Percentage Composition In Chemistry

A compound is made up of more than one element. (E.g. Carbon dioxide is made up of one carbon atom and two oxygen atoms) We can express the composition of the compound in a number of ways:

  • In terms of the number of its constituent atoms
  • In terms of the mass of its constituent elements

Percentage composition of a component (e.g. constituent element) in a compound is defined as the percentage of the total mass of the compound that is due to that component.

$$\begin{aligned} \text{Percentage of an element} \, &= \frac{\text{mass of element in 1 mol. of compound}}{\text{mass of 1 mol. of compound} (M_{r})} \times 100 \\ &= \frac{(\text{atoms of element})\times (A_{r})}{M_{r}} \times 100 \\ &= \frac{\text{Molar mass of component}}{\text{Molar mass of compound}} \times 100 \end{aligned}$$

Steps to calculate percentage composition of a component element in compound

  1. Find molecular mass ($M_{r}$) of compound
  2. Find atomic mass ($A_{r}$) of the component element that you are interested in.
  3. Multiply the atomic mass of the component element with the number of component element in 1 molecule of the compound
  4. Divide the result by the molecular mass of the compound and multiply by 100

Note: Replace $A_{r}$ with $M_{r2}$ for a component compound in a larger compound, where $M_{r2}$ is the molecular mass of the component compound. (E.g. Percentage by mass of water in $\text{CuSO}_{4}.5\text{H}_{2}\text{O}$)


Find the percentage by mass of carbon in carbon dioxide:

  • $M_{r}$ of carbon dioxide: $12 + 2(16) = 44$
  • $A_{r}$ of carbon = 12
  • Number of carbon atoms in carbon dioxide = 1
  • Percentage of C: $\frac{12}{44} \times 100 = 27.3 \%$

Find the percentage by mass of hydrogen in ammonia ($\text{NH}_{3}$):

  • $M_{r}$ of ammonia: $14 + 3(1) = 17$
  • $A_{r}$ of hydrogen = 1
  • Number of hydrogen atoms in ammonia = 3
  • Percentage of H: $\frac{3 \times 1}{17} \times 100 = 17.6 \%$

Find the percentage by mass of carbon in $\text{C}_{6}\text{H}_{12}\text{O}_{6}$:

  • $M_{r}$ of $\text{C}_{6}\text{H}_{12}\text{O}_{6}$: $6(12) + 12(1) + 6(16) = 180$
  • $A_{r}$ of carbon = 12
  • Number of atoms of carbon in $\text{C}_{6}\text{H}_{12}\text{O}_{6}$: 6
  • Percentage of C: $\frac{6\times 12}{180} \times 100 = 40 \%$


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