An experiment was carried out to determine the enthalpy of combustion of propan-1-ol - VCE - SSCE Chemistry - Question 18 - 2020 - Paper 1

First, convert 557 mg of propan-1-ol to grams:

$557 \, \text{mg} = 0.557 \, \text{g}$

Next, calculate the molar mass of propan-1-ol (C₃H₈O):

• C: 12.01 g/mol × 3 = 36.03 g/mol
• H: 1.008 g/mol × 8 = 8.064 g/mol
• O: 16.00 g/mol × 1 = 16.00 g/mol

Total molar mass = 36.03 + 8.064 + 16.00 = 60.094 g/mol

Now, calculate moles of propan-1-ol:

$n = \frac{mass}{molar \, mass} = \frac{0.557 \, \text{g}}{60.094 \, \text{g/mol}} = 0.00926 \, \text{mol}$

Finally, use the heat absorbed by water and the moles of propan-1-ol to find the enthalpy change:

$ΔH = -\frac{q}{n} = -\frac{11.6485 \, \text{kJ}}{0.00926 \, \text{mol}} = -1250 \, \text{kJ/mol}$

Therefore, the enthalpy of combustion is closest to -1250 kJ mol⁻¹.