Question

An organic compound contains $40\%$ C and $6.7\%$ $H_2$. What is the empirical formula of the compound?

• A. $CH_4$
• B. $CH_2O$
• C. $CH_3$
• D. $CH_4O$

Hint:

The percentages $40\%$, $6.7\%$, and $53.3\%$ are the signature ratios for carbohydrates like glucose or acetic acid, which always have the empirical formula $CH_2O$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound.
If the percentages do not add up to $100\%$, the remainder is typically assumed to be Oxygen.

Step 2: Detailed Explanation:
1. Determine the percentage of Oxygen:
$\%$ Oxygen $= 100 - (\% C + \% H) = 100 - (40 + 6.7) = 53.3\%$.
2. Calculate the number of moles of each element (assume $100\text{ g}$ of substance):
Moles of C $= \frac{40}{12} = 3.33 \text{ mol}$.
Moles of H $= \frac{6.7}{1} = 6.7 \text{ mol}$.
Moles of O $= \frac{53.3}{16} = 3.33 \text{ mol}$.
3. Determine the simplest molar ratio by dividing by the smallest value ($3.33$):
C $= \frac{3.33}{3.33} = 1$.
H $= \frac{6.7}{3.33} \approx 2$.
O $= \frac{3.33}{3.33} = 1$.
The empirical formula is $CH_2O$.

Step 3: Final Answer:
The empirical formula of the compound is $CH_2O$.