unit: chemical quantities determining the empirical formula from percent composition day 6 - notes

7
Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - N otes

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Page 1: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

Unit: Chemical QuantitiesDetermining the Empirical

Formula from Percent Composition

Day 6 - Notes

Page 2: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

After today, you will be able to…

•Calculate the empirical formulas for various compounds given their percent makeup

Page 3: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

Empirical Formula Determination from %

CompositionTo determine empirical formulas from % composition follow these steps:1.Assume 100.0g of the compound (change % into grams)2.Convert g moles for each element using molar mass3.Divide each mole answer by the smallest mole answer4.If you do not have a whole number, multiply by the smallest whole # to get a whole # ratio.5.Skip to this step if you already have a whole number ratio from #3. These whole #s are the subscripts for each element.

Page 4: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

To sum up these steps here is a poem for my favorite little

chemists:

“Percent to massMass to MolesDivide by small

Multiply by whole!”

Page 5: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

Example: What is the empirical formula for a compound that is 21.2% N, 6.10% H, 24.3% S, and 48.4%O?

21.2gN

6.10gH

24.3gS

48.4gO

1 mol N 14.01gN=

N2H8SO4

x 1.51

1 mol H 1.01gHx 6.04=

1 mol S 32.07gS

0.758

=x

x 1 mol O 16.00gO

3.03=

0.7580.7580.7580.758

=

=

=

=

2

7.97~

8

1

4

Page 6: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

Example: What is the empirical formula for a compound that is 69.8% Fe and 30.2% O?

69.8gFe

30.92gO

1 mol Fe

55.85gFe

=

Fe2O3

x 1.25

1 mol O 16.00gO

1.89=x

1.25

1.25

=

=

1

1.5

(2)

(2)

=

=

2

3

Page 7: Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition Day 6 - Notes

Time to Practice!

Complete WS 6

HOMEWORK: