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 in a compound.
We assume 100g of the substance to convert percentages directly to grams.
Step 2: Key Formula or Approach:
1. Convert percentage to mass (in grams).
2. Convert mass to moles using the formula: \( \text{moles} = \frac{\text{mass}}{\text{atomic mass}} \).
3. Divide each mole value by the smallest number of moles obtained to get the simplest ratio.
Step 3: Detailed Explanation:
Given percentages: C = 40%, H = 6.7%.
Remaining must be Oxygen (O): \( 100 - (40 + 6.7) = 53.3% \).
Moles of Carbon (C): \( \frac{40}{12} = 3.33 \).
Moles of Hydrogen (H): \( \frac{6.7}{1} = 6.7 \).
Moles of Oxygen (O): \( \frac{53.3}{16} = 3.33 \).
Simplest ratio:
C: \( \frac{3.33}{3.33} = 1 \).
H: \( \frac{6.7}{3.33} \approx 2 \).
O: \( \frac{3.33}{3.33} = 1 \).
The ratio is 1:2:1. Thus, the empirical formula is CH\textsubscript{2}O.
Step 4: Final Answer:
The empirical formula is CH\textsubscript{2}O.