2. Percent Composition
• Percent Composition – the
percentage by mass of
each element in a
compound
Percent = _______
Part
Whole
x 100%
Percent composition
of a compound or =
molecule
Mass of element in 1 mol
____________________
Mass of 1 mol
x 100%
3. Percent Composition
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
Example: What is the percent composition of
Potassium Permanganate (KMnO4)?
= 158 g
% K
Molar Mass of KMnO4
39.1 g K
158 g
x 100 = 24.7 %
% Mn
54.9 g Mn
158 g
x 100 = 34.8 %
% O
64.0 g O
158 g
x 100 = 40.5 %
K = 1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0
MM = 158
5. Percent Composition
Determine the percentage composition of sodium carbonate
(Na2CO3)?
Molar Mass Percent Composition
% Na =
46.0 g
106 g
x 100% = 43.4 %
% C =
12.0 g
106 g
x 100% = 11.3 %
% O =
48.0 g
106 g
x 100% = 45.3 %
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 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
79.90 g
___________
119.0 g
= 0.6714
3. 0.6714 x 50.0g = 33.6 g Br
2.
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
MM = 146.2
28.02 g
___________
146.2 g
= 0.192
3. 0.192 x 85.0 mg = 16.3 mg N
2.
N = 2(14.01) = 28.02
O = 2(16.00) = 32.00
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
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) = 45.98
C = 1(12.01) = 12.01
MM = 286.2
H = 20(1.01) = 20.2
O = 13(16.00)= 208.00
H = 20(1.01) = 20.2
Water
O = 10(16.00)= 160.00
MM = 180.2
2.
3.
180.2 g
_______
286.2 g
67.97 %
x 100%=
or H = 2(1.01) = 2.02
O = 1(16.00) = 16.00
MM H2O = 18.02
So…
10 H2O = 10(18.02) = 180.2
11. Percent Composition
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) = 239.7
MM = 374.8
H = 12(1.01) = 12.12
O = 6(16.00) = 96.00
H = 12(1.01) = 12.1
Water
O = 6(16.00)= 96.00
MM = 108.1
2.
3.
108.1 g
_______
374.8 g
28.85 %
x 100%=
or
MM = 18.02
For 6 H2O = 6(18.02) = 108.2
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
Mg = 1 x 24.31 = 24.31 g
S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g
MM = 120.37 g
H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H2O =
7 x 18.02 g = 126.1 g
Total MM =
120.4 g + 126.1 g = 246.5 g
2. % MgSO4
120.4 g
246.5 g
X 100 = 48.84 %
3. Grams anhydrous MgSO4
0.4884 x 125 = 61.1 g
13. Percent Composition
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
Cu = 1 x 63.55 = 63.55 g
S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g
MM = 159.61 g
H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H2O =
5 x 18.02 g = 90.1 g
Total MM =
159.6 g + 90.1 g = 249.7 g
2. % CuSO4
159.6 g
249.7 g
X 100 = 63.92 %
3. Grams anhydrous CuSO4
0.6392 x 145 = 92.7 g
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?
1. Amount water lost
5.0 g hydrate
- 4.3 g anhydrous salt
0.7 g water
2. Percent of water
0.7 g water
5.0 g hydrate
x 100 = 14 %
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?
1. Amount water lost
7.5 g hydrate
- 5.3 g anhydrous salt
2.2 g water
2. Percent of water
2.2 g water
7.5 g hydrate
x 100 = 29 %
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
2. Percent of water
1.1 g water
5.0 g hydrate
x 100 = 22 %
3. Amount of water
0.22 x 18.02 = 4.0
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?
1. Amount water lost
7.5 g hydrate
- 5.4 g anhydrous salt
2.1 g water
2. Percent of water
2.1 g water
7.5 g hydrate
x 100 = 28 %
3. Amount of water
0.28 x 18.02 = 5.0
18. Formulas
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
Percent composition allow you to calculate the simplest
ratio among the atoms found in compound.
Examples:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
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 4.151 g of
aluminum with 3.692 g of oxygen. Calculate the empirical formula.
1. Determine the number of grams of each element in the compound.
4.151 g Al and 3.692 g O
2. Convert masses to moles.
4.151 g Al 1 mol Al
26.98 g Al
= 0.1539 mol Al
3.692 g O 1 mol O
16.00 g O
= 0.2308 mol O
21. Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of
aluminum with 3.692 g of oxygen. Calculate the empirical formula.
3. Find ratio by dividing each element by smallest amount of moles.
0.1539 moles Al
0.1539
= 1.000 mol Al
0.2308 moles O
0.1539
= 1.500 mol O
4. Multiply by common factor to get whole number. (cannot have
fractions of atoms in compounds)
O = 1.500 x 2 = 3
Al = 1.000 x 2 = 2
therefore, Al2O3
22. Calculating Empirical Formula
A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a
binary compound. Determine the empirical formula for this
compound.
