Photo AI
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 H₃Y·xH₂O. 1.50 g sample of H₃Y·xH₂O 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
-
Calculate moles of CO₂ produced:
moles , CO₂ = \frac{1.89 , g}{44.01 , g/mol} = 0.0428 , mol
- Calculate moles of H₂O produced:
moles , H₂O = \frac{0.643 , g}{18.02 , g/mol} = 0.0357 , mol
- Calculate moles of C from CO₂:
moles , C = moles , CO₂ = 0.0428 , mol
- Calculate moles of H from H₂O:
moles , H = 2 \times 0.0357 , mol = 0.0714 , mol
- Determine the moles of O in H₃Y·xH₂O:
Total , O = \text{(from CO₂ + from H₂O)} = 0.0428 + 0.0357 = 0.0785 , mol
- Calculate empirical formula:
Moles (C:H:O) = 0.0428: 0.0714: 0.0785
Normalize to smallest mole:
C = \frac{0.0428}{0.0428} = 1 \ H = \frac{0.0714}{0.0428} \approx 1.67 \implies 5 \text{ and double the ratio gives approximately } 6 \ O = \frac{0.0785}{0.0428} \approx 1.83 \implies 7.
Therefore, the empirical formula is C₆H₈O₇.
Step 2
Show, using these data, that the value of x = 1
Answer
-
Calculate the mass of H₃Y:
H₃Y = mass , prior - mass , after = 3.00 , g - 2.74 , g = 0.26 , g
- Calculate moles of H₃Y:
moles , H₃Y = \frac{0.26 , g}{210.0 , g/mol} \approx 0.00124 , mol
- Calculate x:
x = \frac{mass , H₂O}{moles , H₃Y} \implies mass , H₂O = 2.74g \implies x = \frac{2.74 , g / 18.02 , g/mol}{0.00124 , mol} = 1.
Step 4