Question

in the compound.

1. a similar chemical analysis is performed on a 500.0 g sample of the sugar isolated from a sample of pure sugar cane. analysis shows this sample contains 211.0 g of carbon, 32.5 g of hydrogen, and 256.5 g of oxygen.

a. determine the percent composition by mass of each element in the sugar cane sample.
b. could the sugar in this sample be sucrose? justify your conclusion.

Explanation:

Step1: Calculate percent composition of Carbon

Percent composition formula: $\% \text{element} = \frac{\text{mass of element}}{\text{mass of sample}} \times 100\%$
For Carbon: $\frac{211.0\ \text{g}}{500.0\ \text{g}} \times 100\% = 42.20\%$

Step2: Calculate percent composition of Hydrogen

For Hydrogen: $\frac{32.5\ \text{g}}{500.0\ \text{g}} \times 100\% = 6.50\%$

Step3: Calculate percent composition of Oxygen

For Oxygen: $\frac{256.5\ \text{g}}{500.0\ \text{g}} \times 100\% = 51.30\%$

(For part b: Sucrose has formula $\text{C}_{12}\text{H}_{22}\text{O}_{11}$. Molar mass of C: $12\times12.01 = 144.12$, H: $22\times1.008 = 22.176$, O: $11\times16.00 = 176.00$. Total molar mass: $144.12 + 22.176 + 176.00 = 342.296\ \text{g/mol}$.
$\%$ C: $\frac{144.12}{342.296} \times 100\% \approx 42.11\%$, $\%$ H: $\frac{22.176}{342.296} \times 100\% \approx 6.48\%$, $\%$ O: $\frac{176.00}{342.296} \times 100\% \approx 51.42\%$. Compare with calculated values: close, so likely sucrose.)

Answer:

(for part a):

• Carbon: $42.20\%$
• Hydrogen: $6.50\%$
• Oxygen: $51.30\%$

(For part b: Yes, the percent compositions are very close to sucrose’s, so it could be sucrose.)