Empirical Formula (Higher Tier)

Definition of Empirical Formula

The empirical formula of a compound shows the simplest whole-number ratio of atoms of each element in that compound. It represents the basic composition of the substance without giving the exact number of atoms in a molecule.

For example, the empirical formula of hydrogen peroxide (H₂O₂) is HO, because the ratio of hydrogen to oxygen atoms is 1:1 when simplified.

Calculating Empirical Formula

To find the empirical formula from experimental data, follow these steps:

• Convert the mass of each element into moles using the formula $\text{moles} = \frac{\text{mass}}{\text{relative atomic mass (Ar)}}$.
• Divide all mole values by the smallest number of moles calculated to get a ratio.
• If the ratio is not a whole number, multiply all ratios by the smallest number that converts them into whole numbers (usually 2, 3, or 4). For example, if a ratio is 1.33, multiply all ratios by 3; if 1.25, multiply by 4.
• Write the empirical formula using these whole-number ratios as subscripts.

For instance, if you have 2.0 g of carbon and 5.3 g of oxygen:

Calculate moles: $\frac{2.0}{12} = 0.167$ mol C, $\frac{5.3}{16} = 0.331$ mol O.

Divide by smallest: $\frac{0.167}{0.167} = 1$, $\frac{0.331}{0.167} = 1.98 \approx 2$.

Empirical formula is $\text{CO}_2$.

Using Percentage Composition

When given percentage composition by mass, assume a 100 g sample. This means the percentage becomes the mass in grams of each element.

Then, convert these masses to moles and follow the same steps as above to find the empirical formula.

For example, if a compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass:

Assume masses: 40 g C, 6.7 g H, 53.3 g O.

Calculate moles: $\frac{40}{12} = 3.33$ mol C, $\frac{6.7}{1} = 6.7$ mol H, $\frac{53.3}{16} = 3.33$ mol O.

Divide by smallest: $\frac{3.33}{3.33} = 1$, $\frac{6.7}{3.33} = 2.01 \approx 2$, $\frac{3.33}{3.33} = 1$.

Empirical formula is $\text{CH}_2\text{O}$.

Applications and Examples

Empirical formulas are useful for identifying substances and understanding their composition. Experimental data from combustion analysis or mass measurements can be used to determine the empirical formula of unknown compounds.

While the empirical formula gives the simplest ratio, it can be related to the molecular formula if the relative molecular mass (Mr) is known (covered in other topics; see the Molecular Formula and Mr section for details).

Example: A compound contains 52.2% carbon, 34.8% oxygen, and 13.0% hydrogen by mass. Find its empirical formula.

Assuming 100 g sample: 52.2 g C, 34.8 g O, 13.0 g H.

Moles: $\frac{52.2}{12} = 4.35$ mol C, $\frac{34.8}{16} = 2.18$ mol O, $\frac{13.0}{1} = 13.0$ mol H.

Divide by smallest (2.18): $\frac{4.35}{2.18} = 2.0$, $\frac{2.18}{2.18} = 1$, $\frac{13.0}{2.18} = 6.0$.

Empirical formula is $\text{C}_2\text{H}_6\text{O}$.