Can you find the value of "n"?

Question

5.00g of hydrated sodium sulphate (Na2SO4 . nH2O) gave 2.20g of anhydrous sodium sulphate on heating to constant mass. Work out the relative molecular mass (Mr) of the hydrated sodium sulphate and find the value of "n". PLEASE CAN YOU WRITE OUT THE STEPS I HAVE TO FOLLOW IN ORDER TO GET THE VALUE OF "n". THANK YOU.

Answer 1

Mass of Na2SO4 produced = 2.20g
Mr of Na2SO4 = 2(23) + 32.1 + 4(16) = 142.1
No. of mole of Na2SO4 produced = (2.20)/(142.1) = 0.01548 mol

Mr of Na2SO4 = 142.1 + n(16+1+1) = 142.1 + 18n
No. of mole of Na2SO4.nH20 = (5.00)/(142.1 + 18n)

Na2SO4.nH2O ---> Na2SO4 + nH2O
From the equation, we know that 1 mole of Na2SO4.nH2O produces 1 mole of Na2SO4.
Therefore, 0.01548 mole of Na2SO4 is produced by 0.01548 mole of Na2SO4.nH2O.

Note: Since the hydrated salt is heated to constant mass, all the hydrated salt is converted into anhydrous salt + water. Thus, no. of mole of Na2SO4 is 0.01548

0.01548 = (5.00) / [142.1 + 18n]

From here, calculate the value of n and you will get answer 10.049 which is ~10. Write your answer as 10.0

Answer: n = 10

Answer 2

Na2SO4.nH2O ---> Na2SO4 + nH2O

Mr for Na2SO4 = (23x2)+32+(16x4) = 142
No. of moles of Na2SO4 = 2.20/142

No. of moles of Na2SO4.nH2O = 2.20/142
[They are the same as the mole ratio in the equation is 1:1.]
Mr for Na2SO4.nH2O = 142+(1+1+16)n = 142+18n

No. of moles = mass/Mr
2.20/142 = 5.00/(142+18n)
2.20(142+18n) = 5.00(142)
312.4 + 39.6n = 710
39.6n = 397.6
n = 10

Answer 3

When the hydrate of sodium sulfate was heated, 2.80 grams of water were removed leaving 2.20 grams of the anhydrous form.

2.80 x 1/18 = moles of water = 0.156 mol H2O
from
2.20 x 1/142 = moles of anhydrous = 0.0155 moles of hydrate

moles of water/ moles of hydrate = 0.156/0.0155 = 10 in whole numbers. n = 10