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 0.355 g CO2(g) (carbon dioxide).

Part A

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?

Express your answer numerically in grams.

mass of CH3OH =

  g  

Answer 1

CH3OH + O2 = H2O + CO2

C2H5OH + O2 = H2O + CO2

balance

CH3OH + 3/2O2 = 2H2O + CO2

C2H5OH + 3O2 = 3H2O + 2CO2

mass of sample = 0.220

let x = methanol

0.22 - x = ethanol

mass of CO2 = 0.355 g of CO2

mol of CO2 = 0.355 /44 = 0.00806 mol of CO2

mol of methanol = mass/MW = x/32.04

mol of ethanol = mass/MW = (0.22 - x)/46

for mol of CO2:

mol of methanol + 2 mol of ethanol = mol of CO2

mol of methanol + 2 mol of ethanol = 0.00806

substitute

x/32.04 + (0.22 - x)/46 = 0.00806

46*32.04x + (0.22-x)*32.04 = 0.00806 *46*32.04

1473.84x + 7.0488 - 32.04x = 11.8791504

x (1473.84-32.04) = 11.8791504-7.0488

x = 4.830/1441.8 = 0.003349

mass of methanol = 0.003349 g