You are here: Home Composition

## Composition

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 four different types of composition problems that you might come across:

1. 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.

2. Problems where you will be given the percentage composition and asked to calculate the formula.

3. 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.

4. Problems where you will be asked to find number of moles of waters of crystallisation.

The following worked examples will show you how to do each of these.

Percentage composition and empirical formula

## Example 1: Calculating the percentage by mass of elements in a compound

### Question

Calculate the percentage that each element contributes to the overall mass of sulphuric acid ($H 2 SO 4$).

#### Use the equation

Hydrogen

(1)

Sulphur

(2)

Oxygen

(3)

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

In other words, in one molecule of sulphuric acid, hydrogen makes up 2,06% of the mass of the compound, sulphur makes up 32,71% and oxygen makes up 65,23%.

## Example 2: Determining the empirical formula of a compound

### Question

A compound contains 52,2% carbon (C), 13,0% hydrogen (H) and 34,8% oxygen (O). Determine its empirical formula.

#### Give the masses

Carbon = 52,2 g, hydrogen = 13,0 g and oxygen = 34,8 g

#### Calculate the number of moles

$n=m M$(4)

Therefore:

(5)
(6)
(7)

#### Find the smallest number of moles

Use the ratios of molar numbers calculated above to find the empirical formula.

$units in empirical formula=moles of this element smallest number of moles$

In this case, the smallest number of moles is 2,175. Therefore:

Carbon

$4,35 2,175=2$(8)

Hydrogen

$12,871 2,175=6$(9)

Oxygen

$2,175 2,175=1$(10)

Therefore the empirical formula of this substance is: $C 2 H 6 O$.

## Example 3: Determining the formula of a compound

### Question

207 g of lead combines with oxygen to form 239 g of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: and ).

(11)

#### Find the moles of oxygen

$n=m M$(12)

(13)

Oxygen

(14)

#### Find the mole ratio

The mole ratio of $Pb:O$ in the product is $1:2$, which means that for every atom of lead, there will be two atoms of oxygen. The formula of the compound is $PbO$.

## Example 4: Empirical and molecular formula

### Question

Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: 39,9% carbon, 6,7% hydrogen and 53,4% oxygen.

1. Determine the empirical formula of acetic acid.

2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is 60,06 g·mol−1.

#### Find the mass

In 100 g of acetic acid, there is , and

#### Find the moles

$n=m M$

(15)

#### Find the empirical formula

 $C$ $H$ $O$ 3,325 6,6337 3,3375 1 2 1

Empirical formula is $CH 2 O$

#### Find the molecular formula

The molar mass of acetic acid using the empirical formula is 30,02 g·mol−1. However the question gives the molar mass as 60,06 g·mol−1. Therefore the actual number of moles of each element must be double what it is in the empirical formula ($60,06 30,02=2$). The molecular formula is therefore $C 2 H 4 O 2$ or $CH 3 COOH$

## Example 5: Waters of crystallisation

### Question

Aluminium trichloride ($AlCl 3$) is an ionic substance that forms crystals in the solid phase. Water molecules may be trapped inside the crystal lattice. We represent this as: $AlCl 3 .nH 2 O$. Carine heated some aluminium trichloride crystals until all the water had evaporated and found that the mass after heating was 2,8 g. The mass before heating was 5 g. What is the number of moles of water molecules in the aluminium trichloride before heating?

#### Find the number of water molecules

We first need to find n, the number of water molecules that are present in the crystal. To do this we first note that the mass of water lost is .

#### Find the mass ratio

The mass ratio is:

 $AlCl 3$ $H 2 O$ 2,8 2,2

#### Find the mole ratio

To work out the mole ratio we divide the mass ratio by the molecular mass of each species:

 $AlCl 3$ $H 2 O$ $0,02099...$ $0,12...$

Next we convert the ratio to whole numbers by dividing both sides by the smaller amount:

 $AlCl 3$ $H 2 O$ 0,020 997 375 0,122 086 57 $0,021 0,021$ $0,122 0,021$ 1 6

The mole ratio of aluminium trichloride to water is: $1:6$

And now we know that there are 6 moles of water molecules in the crystal. The formula is $AlCl 3 .6H 2 O$.

We can perform experiments to determine the composition of substances. For example, blue copper sulphate ($CuSO 4$) crystals contain water. On heating the waters of crystallisation evaporate and the blue crystals become white. By weighing the starting and ending products, we can determine the amount of water that is in copper sulphate. Another example is reducing copper oxide to copper.

## Exercise 1: Moles and empirical formulae

Calcium chloride is produced as the product of a chemical reaction.

