1. Percent Composition, Empirical Formulas, and Molecular Formulas

2. So… 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
158.0 g/mol

4. Percent Composition Molar Mass of KMnO4 = 158.0 g 39.1 g K % K x 100 =
Example: What is the percent composition of Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
= 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 =

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 percent composition of ethanol (C2H5OH)
_______________________________________________
Determine the percent composition of sodium oxalate (Na2C2O4)

7. Percent Composition
Calculate the mass of bromine in 50.0 g of Potassium bromide.
1. Molar Mass of KBr
K = 1(39.10) = 39.1
Br =1(79.90) =79.9
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.

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 of water in a hydrated salt.

10. Percent Composition
Calculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2O
Na = 2(22.99) = 46.0
C = 1(12.0) = 12.0
H = 20(1.01) = 20.2 3.
O = 13(15.99)= 207.9
180.2 g
mm = 285.9
_______
x 100%=
67.97 %
286.2 g 2.
Water
H = 20(1.01) = 20.2
O = 10(15.99)= 159.9
mm = 180.1
H = 2(1.0) = 2.0 or
So…
10 H2O = 10(18.0) = 180.0
O = 1(16.0) = 16.0
mm H2O = 18.0

11. Percent Composition
Calculate the percentage of water in aluminum bromide hexahydrate, AlBr3•6H2O.

12. Percent Composition
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. Percent Composition
If 145 grams of copper(II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain?

14. Percent Composition
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. Percent Composition
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?

16. Percent Composition
A 5.0 gram sample of Cu(NO3)2•nH2O is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?
1. Amount water lost
5.0 g hydrate
3.9 g anhydrous salt
1.1 g water
3. Amount of water
0.22 x =
4.0
2. Percent of water
1.1 g water
x 100 =
22 %
5.0 g hydrate

17. Percent Composition
A 7.5 gram sample of CuSO4•nH2O is heated, and 5.4 g of the anhydrous salt remains. What is the value of n?

18. Formulas
Percent composition allows 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 numbers, if necessary. (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.

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.

25. 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

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 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

27. Calculating Empirical Formula
A borane is a compound containing only boron and hydrogen. If a borane is found to contain 88.45% boron, what is its empirical formula?

28. 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 Compound MM by
Empirical Formula Mass g
= 2
141.94g

29. Calculating Molecular Formula
A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?