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.