Question

The elemental analysis of an oranic compond gave C:38.71%, H: 9.67%. What is the empirical formula of the compound?

• A. CH₂O
• B. CH₃O
• C. CH₄O
• D. CHO
• E. CH₅O

Answer 1

The Correct Option is B

The elemental composition of the compound is given as 38.71% Carbon and 9.67% Hydrogen. To find the empirical formula, we need to determine the simplest ratio of moles of each element in the compound. 1. First, calculate the moles of Carbon (C) and Hydrogen (H) from their percentages: For Carbon: \[ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{38.71}{12} = 3.226 \, \text{mol} \] For Hydrogen: \[ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Molar mass of H}} = \frac{9.67}{1} = 9.67 \, \text{mol} \] 2. Next, calculate the mole ratio between Carbon and Hydrogen: \[ \text{Ratio of C to H} = \frac{3.226}{3.226} : \frac{9.67}{3.226} = 1 : 3 \] 3. Therefore, the empirical formula is CH\(_3\)O.

The correct option is (B) : CH₃O

Answer 2

CH₃O

Explanation:

To find the empirical formula of the compound, we need to determine the mole ratio of carbon (C), hydrogen (H), and oxygen (O) in the compound.

Given the percentage composition:

• C: 38.71%
• H: 9.67%

We assume a 100 g sample, so we have:

• 38.71 g of C, which is 38.71 / 12 = 3.23 mol of C.
• 9.67 g of H, which is 9.67 / 1 = 9.67 mol of H.

Next, we calculate the molar ratio. Since oxygen is not directly given, we subtract the total mass of C and H from 100 g to find the mass of oxygen:

• Mass of O = 100 - (38.71 + 9.67) = 51.62 g
• 51.62 g of O, which is 51.62 / 16 = 3.23 mol of O.

Now, we have the following moles:

• C = 3.23 mol
• H = 9.67 mol
• O = 3.23 mol

The simplest whole-number ratio is approximately 1:3:1 for C:H:O. Thus, the empirical formula is CH₃O.