Q. What is the empirical formula of a compound having 40% carbon, 6.66% hydrogen and 53.34% oxygen?
A) $C_{2}H_{2}O$
B) $C_{2}H_{4}O$
C) $CH_{2}O$
D) $CHO$
Explanation
(c)
Atom Atomic Percentage $\frac{\mathbf{b}}{\mathbf{a}}\mathbf{= x}$ Ratio
Mass $\mathbf{(a)}$ $\mathbf{(b)}$
C 12 40 $\frac{40}{12} = 3.33$ 1
H 1 6.66 $\frac{6.66}{1} = 6.66$ 2
O 16 53.34 $\frac{53.34}{16} = 3.33$ 1
Hence, empirical formula $= CH_{2}O$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.