Empirical Formula

Empirical Formula #1

Wes Prosser 15 Sep 2011 04:00

Q: A white powder has %mass of Ba 69.6%, C 6.09% and O 24.3%. What is the empirical formula? (M(Ba) = 137.3 g mol^-1, M(C) = 12.01 g mol^-1, M(O) = 16.00 g mol^-1)
(Worksheet 6)

In a 100 g sample, m(Ba) = 69.6 g, m(C) = 6.09g, and m(O) = 24.3 g
n(Ba) = m(Ba) ÷ M(Ba) = 69.6 ÷ 137.3 = 0.507 mol
n(C) = m(C) ÷ M(C) = 6.09 ÷ 12.01 = 0.507 mol
n(O) = m(O) ÷ M(O) = 24.3 ÷ 16.00 = 1.52 mol
Mole ratio is 0.507 : 0.507 : 1.52 = $\frac{0.507}{0.507} : \frac{0.507}{0.507} : \frac{1.52}{0.507}$ = 1 : 1 : 3
Therefore, the empirical formula is BaCO₃

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Last edited on 15 Sep 2011 04:08 by Wes Prosser + Show more

Empirical Formula #2

Wes Prosser 15 Sep 2011 04:04

Q: A white solid has %mass of Na 32.4%, S 22.6% and O 45.0%. What is the empirical formula?
(Worksheet 6)

In a 100 g sample, m(Na) = 32.4 g, m(S) = 22.6g, and m(O) = 45.0 g
n(Na) = m(Na) ÷ M(Na) = 32.4 ÷ 22.99 = 1.41 mol
n(S) = m(S) ÷ M(S) = 22.6 ÷ 32.07 = 0.70 mol
n(O) = m(O) ÷ M(O) = 45.0 ÷ 16.00 = 2.81 mol
Mole ratio is 1.41 : 0.70 : 2.81 = $\frac{1.41}{0.70} : \frac{0.70}{0.70} : \frac{2.81}{0.70}$ = 2 : 1 : 4
Therefore, the empirical formula is Na₂SO₄

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Last edited on 15 Sep 2011 04:07 by Wes Prosser + Show more

Empirical Formula comment

Wes Prosser 15 Sep 2011 04:10

It's really gratifying when the empirical formula comes out to be something recognisable… your level of confidence that you got it right goes through the roof :D