[psychkub]

First, here's the question:

A series of sulfate samples is to be analyzed by precipitation as BaSO₄. If it is known that the sulfate content in these samples ranges between 20.0% and 55.0%, what minimum sample mass should be taken to ensure that a precipitate mass no smaller than .300 grams is produced?

At first I thought to find the answer all you had to do was solve
.300g SO₄/mass of sample containing SO₄ = .200
for "mass of sample containing SO₄" to find the answer, but that seemed way too simple. I tried to reason the problem a different way but I couldn't.

Can anyone give me any hints as to how I should approach the problem? Any help or guidance will be greatly appreciated.

[macman104]

You're very close.  It's not .300 grams of Sulfate precipitate, it's BaSO4.  If you need .300 grams of BaSO4 precipitate, can you solve for how much of that mass is going to be from the SO4?

[psychkub]

I think I get it here's what I did:

.300g BaSO₄ * 1mol BaSO₄/233.392g * 1mol SO₄/ 1mol BaSO₄ * 96.065g/ 1mol SO₄ = 0.1235g SO₄

.1235g SO₄/mass of sample = .200

solved for mass of sample and calculated 0.617g

I think that's right. Thank you so much for your help macman104!

[macman104]

Good!  That is the correct way to solve this problem.  Now try checking your other extreme bound of the sulfate percent, make sure you've chosen the correct extrema...

[psychkub]

Ugh! You're right, using the 55% alternative I get 0.225g. Thank you for catching that.

 

[Borek]

You were not wrong - you should calculate for both extreme comopositions. Now reread the question and select correct value of the two