Atomic Mass Question; any short cut to solve this problem out there?


10+ Year Member
5+ Year Member
Sep 5, 2008
Dental Student
Need HELP!!
Thanks in advance!

Question> In a chemical determination of the atomic mass of vanadium, 2.8934g of pure VOCL3 was allowed to undergo a set of reactions as a result of which all the chlorine contained in this compound was converted to AgCL. The weight of the AgCl was 7.18g. Assuming the atomic masses of Ag and Cl are 107.86g and 35.453g, what is the experimental value for the atomic mass of vanadium?

answer> 50.9g/mol


Michael De Coro, DMD - AKA Steve McAwesome
Gold Donor
10+ Year Member
Nov 19, 2008
Visalia, CA
Ok, this is a tricky one with some really crazy calculations if you can't use a calculator, where did you get this problem from? You probably won't see calculations like this on the actual test, but here's how I did it.

Step #1 - So you've got 7.18 grams of AgCl, add the atomic masses togeather and you get 143g per mole of AgCl. Now divide 7.18g/143g = 0.05 moles of AgCl.

Step #2 - Knowing that you have 0.05 moles of AgCl, and that there are 3 moles of Cl in VOCl3 and all of it went to make AgCl, divide 0.05 moles by 3 to get 0.0167 moles of VOCl3.

Step #3 - Since we know there are 2.8934g of pure VOCl3, set up the equation like this:

2.8934g VOCl3 = 0.0167 moles VOCl3, then divide both sides by 0.0167 to get that 1 mole of VOCl3 is 173.26g.

Step #4 - Since we know the molar mass of Cl (35.453) and O (16), just add up what you have, and the molar mass of V is the difference from 173.26.

So, V + 16 + (35.453*3) = 173.26 , solve for V and you get that V = 50.9

Not too bad, but this would take you way too long without a calculator. If you got a question like this on the test, I'd start rounding like crazy.
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