[Addison T]

A 7.96mg sample of a compound containing the elements C, H, and S is burned in oxygen and found to form 16.65mg of CO2. The sulfur in 4.31mg of the compound is converted into sulfate by a series of reactions, and precipitated as Ba(SO4). The Ba(SO4) was found to have a mass of 11.96mg. The molecular weight of the compound was found to be 168g/mol. Using this data, what is the molecular formula of the C, H, S compound?

Okay, so that is the problem and I know the first thing I need is to find the mass of each C, H, and S in the compound in order to get my empirical formula. So i got the mass of S as 4.31mg because it's stated in the problem, but I cannot seem to figure out how to get the masses of C and H. if i got those I know how to solve from there.

Any help would be appreciated, thanks.

[fledarmus]

Actually, that 4.31 mg is not sulfur, it is your C-H-S-containing compound.

What does the burning of 7.96 mg to form 16.65 mg of CO₂ tell you about your original compound?

How about the burning of 4.31 mg to form 11.96 gm of Ba(SO)4?

Can you write chemical equations for these reactions? Perhaps using variables for any values you don't know?

[Addison T]

Okay, I'm not sure if this is what you meant but I think I figured it out. So i get:

7.96mg sample of C-H-S compound forming 16.65mg CO2.
I find the moles of CO2 which gives me my moles for C so then I can find it's weight giving me a mass of 4.544mg for C.

4.31mg sample of C-H-S compound forming 11.96mg Ba(SO4).
I find the moles of Ba(SO4) which gives me the moles for S in 4.31mg sample.
I take the moles of S in the 4.31mg sample times 7.96 then divide by 4.31 to get the moles of S in the 7.96mg sample so I end up with a mass of 3.034mg for S.

I subtract the masses for C and S from 7.96 to get my mass for H. Convert to moles then find the ratio to get my empirical formula. From that I find the molecular formula so I end up with a molecular formula of C8H8S2.

Hopefully that makes sense. Though I'm not sure if I followed what you meant, you helped me figure it out, so thank you.

[Borek]

C8H8S2

Looks OK.