[mattidore]

A metal sulphate has the formula M₂SO₄. 10.99g of the compound was dissolved in water to make 500cm³ of solution. A 25.00cm³ sample was removed and reacted with an excess of BaCl₂ to produce a precipitate of BaSO₄, which, when dried had a mass of 1.167g.

a) Determine the number of moles of BaSO₄
b) Determine the concentration of the M₂SO₄
c) Determine the number of moles of M₂SO₄ solution
d) Determine the molar mass of M₂SO₄
e) Determine the identity of M.

I was able to do (a) using the n=m/M_r and got 0.005mols, but in the answers it said 5mol, I understand it's probably because of the conversion from cm³ to dm³ but why must you take this into account when finding the number of moles?

MAIN QUESTION: I was stumped at (b), but if I knew b I would be able to do solve the rest of the question because the rest are pretty simple. How do I solve (b)?

PS: Also, after trying to find the answer to (b) I gave up and tried a new method
χ₂SO₄ = 10.99g

n(SO₄) = 10.99/96.06 = 0.1144mol

m(S) = 3.668 [m=n x M_r  0.1144 x 32.06]
m(O₂) = 1.830 (same as above)

10.99 - (3.668 + 1.830) = 5.498g [of just x in 10.99g of χ₂SO₄]

then 1 - 0.1144 = 0.8856mol [of just x in 10.99g of χ₂SO₄]

therefore M_r = m/n = 5.498/0.8856 = 6.208gmol^-1
which is ≈ Li.

the answer in the answer sheet was Lithium too. Just wondering, if this was pure coincidence or what not?

[Hunter2]

0.005 mol is correct, because n =m/M.  You have a mass of BaSO₄ divided by the molare Mass ofm it.  It doesnt matter what was the volume before.

Lithium is correct. YOu can calculate more simple.

The 0.005 mol in 25 cm³ calculate to 500 cm³

This value is equal og the unknown substance. Again by using M = m/n you get the molare mass of the substance.

[Borek]

MAIN QUESTION: I was stumped at (b), but if I knew b I would be able to do solve the rest of the question because the rest are pretty simple. How do I solve (b)?

From the precipitation stoichiometry you know there were 5 millimoles of SO₄^2- in the 25 mL - that means there were also 5 millimoles of Li₂SO₄ (dissociated) in these 25 mL. Can you calculate the concentration now?

[Schrödinger]

n(SO₄) = 10.99/96.06 = 0.1144mol

This, and hence the remainder of the method is incorrect. You cannot segregate the moieties of a certain compound and find moles individually according to their atomic/molecular masses.

The actual atomic mass of Li is 6.941 u, but you got 6.208. It's just coincidence, that in this problem the actual answer and your answer have the same GIF value.

Actual method - refer to Hunter2's post