Suppose that samples of methanol and water are mixed so that the mol fraction of methanol is #0.260#, and that the total mass is #"67.0 g"#. From this, determine the mass of water and of methanol that were mixed?

Question

Wyatt Atkins · Accepted Answer

#m_(H_2O) = "41.2 g"#
#m_(MeOH) = "25.8 g"# If you can find the total mols, you can then find the mols of each component and thus their mass. You can use the notion that a two-component mixture allows one to write the mol fraction of one in terms of the other. #chi_(MeOH) = 0.260# #" "" "" "" "" "" "bb((1))# #chi_(H_2O) = 1 - chi_(MeOH) = 0.740# #" "bb((2))# You also know that their masses sum to #"67.0 g"#: #m_(MeOH) + m_(H_2O) = 67.0# or #n_(MeOH)M_(MeOH) + n_(H_2O)M_(H_2O) = 67.0# #" "bb((3))# where #n_i# is the mols and #M_i# is the molar mass in #"g/mol"# of component #i#. If we divide #(3)# by the total mols, we obtain: #chi_(MeOH)M_(MeOH) + chi_(H_2O)M_(H_2O) = 67.0/n_(t ot)# #" "bb((4))# All of these quantities are known except the total mols, and so, we can determine the total mols from solving #(4)# for #n_(t ot)#. #color(green)(n_(t ot)) = 67.0/(chi_(MeOH)M_(MeOH) + (1 - chi_(MeOH))M_(H_2O))# #= "67.0 g"/(0.260cdot "32.04 g/mol" + (1 - 0.260)"18.015 g/mol")# #=# #color(green)ul"3.093 mols total"# Now that we know the total mols, we can solve for the mols of one component and obtain the other by subtraction. The mols of methanol and water are found from #(1)# and #(2)#: #n_(MeOH)/"3.093 mols" = 0.260# #=> color(green)(n_(MeOH) = ul"0.804 mols MeOH")# #=> color(green)(n_(H_2O) = ul("2.289 mols H"_2"O"))# Therefore, from their molar masses, we can then find their masses using #(3)#. #color(blue)(m_(MeOH)) = 0.804 cancel"mols MeOH" xx "32.04 g"/cancel"1 mol" = color(blue)ul("25.8 g MeOH")# #color(blue)(m_(H_2O)) = 2.289 cancel("mols H"_2"O") xx "18.015 g"/cancel"1 mol" = color(blue)ul("41.2 g H"_2"O")# or we could have said #m_(H_2O) = "67.0 g" - m_(MeOH)#. And in fact, we do still satisfy our sets of equations: #"25.8 g MeOH" + "41.2 g H"_2"O" = "67.0 g total"# #color(blue)(sqrt"")# #"0.804 mols MeOH"/"3.093 mols" = 0.260# #color(blue)(sqrt"")# #"2.289 mols MeOH"/"3.093 mols" = 0.740# #color(blue)(sqrt"")# #0.260 + 0.740 = 1.000# #color(blue)(sqrt"")#