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Likurg_2 [28]
3 years ago
6

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B

HT was combusted in an oxygen rich environment to produce 5.995 g of CO 2 ( g ) and 1.963 g of H 2 O ( g ) . Insert subscripts to complete the empirical formula of BHT.
Chemistry
1 answer:
Alborosie3 years ago
3 0

<u>Answer:</u> The empirical formula for the given compound is C_{15}H_{24}O_1

<u>Explanation:</u>

The chemical equation for the combustion of compound having carbon, hydrogen, iron 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=5.995g

Mass of H_2O=1.963g

Mass of sample = 2.001 g

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

So, in 5.995 g of carbon dioxide, \frac{12}{44}\times 5.995=1.635g of carbon will be contained.

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

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

So, in 1.963 g of water, \frac{2}{18}\times 1.963=0.218g of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 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{1.635g}{12g/mole}=0.136moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles

  • <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 0.0092 moles.

For Carbon = \frac{0.136}{0.0092}=14.78\approx 15

For Hydrogen = \frac{0.218}{0.0092}=23.69\approx 24

For Oxygen = \frac{0.0092}{0.0092}=1

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

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is C_{15}H_{24}O_1

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