Can someone please tell me if there is an easier way to answer this question? and if I did it correctly please.

Question

A 2.32g sample of Na2SO4*nH2 yields 1.42g Na2SO4 upon heating. What is the value of n?

Na2SO4*nH20--->Na2SO4(s)+nH20(g)

Molar Mass Na2SO4 =142 g/mol

Molar Mass H2O = 18.016 g/mol

I'm not sure if I did it correctly I kind of guessed but got the right answer in the end?- steps I did:

water loss(nH2O)= (2.32g Na2So4*nH2O) - (1.42g Na2SO4) = 0.9g nH2O

(1.42g Na2SO4)*(1mol/142 g)=0.01mol Na2SO4

(0.9g H2O)*(1mol/18.016 g)H20 = 0.050 mol nH2O

(0.050 mol nH2O)*(1 mol Na2SO4 / n H2O) = 0.01mol Na2SO4

(nH2O)(0.050 mol nH2O)*(1 mol Na2SO4) = (0.01mol Na2SO4)*(nH2O)

(nH2O)*(0.050 mol nH2O) / (0.01)= (nH2O)

[(nH2O)*4.995 mol H20] / (nH2O) = (nH2O)/(nH2O)

=5 mol H2O

Answer 1

Yes, you did it correctly. Nice job. The only comment I would make is that you could shorten the calculations as follows:

Once you have moles of H₂O = 0.05 and moles Na₂SO₄ = 0.01, you can simply divided both by the lowest value (0.01) to get whole numbers. This makes 1 mole Na₂SO₄ and 5 moles H₂O and that is the correct ratio.