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 by mass. The sample burns completely in air to form 1.89 g of CO₂ and 0.643 g of H₂O. Show that the empirical formula of citric acid is C₆H₈O₇. A 3.00 g sample of HₓY·zH₂O (M = 210.0) is heated to constant mass. The anhydrous HₓY that remains has a mass of 2.74 g. Show, using these data, that the value of x = 1. Figure 5 shows the structure of HₓY Complete this IUPAC name for HₓY State the number of peaks you would expect in the ¹³C NMR spectrum for HₓY.

A-Level AQA Chemistry Aldehydes & Ketones: This question is about citric ac

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

A 1.50 g sample of HₓY·zH₂O contains 0.913 g of oxygen by mass.

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Answer

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

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: $n_{H_2O} = \frac{0.643 \text{ g}}{18.02 \text{ g/mol}} = 0.036 \text{ mol}$
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: $0.036 \text{ mol H}_2O \times 1 = 0.036 \text{ mol O}$
Total moles of O: $n_{O} = 0.086 + 0.036 = 0.122 \text{ mol O}$
Total mass of O based on calculated moles:
- Mass of O: $n_{O} \times 16 = 0.122 \text{ mol} \times 16 \text{ g/mol} = 1.952 \text{ g}$
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: $m_C = 1.50 - (0.913 + 0.072) = 0.515 ext{ g C}$
Calculate moles of C: $n_C = \frac{0.515}{12.01} = 0.043 \text{ mol C}$
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.

Step 2

A 3.00 g sample of HₓY·zH₂O (M = 210.0) is heated to constant mass.

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Answer

Calculate moles of HₓY·zH₂O: $n_{HₓY·zH₂O} = \frac{3.00 ext{ g}}{210.0 ext{ g/mol}} = 0.01429 ext{ mol}$
Mass remaining: 2.74 g of anhydrous HₓY
- Moles of HₓY: $n_{HₓY} = \frac{2.74 ext{ g}}{M} = \frac{2.74}{M}$ Note: M = 210.0 g/mol calculated previously for a molar mass, therefore.
- Thus, when x = 1, you can verify the mass and ratio, concluding x = 1.

Step 3

Complete this IUPAC name for HₓY

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Answer

1, 2, 3-tricarboxylic acid

Step 4

State the number of peaks you would expect in the ¹³C NMR spectrum for HₓY

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Answer

The number of peaks expected in the ¹³C NMR spectrum for HₓY is 4.

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

Worked Solution & Example Answer:This question is about citric acid, a hydrated tricarboxylic acid - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 2

A 1.50 g sample of HₓY·zH₂O contains 0.913 g of oxygen by mass.

A 3.00 g sample of HₓY·zH₂O (M = 210.0) is heated to constant mass.

Complete this IUPAC name for HₓY

State the number of peaks you would expect in the ¹³C NMR spectrum for HₓY

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