Question

Ail organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :

• A. \(C_{11}H_{18}O_{12}\)
• B. \(C_{12}H_{20}O_{12}\)
• C. \(C_{14}H_{20}O_{10}\)
• D. \(C_{12}H_{22}O_{11}\)

Answer 1

The Correct Option is D

Only C₁₂H₂₂O₁₁ has 42.1% carbon, 6.4% hydrogen, and 51.5% oxygen.
Thus the correct answer is Option 4.

Answer 2

Step 1: Write the given data clearly
We are told that an organic compound has the following percentage composition:
C = 42.1%, H = 6.4%, and the remainder is O.
Therefore, oxygen percentage = 100 − (42.1 + 6.4) = 51.5%.
The molecular weight (molar mass) of the compound = 342.

Step 2: Convert mass percentages to moles
To find the empirical formula, assume 100 g of the compound:

• Mass of C = 42.1 g → moles of C = 42.1 ÷ 12 = 3.508
• Mass of H = 6.4 g → moles of H = 6.4 ÷ 1 = 6.4
• Mass of O = 51.5 g → moles of O = 51.5 ÷ 16 = 3.219

Step 3: Determine the simplest mole ratio
Divide each by the smallest number (≈ 3.219):
C : H : O = (3.508 ÷ 3.219) : (6.4 ÷ 3.219) : (3.219 ÷ 3.219)
= 1.09 : 1.99 : 1 ≈ 1 : 2 : 1.
Hence, the empirical formula = CH₂O.

Step 4: Calculate the empirical formula mass
Molar mass of CH₂O = 12 + 2 + 16 = 30 g mol⁻¹.

Step 5: Relate molecular formula to empirical formula
Given molecular weight = 342, empirical unit = 30.
\[ \text{Number of empirical units per molecule} = \frac{342}{30} = 11.4 \approx 11. \] Therefore, the molecular formula = (CH₂O)₁₁ = C₁₁H₂₂O₁₁.
But the stable known carbohydrate compound with similar composition is sucrose (C₁₂H₂₂O₁₁), suggesting a rounding effect due to experimental data.

Step 6: Verify by recalculating molar mass
Molar mass of C₁₂H₂₂O₁₁ = (12×12) + (22×1) + (11×16) = 144 + 22 + 176 = 342 g mol⁻¹.
This exactly matches the given molecular weight.

Step 7: Final answer
Hence, the molecular formula of the compound is:

\( C_{12}H_{22}O_{11} \)