Question

An 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

Answer: (D)

$C_{12}H_{22}O_{11}$

Answer 2

To find the molecular formula, we first determine the empirical formula.

The percentage of carbon is 42.1%, hydrogen is 6.4%, and the remainder is oxygen.
Percentage of oxygen = 100% - (42.1% + 6.4%) = 100% - 48.5% = 51.5%.

Assume 100 g of the compound. Then, the mass of each element is:
Mass of carbon = 42.1 g
Mass of hydrogen = 6.4 g
Mass of oxygen = 51.5 g

Convert the mass of each element to moles using their atomic masses (C=12.01, H=1.008, O=16.00). Using integer atomic masses (C=12, H=1, O=16) is usually sufficient for these problems unless high precision is required. Let's use integer masses first.
Moles of carbon = 42.1 g / 12 g/mol ≈ 3.5083 mol
Moles of hydrogen = 6.4 g / 1 g/mol = 6.4 mol
Moles of oxygen = 51.5 g / 16 g/mol ≈ 3.21875 mol

To find the simplest whole-number ratio, divide the number of moles of each element by the smallest number of moles (moles of oxygen):
Ratio of C = 3.5083 / 3.21875 ≈ 1.0899
Ratio of H = 6.4 / 3.21875 ≈ 1.9883
Ratio of O = 3.21875 / 3.21875 = 1

The ratios are approximately 1.09 : 1.99 : 1. These are close to 1.1 : 2 : 1 or 11/10 : 2 : 1.
Multiplying by 10 gives the ratio approximately 11 : 20 : 10.
This suggests an empirical formula of $C_{11}H_{20}O_{10}$.
The empirical formula weight = $11 \times 12 + 20 \times 1 + 10 \times 16 = 132 + 20 + 160 = 312$.
The molecular weight is given as 342.
The ratio of molecular weight to empirical formula weight is $n = 342 / 312 \approx 1.096$. Since n should be an integer, there might be rounding issues with the percentages or the empirical formula derived is not correct.

Alternatively, we can use the molecular weight to directly find the number of atoms of each element in the molecular formula. Let the molecular formula be $C_x H_y O_z$.
The mass of carbon in one molecule is $12x$. The percentage of carbon is $(12x / 342) \times 100 = 42.1$.
$12x = 342 \times (42.1 / 100) = 342 \times 0.421 = 144.002$
$x = 144.002 / 12 \approx 12.000$

The mass of hydrogen in one molecule is $1y$. The percentage of hydrogen is $(1y / 342) \times 100 = 6.4$.
$1y = 342 \times (6.4 / 100) = 342 \times 0.064 = 21.888$
$y \approx 21.888$

The mass of oxygen in one molecule is $16z$. The percentage of oxygen is $(16z / 342) \times 100 = 51.5$.
$16z = 342 \times (51.5 / 100) = 342 \times 0.515 = 176.13$
$z = 176.13 / 16 \approx 11.008$

Rounding the number of atoms to the nearest integer, we get x=12, y=22, z=11.
So, the molecular formula is $C_{12}H_{22}O_{11}$.

Let's verify the molecular weight of $C_{12}H_{22}O_{11}$:
Molecular weight = $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342$.
This matches the given molecular weight.

Let's verify the percentages for $C_{12}H_{22}O_{11}$:
Percentage of C = $(12 \times 12 / 342) \times 100 = (144 / 342) \times 100 \approx 42.105 \%$
Percentage of H = $(22 \times 1 / 342) \times 100 = (22 / 342) \times 100 \approx 6.433 \%$
Percentage of O = $(11 \times 16 / 342) \times 100 = (176 / 342) \times 100 \approx 51.462 \%$

These calculated percentages are very close to the given percentages (42.1%, 6.4%, 51.5%).

The final answer is $C_{12}H_{22}O_{11}$.