heinemann student workbook

[nooshnoosh95]

if anyone has the solutions could you please upload them
or just any help with this (pg47) would be appreciated
-barium sulfate precipitate has a mass of 0.413g

1. find the mass of sulfate ions in the barium sulfate precipitate...
2. determine the percentage by mass of sulfate in the fertiliser (mass of fertiliser is 1g)
3. determine the percentage by mass of sulfur (as sulfate) in the fertiliser

thanks in advance

[b^3]

Solutions for the workbook are on the Heinemann page http://www.hi.com.au/chemistry/chem1.asp
Remember to not rely on the solutions, just use them as the last resort when you get really stuck so that you can look though and understand how to do the question.

As for your question
1) Barium Sulfate is the compond BaSO₄, so that means that the sulfate ions will become the precipitate in a 1:1 ratio, so the mol of the precipitate will be equal to the mol of the sulfate ions. n=m/M
n(SO₄^-)=n(BaSO₄)

2) % by mass of sulfate ions=(mass of sulfate ions)/(mass of precipitate)*100 (m=nM) - refer to mihir94's post

3) I remember getting tripped up on this at one stage, and since I'm a bit rusty with chem, not sure which way to go. Someone else may be able to help you out here (or refer to the workbook solutions)

[nooshnoosh95]

ohh okay
thanks for the help
answers for this prac arent in the solutions so if anyone else knows how to do the last one??
answers i got so far
1. 0.17g
2. 41.16%
3. ?

[mihir94]

I thought to work out (b) you would have to divide the mass of slufate ions by 1 as it says percentage by mass of sulfate in the fertiliser. Therefore givng you 17%

To work out (c) you use the same approach:
n(SO4-) = n(S2-) =0.00177mol
therefore m(S2-)= n x M = 0.00177 x 32.1 =0.0568g
therefore  percentage by mass of sulfur (as sulfate) in the fertiliser= m(S2-)/m(fertiliser) x100 = .0568/1 x 100 = 5.68%

sorry S2- is the sulfate ion. Don't know how to use the subscript thingy.

[nooshnoosh95]

yea! the answer was 17% but i thought maybe it was wrong
lols
thanks heaps

[mihir94]

No problem 

[b^3]

I thought to work out (b) you would have to divide the mass of slufate ions by 1 as it says percentage by mass of sulfate in the fertiliser. Therefore givng you 17%

To work out (c) you use the same approach:
n(SO4-) = n(S2-) =0.00177mol
therefore m(S2-)= n x M = 0.00177 x 32.1 =0.0568g
therefore  percentage by mass of sulfur (as sulfate) in the fertiliser= m(S2-)/m(fertiliser) x100 = .0568/1 x 100 = 5.68%

sorry S2- is the sulfate ion. Don't know how to use the subscript thingy.

Ah yeh my bad, haven't done this kind of chem in a while, sorry guys.