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romanna [79]
3 years ago
14

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

2 collected is 22.0 g. What is the empirical formula of the compound?
Chemistry
1 answer:
Akimi4 [234]3 years ago
4 0

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

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

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: 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 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

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