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.

Example: Calculating the Percentage by Mass of Elements in a Compound


Calculate the percentage that each element contributes to the overall mass of sulphuric acid (\(\text{H}_{2}\text{SO}_{4}\)).

Step 1: Calculate the molar masses

\[\text{Hydrogen } = 2 \times \text{1.01} = \text{2.02}\text{ g·mol$^{-1}$} \\ \text{Sulfur } = \text{32.1}\text{ g·mol$^{-1}$} \\ \text{Oxygen } = 4 \times \text{16.0} = \text{64.0}\text{ g·mol$^{-1}$}\]

Step 2: Use the calculations in the previous step to calculate the molecular mass of sulphuric acid.

\[\text{Mass } = \text{2.02}\text{ g·mol$^{-1}$} + \text{32.1}\text{ g·mol$^{-1}$} + \text{64.0}\text{ g·mol$^{-1}$} = \text{98.12}\text{ g·mol$^{-1}$}\]

Step 3: Use the equation

\[\text{Percentage by mass } = \frac{\text{atomic mass}}{\text{molecular mass of H}_{2}\text{SO}_{4}} \times 100\]


\[\frac{\text{2.02}\text{ g·mol$^{-1}$}}{\text{98.12}\text{ g·mol$^{-1}$}} \times \text{100}\% = \text{2.0587}\%\]


\[\frac{\text{32.1}\text{ g·mol$^{-1}$}}{\text{98.12}\text{ g·mol$^{-1}$}} \times \text{100}\% = \text{32.7150}\%\]


\[\frac{\text{64.0}\text{ g·mol$^{-1}$}}{\text{98.12}\text{ g·mol$^{-1}$}} \times \text{100}\% = \text{65.2263}\%\]

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

In other words, in one molecule of sulphuric acid, hydrogen makes up \(\text{2.06}\%\) of the mass of the compound, sulfur makes up \(\text{32.71}\%\) and oxygen makes up \(\text{65.23}\%\).

Example: Determining the Empirical Formula of a Compound


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

Step 1: Give the masses

Carbon = \(\text{52.2}\) \(\text{g}\), hydrogen = \(\text{13.0}\) \(\text{g}\) and oxygen = \(\text{34.8}\) \(\text{g}\)

Step 2: Calculate the number of moles

\[n = \frac{m}{M}\]


\begin{align*} n_{\text{carbon}} & = \frac{\text{52.2}\text{ g}}{\text{12.0}\text{ g·mol$^{-1}$}} \\ & = \text{4.35}\text{ mol} \\\\ n_{\text{hydrogen}} & = \frac{\text{13.0}\text{ g}}{\text{1.01}\text{ g·mol$^{-1}$}} \\ & = \text{12.871}\text{ mol} \\\\ n_{\text{oxygen}} & = \frac{\text{34.8}\text{ g}}{\text{16.0}\text{ g·mol$^{-1}$}} \\ & = \text{2.175}\text{ mol} \end{align*}

Step 3: Find the smallest number of moles

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

\[\text{units in empirical formula } = \frac{\text{moles of this element}}{\text{smallest number of moles}}\]

In this case, the smallest number of moles is \(\text{2.175}\). Therefore:


\[\frac{\text{4.35}}{\text{2.175}} = \text{2}\]


\[\frac{\text{12.87}}{\text{2.175}} = \text{6}\]


\[\frac{\text{2.175}}{\text{2.175}} = \text{1}\]

Therefore the empirical formula of this substance is: \(\text{C}_{2}\text{H}_{6}\text{O}\).

Example: Determining the Formula of a Compound


\(\text{207}\) \(\text{g}\) of lead combines with oxygen to form \(\text{239}\) \(\text{g}\) of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: \(\text{Pb } = \text{202.7}\text{ u}\) and \(\text{O } = \text{16.0}\text{ u}\)).

