Determining empirical and molecular formulae

To turn a molecular formula into an empirical formula, divide all its subscripts by their highest common factor. Sometimes there is no common factor, so the two formulae are identical: methane is CH₄ in both cases. Ionic compounds such as magnesium oxide form lattices rather than separate molecules; their formulae express the simplest ratio of ions.

From reacting masses to an atom ratio

Different elements have different masses per atom. Equal masses of two elements therefore do not usually contain equal numbers of atoms. Relative atomic mass, $A_r$, allows us to account for this difference.

For each element, calculate:

$$\text{relative number of atoms} \propto \frac{\text{mass of element}}{A_r}$$

Use masses in the same unit. Divide each result by the smallest result to find the simplest ratio, then write that ratio as subscripts in the formula.

For example, a compound contains 10 g of hydrogen and 80 g of oxygen. Using $A_r(\text{H})=1$ and $A_r(\text{O})=16$:

The atom ratio is H:O = 2:1, so the empirical formula is H₂O. The original mass ratio, 1:8, would not give the correct formula.

The final ratio must contain whole numbers. If dividing by the smallest result gives 1:1.5, multiply both numbers by 2 to obtain 2:3. This preserves the ratio rather than changing it by rounding 1.5 to an integer.

Using percentage composition

Percentage by mass tells you how much of each element is present in every 100 g of a compound. You can therefore assume a 100 g sample and use the percentages as masses in grams. This does not change the atom ratio: a larger or smaller sample of the same compound has the same composition.

For example, a compound contains 27.29% carbon and 72.71% oxygen by mass. A 100 g sample would contain 27.29 g of carbon and 72.71 g of oxygen. Using $A_r(\text{C})=12$ and $A_r(\text{O})=16$:

• Carbon: $27.29/12\approx2.274$.
• Oxygen: $72.71/16\approx4.544$.
• Dividing both by 2.274 gives approximately 1:2.

The empirical formula is CO₂. Percentage data and reacting-mass data lead to the same calculation method.

Finding the molecular formula

An empirical formula alone cannot tell you how many atoms are in a molecule. For example, CH₂O and C₆H₁₂O₆ both have the simplest ratio C:H:O = 1:2:1. To identify the molecular formula, you also need the compound’s relative molecular mass, $M_r$.

First add the relative atomic masses represented by the empirical formula. Then calculate the whole-number multiplier:

$$\text{multiplier}=\frac{M_r}{\text{relative mass of the empirical formula}}$$

For a compound with empirical formula CH₂O and $M_r=180$, using $A_r$ values of C = 12, H = 1 and O = 16:

$$\text{empirical formula mass}=12+(2\times1)+16=30$$

The multiplier is $180/30=6$. Multiply every subscript in CH₂O by 6 to give C₆H₁₂O₆. Each molecule has six times the numbers of atoms represented by the empirical formula, but the same ratio.

Measuring the empirical formula of magnesium oxide

Magnesium reacts with oxygen from the air when heated. By measuring the magnesium mass and the mass gained during the reaction, you can find the masses of both elements in the oxide.

Use a balance, magnesium ribbon, a crucible with a lid, a Bunsen burner, suitable crucible support and tongs.

1. Weigh the empty crucible and lid.
2. Add magnesium ribbon and weigh the crucible, lid and magnesium together. Subtract the first reading to obtain the magnesium mass.
3. Heat the crucible strongly with the lid on. Briefly lift the lid at intervals to admit oxygen, replacing it promptly to minimise the escape of magnesium oxide smoke.
4. Allow the crucible to cool, then weigh it with its lid and contents. Use tongs to handle the hot apparatus.
5. Repeat heating, cooling and weighing until a constant mass is reached. This indicates that no further oxygen is being taken in and the reaction is complete.

The lid helps retain the product, but must not prevent oxygen entering. Losing product or leaving magnesium unreacted would make the measured mass gain too small.

Turning the practical results into a formula

Consider these recorded masses:

The magnesium mass is $25.07-24.83=0.24\text{ g}$. The oxygen mass is the gain in mass: $25.23-25.07=0.16\text{ g}$.

Using $A_r(\text{Mg})=24$ and $A_r(\text{O})=16$, divide each element’s mass by its relative atomic mass:

$$\text{Mg:O}=\frac{0.24}{24}:\frac{0.16}{16}=0.01:0.01=1:1$$

The empirical formula is therefore MgO. The elements have different masses in the product, but equal relative numbers of atoms.

1. Divide each element’s mass by its $A_r$.
2. Divide all results by the smallest.
3. If necessary, multiply the whole ratio to obtain whole numbers.
4. Use the ratio as formula subscripts.

For percentages, assume 100 g: each percentage becomes that element’s mass in grams.

Empirical → molecular formula

$$\text{multiplier}=\frac{M_r}{\text{empirical formula mass}}$$

Multiply every empirical-formula subscript by this factor.

Magnesium oxide experiment

• Weigh crucible + lid; then crucible + lid + magnesium.
• Heat strongly. Briefly lift the lid to admit oxygen while minimising product loss.
• Cool and weigh; repeat to constant mass.
• Magnesium mass = initial loaded mass − empty crucible-and-lid mass.
• Oxygen mass = final loaded mass − initial loaded mass.
• Divide these masses by their respective $A_r$ values and simplify. Mg:O = 1:1 gives MgO.