Question

Question #74a4c

Answer 1

`41.9%`

Explanation:

Your starting point here will be to determine how many grams of zinc are present in `"5.00 g"` of zinc sulfide, `"ZnS"`.

You know that zinc sulfide has a molar mass of `"97.474 g mol"^(-1)`, which basically means that `1` mole of zinc sulfide has a mass of `"97.474 g"`.

Elemental zinc, `"Zn"`, has a molar mass of `"65.38 g mol"^(-1)`, which means that `1` mole of zinc has a mass of `"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 `"97.474 g"` of zinc sulfide, the equivalent of `1` mole of zinc sulfide, contains `"65.38 g"` of zinc, the equivalent of `1` mole of zinc.

This means that `"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 `"3.354 g"` of zinc in a total mass of `"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 `"5.00 g"` as opposed to `"8.00 g"`.