Question

A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is its empirical formula?

• A. \( \text{CH}_2\text{O} \)
• B. \( \text{C}_2\text{H}_4\text{O} \)
• C. \( \text{CH}_3\text{O} \)
• D. \( \text{C}_3\text{H}_6\text{O}_3 \)

Hint:

To find the empirical formula: \begin{itemize} \item Assume a 100 g sample. \item Convert mass % to moles. \item Divide all moles by the smallest mole value. \item Express as whole-number ratios. \end{itemize}

Answer 1

The Correct Option is A

Steps to Calculate the Empirical Formula

Step 1: Assume 100 g of the compound.
Then the masses of the elements are:
C = 40 g, H = 6.7 g, O = 53.3 g

Step 2: Convert masses to moles using atomic masses.

• Moles of C = 40 / 12 = 3.33
• Moles of H = 6.7 / 1 = 6.7
• Moles of O = 53.3 / 16 = 3.33

Step 3: Divide all mole values by the smallest number of moles.

• C: 3.33 / 3.33 = 1
• H: 6.7 / 3.33 ≈ 2
• O: 3.33 / 3.33 = 1

Step 4: Write the empirical formula based on whole-number mole ratios.

Empirical formula = CH₂O