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

We are given the following percentages by mass:

• Sulfur (S) = 40%
• Oxygen (O) = 60%

Step 1: Assume 100 g of the compound

This simplifies the calculation, as the percentage directly translates to grams:

• Sulfur: 40 g
• Oxygen: 60 g

Step 2: Convert mass to moles

The molar mass of sulfur (S) is 32 g/mol, and for oxygen (O) it is 16 g/mol:

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

Step 3: Find the mole ratio

To find the empirical formula, divide the moles of each element by the smallest number of moles:

• For sulfur: \[ \frac{1.25}{1.25} = 1 \]
• For oxygen: \[ \frac{3.75}{1.25} = 3 \]

Thus, the ratio of sulfur to oxygen is 1:3.

Step 4: Write the empirical formula

From the mole ratio of 1:3, the empirical formula of the compound is \( \text{SO}_3 \).