Question

A solution is made by moving equal masses of methanol CH4O and ethanol C2H6O determine the mole fraction of each component to at least three sig. figs

Answer 1

Since mass is equal for both
Lets say both are 100 g each

Molar mass of CH4O,
MM = 1*MM(C) + 4*MM(H) + 1*MM(O)
= 1*12.01 + 4*1.008 + 1*16.0
= 32.042 g/mol


Molar mass of C2H6O,
MM = 2*MM(C) + 6*MM(H) + 1*MM(O)
= 2*12.01 + 6*1.008 + 1*16.0
= 46.068 g/mol

n(CH4O) = mass/molar mass
= 100.0/32.042
= 3.1209

n(C2H6O) = mass/molar mass
= 100.0/46.068
= 2.1707

n(CH4O),n1 = 3.1209 mol
n(C2H6O),n2 = 2.1707 mol

Total number of mol = n1+n2
= 3.1209 + 2.1707
= 5.2916 mol

Mole fraction of each components are
X(CH4O) = n1/total mol
= 3.1209/5.2916
= 0.590

X(C2H6O) = n2/total mol
= 2.1707/5.2916
= 0.410

Answer:
Mole fraction of CH4O = 0.590
Mole fraction of C2H6O = 0.410