6.1 Stoichiometry and composition (Page 7/5)

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The composition of substances

The empirical formula of a chemical compound is a simple expression of the relative number of each type of atom in that compound. In contrast, the molecular formula of a chemical compound gives the actual number of atoms of each element found in a molecule of that compound.

Empirical formula

The empirical formula of a chemical compound gives the relative number of each type of atom in that compound.

Molecular formula

The molecular formula of a chemical compound gives the exact number of atoms of each element in one molecule of that compound.

The compound ethanoic acid for example, has the molecular formula CH $_{3}$ COOH or simply C $_{2}$ H $_{4}$ O $_{2}$ . In one molecule of this acid, there are two carbon atoms, four hydrogen atoms and two oxygen atoms. The ratio of atoms in the compound is 2:4:2, which can be simplified to 1:2:1. Therefore, the empirical formula for this compound is CH $_{2}$ O. The empirical formula contains the smallest whole number ratio of the elements that make up a compound.

Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula. There are three different types of composition problems that you might come across:

Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance.
Problems where you will be given the percentage composition and asked to calculate the formula.
Problems where you will be given the products of a chemical reaction and asked to calculate the formula of one of the reactants. These are often referred to as combustion analysis problems.

Khan academy video on molecular and empirical formulae - 1

Khan academy video on mass composition - 1

Calculate the percentage that each element contributes to the overall mass of sulfuric acid ( $H_{2} S O_{4}$ ).

Write down the relative atomic mass of each element in the compound.
Hydrogen = 1.008 $\times$ 2 = 2.016 u

Sulfur = 32.07 u

Oxygen = 4 $\times$ 16 = 64 u
Calculate the molecular mass of sulfuric acid.
Use the calculations in the previous step to calculate the molecular mass of sulfuric acid.

$M a s s = 2.016 + 32.07 + 64 = 98.09 u$
Convert the mass of each element to a percentage of the total mass of the compound
Use the equation:

Percentage by mass = atomic mass / molecular mass of H $_{2}$ SO $_{4}$ $\times$ 100%

Hydrogen

$\frac{2.016}{98.09} \times 100 % = 2.06 %$

Sulfur

$\frac{32.07}{98.09} \times 100 % = 32.69 %$

Oxygen

$\frac{64}{98.09} \times 100 % = 65.25 %$

(You should check at the end that these percentages add up to 100%!)

In other words, in one molecule of sulfuric acid, hydrogen makes up 2.06% of the mass of the compound, sulfur makes up 32.69% and oxygen makes up 65.25%.

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A compound contains 52.2% carbon (C), 13.0% hydrogen (H) and 34.8% oxygen (O). Determine its empirical formula.

If we assume that we have 100 g of this substance, then we can convert each element percentage into a mass in grams.
Carbon = 52.2 g, hydrogen = 13.0 g and oxygen = 34.8 g
Convert the mass of each element into number of moles
$n = \frac{m}{M}$

Therefore,

$n (c a r b o n) = \frac{52.2}{12.01} = 4.35 m o l$

$n (h y d r o g e n) = \frac{13.0}{1.008} = 12.90 m o l$

$n (o x y g e n) = \frac{34.8}{16} = 2.18 m o l$
Convert these numbers to the simplest mole ratio by dividing by the smallest number of moles
In this case, the smallest number of moles is 2.18. Therefore...

Carbon

$\frac{4.35}{2.18} = 2$

Hydrogen

$\frac{12.90}{2.18} = 6$

Oxygen

$\frac{2.18}{2.18} = 1$

Therefore the empirical formula of this substance is: $C_{2} H_{6} O$ . Do you recognise this compound?

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Source: OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2

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