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Stoichiometry - Empirical and molecular formulae

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Empirical Formula is defined as the simplest whole-number ratio of the atoms of each element present in a compound.

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The Molecular Formula specifies the actual number of atoms of each element in one molecule of a substance. It is always a multiple of the empirical formula.

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To calculate the empirical formula from mass or percentage composition: 1. Divide the mass/percentage of each element by its relative atomic mass (ArA_r) to find the number of moles. 2. Divide all results by the smallest number of moles obtained. 3. If necessary, multiply to reach the nearest whole-number ratio.

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The relationship between the molecular formula and the empirical formula is given by the multiplier nn, where Molecular Formula=n×(Empirical Formula)\text{Molecular Formula} = n \times (\text{Empirical Formula}).

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Relative atomic masses (ArA_r) used for common elements: C=12C = 12, H=1H = 1, O=16O = 16, N=14N = 14, S=32S = 32, Cl=35.5Cl = 35.5.

📐Formulae

moles(n)=mass (g)Relative Atomic Mass(Ar)\text{moles} (n) = \frac{\text{mass (g)}}{\text{Relative Atomic Mass} (A_r)}

n=Relative Molecular Mass(Mr)Relative Empirical Formula Massn = \frac{\text{Relative Molecular Mass} (M_r)}{\text{Relative Empirical Formula Mass}}

Molecular Formula=n×(Empirical Formula)\text{Molecular Formula} = n \times (\text{Empirical Formula})

💡Examples

Problem 1:

A compound is found to contain 40.0%40.0\% Carbon (CC), 6.7%6.7\% Hydrogen (HH), and 53.3%53.3\% Oxygen (OO) by mass. Determine its empirical formula.

Solution:

  1. Moles of C=40.012=3.33C = \frac{40.0}{12} = 3.33 mol.
  2. Moles of H=6.71=6.7H = \frac{6.7}{1} = 6.7 mol.
  3. Moles of O=53.316=3.33O = \frac{53.3}{16} = 3.33 mol.
  4. Dividing by the smallest value (3.333.33): C=3.333.33=1C = \frac{3.33}{3.33} = 1; H=6.73.33≈2H = \frac{6.7}{3.33} \approx 2; O=3.333.33=1O = \frac{3.33}{3.33} = 1. Result: CH2OCH_2O.

Explanation:

The percentage composition is treated as mass in 100 g100\text{ g}. We convert mass to moles using n=mArn = \frac{m}{A_r} and find the simplest integer ratio.

Problem 2:

The empirical formula of a compound is CH2CH_2. If its relative molecular mass (MrM_r) is 5656, what is its molecular formula?

Solution:

  1. Calculate empirical formula mass: 12+(2×1)=1412 + (2 \times 1) = 14.
  2. Find multiplier nn: n=MrEmpirical Mass=5614=4n = \frac{M_r}{\text{Empirical Mass}} = \frac{56}{14} = 4.
  3. Molecular formula =4×(CH2)=C4H8= 4 \times (CH_2) = C_4H_8.

Explanation:

The molecular formula is always a whole-number multiple of the empirical formula. We find the ratio of the actual molar mass to the mass of the empirical unit.

Empirical and molecular formulae Revision - Grade 11 Chemistry IGCSE