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Esters are common components of artificial flavours - VCE - SSCE Chemistry - Question 5 - 2002 - Paper 1
Question 5
Esters are common components of artificial flavours. An ester, known to contain only the elements carbon, hydrogen and oxygen, was isolated and its composition analy... show full transcript
Worked Solution & Example Answer:Esters are common components of artificial flavours - VCE - SSCE Chemistry - Question 5 - 2002 - Paper 1
Step 1
mass of carbon in 1.02 g of the compound
Answer
To find the mass of carbon, we first calculate the amount of carbon produced from the carbon dioxide formed:
- Molar mass of CO₂ = 12 (for C) + 16×2 (for O) = 44 g/mol.
- Moles of CO₂ produced = ( \frac{2.20 \text{ g}}{44 \text{ g/mol}} = 0.050 \text{ mol} ).
- Since each mole of CO₂ contains one mole of C, moles of C = 0.050 mol.
- Mass of C = moles × molar mass = ( 0.050 \text{ mol} \times 12 ext{ g/mol} = 0.60 ext{ g} ).
Step 2
mass of hydrogen in 1.02 g of the compound
Answer
To find the mass of hydrogen, we first calculate the amount of water produced:
- Molar mass of H₂O = 2 (for H) + 16 (for O) = 18 g/mol.
- Moles of H₂O produced = ( \frac{0.90 ext{ g}}{18 ext{ g/mol}} = 0.050 \text{ mol} ).
- Each mole of H₂O contains 2 moles of H, hence moles of H = 2 × 0.050 mol = 0.10 mol.
- Mass of H = moles × molar mass = ( 0.10 ext{ mol} \times 1 ext{ g/mol} = 0.10 ext{ g} ).
Step 3
Step 4
empirical formula of the compound
Answer
To find the empirical formula, we need to determine the mole ratio of C, H, and O:
- Moles of C = ( \frac{0.60 ext{ g}}{12 ext{ g/mol}} = 0.050 ext{ mol} ).
- Moles of H = ( \frac{0.10 ext{ g}}{1 ext{ g/mol}} = 0.10 ext{ mol} ).
- Moles of O = ( \frac{0.32 ext{ g}}{16 ext{ g/mol}} = 0.020 ext{ mol} ).
Next, to find the simplest ratio:
- C: 0.050 / 0.020 = 2.5
- H: 0.10 / 0.020 = 5
- O: 0.020 / 0.020 = 1
To convert to whole numbers, multiply each by 2:
- C: 2 × 2.5 = 5
- H: 5 × 2 = 10
- O: 1 × 2 = 2
Thus, the empirical formula is ( C_5H_{10}O_2 ).