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An organic liquid is a mixture of methyl alcohol (CH3OH m CH_3OH) and ethyl alcohol (C2H5OH m C_2H_5OH ). A 0.220-g m g sample of the liquid is burned in an excess of O2(g) m O_2(g) and yields 0.367g g CO2(g) m CO_2(g) (carbon dioxide). Set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.

What is the mass of methyl alcohol (CH3OH m CH_3OH) in the sample?

Respuesta :

Answer: The mass of methanol in the sample is 0.1002 grams

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

  • For carbon dioxide:

Given mass of carbon dioxide = 0.367 g

Molar mass of carbon dioxide = 44.0 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of carbon dioxide}=\frac{0.367g}{44.0g/mol}=0.00834mol[/tex]

Let the mass of methanol in the sample be 'x' grams.

We are given:

Mass of the mixture of methanol and ethanol = 0.220 g

Mass of methanol = x g

Mass of ethanol = (0.220 - x) g

  • For methanol:

Given mass of methanol = x g

Molar mass of methanol = 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of methanol}=\frac{xg}{32g/mol}=\frac{x}{32}mol[/tex]

  • For ethanol:

Given mass of ethanol = (0.220 - x) g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ethanol}=\frac{(0.220-x)g}{46g/mol}=\frac{(0.220-x)}{46}mol[/tex]

  • The chemical equation for the combustion of methanol follows:

[tex]CH_3OH+\frac{3}[2}O_2\rightarrow CO_2+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of methanol produces 1 mole of carbon dioxide

So, [tex]\frac{x}{32} moles of methanol will produce = [tex]\frac{1}{1}\times \frac{x}{32}=\frac{x}{32}[/tex] moles of carbon dioxide

  • The chemical equation for the combustion of ethanol follows:

[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, [tex]\frac{(0.220-x)}{46}[/tex] moles of ethanol will produce = [tex]\frac{1}{2}\times \frac{(0.220-x)}{46}=\frac{(0.220-x)}{23}[/tex] moles of carbon dioxide

Total moles of carbon dioxide:

[tex]n_{CO_2}=\frac{x}{32}+\frac{(0.220-x)}{23}\\\\0.00834=\frac{x}{32}+\frac{(0.220-x)}{23}\\\\23x+7.04-32x=6.1382\\9x=0.9018\\x=0.1002[/tex]

Hence, the mass of methanol in the sample is 0.1002 grams

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