[claudia]

Hello...i was just wondering if anyone could help me out with this problem?

(a)A sample of metal oxide (0.4150g) with formula M2O3is quantitatively converted to a metal sulfide MS yielding 0.457g. What is the atomic weight of metal M?

(b) A thoroughly dry sample of Na2SO4 (1.0000g)nis exposed to the atmosphere for a period of time and is observed to have gained 0.4250g. What is the percent by mass of Na2SO4 10H2O in the resulting mixture of hydrate and anhydrous salt?
(f. wt. Na2SO4=141.92 and f.wt.Na2SO4 10H2O=322.07)

-->for (a) i came up with:
atomic weight of O=16
formula weight of metal oxide tells us: moles M=moles S
moles S= (0.4570-0.4150)/32.065=0.001309839
0.001309839=molesM=0.4150/at weight of M
so at weight of M= 0.4150/0.001309839=316.8

i am really not sure of what i did, and for question (b)i am extremely confused....
i would really appreciate it if someone could help me out 
thank you very much in advance...

[Borek]

For a - you have to set up system of two equations. One will tell you mass of oxide for n moles of M, second mass of sulfide for n moles of M. You will have two unknowns (n and molar mass). Solve.

For b - what is the mass of Na₂SO₄^.10H₂O produced form 0.4250 g of H₂O?