chemistry homework

Question

In a 74.0-g  aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.160. 

What is the mass of each component?

Mass of CH4O (g) ?

 

Mass of water (g) ?

Answer 1

Good evening.

We need to be sure we understand what is in our system (its components). Here we have methanol, CH₄O, in an aqueous solution. That means the solvent is water, H₂O.

We now need to calculate the molar mass of both our solute and our solvent.

The molar mass of methanol is 32.04 g/mol. The molar mass of water is 18.01528 g/mol.

Mole fractions have to add up to one. So do the mass fractions.

A mole fraction is defined as the number of moles of a component of the solution divided by the total number of moles in the solution.

A mass fraction is defined as the mass of a component of the solution divided by the total mass of the solution.

The problem tells us that the mole fraction of methanol is 0.160. That means the mole fraction of water must be 0.840 (1 - 0.160). We're going to assume that we have a 100-mole solution, which will make our mass fraction calculations easier. In a 100-mole solution, we'd have 16.0 moles of methanol and 84.0 moles of water.

16.0 mol CH₄O * 32.04 g/mol = 512.64 g CH₄O

84.0 mol H₂O * 18.01528 g/mol = 1513.28352 g H₂O

We're going to adjust those mass calculations for significant figures, so we end up with 513 g CH₄O and 1513 g H₂O.

Our total mass = 1513 + 513 = 2,026 g.

We can now compute our mass fractions:

CH₄O: 513 / 2,026 = 0.2532

H2O: 1,513 / 2,026 = 0.7468

Now we can calculate the actual mass of both components.

Mass of CH₄O = (Mass Percent CH₄O) (overall mass of actual solution) = .2532 x 74.0 g = 18.7368 g

Mass of H₂O = (Mass Percent H₂O) (overall mass of actual solution) = 0.7468 x 74.0 g = 55.2632 g.

We're going to round those off to take account of significant figures. They end up as 18.7 g CH₄O and 55.3 g H₂O.

I hope this helps.