Question

question 9 of 10
what is the percent composition of sulfur in so₂?

a. \\(\frac{(32.1\\ g + 16.0\\ g + 16.0\\ g)}{32.1\\ g} \times 100\\%\\)

b. \\(\frac{32.1\\ g}{(32.1\\ g + 16.0\\ g + 16.0\\ g)} \times 100\\%\\)

c. \\(\frac{32.1\\ g}{(16.0\\ g + 16.0\\ g)} \times 100\\%\\)

d. \\(\frac{(32.1\\ g)(6.02 \times 10^{23})}{100\\ g}\\)

Explanation:

Step1: Recall percent composition formula

Percent composition of an element in a compound is (mass of element / molar mass of compound) × 100%.
For \( SO_2 \), molar mass of \( S \) is 32.1 g, molar mass of \( O \) is 16.0 g. Molar mass of \( SO_2 = 32.1 + 16.0 + 16.0 \) g.

Step2: Analyze each option

• Option A: Numerator is molar mass of \( SO_2 \), denominator is mass of \( S \). Incorrect.
• Option B: Numerator is mass of \( S \), denominator is molar mass of \( SO_2 \). Correct formula.
• Option C: Denominator is mass of \( O_2 \), not \( SO_2 \). Incorrect.
• Option D: Involves Avogadro's number, not for percent composition. Incorrect.

Answer:

B. \(\frac{32.1\ \text{g}}{(32.1\ \text{g} + 16.0\ \text{g} + 16.0\ \text{g})} \times 100\%\)