Question

An organic liquid is a mixture of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH). A 0.220-gsample of the liquid is burned in an excess of O2(g) and yields 0.348 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

Let mass of CH3OH be m1 g

  mass of C2H5OH be m2 g

CH3OH + 1.5 O2 ---> CO2 + 2H2O
1 mol of CH3OH gives 1 mol of CO2
number of moles of CH3OH = mass/molar mass = m1 / 32
number of moles of CO2 = m1/32
mass of CO2 = molar mass of CO2 * number of moles of CO2
                          = 44*m1/32
                           =1.375*m1
C2H5OH + 3O2 ---> 2CO2 + 3H2O
1 mol of C2H5OH gives 2 mol of CO2
number of moles of C2H5OH = mass/molar mass

= m2 / 46
number of moles of CO2 = 2*m1/46 = m2/23
mass of CO2 = molar mass of CO2 * number of moles of CO2
                          = 44*m2/23
                           =1.913*m2

2 equation are:
m1+m2 = 0.22 ------>1
1.375* m1 + 1.913* m2 = 0.348 ----->2

Multiply equation 1 by 1.375,

solving above 2 equation we get,

m1 = 0.135 g and m2 = 0.08 g

So, mass of methyl alcohol = 0.13 g