1. What is the formula of calcium chloride?

2. What is the percentage mass of each of the elements in a molecule of calcium chloride?

3. If the sample contains 5 g of calcium chloride, what is the mass of calcium in the sample?

4. How many moles of calcium chloride are in the sample?

a) ${\text{CaCl}}_{2}$

b) The percentage by mass is the atomic mass of the element divided by the molecular mass of the compound:

Ca: $\text{% mass}=\frac{40,08}{110,98}×100=36,15\text{%}$

Cl: $\text{% mass}=\frac{\left(2\left(35,45\right)\right)}{110,98}×100=63,89\text{%}$

If we add these up we get 100%: $36,15+63,89=100$

c) Calcium makes up 36,15% of the calcium chloride, so the mass of calcium in the sample must be 36,15% of 5g:

d) The number of moles is:

 $n=\frac{m}{M}$ $n=\frac{1,808}{40,08}$

(Notice that this is the same as the number of moles in 5g of calcium chloride: )

13 g of zinc combines with 6,4 g of sulphur.

1. What is the empirical formula of zinc sulphide?

2. What mass of zinc sulphide will be produced?

3. What is the percentage mass of each of the elements in zinc sulphide?

4. The molar mass of zinc sulphide is found to be 97,44 g·mol−1. Determine the molecular formula of zinc sulphide.

a) We work out the number of moles of each reactant:

Zinc:

 $n=\frac{m}{M}$ $n=\frac{13}{65,38}$

Sulphur:

 $n=\frac{m}{M}$ $n=\frac{6,4}{32,06}$

Zinc and sulphur combine in a 1:1 ratio, i.e. there 1 atom of zinc reacts with 1 atom of sulphur. The empirical formula of zinc sulphide is therefore:

ZnS

b) The mass of zinc sulphide produced is:

 $m=nM$ $m=\left(0,20\right)\left(97,44\right)$

c) Zinc: $\frac{65,38}{97,44}×100=67\text{%}$

Sulphur: $\frac{32,06}{97,44}×100=33\text{%}$

d) The molecular mass from the empirical formula is the same as the molecular mass of the molecular formula and so the molecular formula is:

ZnS

A calcium mineral consisted of 29,4% calcium, 23,5% sulphur and 47,1% oxygen by mass. Calculate the empirical formula of the mineral.

In 100g of the mineral there will be 29,4 g calcium, 23,5 g sulphur and 47,1 g oxygen. We can calculate the number of moles for each of these:

Calcium:

 $n=\frac{m}{M}$ $n=\frac{29,4}{40,08}$

Sulphur:

 $n=\frac{m}{M}$ $n=\frac{23,5}{32,06}$

Oxygen:

 $n=\frac{m}{M}$ $n=\frac{47,1}{16,00}$

The smallest number is 0,73 so we divide all the values by 0,73.

This gives 1 for calcium and sulphur and $\frac{2,95}{0,73}=4$ for oxygen.

So the molar ratio of the mineral is: 1:1:4 and the empirical formula is:

${\text{CaSO}}_{4}$

A chlorinated hydrocarbon compound was analysed and found to consist of 24,24% carbon, 4,04% hydrogen and 71,72% chlorine. From another experiment the molecular mass was found to be 99 g·mol−1. Deduce the empirical and molecular formula.

In 100g of the mineral there will be 24,24 g carbon, 4,04 g hydrogen and 71,72 g chlorine. We can calculate the number of moles for each of these:

Carbon:

 $n=\frac{m}{M}$ $n=\frac{24,24}{12,011}$

Hydrogen:

 $n=\frac{m}{M}$ $n=\frac{4,04}{1,01}$

Chlorine:

 $n=\frac{m}{M}$ $n=\frac{71,72}{35,45}$

The smallest number is 2 so we divide all the values by 2.

This gives 1 for carbon and chlorine and 2 for hydrogen.

So the molar ratio of the mineral is: 1:2:1 and the empirical formula is:

${\text{CH}}_{2}\text{Cl}$

The molecular mass is:

The molecular mass of the compound is found to

The empirical formula must be multiplied by: $\frac{99}{48,471}=2$

The molecular formula is therefore: ${\text{C}}_{2}{\text{H}}_{4}{\text{Cl}}_{2}$

Magnesium sulphate has the formula $MgSO 4 .nH 2 O$. A sample containing 5,0 g of magnesium sulphate was heated until all the water had evaporated. The final mass was found to be 2,6 g. How many water molecules were in the original sample?

We first need to find n, the number of water molecules that are present in the crystal. To do this we first note that the mass of water lost is $5-2,6=2,4\text{g}$.

The mass ratio is: $2,6:2,4$

To work out the mole ratio we divide the mass ratio by the molecular mass of each species:

$\frac{2,6}{120,4}:\frac{2,4}{18,02}=0.021594684:0.13318535$

$1:6$

So the number of water molecules is 6 and the formula is: ${\text{MgSO}}_{4}\cdot 6{\text{H}}_{2}\text{O}$