Percentage Composition: Definition, Common Mistakes to Avoid & Sample Questions

Percentage composition is the mass percentage of each element in a compound, calculated as the mass of the element divided by the molar mass of the compound, multiplied by 100 . For a compound with molar mass $M$ and element $E$ contributing mass $m_E$ : $\mathrm{\%}\text{of}E=\frac{m_E}{M}\times 100$ Do note that percentage composition quantifies the relative contribution of each element in a compound, a critical concept for JEE Main and NEET. Let us now take a look at the example:

Example: For ${\mathrm{H}}_2\mathrm{O}$ (molar mass $=18\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ): $\mathrm{\%}\mathrm{H}=\frac{2\times 1}{18}\times 100=11.11\mathrm{\%},\mathrm{}\mathrm{\%}\mathrm{O}=\frac{16}{18}\times 100=88.89\mathrm{\%}$

Percentage composition is calculated using:

1. Molar Mass Approach: Sum atomic masses of elements in one mole of the compound, then compute each element's percentage.

2. Experimental Data: Use given mass percentages (e.g., from combustion analysis) to find element ratios.

Example: For ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}$ (molar mass $=46\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ): $\mathrm{\%}\mathrm{C}=\frac{2\times 12}{46}\times 100=52.17\mathrm{\%},\mathrm{}\mathrm{\%}\mathrm{H}=\frac{6\times 1}{46}\times 100=13.04\mathrm{\%},\mathrm{}\mathrm{\%}\mathrm{O}=\frac{16}{46}\times 100=34.79\mathrm{\%}$

 

The empirical formula is the simplest whole-number ratio of atoms in a compound, derived from percentage composition:

1. Assume 100 g of compound, converting percentages to grams.

2. Calculate moles of each element: $\text{Moles}=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\left(\mathrm{g}\right)}{\text{atomic mass}(\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l})}$

3. Divide by the smallest mole value to get the ratio.

4. Multiply by integers if needed to obtain whole numbers.

Example: A compound with 40 $\text{Moles}\mathrm{C}=\frac{40}{12}=3.33,\mathrm{}\mathrm{H}=\frac{6.67}{1}=6.67,\mathrm{}\mathrm{O}=\frac{53.33}{16}=3.33$

Ratio: $3.33:6.67:3.33=1:2:1$ . Empirical formula: ${\mathrm{C}\mathrm{H}}_2\mathrm{O}$ . JEE/NEET frequently test empirical formula determination.

The molecular formula is a multiple of the empirical formula: $\text{Molecular formula}=n\times \text{empirical formula},\mathrm{}n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$

Example: If the above compound has molecular mass $180\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ , empirical mass of ${\mathrm{C}\mathrm{H}}_2\mathrm{O}=$ $30\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ : $n=\frac{180}{30}=6,\mathrm{}\text{Molecular formula}={\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

NEET questions often combine percentage composition with molecular mass data.

Percentage composition aids in:

1. Calculating reactant/product masses in reactions.

2. Determining purity of compounds (e.g., percentage of active ingredient).

3. Analyzing combustion products (e.g., ${\mathrm{C}\mathrm{O}}_2$ and ${\mathrm{H}}_2\mathrm{O}$ from hydrocarbons).

Example: To find carbon content in $11\mathrm{g}{\mathrm{C}\mathrm{O}}_2$ : $\mathrm{\%}\mathrm{C}=\frac{12}{44}\times 100=27.27\mathrm{\%},\mathrm{}\text{Mass}\mathrm{C}=0.2727\times 11=3\mathrm{g}$

 

Percentage purity measures the mass percentage of a desired compound in an impure sample: $\mathrm{\%}\text{purity}=\frac{\text{mass of pure compound}}{\text{total mass of sample}}\times 100$

Example: A 10 g sample of limestone $\left({\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3\right)$ produces $4.4\mathrm{g}{\mathrm{C}\mathrm{O}}_2$ . Moles ${\mathrm{C}\mathrm{O}}_2$ : $n=\frac{4.4}{44}=0.1,\mathrm{}\text{Moles}{\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3=0.1$

Mass ${\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3$ : $0.1\times 100=10\mathrm{g},\mathrm{}\mathrm{\%}\text{purity}=\frac{10}{10}\times 100=100\mathrm{\%}$

NEET tests purity calculations in analytical chemistry.

Example 1: A compound contains 26.67Assume 100 g : $\text{Moles}\mathrm{C}=\frac{26.67}{12}=2.22,\mathrm{}\mathrm{H}=\frac{2.22}{1}=2.22,\mathrm{}\mathrm{O}=\frac{71.11}{16}=4.44$

Ratio: $2.22:2.22:4.44=1:1:2$ . Empirical formula: ${\mathrm{C}\mathrm{H}\mathrm{O}}_2$ .

 

Example 2: 0.92 g of a hydrocarbon produces $2.76\mathrm{g}{\mathrm{C}\mathrm{O}}_2$ and $1.13\mathrm{g}{\mathrm{H}}_2\mathrm{O}$ . Find its empirical formula. (JEE Main) Mass C: $\frac{12}{44}\times 2.76=0.753\mathrm{g}$

Mass H: $\frac{2\times 1}{18}\times 1.13=0.126\mathrm{g}$

Moles: $\mathrm{C}=\frac{0.753}{12}=0.0628,\mathrm{}\mathrm{H}=\frac{0.126}{1}=0.126$

Ratio: 0.0628 : $0.126=1:2$ . Empirical formula: ${\mathrm{C}\mathrm{H}}_2$ .