[John623]

A mixture of salt (NaCl) and Sugar(C₁₂H₂₂O₁₁) contains 5% chlorine by mass. Calculate the % of salt in the mixture

Let X be the % of salt
Let Y be the % of sugar

RAM of Cl = 35.5g
RMM of NaCl =58.5g
RMM of C₁₂H₂₂O₁₁) = 342g

1) 35.5x/(58.5x + 342y) = 0.05
2) x + y = 1
x = 1-y

35.5(1-y)/(58.5(1-y) + 342y) = 0.05
35.5-35.5y/(58.5-58.5y + 342y) = 0.05
35.5-35.5y = 0.05(58.5-58.5y + 342y)
35.5-35.5y = 2.925 - 2.925y + 17.1y
32.575 = 49.675y
y = 0.65572 = 65.572%
which means x = 0.34428 = 34.428%

checking:
35.5x/(58.5x + 342y) = 0.05
35.5*0.34428/(58.5*0.34428 + 342*0.65572) = 0.05
0.05 = 0.05

My teacher gave us this answer:
Moles Cl =5.0/35.5=0.14=moles Na
mass(%) Na = 23x0.14=3.22
total % NaCl =5.0+ 3.22=8.22

I don't understand why my way of doing it is wrong. Is there a way to modify my method to make it work?

[Hunter2]

1) 35.5x/(58.5x + 342y) = 0.05

This equation makes no sense. You multiply molar weight of chlorine with x (mass% of NaCl). This x should be the 5%.

[Borek]

Your approach is perfectly valid, but incomplete.

Let X be the % of salt
Let Y be the % of sugar

In your solution these are not mass percentages, but molar fractions. That's OK, and the equation you used is valid. However, your final answer is also given as a molar fraction, when the question asks for mass fraction - hence the discrepancy.