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. Hydrates
Hydrated salt – salt that has water molecules trapped within the crystal lattice
Examples: CuSO4•5H2O , CuCl2•2H2O
Anhydrous salt – salt without water molecules
Examples: CuCl2
Can calculate the percentage by mass of water in a hydrated salt.

10. Hydrates
Calculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2O
Na = 2(22.99) = 45.98
C = 1(12.01) = 12.01
H = 20(1.01) = 20.2 3.
O = 13(16.00)=
180.2 g
MM = 286.2
_______
x 100%=
62.96 %
286.2 g 2.
Water
H = 20(1.01) = 20.2
O = 10(16.00)=
MM = 180.2
H = 2(1.01) = 2.02 or
So…
10 H2O = 10(18.02) = 180.2
O = 1(16.00) = 16.00
MM H2O = 18.02

11. Hydrates
Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr3•6H2O.
1. Molar Mass of AlBr3•6H2O
Al = 1(26.98) = 26.98
Br = 3(79.90) =
H = 12(1.01) = 12.12
O = 6(16.00) = 96.00
MM = 3. 2.
Water g
_______
x 100%= % g
H = 12(1.01) = 12.12
O = 6(16.00)= 96.00
MM =
MM = 18.02
For 6 H2O = 6(18.02) = 108.2 or

12. Hydrates
If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?
MgSO4 . 7 H2O
1. Molar Mass
2. % MgSO4
Mg = 1 x = g
S = 1 x = g
O = 4 x = g
MM = g
120.4 g
X 100 =
48.84 %
246.5 g
3. Grams anhydrous MgSO4
H = 2 x = 2.02 g
O = 1 x = g
MM = g
x 125 =
61.1 g
MM H2O =
7 x g = g
Total MM =
120.4 g g = g

13. Hydrates
If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain?
CuSO4 . 5 H2O
1. Molar Mass
2. % CuSO4
Cu = 1 x = g
S = 1 x = g
O = 4 x = g
MM = g
159.6 g
X 100 =
63.92 %
249.7 g
3. Grams anhydrous CuSO4
H = 2 x = 2.02 g
O = 1 x = g
MM = g
x 145 =
92.7 g
MM H2O =
5 x g = 90.1 g
Total MM =
159.6 g g = g

14. Hydrates
A 5.0 gram sample of a hydrate of BaCl2 was heated, and only 4.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate?
2. Percent of water
1. Amount water lost
0.7 g water
5.0 g hydrate
4.3 g anhydrous salt
0.7 g water
x 100 =
14 %
5.0 g hydrate

15. A 7. 5 gram sample of a hydrate of CuCl2 was heated, and only 5
A 7.5 gram sample of a hydrate of CuCl2 was heated, and only 5.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate?
2. Percent of water
1. Amount water lost
2.2 g water
7.5 g hydrate
5.3 g anhydrous salt
2.2 g water
x 100 =
29 %
7.5 g hydrate

16. Hydrates
A 5.0 gram sample of Cu(NO3)2•xH2O is heated, and 3.9 g of the anhydrous salt remains. What is the value of x?
1. Amount water lost in g and mols
5.0 g hydrate
3.9 g anhydrous salt
1.1 g water
1.1 g/ g/mol = mols of H20
3. Ratio of mols
0.0610/ =2.9
N=3
4. Formula
2. Amount of anhydrous salt in mols
Cu(NO3)2•3H2O
Molar mass ( x x 16)= g/mol
3.9 g / g/mol= mols of anhydrous salt

17. Hydrates
A 2.54 gram sample of CuSO4•nH2O is heated, and 1.61 g of the anhydrous salt remains. What is the value of n?
1. Amount water lost
2.54 g hydrate
1.61 g anhydrous salt
0.93 g water
0.93/18.02 g/mol= mols water
3. Ratio of mols
0.0517/ = 5.1
2. Mols of anhydrous salt
4. Formula
CuSO4•5H2O
1.61 g anhydrous / g/mol= mols

18. 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

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

20. 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

21. 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

22. 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

23. 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 23

24. 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

25. 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

26. 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

27. 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
What is the
empirical formula mass?
6x =
g/mol
mol O
= mol O
0.8837

28. Molecular Formula
But what if the molar mass is 339g/mol what is the molecular formula? What is the mathematical relationship between the empirical formula mass versus the molar mass? The molar mass is greater by 3X
339 g/mol/ g/mol= 3
Molecule is 3 X bigger than C6N1H11O1
Molecule = C18N3H33O3

29. 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: Empirical formula 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

30. 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
13.02 g
C6H6
78 g/mol
= 6
13.01 g/mol