Question

An organic compound containing C, H and N has 72.7% C, 13.1% H and 14.1% N. Its empirical formula is

• A. C$_6$H$_{11}$N
• B. C$_4$H$_{13}$N
• C. C$_3$H$_7$N
• D. C$_2$H$_{13}$N$_2$

Hint:

For empirical formula: convert % to mole ratios and simplify to the smallest whole numbers.

Answer 1

The Correct Option is B

Convert percentages to moles:
C: $\dfrac{72.7}{12} = 6.06$, H: $\dfrac{13.1}{1} = 13.1$, N: $\dfrac{14.1}{14} = 1.01$
Now divide each by the smallest (1.01):
C: 6.06, H: 13.1, N: 1.01 ⇒ ≈ C$_6$H$_{13}$N
But this must be simplified: divide all values by 1.5 to get small whole numbers ⇒ C$_4$H$_{13}$N