Question

An organic compound was found to contain 40.0% C and 6.66% H. Find its molecular formula (molar mass = 180).

• A. C\(_{22}\)H\(_{24}\)O\(_{11}\)
• B. C\(_2\)H\(_4\)O\(_2\)
• C. CH\(_2\)O
• D. C\(_6\)H\(_{12}\)O\(_6\)

Hint:

To determine the molecular formula, first calculate the empirical formula and then use the molar mass to find the ratio between the empirical and molecular formula.

Answer 1

The Correct Option is D

Step 1: Understanding the mass percentages.
Let the molar mass of the compound be 180 g/mol. From the given percentages, the mass of carbon (C) in 180 g is: \[ \text{Mass of C} = \frac{40.0}{100} \times 180 = 72 \, \text{g} \] The mass of hydrogen (H) is: \[ \text{Mass of H} = \frac{6.66}{100} \times 180 = 12 \, \text{g} \]
Step 2: Determining moles of C and H.
- Moles of C = \(\frac{72}{12} = 6\) moles. - Moles of H = \(\frac{12}{1} = 12\) moles.
Step 3: Finding the empirical formula.
The empirical formula is C\(_6\)H\(_{12}\), and since the molar mass is 180 g/mol, we can find the molecular formula by dividing the molar mass by the empirical formula mass (which is \(6 \times 12 + 12 \times 1 = 72 + 12 = 84\)). \[ \frac{180}{84} = 2.14 \quad \text{(approximately 2)} \] Thus, the molecular formula is C\(_6\)H\(_{12}\)O\(_6\).

Step 4: Conclusion.
The correct molecular formula is (D) C\(_6\)H\(_{12}\)O\(_6\).