STOICHIOMETRIC CALCULATIONS

1. What is "stoichiometry"?

The word STOICHIOMETRY simply means the determination of the numerical proportions in which chemical substances react. More specifically, it involves the relationships BY MASS that are involved in chemical reactions. Stoichiometric calculations are used routinely by chemists, and form the basis of a very important branch of chemistry known as QUANTITATIVE ANALYSIS.

In this topic, we will look at typical examples of stoichiometric calculations.

2. Calculation of the percentage composition of a compound

If the formula for a compound is known, it is easy to calculate the proportions by mass of each element in that compound. For this, however, one must know the relative atomic masses of each element. In most cases, these are known with great accuracy, and are either given as part of the problem, or they can be obtained from tables of atomic masses.

The basis for all stoichiometric calculations are the Law of conservation of mass () and the Law of constant composition ().

3. Calculation of empirical formulae

If one knows the percentage composition of a compound, as well as the relative atomic masses of the elements which make up the compound, one can calculate the simplest possible formula for that compound.

A formula that expresses the simplest whole-number ratio of its constituent atoms is known as an EMPIRICAL FORMULA.

4. Mass relations from chemical equations

Given a BALANCED chemical equation, and the relative atomic masses of all the atoms that participate in the reaction, it is possible to determine the mass relationships of all the molecules involved.

5. Additional questions












Calculation of the percentage composition of a compound: Worked example

What is the percentage of sulphur in sulphuric acid, H2SO4, given that the relative atomic masses of the constituent atoms are H = 1.008, S= 32.064, O = 16.000?

Answer:

The relative molar mass is (2 x 1.008) + (1 x 32.064) + (4 x 16.000) = 98.080.

Thus, in 98.080 g of H2SO4, there will be 32.064 g sulphur, since one molecule of sulphuric acid contains one atom of sulphur.

The percentage sulphur in H2SO4 is 100 x 32.064/98.080 = 32.63%












Calculation of empirical formulae: Worked example

A certain oxide of iron contains 72.36% Fe and 27.64% O. What is the simplest formula that can be assigned to this oxide? (Ar: Fe = 55.847, O = 16.00.)

Answer

Knowing the percentage composition of each of the elements and the relative atomic masses of these elements, we can calculate the ATOMIC RATIO X : Y in the formula FeXOY.

X = 72.36/55.847 = 1.296, and Y = 27.64/16 = 1.728.

We can write the oxide as Fe1.296O1.728, or (by dividing each ratio by 1.296) Fe1O1.33

Since the formula must have a whole number of each of its constitutent atoms, the empirical formula must be Fe3O4.












Mass relations from chemical equations: Worked example

Calculate the mass of quicklime, (calcium oxide, CaO), that can be obtained by heating 2.0 kg of pure calcium carbonate, CaCO3. The reaction is:

CaCO3 → CaO + CO2

Answer

The relevant atomic masses are: Ca = 40.08, C = 12.01, O = 16.00. From these we calculate the relative formula masses: CaCO3 = 40.08 + 12.01 + 3 x 16 = 100.1, CaO = 40.08 + 16.00 = 56.08, CO2 = 44.01 (this is not required to solve the problem).

From the reaction equation we see that since one molar mass CaCO3 (100.1 g) gives, upon heating, one molar mass of CaO (56.08 g), then 2000 g CaCO3 will give 2.0 (kg) x 1000 (g·kg-1)x 56.08/100.1 = 1120 g = 1.120 kg CaO/b>.