A pure compound has one Percent Composition. A mixture does not: a spoonful of a salt mixture can be mostly one salt or mostly the other. But each compound inside still has its own fixed composition, and that is enough to work out how much of each is there, using a system of two equations.
Key idea
Write one unknown for each component's mass. Each fact you measured gives one equation: the total mass, and the mass of one element, which is the sum of (mass of each component) × (that element's mass fraction in it). Two unknowns need two equations.
Why it works
Mass fractions are fixed inside each pure compound, so the chlorine in $x$ grams of NaCl is $0.6066x$ no matter what else is mixed in. The mixture's measured chlorine is the sum of the chlorine from each part. A mixture's percent of an element therefore always lies between the percents of its pure components, nearer to the one there is more of, like a weighted average.
Worked example: One component holds the element
A $5.00\ \text{g}$ sample of impure limestone contains $1.60\ \text{g}$ of calcium. The impurities contain no calcium. What percent of the sample is $\mathrm{CaCO_3}$ ($M = 100.09\ \text{g/mol}$, Ca $40.08$)?
- Calcium's mass fraction in $\mathrm{CaCO_3}$: $40.08/100.09 = 0.4004$.
- All the calcium is in the $\mathrm{CaCO_3}$, so its mass is $1.60/0.4004 = 3.996\ \text{g}$.
- Percent: $3.996/5.00 \times 100 = 79.9\%$.
Answer: $79.9\%$ $\mathrm{CaCO_3}$.
Worked example: Both components hold the element
A $10.00\ \text{g}$ mixture of NaCl and KCl contains $5.50\ \text{g}$ of chlorine. How many grams of NaCl does it hold? (Mass fractions of Cl: NaCl $0.6066$, KCl $0.4755$.)
- Let $x$ = grams of NaCl and $y$ = grams of KCl. Total mass: $x + y = 10.00$.
- Chlorine: $0.6066x + 0.4755y = 5.50$.
- Substitute $y = 10.00 - x$: $0.6066x + 4.755 - 0.4755x = 5.50$, so $0.1311x = 0.745$.
- $x = 5.68\ \text{g}$ NaCl (and $y = 4.32\ \text{g}$ KCl). Check: $$0.6066(5.68) + 0.4755(4.32) = 3.45 + 2.05 = 5.50\ \text{g}$$
Answer: $5.68\ \text{g}$ of NaCl.
Common mistake
Treating the mixture as one compound, for instance dividing $5.50\ \text{g}$ of Cl by NaCl's fraction (0.6066) to get "9.07 g of NaCl", or using either percent for the whole sample. That ignores the other component's chlorine. Check: the mixture's percent (55.0%) must lie between the pure percents (47.55% and 60.66%), and your two masses must add to the total.
The same idea finds the purity of an ore, the makeup of an alloy from its density, or the share of each salt in a fertilizer: one equation for the total and one for each measured element.