Question

Please help!!!
How many grams of each element are present in the 8.7 g sample of iron II sulfide, FeS:
Mass of Fe in the sample: ?? grams
Mass of S in the sample: ?? grams

Answer 1

you could calculate the molar mass of each sample first

Answer 2

Fe- 55.85
S-32
FeS-87.85
@greatlife44

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Dimpalk104
Fe- 55.85
S-32
FeS-87.85
@greatlife44
\(\color{#0cbb34}{\text{End of Quote}}\)

Yeah, I would calculate the % of each first like this:

we take the mass of sulfur and put it over the total mass of the compound to get the %

FeS IS 36% Sulfur so we subtract that by 100 and get 64% Fe by mass

\[%Sulfur \frac{ 32 }{ 87.85 } = 36% sulfur S ; 64% Fe \]

then we take the % and multiply it by the total mass of the sample

0.64*8.75g = 5.6g Iron

0.36*8.75g =3.15g Sulfur

Answer 4

Hint:
the mass of \(\text{FeS}\) is: \(55.85+32=87.85\)
so we have:
\(8.7/87.85=0.099\) mol of \(\text{FeS}\), and \(0.099\) mol of \(\text{FeS}\) contains 0.099 mol of \(\text{Fe}\), and 0.099 mol of \(\text{S}\)
so we have:
\(0.099⋅55.85=...?\) grams of \(\text{Fe}\), and
\(0.099⋅32=...?\) grams of \(\text{S}\)