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Ostrovityanka [42]
3 years ago
7

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:
Kisachek [45]3 years ago
6 0

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_9H_{12}O and C_{18}H_{24}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.01 g of carbon dioxide, \frac{12}{44}\times 39.01=10.64g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, \frac{2}{18}\times 10.65=1.18g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given 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

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

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

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

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

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

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

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

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

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

Multiplying this valency by the subscript of every element of empirical formula, we get:

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

Hence, the empirical and molecular formula for the given organic compound is C_9H_{12}O and C_{18}H_{24}O_2

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