[Squall14]

OK, I'm stuck on how to proceed with this problem that I have:

When M2S3(s) is heated in air, it is converted to MO2(s). A 4.000 g sample of M2S3(s) shows a decrease of 0.218g when heated in air. What is the atomic mass of M.

One hint given was to write the balanced equation which I think is:

2M2S3 (s) + 4O2 (g) -----> 4MO2 (s) + 3S2 (g)  (I'm assuming that sulfur gas is a product as well)

My main thing would be what the decrease in .218g is when heating M2S3, since I think that once I figure that out, I can figure out the problem.  Does it make sense that the decrease means that .218g of S3 has resulted in 3.782g of MO2?

Thanks in Advance!

[AWK]

M2S3 + 5O2 = 2MO2 + 3SO2(g)

[Squall14]

Thanks for the correction in my formula.  Helps me make more sense of the problem.  So based on this, I now assume that the .218g decrease is the SO2 escaping after combustion.  Given this assumption, I get:

.218g SO2 X (1mol/64.07g SO2) = 3.40e-3 mol of SO2

Based on the new formula, I can multiply this by the 5 mol O2:3 mol SO2 ratio to get 5.67e-3 mol O2 which equals .181 g O2.  This now gives me that 4.000g of M2S3 reacts with .181g of O2 to create .218g of SO2 and 3.963g of MO2.

However, I'm not kinda stuck after this part since I'm not sure how to proceed.  I've tried doing this

.218g SO2 X (32.07g S/64.07g SO2) = .109g S

And then using this figure to assume that we have 3.891g M2 since .109g of the compound reacts completely with O2 to form SO2.  However, this ended up giving me a ridiculously large atomic number when I divided it by the mol of O2 based on the .181g O2 I calculated above.

Am I wrong in my assumptions?

[jdurg]

With the information given, this problem cannot be solved.  One must remember that the mass given on the left hand side is the mass of the M2S3 and that on the right hand side is MO2.  You were never given the mass of O2 that reacted, and even though the total difference is 0.218 grams, that doesn't tell you how much SO2 was lost.  (Because the O2 on the left would have combined with the sulfur to produce SO2 gas.  As a result, we do not know the mass of SO2 gas that was produced).

[Squall14]

If only I could tell my professor that the problem can't be solved...  I think that the problem can be solved and that we have to make assumptions in order to determine the amount of O2 gas reacted with the M2S3.  In any case, does anyone else have any ideas that I can use?

[Borek]

jdurg: you are wrong, problem can be solved
Squall14: there is no need for any asumptions.

Decrease of the mass means that M₂S₃ weights 4.00 g, while mass of produced MO₂ is 4-0.218 = 3.782 g.

M₂S₃ + 5O₂ -> 2MO₂ + 3SO₂

Oxygen and sulfur dioxide are not interesting, what is important is this:

M₂S₃  -> 2MO₂

1 mole of M₂S₃ gives 2 moles of MO₂.

In other words 2*M+3*32 g of sulfide gives 2*(M+2*16) g of oxide (M stands now for molar mass of M).

At the same time we know that 4 g of sulfide gives 3.782 g of oxide after roasting.

We can combine all that using proportions into one equation and solve for M.

While I am 100% sure about the method, there is either some problem with my math or with the data in question as there is no element with such high molar mass.

[jdurg]

Hmmm.  You know, I think I just needed to get away from the problem for a bit as I have come up with a 'reasonable' answer.

Set up the following ratio with 'M' being the molar mass:

(2M+96)/4.00 = (2M + 64)/3.782

Rearrange everything by cross-multiplying and you'll get:

7.564M + 363.072 = 256 + 8M

Simplify all that and you'll get 107.072 = 0.436M.

M = 245.58 grams per mole.  That's about the molar mass of an isotope of Curium.  Though I still think that there's something wrong with the question or the data.

[Squall14]

Thanks for the help jdurg and everyone else!  I agree that there's something wrong with the question in general as I've asked 3 different people who do these kind of calculations for their job and all 3 of them were stumped!  (One of them even TA'd for freshman chem and she couldn't figure it out!!)  In any case, the answer that you got was unfortunately incorrect as I submitted my homework online (it's an online homework) and it told me that it was incorrect.  Fortunately however, I get 1 more shot to get the right answer.  Any other ideas?

[mike]

M₂S₃ + O₂ --> 2MO₂

m(M2S3) = 4.00g
M(M2S3) = 2M+3S = 2M+96
n(M2S3) = 4.00/(2M+96)

m(M02) = 4.00g-0.218g = 3.782g
M(MO2) = M+2O = M+32
n(MO2) = 3.782/(M+32)

now: n(M2S3) : n(MO2) = 1 : 2

therefore: 2n(M2S3) = n(MO2)

and:

2x[4.00/(2M+96)] = 3.782/(M+32)

8.00/(2M+96) = 3.782/(M+32)

8M+ 256 = 7.564M + 363.072

0.436M = 107.072

M = 245.6

This has to be right, the computer that is checking your answer must be wrong, I have seen this question before and the mass loss is different so if your teacher has copied it incorrectly but kept the original answer then this could be the problem.

http://mailer.uwf.edu/listserv/wa.exe?A2=ind0110&L=chemed-l&D=1&P=15910
http://69.65.19.208/~apstuden/forum/viewtopic.php?t=1491&sid=c14519d1c22463ac8cb9c375568af8bd

[Borek]

I agree that there's something wrong with the question in general

Sorry, but there is nothing wrong with the question in general. Perhaps it is not an easy one, but it is perfectly doable. There is something wrong with numbers, but the idea behind the solution is clear.

It is very similar to the question like: 10 g of iron filings were dropped into solution of copper sulfate. How much copper was in the solution if the weight of the filings after reaction was 11 g. I suppose the same people that had problems with your question will have problem with this one.

[mike]

Borek is correct there is nothing wrong with the question, but there is something wrong with the answer you are being given.