Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 40.1 g
mass of calorimeter and water: 340.1 g
initial temperature of water: 23.2 ºc
final temperature of water: 31.5 ºc
mass of fuel burned: 0.211 g
type of fuel: butane c4h10

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \( 340.1\ g - 40.1\ g = 300.0\ g \)

Step2: Calculate heat absorbed by water

Using \( q = mc\Delta T \), where \( c = 4.184\ J/g^\circ C \), \( m = 300.0\ g \), \( \Delta T = 31.5^\circ C - 23.2^\circ C = 8.3^\circ C \)
\( q_{water} = 300.0\ g \times 4.184\ J/g^\circ C \times 8.3^\circ C = 300.0 \times 4.184 \times 8.3\ J \)
\( q_{water} = 10395.12\ J \approx 10400\ J = 10.4\ kJ \) (assuming calorimeter heat capacity is negligible or not given, so we consider only water's heat)

Step3: Moles of butane (\( C_4H_{10} \))

Molar mass of \( C_4H_{10} = 4\times12.01 + 10\times1.008 = 48.04 + 10.08 = 58.12\ g/mol \)
Moles (\( n \)) = \( \frac{0.211\ g}{58.12\ g/mol} \approx 0.00363\ mol \)

Step4: Molar enthalpy of combustion (\( \Delta H \))

Combustion is exothermic, so \( \Delta H = -\frac{q}{n} \) (heat released by fuel is absorbed by water, so \( q_{fuel} = -q_{water} \))
\( \Delta H = -\frac{10395.12\ J}{0.00363\ mol} \approx -2863669\ J/mol = -2860\ kJ/mol \) (approximate, more precise calculation: \( \frac{10395.12}{0.00363} \approx 2863669\ J/mol = 2860\ kJ/mol \), so \( \Delta H \approx -2860\ kJ/mol \))

Answer:

The molar enthalpy of combustion of butane is approximately \(\boldsymbol{-2860\ kJ/mol}\) (or more precisely around \(-2860\ kJ/mol\) depending on calculation precision).