Question

An organic compound is found to contain $ \mathrm{C} = 54.5% $, $ \mathrm{O} = 36.4% $, and $ \mathrm{H} = 9.1% $ by weight. Its empirical formula is:

• A. \(\mathrm{CHO}_2\)
• B. \(\mathrm{C_2H_4O}\)
• C. \(\mathrm{C_2H_6O}\)
• D. \(\mathrm{C_3H_4O}\)

Hint:

Always convert mass percentages to moles by dividing with atomic masses and then normalize using the smallest mole value.

Answer 1

The Correct Option is B

Step 1: Divide each percentage by atomic mass to get mole ratio. 

C: \( \frac{54.5}{12} = 4.54 \),  H: \( \frac{9.1}{1} = 9.1 \),  O: \( \frac{36.4}{16} = 2.275 \) 

Step 2: Divide all by smallest value (2.275). 

C: \( \frac{4.54}{2.275} = 2 \),  H: \( \frac{9.1}{2.275} = 4 \),  O: \( \frac{2.275}{2.275} = 1 \) 

Empirical formula: \( \mathrm{C_2H_4O} \)