Questions: A chemist carefully measures the amount of heat needed to raise the temperature of a 868.0 mg sample of C9H10O2 from 43.4 °C to 58.8 °C. The experiment shows that 22.9 J of heat are needed. What can the chemist report for the molar heat capacity of C9H10O2? Be sure your answer has the correct number of significant digits. J · mol^-1 · K^-1

Solution Steps

First, we need to convert the mass of the sample from milligrams to grams:

\[ 868.0 \, \text{mg} = 0.8680 \, \text{g} \]

Next, we calculate the number of moles of \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}\) using its molar mass. The molar mass is calculated as follows:

• Carbon (C): \(9 \times 12.01 \, \text{g/mol} = 108.09 \, \text{g/mol}\)
• Hydrogen (H): \(10 \times 1.008 \, \text{g/mol} = 10.08 \, \text{g/mol}\)
• Oxygen (O): \(2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}\)

Adding these together gives the molar mass of \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}\):

\[ 108.09 + 10.08 + 32.00 = 150.17 \, \text{g/mol} \]

Now, calculate the number of moles:

\[ \text{moles} = \frac{0.8680 \, \text{g}}{150.17 \, \text{g/mol}} = 0.005780 \, \text{mol} \]

The temperature change \(\Delta T\) is given by:

\[ \Delta T = 58.8^\circ \text{C} - 43.4^\circ \text{C} = 15.4 \, \text{K} \]

The molar heat capacity \(C\) is calculated using the formula:

\[ q = nC\Delta T \]

where \(q\) is the heat absorbed (22.9 J), \(n\) is the number of moles (0.005780 mol), and \(\Delta T\) is the temperature change (15.4 K). Solving for \(C\):

\[ C = \frac{q}{n \Delta T} = \frac{22.9 \, \text{J}}{0.005780 \, \text{mol} \times 15.4 \, \text{K}} \]

\[ C = \frac{22.9}{0.089012} = 257.2 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \]

Final Answer