Percentage Composition

Full Notes:

What is Percentage Composition?

Percentage composition refers to the percentage by mass of each element in a compound. It tells us how much of each element is present in a 100 g sample of the compound.

This is especially useful for:

• Determining chemical formulae
• Verifying compound purity
• Carrying out stoichiometric calculations

Formula for Percentage Composition

To calculate the % of an element:

% Element =
(Total mass of element in 1 mole of compound / Molar mass of compound) × 100

Example Calculate % composition of H₂SO₄

Molar mass of H₂SO₄ = 98.08 g/mol
H: 2 × 1.008 = 2.016
S: 1 × 32.06 = 32.06
O: 4 × 16.00 = 64.00
Total = 98.08 g
% H = (2.016 / 98.08) × 100 = 2.06%
% S = (32.06 / 98.08) × 100 = 32.68%
% O = (64.00 / 98.08) × 100 = 65.26%
Total = 100% (confirming accuracy)

1.9.1 Empirical Formula and Molecular Formula

Empirical Formula

This is the simplest whole number ratio of atoms of different elements in a compound.

It gives only the relative proportions, not the actual number of atoms.

Molecular Formula

This represents the actual number of atoms of each element in one molecule of the compound.

It is a whole number multiple of the empirical formula.

Relationship Between the Two

Molecular formula = (Empirical formula) × n
Where n = (Molar mass of compound / Empirical formula mass)

Steps to Calculate Empirical Formula (NCERT Method)

To Calculate Empirical Formula using the NCERT Method:

1. Convert % composition to grams → Assume 100 g sample → % becomes g
2. Convert grams to moles → Moles = Mass / Atomic mass
3. Divide all mole values by the smallest → This gives relative atom ratio
4. Round to nearest whole numbers → If needed, multiply all values by 2, 3 etc. to get integers

Example Finding Empirical and Molecular Formula

A compound has: C = 40%, H = 6.7%, O = 53.3%

Step 1: Assume 100 g sample → 40 g C, 6.7 g H, 53.3 g O
Step 2: Convert to moles
C: 40 / 12.01 = 3.33 mol
H: 6.7 / 1.008 = 6.65 mol
O: 53.3 / 16.00 = 3.33 mol

Step 3: Divide by smallest
C: 3.33 / 3.33 = 1
H: 6.65 / 3.33 ≈ 2
O: 3.33 / 3.33 = 1
Empirical formula = CH₂O

Step 4: Given molar mass = 180 g/mol
Empirical formula mass = 30 g/mol
n = 180 / 30 = 6
Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆

Matt’s exam tip

Always check: if mole ratios are not exact integers, multiply to get whole numbers. Molar mass must be given or calculated to find molecular formula. A common NEET trap is when students calculate an empirical formula (i.e CH₂O) but forget to find n to get the molecular formula.

Summary

• Percentage composition gives the mass percentage of each element in a compound.
• Empirical formula gives the simplest ratio of atoms.
• Molecular formula shows actual atom numbers and equals the empirical formula multiplied by n.
• Use given molar mass to find n and confirm correct molecular formula.