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Stoichiometry - Empirical and molecular formulae

Grade 12IGCSEChemistry

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 atoms of each element present in a compound. For example, the empirical formula of benzene (C6H6C_6H_6) is CHCH.

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The Molecular Formula provides the actual number of atoms of each element in one molecule of a compound. It is always an integer multiple (nn) 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 (n=mArn = \frac{m}{A_r}). 2. Divide all mole values by the smallest number of moles calculated. 3. If necessary, multiply by a small integer to obtain a whole-number ratio.

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The relationship between the molecular formula and the empirical formula is given by the factor nn, calculated as n=Relative Molecular Mass (Mr)Empirical Formula Massn = \frac{\text{Relative Molecular Mass (} M_r \text{)}}{\text{Empirical Formula Mass}}.

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In combustion analysis, the mass of CC is derived from the mass of CO2CO_2 produced, and the mass of HH is derived from the mass of H2OH_2O produced using their respective stoichiometric ratios.

📐Formulae

n=massArn = \frac{\text{mass}}{A_r}

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

n=Molar Mass of CompoundMolar Mass of Empirical Formulan = \frac{\text{Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}}

% of element=Total Ar of element in formulaMr of compound×100\% \text{ of element} = \frac{\text{Total } A_r \text{ of element in formula}}{M_r \text{ of compound}} \times 100

💡Examples

Problem 1:

A sample of a compound contains 2.4 g2.4\ g of Carbon and 0.6 g0.6\ g of Hydrogen. The relative molecular mass of the compound is 30.030.0. Calculate its empirical and molecular formulae.

Solution:

  1. Calculate moles of each element: n(C)=2.4 g12.0 g/mol=0.2 moln(C) = \frac{2.4\ g}{12.0\ g/mol} = 0.2\ mol n(H)=0.6 g1.0 g/mol=0.6 moln(H) = \frac{0.6\ g}{1.0\ g/mol} = 0.6\ mol
  2. Find the simplest ratio: C:H=0.20.2:0.60.2=1:3C : H = \frac{0.2}{0.2} : \frac{0.6}{0.2} = 1 : 3 Empirical Formula = CH3CH_3
  3. Calculate Empirical Formula Mass: 12.0+(3×1.0)=15.012.0 + (3 \times 1.0) = 15.0
  4. Find the multiplier nn: n=30.015.0=2n = \frac{30.0}{15.0} = 2
  5. Molecular Formula = 2×(CH3)=C2H62 \times (CH_3) = C_2H_6

Explanation:

We first determined the molar ratio of the elements to find the simplest integer ratio (1:31:3). Then, by comparing the empirical mass (15.015.0) to the given molecular mass (30.030.0), we found the scaling factor n=2n=2 to determine the true molecular formula.

Problem 2:

Determine the empirical formula of a compound that consists of 40.0%40.0\% Calcium, 12.0%12.0\% Carbon, and 48.0%48.0\% Oxygen by mass.

Solution:

Assume a 100 g100\ g sample: n(Ca)=40.040.1≈1.0 moln(Ca) = \frac{40.0}{40.1} \approx 1.0\ mol n(C)=12.012.0=1.0 moln(C) = \frac{12.0}{12.0} = 1.0\ mol n(O)=48.016.0=3.0 moln(O) = \frac{48.0}{16.0} = 3.0\ mol Ratio Ca:C:O=1:1:3Ca:C:O = 1:1:3. Empirical Formula = CaCO3CaCO_3

Explanation:

By assuming a 100 g100\ g sample, we treat the percentages as masses. Dividing by the relative atomic masses gives a molar ratio of 1:1:31:1:3, which corresponds to Calcium Carbonate.

Empirical and molecular formulae Revision - Grade 12 Chemistry IGCSE