# Question 74a4c

Feb 8, 2018

41.9%

#### Explanation:

Your starting point here will be to determine how many grams of zinc are present in $\text{5.00 g}$ of zinc sulfide, $\text{ZnS}$.

You know that zinc sulfide has a molar mass of ${\text{97.474 g mol}}^{- 1}$, which basically means that $1$ mole of zinc sulfide has a mass of $\text{97.474 g}$.

Elemental zinc, $\text{Zn}$, has a molar mass of ${\text{65.38 g mol}}^{- 1}$, which means that $1$ mole of zinc has a mass of $\text{65.38 g}$.

Now, the chemical formula of zinc sulfide tells you that $1$ mole of zinc sulfide contains $1$ mole of zinc, so you can use the molar masses to say that $\text{97.474 g}$ of zinc sulfide, the equivalent of $1$ mole of zinc sulfide, contains $\text{65.38 g}$ of zinc, the equivalent of $1$ mole of zinc.

This means that $\text{5.00 g}$ of zinc sulfide will contain

5.00 color(red)(cancel(color(black)("g ZnS"))) * "65.38 g Zn"/(97.474 color(red)(cancel(color(black)("g ZnS")))) = "3.354 g Zn"

So, you know that your crude sample contains $\text{3.354 g}$ of zinc in a total mass of $\text{8.00 g}$. To find the percent composition of zinc in the crude sample, simply divide the mass of zinc by the total mass of the sample and multiply the result by 100%.

(3.354 color(red)(cancel(color(black)("g"))))/(8.00color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(41.9%)))

The answer is rounded to three sig figs.

Notice that the percent composition of zinc in zinc sulfide

(3.354 color(red)(cancel(color(black)("g"))))/(5.00color(red)(cancel(color(black)("g")))) * 100% = 67.1%#

if higher than the percent composition of zinc in the crude sample because you're dealing with the same mass of zinc in a smaller mass, i.e. in $\text{5.00 g}$ as opposed to $\text{8.00 g}$.