Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Question

3 months ago

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

ced 39.01 g CO2 and
10.65 g H2O. The molar mass of the unknown compound is 272.38 g/mol.
Find the molecular formula of the unknown compound.

Chemistry

1 answer:

lions [2.9K] · Answer 1 · 2026-03-03T02:05:20+03:00

Answer: The empirical formula and molecular formula for the analyzed organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

Explanation:

The combustion chemical equation for a hydrocarbon containing carbon, hydrogen, and oxygen is:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' represent the subscripts for Carbon, Hydrogen, and Oxygen.

We have the following data:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For carbon mass calculation:

12 grams of carbon are found in 44 grams of carbon dioxide.

Thus, in 39.01 grams of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ grams of carbon will be present.

For hydrogen mass calculation:

In 18 grams of water, 2 grams of hydrogen are contained.

Therefore, in 10.65 grams of water, $\frac{2}{18}\times 10.65=1.18g$ grams of hydrogen will be present.

The oxygen mass in the compound = (13.42) - (10.64 + 1.18) = 1.6 grams.

To derive the empirical formula, you need to perform a few steps:

Step 1: Convert the provided masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Determine the mole ratio of the elements.

Each mole value is divided by the smallest mole value, which is 0.1 moles, to find the mole ratio.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Use the mole ratio values as subscripts.

The ratio of C: H: O = 9: 12: 1

The empirical formula for the compound is $C_9H_{12}O$

To find the molecular formula, it’s necessary to ascertain the valency, which is then multiplied by each elemental subscript of the empirical formula.

The formula used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

Given:

Mass of the molecular formula = 272.38 g/mol

Mass of the empirical formula = 136 g/mol

Substituting values into the equation gives:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying the determined valency with the empirical formula’s element subscripts results in:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Therefore, the forms of the organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$