1. Percent Composition, Empirical Formulas, Molecular Formulas

2. So… Percent Composition Part _______ Percent = x 100% Whole
Percent Composition – the percentage by mass of each element in a compound
Part
_______
Percent =
x 100%
Whole
So…
Percent composition
of a compound or =
molecule
Mass of element in 1 mol
____________________
x 100%
Mass of 1 mol

3. Percent Composition Molar Mass of KMnO4 K = 1(39.1) = 39.1
Example: What is the percent composition of Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
K = 1(39.1) = 39.1
Mn = 1(54.9) = 54.9
O = 4(16.0) = 64.0
MM = 158 g

4. Percent Composition Molar Mass of KMnO4 = 158 g 39.1 g K % K x 100 =
Example: What is the percent composition of Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
= 158 g
39.1 g K
% K
x 100 =
24.7 %
158 g
54.9 g Mn
34.8 %
x 100 =
% Mn
158 g
K = 1(39.10) = 39.1
64.0 g O
x 100 =
40.5 %
Mn = 1(54.94) = 54.9
% O
158 g
O = 4(16.00) = 64.0
MM = 158

5. Percent Composition
Determine the percentage composition of sodium carbonate (Na2CO3)?
Molar Mass
Percent Composition
46.0 g
x 100% =
43.4 %
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= g
% Na =
106 g
12.0 g
x 100% =
11.3 %
% C =
106 g
48.0 g
x 100% =
45.3 %
% O =
106 g

6. Percent Composition
Determine the percentage composition of ethanol (C2H5OH)?
% C = 52.13%, % H = 13.15%, % O = 34.72%
_______________________________________________
Determine the percentage composition of sodium oxalate
(Na2C2O4)?
% Na = 34.31%, % C = 17.93%, % O = 47.76%

7. Percent Composition
Calculate the mass of bromine in 50.0 g of Potassium bromide.
1. Molar Mass of KBr
K = 1(39.10) = 39.10
Br =1(79.90) =79.90
MM = 119.0 2.
79.90 g
___________ =
119.0 g
x 50.0g = 33.6 g Br

8. Percent Composition
Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2.
1. Molar Mass of C6H14N2O2
C = 6(12.01) = 72.06
H =14(1.01) = 14.14
N = 2(14.01) = 28.02
O = 2(16.00) = 32.00
MM = 146.2 2.
28.02 g
___________
= 0.192
146.2 g
x 85.0 mg = 16.3 mg N

9. Formulas
Percent composition allow you to calculate the simplest ratio among the atoms found in compound.
Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms.
Molecular Formula – actual formula of a compound showing the number of atoms present
Examples:
C6H12O6
- molecular
C4H10
- molecular
C2H5
- empirical
CH2O
- empirical

10. Formulas Is H2O2 an empirical or molecular formula?
Molecular, it can be reduced to HO
HO = empirical formula

11. Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula.
1. Determine the number of grams of each element in the compound.
4.151 g Al and g O
2. Convert masses to moles.
1 mol Al
4.151 g Al =
mol Al
26.98 g Al
1 mol O
3.692 g O =
mol O
16.00 g O

12. Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula.
3. Find ratio by dividing each element by smallest amount of moles.
moles Al
= mol Al
0.1539
moles O
= mol O
0.1539
4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)
O = x 2 = 3
Al = x 2 = 2
therefore,
Al2O3

13. Calculating Empirical Formula
A g sample of cobalt reacts with g chlorine to form a binary compound. Determine the empirical formula for this compound.
4.550 g Co
1 mol Co
= mol Co
58.93 g Co
1 mol Cl
5.475 g Cl
= mol Cl
35.45 g Cl
mol Co
mol Cl
= 1
= 2
CoCl2

14. Calculating Empirical Formula
When a g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of g. Determine the empirical formula.
Fe = g
O = g – g = g
1 mol Fe
2.000 g Fe
= mol Fe
55.85 g Fe
1 mol O
0.573 g O
= mol Fe
16.00 g
1 : 1
FeO 14

15. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate.
1 mol Pb
g Pb
= mol Pb
207.2 g Pb gH
1 mol H
= mol H
1.008 g H
1 mol As
g As
= mol As
74.92 g As
1 mol O
0.4267g Fe
= mol O
16.00 g O

16. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate.
mol Pb
= mol Pb
mol H
= 1.00 mol H
PbHAsO4
mol As
= mol As
mol O
= mol O

17. Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 1:
In g of Nylon-6 the masses of elements present are g C, g n, 9.80 g H, and g O.
Step 2:
63.38 g C
1 mol C
9.80 g H
1 mol H
= mol C
= 9.72 mol H
12.01 g C
1.01 g H
12.38 g N
1 mol N
= mol N
14.01 g N
14.14 g O
1 mol O
= mol O
16.00 g O

18. Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 3:
5.302 mol C
= mol C
6:1:11:1
0.8837
mol N
= mol N
0.8837
C6NH11O
9.72 mol H
= 11.0 mol H
0.8837
mol O
= mol O
0.8837

19. Calculating Molecular Formula
A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of g. What is the compound’s molecular formula?
Step 3: Multiply
Step 1: Molar Mass
P = 2 x g = 61.94g
O = 5 x 16.00g = g g
(P2O5)2 =
P4O10
Step 2: Divide MM by
Empirical Formula Mass g
= 2
141.94g

20. Calculating Molecular Formula
A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?
(CH)6 =
C = g
H = g
13.01 g
C6H6
78 g/mol
= 6
13.01 g/mol