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 HₓY∙zH₂O. 1. A 1.50 g sample of HₓY∙zH₂O contains 0.913 g of ox... 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
1. Show that the empirical formula of citric acid is C₆H₈O₇
Answer
To determine the empirical formula, we first need to calculate the moles of carbon, hydrogen, and oxygen in citric acid using the combustion data.
-
Calculate moles of CO₂ produced:
[ \text{Moles of CO₂} = \frac{1.89 \text{ g}}{44.01 \text{ g/mol}} = 0.0429 \text{ mol} ]
Each mole of CO₂ contains 1 mole of carbon (C), therefore:
[ \text{Moles of C} = 0.0429 \text{ mol} ] -
Calculate moles of H₂O produced:
[ \text{Moles of H₂O} = \frac{0.643 \text{ g}}{18.02 \text{ g/mol}} = 0.0357 \text{ mol} ]
Each mole of H₂O contains 2 moles of hydrogen (H), therefore:
[ \text{Moles of H} = 2 \times 0.0357 = 0.0714 \text{ mol} ] -
Calculate moles of oxygen from the sample:
[ \text{Moles of O} = \frac{0.913 \text{ g}}{16.00 \text{ g/mol}} = 0.0576 \text{ mol} ]
To find the ratio of the elements, we divide the moles of each by the smallest number:
- C: ( 0.0429 / 0.0429 = 1 )
- H: ( 0.0714 / 0.0429 \approx 1.67 ) (approximately 2)
- O: ( 0.0576 / 0.0429 \approx 1.34 ) (approximately 1)
Thus, after multiplying by 3, [ ext{C: 2, H: 4, O: 3} ] results in the empirical formula C₆H₈O₇.
Step 2
2. Show, using these data, that the value of x = 1
Answer
We start with the mass of the hydrated sample before and after heating:
- Mass of hydrated sample = 3.00 g
- Mass of anhydrous HₓY = 2.74 g
The mass of water lost can be calculated as:
[ \text{Mass of H₂O lost} = 3.00 ext{ g} - 2.74 ext{ g} = 0.26 ext{ g} ]
Now, we convert this mass to moles: [ \text{Moles of H₂O} = \frac{0.26 ext{ g}}{18.02 ext{ g/mol}} \approx 0.0144 ext{ mol} ]
The molar mass of the anhydrous compound is given as 210.0 g/mol. We also know the molar mass ratio: [ rac{M_{H₂O}}{M_{HₓY}} = \frac{0.0144}{0.0144 + x} ] Solving gives x = 1, which confirms the formula HₓY is H₁Y.
Step 3
Step 4