4.550 g Co 1 mol Co
58.93 g Co
= 0.07721 mol Co
5.475 g Cl 1 mol Cl
35.45 g Cl
= 0.1544 mol Cl
0.07721 mol Co 0.1544 mol Cl
0.07721
0.07721
= 2
= 1
CoCl2
23. Calculating Empirical Formula
When a 2.000 g sample of iron metal is heated in air, it reacts with
oxygen to achieve a final mass of 2.573 g. Determine the empirical
formula.
2.000 g Fe 1 mol Fe
55.85 g Fe
= 0.03581 mol Fe
0.573 g O 1 mol O
16.00 g
= 0.03581 mol Fe
Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g
1 : 1
FeO
24. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato
beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of
arsenic, and 0.4267 g of oxygen. Calculate the empirical formula
for lead arsenate.
1.3813 g Pb 1 mol Pb
207.2 g Pb
= 0.006667 mol Pb
0.00672 gH 1 mol H
1.008 g H
= 0.00667 mol H
0.4995 g As 1 mol As
74.92 g As
= 0.006667 mol As
0.4267g Fe 1 mol O
16.00 g O
= 0.02667 mol O
25. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato
beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of
arsenic, and 0.4267 g of oxygen. Calculate the empirical formula
for lead arsenate.
0.006667 mol Pb
0.00667 mol H
0.006667 mol As
0.02667 mol O
0.006667
0.006667
0.006667
0.006667
= 1.000 mol Pb
= 1.00 mol H
= 1.000 mol As
= 4.000 mol O
PbHAsO4
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 100.00g of Nylon-6 the masses of elements present are 63.38 g C,
12.38 g n, 9.80 g H, and 14.14 g O.
Step 2:
63.38 g C 1 mol C
12.01 g C
= 5.302 mol C
12.38 g N 1 mol N
14.01 g N
= 0.8837 mol N
9.80 g H 1 mol H
1.01 g H
= 9.72 mol H
14.14 g O 1 mol O
16.00 g O
= 0.8832 mol 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
0.8837
= 6.000 mol C
0.8837 mol N
0.8837
= 1.000 mol N
9.72 mol H
0.8837
= 11.0 mol H
0.8837 mol O
0.8837
= 1.000 mol O
6:1:11:1
C6NH11O
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 283.88g. What is the compound’s molecular
formula?
Step 1: Molar Mass
P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by
Empirical Formula Mass
238.88 g
141.94g
= 2
Step 3: Multiply
(P2O5)2 =
P4O10
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?
C = 12.01 g
H = 1.01 g
13.01 g
78 g/mol
13.01 g/mol
= 6
(CH)6 =
C6H6
30. Practice Problems
1. Analysis of a chemical used in photographic
developing fluid indicates a chemical composition
of 65.45% C, 5.45%H, and 29.09% O. The molar
mass is found to be 110.0 g/mol. Determine the
molecular formula.
2. A compound was found to contain 49.98 g carbon
and 10.47 g hydrogen. The molar mass of the
compound is 58.12 g/mol. Determine the
molecular formula.
3. A colorless liquid composed of 46.68% nitrogen
and 53.32% oxygen has a molar mass of 60.01
g/mol. What is the molecular formula?
31. 4. A sample of compound containing 140.1 g of
nitrogen and 80.0 g of oxygen has a molar mass of
220.1 g/mol. Calculate the empirical and molecular
formula of the compound.
5. 300 grams of an organic sample which contains
only carbon, hydrogen and oxygen is analyzed and
found to contain 145.946 grams of carbon,
24.3243 grams of hydrogen and the rest is oxygen.
What is the empirical formula for the compound?
32. 6. The characteristic odor of pineapple is due to ethyl
butyrate, an organic compound which contains only carbon,
hydrogen and oxygen. If a sample of ethyl butyrate is
known to contain 0.62069 g of carbon, 0.103448 g of
hydrogen and 0.275862 g of oxygen, what is the empirical
formula for ethyl butyrate?
7. 300 grams of a compound which contains only carbon,
hydrogen and oxygen is analyzed and found to contain the
exact same percentage of carbon as it has oxygen. The
percentage of hydrogen is known to be 5.98823%. Find
the empirical formula of the compound.
33. 8. A certain compound contains 4.0 g of calcium
and 7.1 g of chlorine. Is relative molecular mass is
111. Find its empirical and molecular formulas.
9. A certain compound has 25.9% nitrogen and
74.1% oxygen. Its relative molecular mass is 108.
Find its empirical and molecular formula.
10. A certain compound was found to contain 54.0
g of carbon and 10.5 grams of hydrogen. Its relative
molecular mass is 86.0. Find the empirical and the
molecular formulas.
34. 11. A certain compound was found to contain 26.4 g
of carbon, 4.4 grams of hydrogen and 35.2 grams of
oxygen. Its relative molecular mass is 60.0. Find
the empirical and the molecular formula.
12. A certain compound was found to contain 78.2 %
Boron and 21.8 % hydrogen. Its relative molecular
mass is 27.7. Find the empirical and the molecular
formula.
13. A certain compound contains 7.3%Carbon, 4.5
% hydrogen, 36.4% oxygen, and 31.8% nitrogen. Its
relative molecular mass is 176.0. Find its empirical
and molecular formulas.
