percentage compositions (1 Viewer)

[clicker]

a sample of impure Ba(NO3)2 is analysed by dissolving 0.222g of the mixture completely in water. excess Na2SO4 solution is added and a precipitate of BaSO4 is formed. the mass of the precipitate, after filtering and drying, is 0.182g .

calculate the percent of purity of the original Ba(NO3)2 sample.

 

[jamesy_1988]

The percentage of Ba(NO3)2 in the sample is 70.45%, as per my calculations.

I will post the calculations when I have a bit more spare time

Jamesy

 

[thunderdax]

Ba(NO₃)₂+Na₂SO₄ -----> 2NaNO₃+BaSO₄
Then, work out the mass that should be created:
n=m/M
=.222/261.3
=.000850 moles
Now, for the barium sulfate,
n=m/M
.000850=m/233.4
m=.198g

Impurity=.198/.259*100
=76.6%

EDIT: Happy?

 

[Pace_T]

isnt molar mass of Ba(NO3)2 = 137.3 + (14 + 48)*2 = 261.3?

 

[currysauce]

that would be molar mass

 

[Pace_T]

oh yep thats what i meant hehe