This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2

To find the empirical formula, we first need to determine the moles of carbon (C), hydrogen (H), and oxygen (O) in the citric acid.

1. Calculate moles of CO₂ and H₂O:

   • Moles of CO₂: $n_{CO_2} = \frac{1.89 \text{ g}}{44.01 \text{ g/mol}} = 0.043 \text{ mol}$
   • Moles of H₂O: $n_{H_2O} = \frac{0.643 \text{ g}}{18.02 \text{ g/mol}} = 0.036 \text{ mol}$

2. Calculate moles of O from products: For CO₂, each mole contains 2 moles of O:

   • Moles of O from CO₂: $0.043 \text{ mol CO}_2 \times 2 = 0.086 \text{ mol O}$ For H₂O, each mole contains 1 mole of O:
   • Moles of O from H₂O: $0.036 \text{ mol H}_2O \times 1 = 0.036 \text{ mol O}$

3. Total moles of O: $n_{O} = 0.086 + 0.036 = 0.122 \text{ mol O}$

4. Total mass of O based on calculated moles:

   • Mass of O: $n_{O} \times 16 = 0.122 \text{ mol} \times 16 \text{ g/mol} = 1.952 \text{ g}$

5. Finding mass of C and H:

   • Mass of H from H₂O: $0.036 \text{ mol H}_2O \times 2 \text{ g/mol} = 0.072 \text{ g H}$
   • Total mass (1.50 g) = mass of C + mass of H + mass of O: $1.50 = m_C + (0.072 + (0.913))$
   • Therefore, mass of C: $m_C = 1.50 - (0.913 + 0.072) = 0.515 ext{ g C}$

6. Calculate moles of C: $n_C = \frac{0.515}{12.01} = 0.043 \text{ mol C}$

7. Determine the molar ratios: Convert to simplest ratio:

   • C: 0.043
   • H: 0.072 / 2 = 0.036
   • O: 0.122
   • Empirical ratio C:H:O = 1:2:4 thus empirical formula is C₆H₈O₇.