Specifically chemistry stemlords A 10.00 g sample of an unknown water soluble barium salt is treated with an excess of Na2SO4 to precipitate 11.21 g BaSO4. Which barium salt is it?
A. BaCl2
B. Ba(O2CH)2
C. Ba(NO3)2
D. BaBr2
Help
A, BaCl2.
Its a density problem
Weight of non-barium part of barium salt = X
Br + X = BR + SO4
Br + X = 10g
Br + SO4 = 11.21 g
Molar mass of BrSO4 = 233.39 g/mol
So you have 11.21/233.39= .048 mols of BrSO4
10g/.064m = 208.2 g/mol
Which of the answers has that molar mass?
(its BaCl2, at 208.233g/mol)
I did that and I got BaCl2, I was just wondering if there was an easier way than doing trial and error with all the possible answers.
Thanks!
Subtracting barium from the final molar mass.get you ~70 g/mol, ie the mm of Cl x 2. I double checked the other options but you can definintely rule out the bromide salt based on that
Take the mass of precipitate produced and multiply it by the molar mass of BaSO4 (233.4 g/mol)to get the moles of precipitate. Then, divide the mass of the unknown by that amount of moles to find the molar mass and see which of the answers it corresponds to.
spoiler
11.21 g BaSO4 / 233.4 g/mol BaSO4 = 0.04804 mol BaSO4
10.00 g unknown / 0.04804 mol = 208.15 g/mol unknown
Going through the list we can see that this corresponds to the mass of barium chloride (BaCl2)
BaCl2 because that Na2 in front of the SO4 looks like it would want to have sex with the Cl2 to make 2NaCl aka table salt (i forgot all my highschool chem knowledge)
Mass = moles * Mr
Moles = mass/Mr
BaSO4 Mr = 233.43
Moles BaSO4 = 11.21g / 233.43 = 0.048 mol
Reaction is 1:1, thus 0.048 mol of initial salt produces 0.048 mol of resultant saltMr = Mass / moles
10.00 g / 0.048 = 208. 33
Since Barium Mr is 137.33, 208.33 - 137.33 = 71Cl2 = 71
(O2CH)2 = 90
(NO3)2 = 124
Br2 = 160Therefore BaCl2
