Let's calculate the percent composition of acetic acid: C _{2} H _{4} O _{2}

- Determining the mass of a mole of substance

m(C _{2} H _{4} O _{2} ) = 12.01, x 2 + 1.008 x 4 + 15.99 x 2 = 60.04 g/mol

- Calculation of the percentage of each element in the compound, dividing the mass of the element by the mass of one mole of the compound

$$\%C=\frac{12.01 \cdot 2}{60.04}\cdot 100 =40.00\%$$

$$\%H=\frac{1.008 \cdot 4}{60.04}\cdot 100 =6.728\%$$

$$\%O=\frac{15.99 \cdot 2}{60.04}\cdot 100 =53.26\%$$

The experimental determination of the percent composition of a compound and its comparison with the theoretical calculation allows the identification of said compound. This type of analysis is frequently carried out after carrying out a chemical reaction to verify that the product obtained is as expected.

Note that the sum of percentages equals 100.