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 0.334 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

Molar mass of CH3OH =32.0419 g / mole

Molar mass of C2H5OH = 46.0684 g/ mole

Molar mass of CO2 = 44.00964 g/ mole

Now calculate the amount of CO2 from CH3OHand C2H5OH as follows:

CH3OH + 2 O2 --> CO2 + 2 H2O
0.220 g / 32.043 g/mole = 0.006866 mole CH3OH
0.006866 mole CH3OH = 0.006866 mole CO2

C2H5OH + 2.5 O2 -->2CO2 + 3 H2O
0.220 g / 46.502 g/mole = 0.004731 mole of C2H5OH
0.004731 mole C2H5OH = 0.00946 mole CO2

0.334 g CO2 / 44.011 g/mole = 0.00759 mole CO2

let X = fraction of mass that was CH3OH
let (1-X) = fraction of mass that was C2H5OH
mole of CO2 from CH3OH + moles of CO2 from C2H5OH = total moles of CO2
X x 0.006866 + (1 - X) x 0.00946 = 0.00759