Question

A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is its empirical formula?
(Atomic masses: C = 12, H = 1, O = 16)

• A. \(CH_{2}O\)
• B. \(C_{2}H_{4}O_{2}\)
• C. \(C_{3}H_{6}O_{3}\)
• D. \(C_{2}H_{2}O\)

Hint:

To find the empirical formula, convert percentages to grams, grams to moles, divide by the smallest mole value, and round to nearest whole numbers to get the simplest ratio.

Answer 1

The Correct Option is A

Step 1: Assume 100 g of the compound. So, mass of each element is: 
Carbon = 40 g, Hydrogen = 6.7 g, Oxygen = 53.3 g 
Step 2: Convert mass to moles: 
Moles of C = \( \frac{40}{12} = 3.33 \) 
Moles of H = \( \frac{6.7}{1} = 6.7 \) 
Moles of O = \( \frac{53.3}{16} = 3.33 \) 
Step 3: Divide each mole value by the smallest number of moles (which is 3.33): 
\[ \text{C: } \frac{3.33}{3.33} = 1, \quad \text{H: } \frac{6.7}{3.33} \approx 2, \quad \text{O: } \frac{3.33}{3.33} = 1 \] Step 4: The simplest ratio is C:H:O = 1:2:1, so the empirical formula is \(CH_{2}O\).