Question

An organic liquid is a mixture of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH). A 0.220-g sample of the liquid is burned in an excess of O2(g) and yields 0.380 g CO2(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) in the sample?

Answer 1

the balanced equations will be

CH₃OH +1.5O₂   --> CO₂ + 2H₂O

C₂H₅OH + 3O₂ -> 2CO₂ + 3H₂O

Molecular weight of methanol = 36g / mole

Molecular weight of ethanol = 46g / mole

Mass of CO2 produced = 0.380 grams

Let the mass of CH4 in the mixture = xgrams

As per balanced combustion equation , one mole of methane will give one mole of CO2

This means that 32grams of methanol will give 44grams of CO2

Hence x grams of methanol will give = 44 x/ 32 grams of CO2 = 1.375x grams of CO2

The mass of ethanol = 0.22-x grams

Again by balanced equation, one mole of ethanol will give two moles of CO2

Therefore 46grams of ethanol will give 2 X 44 grams of CO2

0.22-x grams of ethanol will give = 88(0.22-x) / 46 grams of CO2 = 1.91(0.22-x)

total mass of CO2 produced by the two fuels = 1.375x + 1.91(0.22-x) = 0.380 grams

1.375x + 0.4202 - 1.91x = 0.380

0.535x = 0.0402

x = 0.0751 grams = mass of methanol

Mass of ethanol = 0.220 - 0.0751 = 0.1449 grams