A rock sample contains only Sb2S3 and FeS. If 835g of the sample contains 11.2% antimony, what is the mass percent of the Sb2S3 in the sample? What is the mass percent of the sulfur in the sample?

Question

So far, I have determined that 11.2% of an 835g sample is 93.52g Sb, the molar mass of FeS is 87.92 g/mol and the molar mass of Sb2S3 is 339.73 g/mol. I have no idea where to go from here.

Answer 1

pretend you have 100 g of the sample
Then, you have 11.2 grams Sb. That is

moles Sb=11.2/molemassSb.

But you know the compound is Sb2S3, which means there are 1/2 * moles Sb of Sb2S3 (reason, if you have ten moles SB, you can make 5 moles Sb2S3).

Now, you know the moles Sb2S3. That means you have a mass of Sb2S3 of molesSb2S3*molmassSb2S3. So the percent of antomony sulide is then that mass divided by 100grams.

The remaining mass is Ferrous Sulfide, and you can determine its percent by dividing by 100