Question

What is the empirical formula of a compound containing 40% sulfur and 60% oxygen by mass?

• A. \( \text{SO}_3 \)
• B. \( \text{SO}_2 \)
• C. \( \text{S}_2\text{O}_3 \)
• D. \( \text{SO} \)

Hint:

Remember: To determine the empirical formula, convert the mass percentages of elements to moles, find the simplest ratio, and write the formula using whole numbers.

Answer 1

The Correct Option is A

The compound's composition by mass is:

• Sulfur (S) = 40%
• Oxygen (O) = 60%

Step 1: Assume a 100 g sample

This allows percentages to be directly converted to grams:

• Sulfur: 40 g
• Oxygen: 60 g

Step 2: Convert mass to moles

Using the molar masses: Sulfur (S) = 32 g/mol, Oxygen (O) = 16 g/mol:

• Sulfur moles: \[ \text{moles of S} = \frac{40}{32} = 1.25 \, \text{mol} \]
• Oxygen moles: \[ \text{moles of O} = \frac{60}{16} = 3.75 \, \text{mol} \]

Step 3: Determine the simplest mole ratio

Divide each element's mole count by the smallest mole value:

• Sulfur ratio: \[ \frac{1.25}{1.25} = 1 \]
• Oxygen ratio: \[ \frac{3.75}{1.25} = 3 \]

The sulfur to oxygen mole ratio is 1:3.

Step 4: State the empirical formula

Based on the 1:3 mole ratio, the empirical formula is \( \text{SO}_3 \).