Processing data help

1. Processing your data is like prelab question number 3 but instead of using a 0.100 g of compound use what you weighed for each trial. Be sure to have a clear table set up with mass of original sulfate, mass of barium sulfate produced, mole of barium sulfate calculated, moles of sulfate ions in the reactant, mass of sulfate ions in the reactant, mass percent.

2. As they mentioned the unknown was a mixture of K₂SO₄ and Na₂SO₄. Read the “here is what you” part the processing data and we will use this to calculate the percent of each.

K₂SO₄ + Na₂SO₄ + BaCl₂ → BaSO₄ + KCl + NaCl

As it says you need a little algebra combined with what is going stiochiometrically.

Using an example that has a mass of sample of 0.132g of mixture and mass of precipitate barium sulfate of 0.195 g we can set up the problem.

Lets call

X = moles of K2SO4 and converting 0.195g of BaSO4 to mole = 8.35x10-4 mol BaSO4

So

(8.35x10-4 mol – X)= mole of Na2SO4

And multiplying by molar mass for Na2SO4 and K2SO4 you get

[(8.35x10-4 mol – X)·142.05g/mol] + (X·174.27g/mol) = 0.132 g solve for X

X = 4.03x10-4 mol of K2SO4

Calculating the number of grams

4.03x10-4 mol K2SO4 x 174.27 g/mol = 0.070 g K2SO4

0.132g K2SO4-Na2SO4 mix - 0.070 g K2SO4 = 0.062 g Na2SO4

Finally dividing and multiplying by a 100% you end up with 53% K2SO4 and 47% Na2SO4