# In a 56.0 g aqueous solution of methanol, CH_4O, the mole fraction of methanol is .270. What is the mass of each component?

May 21, 2016

$\text{22.2 g CH"_3"OH}$

$\text{33.8 g H"_2"O}$

#### Explanation:

The idea here is that you need to use the definition of mole fraction to find a relationship between the number of moles of methanol, $\text{CH"_3"OH}$, and the number of moles of water.

As you know, mole fraction is defined as the ratio between the number of moles of a component $i$ of a solution and the total number of moles present in that solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\chi}_{i} = \text{number of moles of i"/"total number of moles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, you solution contains methanol, your solute, and water, your solvent. If you take $x$ to be the number of moles of methanol and $y$ to be the number of moles of water, the mole fraction of methanol will be

chi_(CH_3OH) = (xcolor(white)(a)color(red)(cancel(color(black)("moles"))))/((x + y)color(red)(cancel(color(black)("moles")))) = x/(x + y) = 0.270" " " "color(orange)((1))

You know that your solution has a total mass of $\text{56.0 g}$. You can use the number of moles of methanol and the compound's molar mass to determine the mass present in solution

x color(red)(cancel(color(black)("moles CH"_3"OH"))) * "32.042g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = (32.042 * x)color(white)(a)"g"

Do the same for water

y color(white)(a)color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(white)(a)color(red)(cancel(color(black)("mole H"_2"O")))) = (18.015 * y)color(white)(a)"g"

You thus have

32.042 * xcolor(white)(a)color(red)(cancel(color(black)("g"))) + 18.015 * ycolor(white)(a)color(red)(cancel(color(black)("g"))) = 56.0color(red)(cancel(color(black)("g")))" " " "color(orange)((2))

You now have two equations with two unknowns. Use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to find

$x = 0.270 \cdot \left(x + y\right)$

$0.730 \cdot x = 0.270 \cdot y \implies x = \frac{0.270}{0.730} y$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find

$32.042 \cdot \frac{0.270}{0.730} y + 18.015 y = 56.0$

$11.85 y + 18.015 y = 56.0$

$y = \frac{56.0}{11.85 + 18.015} = 1.875$

This will give you

$x = \frac{0.270}{0.730} \cdot 1.875 = 0.6935$

Use the molar masses of the two species to convert the number of moles to grams

$0.6935 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CH"_3"OH"))) * "32.042 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = color(green)(|bar(ul(color(white)(a/a)"22.2 g CH"_3"OH} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$1.875 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(green)(|bar(ul(color(white)(a/a)"33.8 g H"_2"O} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answers are rounded to three sig figs.