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This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2
Question 8
This question is about citric acid, a hydrated tricarboxylic acid. Its formula can be represented as HₓY·zH₂O A 1.50 g sample of HₓY·zH₂O contains 0.913 g of oxygen... show full transcript
Worked Solution & Example Answer:This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2
Step 1
Show that the empirical formula of citric acid is C₆H₈O₇.
Answer
To determine the empirical formula, we start with the mass percentages from the combustion products. From the data:
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Calculate moles of CO₂:
The molar mass of CO₂ = 44.01 g/mol.
Moles of CO₂ = rac{1.89 ext{ g}}{44.01 ext{ g/mol}} = 0.043 ext{ mol}.
Each mole of CO₂ contains one mole of carbon. So, moles of C = 0.043 mol.
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Calculate moles of H₂O:
The molar mass of H₂O = 18.02 g/mol.
Moles of H₂O = rac{0.643 ext{ g}}{18.02 ext{ g/mol}} = 0.036 ext{ mol}.
Each mole of H₂O contains two moles of hydrogen. So, moles of H = 0.036 mol × 2 = 0.072 mol.
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Calculate moles of oxygen:
Total mass = 1.50 g. Mass of H + mass of C = 0.072 mol × 1.01 g/mol + 0.043 mol × 12.01 g/mol = 0.072 + 0.515 = 0.587 g.
Therefore, moles of O = rac{(mass ext{ of } O)}{16.00 ext{ g/mol}} = rac{0.913 ext{ g}}{16.00 ext{ g/mol}} = 0.057 ext{ mol}.
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Ratio of moles:
C : H : O = 0.043 : 0.072 : 0.057.
To simplify, divide each by 0.043:
C : H : O = 1 : 1.67 : 1.33.
Multiplying through by 3 gives:
C : H : O = 3 : 5 : 4.
So, empirical formula = C₆H₈O₇.
Step 2
Show, using these data, that the value of x = 1.
Answer
Given that the initial mass of HₓY·zH₂O is 3.00 g and the mass of the anhydrous HₓY remaining is 2.74 g, we can find the mass of water lost:
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Calculate mass of water lost:
Mass of water lost = 3.00 g - 2.74 g = 0.26 g.
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Calculate moles of anhydrous compound:
Molar mass of HₓY = 210.0 g/mol.
Moles of HₓY = rac{2.74 ext{ g}}{210.0 ext{ g/mol}} = 0.0130 ext{ mol}.
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Calculate moles of water lost:
Molar mass of H₂O = 18.02 g/mol.
Moles of water lost = rac{0.26 ext{ g}}{18.02 ext{ g/mol}} ext{ = 0.0144 mol}.
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Ratio Determination:
From the mass of water (2 per molecule of H₂O), find x:
Moles of H₂O lost = 0.0144 mol and ratio of H₂O to HₓY = 1 : 1. Hence x = 1.
Step 3
Step 4