[deepbreathofsolvents]

Hi

Al2 (SO4)3
If Al = 1.4 mg/L
What is the Concentration of the Al2 (SO4)3
Find concentration of Al2 (SO4) 3

Moles (Al) = 1.4/1000/53.96=0.0000259
0.0000259 x 342.14 = 0.008877 g/L
Is this correct?  Is there anyway the answer could be 0.0044 g/L? Which I have been told is the correct answer.
I know that there are two moles of Al for every mole of Al2SO4 3 but I thought that is already accounted for in the calculation.
Thanks for any help/advice.

[UG]

By the way you have worded the question, your answer looks correct.

[fledarmus]

Hi

Al2 (SO4)3
If Al = 1.4 mg/L
What is the Concentration of the Al2 (SO4)3
Find concentration of Al2 (SO4) 3

Moles (Al) = 1.4 (mg Al/L)/1000 (mg/g)/53.96 g Al/mole Al)=0.0000259 (units?)
0.0000259 (units?) x 342.14 (units?) = 0.008877 g (of what?)/L
Is this correct?  Is there anyway the answer could be 0.0044 g/L? Which I have been told is the correct answer.
I know that there are two moles of Al for every mole of Al2SO4 3 but I thought that is already accounted for in the calculation. (Where?)
Thanks for any help/advice.

Be very careful (possibly even obsessive) about your units - I've done some markup of your equation, see if you can fill in the question marks...

[deepbreathofsolvents]

Moles (Al) = 1.4 (mg Al/L)/1000 (mg/g)/53.96g (Al/mole Al)=0.0000259 moles Al/L
0.0000259 (moles/Al ) x 342.14g Al2(SO4)3/L = 0.008877g Al2(SO4)3/L
Is this correct?  Is there anyway the answer could be 0.0044 g/L? Which I have been told is the correct answer.
I know that there are two moles of Al for every mole of Al2SO4 3 but I thought that is already accounted for in the calculation. (Where? 0.0014g Al/L / 53.96g Al/mole Al)
The answer seems to be right as the original Al weight is 15.7% my calculated Al2SO4 3 weight, I think the original question may have been trying to get at something else with the number of Al3+ ions in the dissociation reaction but I'm not sure.
Anyway thankyou.

[fledarmus]

Moles (Al) = 1.4 (mg Al/L)/1000 (mg/g)/53.96g (Al/mole Al)=0.0000259 moles Al/L
0.0000259 (moles Al/L ) x 342.14g Al2(SO4)3/Lmole Al2(SO4)3 = 0.008877 g Al2(SO4)3/L

This is what I mean about being obsessive about your units. You have moles of Al magically going to grams of Al2(SO4)3 with no way of cancelling it, and your liters are drifting.

Being obsessive about units isn't just a way to get full credit on the problem, it is a way to satisfy yourself that the arithmetic is really doing what you think it is. In your case, right now, it isn't.

[deepbreathofsolvents]

Thanks UG I’ve done it a couple of ways now and I got the same answer, thanks for going to all the trouble to reply Fled.   I didn’t realise you wanted me to identify all the units to do the factor label method.  Where I’m from in Australia that method is not really taught at the High School or technician level, though it probably is at university level I’m not sure. 
Actually I’m not a high school student I just wanted someone to verify an answer from a peculiar situation that came up at work so I may have put this in the wrong section.
How about this:
Al = 1.4mg/L
Al = 0.0014g/L
n moles (Al):   0.0014g/L  / 26.98g/Mole  = 0.00005189 M/L
2 moles Al : 1 mole Al2(SO4)3
No moles Al2(SO4)3 = 0.00005189/2
              = 0.000025945 M/L
Conc Al2 (SO4)3 = 0.000025945 x 342.14g
         = 0.00877g/L
I think my issue is similar to this problem: http://au.answers.yahoo.com/question/index?qid=20110327231539AAJJ5RE
I’m getting an answer of 1.026g/50ml for that one so I think I'm now on the right track.
Just to recap the question
Al = 1.4 mg/L
What is the conc of Al2(SO4)3
If anyone wants to give it a go to verify.

[Borek]

Moles (Al) = 1.4/1000/53.96=0.0000259
0.0000259 x 342.14 = 0.008877 g/L

That was correct from the very beginning.