Question #74a4c

1 Answer
Feb 8, 2018

Answer:

#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"#.