Question

a chemist carefully measures the amount of heat needed to raise the temperature of a 644.0 mg sample of ( c_{5}h_{12}s ) from ( 2.8^{circ}c ) to ( 19.4^{circ}c ). the experiment shows that 21.4 j of heat are needed. what can the chemist report for the molar heat capacity of ( c_{5}h_{12}s )? be sure your answer has the correct number of significant digits.
( j mol^{-1} k^{-1} )

Explanation:

Step1: Calculate moles of \( C_5H_{12}S \)

Molar mass of \( C_5H_{12}S \): \( C = 12.01, H = 1.008, S = 32.07 \)
Molar mass \( M = 5\times12.01 + 12\times1.008 + 32.07 = 60.05 + 12.096 + 32.07 = 104.216 \, g/mol \)
Mass \( m = 644.0 \, mg = 0.6440 \, g \)
Moles \( n = \frac{m}{M} = \frac{0.6440}{104.216} \approx 0.006179 \, mol \)

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

\( \Delta T = 19.4 - 2.8 = 16.6 \, ^\circ C \) (change in Kelvin is same as Celsius)

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

Rearrange: \( C = \frac{q}{n\Delta T} \)
\( q = 21.4 \, J \), \( n \approx 0.006179 \, mol \), \( \Delta T = 16.6 \, K \)
\( C = \frac{21.4}{0.006179\times16.6} \approx \frac{21.4}{0.1026} \approx 208.6 \, J \, mol^{-1}K^{-1} \) (Wait, no, recalculate: 0.006179*16.6=0.1025714; 21.4/0.1025714≈208.6? Wait, no, maybe miscalculation. Wait, 0.6440g / 104.216g/mol = 0.006179mol. Then 21.4J = 0.006179mol * C * 16.6K. So C = 21.4 / (0.006179*16.6) = 21.4 / (0.10257) ≈ 208.6? But let's check significant digits. Mass is 644.0 (4 sig figs), temp change 16.6 (3 sig figs), heat 21.4 (3 sig figs). Molar mass: 104.216 (from precise values, but mass is 4 sig figs, so moles: 0.6440 / 104.216 = 0.006179 (4 sig figs? Wait, 0.6440 is 4, 104.216 is more, so moles is 4 sig figs? Wait, no: 0.6440 / 104.216 = 0.006179 (let's calculate: 0.6440 ÷ 104.216 ≈ 0.006179. Then 0.006179 * 16.6 = 0.10257. 21.4 / 0.10257 ≈ 208.6. But wait, maybe I made a mistake in molar mass. Wait, \( C_5H_{12}S \): 5C=60.05, 12H=12.096, S=32.07. Sum: 60.05+12.096=72.146+32.07=104.216 g/mol. Correct. Then 0.6440g is 0.6440/104.216 = 0.006179 mol. Then 21.4J = n*C*ΔT. So C = 21.4 / (n*ΔT) = 21.4 / (0.006179*16.6) = 21.4 / 0.10257 ≈ 208.6. But wait, maybe I messed up the moles. Wait, 644.0 mg is 0.6440 g. 0.6440 / 104.216 = 0.006179 mol. Then 0.006179 * 16.6 = 0.10257. 21.4 / 0.10257 ≈ 208.6. But let's check again. Wait, maybe the molar mass is 104.22 g/mol (rounded). Then 0.6440 / 104.22 = 0.006179 mol. Then 21.4 / (0.006179*16.6) = 21.4 / 0.10257 ≈ 208.6. But the answer should have 3 significant digits? Wait, heat is 21.4 (3), temp change 16.6 (3), mass 644.0 (4). So the least is 3, so 209? Wait, no, 208.6 rounds to 209? Wait, no, 208.6 is approximately 209? Wait, no, let's do the calculation more accurately.

Wait, 0.6440 g / 104.216 g/mol = 0.006179 mol (exact: 0.6440 ÷ 104.216 = 0.00617903 mol)

ΔT = 16.6 K

So nΔT = 0.00617903 * 16.6 = 0.1025719 K·mol

Then C = 21.4 J / 0.1025719 K·mol ≈ 208.6 J·mol⁻¹K⁻¹. Wait, but maybe I made a mistake in the formula. The formula is q = nCΔT, so C = q/(nΔT). Yes. So that's correct. Wait, but the initial answer was wrong, maybe the molar mass was miscalculated? Wait, \( C_5H_{12}S \): 5 carbons (12.01*5=60.05), 12 hydrogens (1.008*12=12.096), 1 sulfur (32.07). Sum: 60.05+12.096=72.146+32.07=104.216 g/mol. Correct. So 0.6440g is 0.006179mol. Then 21.4J = 0.006179mol * C * 16.6K. So C = 21.4 / (0.006179*16.6) = 21.4 / 0.10257 ≈ 208.6. So approximately 209 J·mol⁻¹K⁻¹? Wait, no, 208.6 is about 209, but let's check with more precise calculation. Wait, 0.006179*16.6=0.1025714. 21.4 divided by 0.1025714: 21.4 ÷ 0.1025714 ≈ 208.6. So maybe the correct answer is approximately 209? Wait, no, maybe I messed up the mass. Wait, 644.0 mg is 0.6440 g. Correct. Molar mass 104.216 g/mol. Correct. So moles 0.006179. Correct. Then 21.4 / (0.006179*16.6) = 21.4 / 0.10257 ≈ 208.6. So the molar heat capacity is approximately 209 J·mol⁻¹K⁻¹? Wait, but let's check the significant figures. The heat is 21.4 (3 sig figs), temperature change 16.6 (3 sig figs), mass 644.0 (4 sig figs). So the limiting is 3 sig figs, so 209? Wait, 208.6 rounds to 209? Wait, no, 208.6 is closer to 209, but maybe the calculation is wrong. Wait, wait, maybe I used the wrong formula. Wait, molar heat capacity is per mole, so yes, q = nCΔT. So that's correct. Alternatively, maybe the molar mass is 104.22 g/mol, so 0.6440 / 104.22 = 0.006179 mol. Then 21.4 / (0.006179*16.6) = 208.6. So the answer should be approximately 209 J·mol⁻¹K⁻¹? Wait, but let's do the calculation again:

0.6440 g / 104.216 g/mol = 0.006179 mol (exact value: 0.6440 ÷ 104.216 = 0.00617903 mol)

ΔT = 19.4 - 2.8 = 16.6 K

nΔT = 0.00617903 mol * 16.6 K = 0.1025719 K·mol

C = 21.4 J / 0.1025719 K·mol ≈ 208.6 J·mol⁻¹K⁻¹

So rounding to 3 significant figures, it's 209 J·mol⁻¹K⁻¹? Wait, but 208.6 is 209 when rounded to 3 sig figs? Wait, 208.6: the third sig fig is 8, next digit 6, so round up: 209. Yes.

Answer:

\boxed{209}