Question

a chemist carefully measures the amount of heat needed to raise the temperature of a 306.0 mg sample of $c_{4}h_{8}o_{2}$ from $55.7^{circ}c$ to $66.8^{circ}c$. the experiment shows that 5.41 j of heat are needed. what can the chemist report for the molar heat capacity of $c_{4}h_{8}o_{2}$? be sure your answer has the correct number of significant digits.
$\square j\cdot mol^{-1}\cdot k^{-1}$

Explanation:

Step1: Calculate the moles of \( C_4H_6O_2 \)

The molar mass of \( C_4H_6O_2 \): \( C = 12.01 \, g/mol \), \( H = 1.008 \, g/mol \), \( O = 16.00 \, g/mol \)
Molar mass \( M = 4\times12.01 + 6\times1.008 + 2\times16.00 = 48.04 + 6.048 + 32.00 = 86.088 \, g/mol \)
Mass of sample \( m = 306.0 \, mg = 0.3060 \, g \)
Moles \( n = \frac{m}{M} = \frac{0.3060 \, g}{86.088 \, g/mol} \approx 0.003554 \, mol \)

Step2: Calculate the temperature change \( \Delta T \)

\( \Delta T = 66.8^\circ C - 55.7^\circ C = 11.1^\circ C \). Since \( \Delta T \) in Kelvin is the same as in Celsius, \( \Delta T = 11.1 \, K \)

Step3: Use the heat formula \( q = nC\Delta T \) to find molar heat capacity \( C \)

Rearranging for \( C \): \( C = \frac{q}{n\Delta T} \)
Given \( q = 5.41 \, J \), \( n \approx 0.003554 \, mol \), \( \Delta T = 11.1 \, K \)
\( C = \frac{5.41 \, J}{0.003554 \, mol \times 11.1 \, K} \approx \frac{5.41}{0.03945} \approx 137.1 \, J \cdot mol^{-1} \cdot K^{-1} \) (rounded to correct significant figures: 137 \( J \cdot mol^{-1} \cdot K^{-1} \) or more precisely, let's recalculate with more precision)

Wait, let's do the calculation more accurately:

\( n = \frac{0.3060}{86.088} = 0.3060 \div 86.088 \approx 0.003554 \)

\( n\Delta T = 0.003554 \times 11.1 = 0.03945 \)

\( C = 5.41 \div 0.03945 \approx 137.1 \). Checking significant figures: \( q \) has 3, \( m \) has 4, \( \Delta T \) has 3, so result should have 3 significant figures. So \( C \approx 137 \, J \cdot mol^{-1} \cdot K^{-1} \) (wait, let's check again: 5.41 (3 sig figs), 0.3060 (4), 86.088 (5), 11.1 (3). The least number of sig figs in multiplication/division steps: 3 (from 5.41 and 11.1). So let's recalculate with more precise n:

\( n = 0.3060 / 86.088 = 0.0035544 \)

\( n\Delta T = 0.0035544 * 11.1 = 0.03945384 \)

\( C = 5.41 / 0.03945384 ≈ 137.1 \). Rounding to 3 sig figs: 137 \( J \cdot mol^{-1} \cdot K^{-1} \)? Wait, 5.41 is 3, 0.3060 is 4, 11.1 is 3. So the result should have 3 sig figs. Wait, 137.1 rounds to 137? Wait, 5.41 / (0.3060/86.088 11.1) = 5.41 86.088 / (0.3060 * 11.1)

Calculate numerator: 5.41 86.088 ≈ 5.41 86.088 ≈ 465.736

Denominator: 0.3060 * 11.1 ≈ 3.3966

Then 465.736 / 3.3966 ≈ 137.1. So 137 \( J \cdot mol^{-1} \cdot K^{-1} \) (3 sig figs)

Wait, maybe I made a mistake in molar mass? Let's recalculate molar mass:

4C: 4*12.01 = 48.04

6H: 6*1.008 = 6.048

2O: 2*16.00 = 32.00

Sum: 48.04 + 6.048 = 54.088 + 32.00 = 86.088 g/mol. Correct.

Mass: 306.0 mg = 0.3060 g. Correct.

Moles: 0.3060 / 86.088 ≈ 0.003554 mol. Correct.

Delta T: 11.1 K. Correct.

q = 5.41 J. Correct.

So C = 5.41 / (0.003554 * 11.1) = 5.41 / 0.03945 ≈ 137.1. So with 3 sig figs, 137 J·mol⁻¹·K⁻¹? Wait, 5.41 has 3, 0.3060 has 4, 11.1 has 3. The limiting is 3, so 137 (or 1.37×10²? Wait, 137 is three sig figs. Wait, let's check the calculation again:

0.003554 mol * 11.1 K = 0.03945 mol·K

5.41 J / 0.03945 mol·K ≈ 137.1 J·mol⁻¹·K⁻¹. So the answer is approximately 137 \( J \cdot mol^{-1} \cdot K^{-1} \) (or 1.37 × 10², but 137 is three sig figs which matches 5.41's three sig figs).

Wait, maybe I miscalculated the moles. Let's do 0.3060 g / 86.088 g/mol:

86.088 * 0.0035 = 0.301308

0.3060 - 0.301308 = 0.004692

0.004692 / 86.088 ≈ 0.0000545

So total moles ≈ 0.0035 + 0.0000545 ≈ 0.0035545, which matches. So that's correct.

So the molar heat capacity is approximately 137 \( J \cdot mol^{-1} \cdot K^{-1} \) (or more precisely, 137 when rounded to three significant figures).

Answer:

\boxed{137} (or more accurately, if we consider the calculation more precisely, maybe 1.37×10², but 137 is three sig figs. Wait, let's check the exact calculation:

5.41 / ( (0.3060 / 86.088) 11.1 ) = 5.41 86.088 / (0.3060 * 11.1)

5.41 86.088 = 5.41 86 + 5.41 * 0.088 = 464.26 + 0.47608 = 464.73608

0.3060 * 11.1 = 3.3966

464.73608 / 3.3966 ≈ 136.8, which rounds to 137. So yes, 137.