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?

1 Answer
May 21, 2016

Answer:

#"22.2 g CH"_3"OH"#

#"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, #"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.

#color(blue)(|bar(ul(color(white)(a/a)chi_i = "number of moles of i"/"total number of moles"color(white)(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 #"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 #color(orange)((1))# to find

#x = 0.270 * (x+y)#

#0.730 * x = 0.270 * y implies x = 0.270/0.730y#

Plug this into equation #color(orange)((2))# to find

#32.042 * 0.270/0.730y + 18.015y = 56.0#

#11.85y + 18.015y = 56.0#

#y = 56.0/(11.85 + 18.015) = 1.875#

This will give you

#x = 0.270/0.730 * 1.875 = 0.6935#

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

#0.6935 color(red)(cancel(color(black)("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"color(white)(a/a)|)))#

#1.875 color(red)(cancel(color(black)("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"color(white)(a/a)|)))#

The answers are rounded to three sig figs.