Anyone good with chem?

[Tim]

I'm having a bit of trouble on this one problem. It doesn't sound like a combustion problem, because it would normally mix with 02 to make CO2 and H20 but I cant figure out how to write a balanced chemical equation out of it, because it doesnt specify what happens to the sulfer:

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 0.277g when heated in air. What is the atomic mass of M. Report your answer to 1 decimal place.


Any help would be greatly appreciated

p.s. M is an unspecified element

[Landshark]

i don't know, but i do know that:

6pack + 12hotwings = CH4

[Earthworm]

I haven't done this in a while (11 years maybe) so I'm just looking at the numbers.

M2S3=4.0g
M2 = .277g
M = .138
S3 = 3.723g
S = 1.274
Atomic weight of S = 32.065

M is 10.8% the weight of S
Atomic weight of M = 3.473??

Probably not right but there is an S3CL2 compound so that's probably the right answer.

[Tim]

The oxygen is missing in your calculations, isnt it?

[lee]

found out an answer yet?

Here's my thought process (not a chemist BTW). And the lack of subscripts makes this look ugly, anyway....

set X=number of mole (fractional it turns out in this case) of each of the chemical elements - so 2units for M and 3 of S in the original state. Also, the units for M, S and O are atomic weights e.g. S=32.065g/mole. Note, we have two unknowns and two equations: a) unknown #1 is the atomic weight of M, and b) the number of moles used to form the 4g sample.

Mathematically, M*2*X + S*3*X = 4g (where X=the part of a mole present). To verify, look at the units, i.e., S=32.065g/mole, 3 is dimensionless and X=number of moles - so left side of equation reduces to grams - which matches right side.

using same thought process, 2*(M*X +O2*X) = 4-0.277 or m2*X + O4*X=4-0.277, or M2*X + O4*X + 0.277 = 4
Why? Conserving the number of M molecules in the reaction means 4 oxygen molecules form assuming reaction went to completion. Also, we were given the compound got 0.277g lighter during the reaction.

Now, since the number of M atoms remains constant (remember, we are looking for the number of moles used in the oxidation reaction) and last time I checked, 4=4, so we can set one side equal to the other:
M*2*X + S*3*X = M*2*X + O*4*X + 0.277, then subtracting M2*X from both sides you get 3*S*X=4*O*X+.277

then putting in weights for sulpher and oxygen: 3*32.065*X=4*15.994*X+.277 and solving for X -> X=0.0086 moles

Since we now know the number of moles in the original equation, the only unknown is the atomic weight for M, so going back to the original equation

2*M*0.0086 + 3*32.065*0.0086 = 4, so 2*M*0.0086=3.1727, or M=184.5 g/mole (atomic weight)

plugging the number for the atomic weight for M into either equation yields correct results, so a chemist might cringe, but this is how a physicist approached the problem.

[upnygimp]

Originally Posted by Landshark

i don't know, but i do know that:

6pack + 12hotwings = CH4

I'm writing my thesis on this. If I ever have to write a chemistry thesis, which is unlikely.

[Landshark]

Originally Posted by upnygimp

I'm writing my thesis on this. If I ever have to write a chemistry thesis, which is unlikely.

you know what CH4 is, right?

[Noir]

Originally Posted by Landshark

you know what CH4 is, right?

i'm sure it tastes pretty good whatever it is.

[Landshark]

Originally Posted by Noir

i'm sure it tastes pretty good whatever it is.

you probably would like the taste - its methane.

[Manarius]

I think he'd like C4-H8 better myself...

[upnygimp]

Originally Posted by Landshark

you know what CH4 is, right?

Of course I know what CH4 is.

[Landshark]

Originally Posted by upnygimp

Of course I know what CH4 is.

i didn't - i Googled it.


i frickin' hated chemistry in school.

[Tim]

Ahhh this problem is so confusing! Not quite figured out yet but i'll keep you posted. Thanks for your effors so far

[lee]

I'll try to better explain my steps (probably had too much of a "it ca be shown that" flavor), see edited text above.

[oab_au]

Originally Posted by lee

I'll try to better explain my steps (probably had too much of a "it ca be shown that" flavor), see edited text above.

Don't change a thing Lee, that looks lovely. Don't understand a word, but all those numbers and words,,,,,, just fit the page, so well.

Harvey.