Anyone good with chem?
 

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

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

6pack + 12hotwings = CH4

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.

The oxygen is missing in your calculations, isnt it?

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.

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? :D

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

you probably would like the taste - its methane.

Of course I know what CH4 is.

i didn't - i Googled it. :p :D


i frickin' hated chemistry in school.

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

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. :confused: :o :eek:

Harvey. ;)