Question
How do I get the formula of a hydrocarbon relative formula mass 56 .on complete combustion 3.52g of carbon iv oxide and 1.44g of water is formed
Answers
Answered by
GPT-4o mini
To determine the formula of the hydrocarbon based on the combustion data given, we can follow these steps:
1. **Determine moles of carbon dioxide (CO₂):**
- Molar mass of CO₂ = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol
- Moles of CO₂ produced = mass of CO₂ / molar mass of CO₂
\[
\text{Moles of CO₂} = \frac{3.52 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0800 \, \text{mol}
\]
2. **Determine moles of water (H₂O):**
- Molar mass of H₂O = 2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H₂O produced = mass of H₂O / molar mass of H₂O
\[
\text{Moles of H₂O} = \frac{1.44 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0800 \, \text{mol}
\]
3. **Determine moles of carbon (C) and hydrogen (H) in the hydrocarbon:**
- From the moles of CO₂ produced, we know that each mole of CO₂ contains 1 mole of carbon. Therefore, moles of carbon:
\[
\text{Moles of C} = 0.0800 \, \text{mol}
\]
- From the moles of H₂O produced, we know that each mole of H₂O contains 2 moles of hydrogen. Therefore, moles of hydrogen:
\[
\text{Moles of H} = 2 \times 0.0800 \, \text{mol} = 0.1600 \, \text{mol}
\]
4. **Express the ratio of moles of carbon to moles of hydrogen:**
- Carbon to hydrogen ratio is:
\[
\text{C} : \text{H} = 0.0800 : 0.1600 = 1 : 2
\]
5. **Write the empirical formula:**
- The empirical formula based on the mole ratio is \( \text{CH}_2 \).
6. **Determine the molecular formula:**
- The relative molar mass of the empirical formula \( \text{CH}_2 \):
\[
\text{C: } 12.01 \, \text{g/mol} + \text{H: } (2 \times 1.01 \, \text{g/mol}) = 12.01 + 2.02 = 14.03 \, \text{g/mol}
\]
- The molecular formula mass (given) = 56 g/mol. Therefore, we can find how many times the empirical formula mass fits into the molecular mass:
\[
n = \frac{56 \, \text{g/mol}}{14.03 \, \text{g/mol}} \approx 4
\]
7. **Determine the molecular formula:**
- Multiply subscripts in the empirical formula \( \text{CH}_2 \) by 4 to find the molecular formula:
\[
\text{C}_4\text{H}_8
\]
Thus, the molecular formula of the hydrocarbon is \( \text{C}_4\text{H}_8 \) (butene).
1. **Determine moles of carbon dioxide (CO₂):**
- Molar mass of CO₂ = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol
- Moles of CO₂ produced = mass of CO₂ / molar mass of CO₂
\[
\text{Moles of CO₂} = \frac{3.52 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0800 \, \text{mol}
\]
2. **Determine moles of water (H₂O):**
- Molar mass of H₂O = 2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- Moles of H₂O produced = mass of H₂O / molar mass of H₂O
\[
\text{Moles of H₂O} = \frac{1.44 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0800 \, \text{mol}
\]
3. **Determine moles of carbon (C) and hydrogen (H) in the hydrocarbon:**
- From the moles of CO₂ produced, we know that each mole of CO₂ contains 1 mole of carbon. Therefore, moles of carbon:
\[
\text{Moles of C} = 0.0800 \, \text{mol}
\]
- From the moles of H₂O produced, we know that each mole of H₂O contains 2 moles of hydrogen. Therefore, moles of hydrogen:
\[
\text{Moles of H} = 2 \times 0.0800 \, \text{mol} = 0.1600 \, \text{mol}
\]
4. **Express the ratio of moles of carbon to moles of hydrogen:**
- Carbon to hydrogen ratio is:
\[
\text{C} : \text{H} = 0.0800 : 0.1600 = 1 : 2
\]
5. **Write the empirical formula:**
- The empirical formula based on the mole ratio is \( \text{CH}_2 \).
6. **Determine the molecular formula:**
- The relative molar mass of the empirical formula \( \text{CH}_2 \):
\[
\text{C: } 12.01 \, \text{g/mol} + \text{H: } (2 \times 1.01 \, \text{g/mol}) = 12.01 + 2.02 = 14.03 \, \text{g/mol}
\]
- The molecular formula mass (given) = 56 g/mol. Therefore, we can find how many times the empirical formula mass fits into the molecular mass:
\[
n = \frac{56 \, \text{g/mol}}{14.03 \, \text{g/mol}} \approx 4
\]
7. **Determine the molecular formula:**
- Multiply subscripts in the empirical formula \( \text{CH}_2 \) by 4 to find the molecular formula:
\[
\text{C}_4\text{H}_8
\]
Thus, the molecular formula of the hydrocarbon is \( \text{C}_4\text{H}_8 \) (butene).