posted by London on .
When a0.269g sample of a barium compound was treated with excess H2SO4, 0.0891g of BaSO4 formed. What is the percentage of Barium in the compound?
% Ba = (mass Ba/mass sample)*100
mass sample = 0.269 g.
mass BaSO4 = 0.0891 g.
moles BaSO4 = grams/molar mass
There is one mole Ba per mole BaSO4; therefore, that, also, will be the moles of Ba.
g Ba = moles Ba x molar mass Ba. Plug into the top formula and calculate percent Ba. Check my work.
[Note: I can't help but mention that there is an easier way to do this but "they don't teach that method anymore."]
Please do tell...
What the IUPAC calls and obsolete factor, the gravimetric factor. It will convert grams of one substance into another without going through all the mole business.
For example, for BaSO4, the mass was 0.0891 g. We want to know how much Ba was there.
mass BaSO4 * (atomic mass Ba/molar mass BaSO4) = 0.0891 x (137.33/233.39) = 0.05243
and %Ba = (0.05243/0.269)*100 =19.49% which rounds to 19.5% to three significant figures (determined by the 0.0891 g sample). Check my arithmetic and look up the atomic mass and molar mass. I'm just recalling them from memory and they may have changed. Check my answer against your answer.
Thank you, that was so helpful