Step 1: Find the mass of oxygen

\[\text{239}\text{ g} – \text{207}\text{ g} = \text{32}\text{ g}\]

Step 2: Find the moles of oxygen

\[n = \frac{m}{M}\]


\[n = \frac{\text{207}\text{ g}}{\text{207.2}\text{ g·mol$^{-1}$}} = \text{1}\text{ mol}\]


\[n = \frac{\text{32}\text{ g}}{\text{16.0}\text{ g·mol$^{-1}$}} = \text{2}\text{ mol}\]

Step 3: Find the mole ratio

The mole ratio of \(\text{Pb }: \text{ O}\) 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 \(\text{PbO}_{2}\).

Example: Empirical and Molecular Formula


Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: \(\text{39.9}\%\) carbon, \(\text{6.7}\%\) hydrogen and \(\text{53.4}\%\) 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 \(\text{60.06}\) \(\text{g·mol$^{-1}$}\).

Step 1: Find the mass

In \(\text{100}\) \(\text{g}\) of acetic acid, there is \(\text{39.9}\) \(\text{g}\) \(\text{C}\), \(\text{6.7}\) \(\text{g}\) \(\text{H}\) and \(\text{53.4}\) \(\text{g}\) \(\text{O}\).

Step 2: Find the moles

\[n = \frac{m}{M}\]\begin{align*} n_{\text{C}} & = \frac{\text{39.9}\text{ g}}{\text{12.0}\text{ g·mol$^{-1}$}} \\ & = \text{3.325}\text{ mol} \\\\ n_{\text{H}} & = \frac{\text{6.7}\text{ g}}{\text{1.01}\text{ g·mol$^{-1}$}} \\ & = \text{6.637}\text{ mol} \\\\ n_{\text{O}} & = \frac{\text{53.4}\text{ g}}{\text{16.0}\text{ g·mol$^{-1}$}} \\ & = \text{3.375}\text{ mol} \end{align*}

Step 3: Find the empirical formula










Empirical formula is \(\text{CH}_{2}\text{O}\)

Step 4: Find the molecular formula

The molar mass of acetic acid using the empirical formula is \(\text{30.02}\) \(\text{g·mol$^{-1}$}\). However the question gives the molar mass as \(\text{60.06}\) \(\text{g·mol$^{-1}$}\). Therefore the actual number of moles of each element must be double what it is in the empirical formula (\(\frac{\text{60.06}}{\text{30.02}} = 2\)). The molecular formula is therefore \(\text{C}_{2}\text{H}_{4}\text{O}_{2}\) or \(\text{CH}_{3}\text{COOH}\)

Example: Waters of Crystallization


Aluminium trichloride (\(\text{AlCl}_{3}\)) is an ionic substance that forms crystals in the solid phase. Water molecules may be trapped inside the crystal lattice. We represent this as: \(\text{AlCl}_{3}.\text{nH}_{2}\text{O}\). Carine heated some aluminium trichloride crystals until all the water had evaporated and found that the mass after heating was \(\text{2.8}\) \(\text{g}\). The mass before heating was \(\text{5}\) \(\text{g}\). What is the number of moles of water molecules in the aluminium trichloride before heating?

Step 1: 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 \(\text{5}\text{ g} – \text{2.8}\text{ g} = \text{2.2}\text{ g}\).

Step 2: Find the mass ratio

The mass ratio is:





Step 3: Find the mole ratio

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



\(\frac{\text{2.8}\text{ g·mol$^{-1}$}}{\text{133.35}\text{ g}}\)

\(\frac{\text{2.2}\text{ g·mol$^{-1}$}}{\text{18.02}\text{ g}}\)



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









The mole ratio of aluminium trichloride to water is: \(1:6\)

Step 4: Write the final answer

And now we know that there are 6 moles of water molecules in the crystal. The formula is \(\text{AlCl}_{3}.6\text{H}_{2}\text{O}\).

We can perform experiments to determine the composition of substances. For example, blue copper sulphate (\(\text{CuSO}_{4}\)) 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.

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This is a lesson from the tutorial, Quantitative Aspects of Chemical Change and you are encouraged to log in or register, so that you can track your progress.

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