrogen, and ammonia is 3:1:2 which is a simple whole-number ratio that illustrates the gaseous volume.


Atomic mass

The atomic mass of an element is defined as the mass which shows how heavy is an atom of an element as compared to the one-twelfth mass of carbon-12 isotope.

Atomic\ mass = \frac{Mass\ of\ an\ atom}{1/12 \times mass\ of\ 1\ atom\ of\ C-12\ isotopes}

Atomic mass unit (amu)

One-twelfth the mass of one atom of carbon-12 isotope is called 1 amu.

Gram atomic mass

It is the atomic mass expressed in grams.
Atomic weight of hydrogen = 1.008 amu
1 gm atom of hydrogen = 1.008 gm
Atomic weight of oxygen = 16 amu
1 gm atom of oxygen = 16 gm


Molecular mass

The molecular mass of a molecule is defined as the mass of that molecule compared to one amu.

Molecular\ mass = \frac{Mass\ of\ a\ molecule}{1/12 \times mass\ of\ 1\ atom\ of\ C-12\ isotope}

Since both atomic mass and molecular mass are compared to the same amu, the molecular mass can be calculated by adding the atomic mass of all atoms present in the molecule.

For example:
Molecular mass of CO2 = 12 + 2 6 = 44 amu
Molecular mass of CaCO3 = 40 + 12 + 3 x 16 = 100 amu.

Some Important Questions
  1. State a. Law of conservation of mass b. Law of constant or definite proportion c. Law of multiple proportion d. Law of reciprocal proportion.
  2. Nitrogen reacts with oxygen to give N2O3 and N2O5. Which law is illustrated? State the law.
  3. 1.81 gm of copper was obtained by heating 2.26 gm of copper oxide. In a second experiment, 2.13 gm of copper was dissolved in nitric acid. Copper nitrate obtained was ignited and 2.67 gm of copper oxide was obtained. Show that the data explains the law of definite proportion.
  4. In an experiment, 0.936 gm of zinc was converted into oxide and the weight of oxide formed was 1.165 gm. In a second experiment, 1.236 gm of the metal was converted into oxide and the weight of oxide formed was 1.538 gm. Show that these results illustrate the law of constant proportion.
  5. A metal X forms two oxides, A and B. 3 gm of A and B contains 0.720 gm and 1.160 gm of oxygen, resp. Calculate the masses of metal in gm which combine with one gm of oxygen in each case. What chemical law do these data illustrate? State the law.
  6. Carbon combines with hydrogen to form three binary compounds X, Y and Z. The percentage of hydrogen in X, Y and Z is 25%, 14.3% and 7.7% resp. Which law of chemical combination is illustrated by this data?
  7. Three oxides of Pb named red lead, lead peroxide and litharge contain 90.6%, 86.6% and 92.83% of Pb respectively. Illustrate the law of multiple proportions from these data.
  8. Phosphine contains 91.1% of P. Water contains 88.8% of O and phosphorous tetraoxide contains 56.4% of P. Show that these data illustrate the law of reciprocal proportion.
  9. Two oxides of the same metal contain 7.18% and 13.39% of oxygen in their metal oxide. Find the formula of the second oxide if the formula of the first oxide is MO. (MO2)
  10. Hydrocarbons x, y and z are formed from carbon and hydrogen where the percentage of carbon is 75, 85.7 and 92.3 resp. Find the molecular formula of x, y and z. (CH4, C2H6, C2H2)

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