Percent Composition
• Percent Composition – the
percentage by mass of
each element in a
compound
Part
_______
Percent =
x 100%
Whole
Percent composition Mass of element in 1 mol
of a compound or = ____________________ x 100%
Mass of 1 mol
molecule
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
Percent Composition
Example: What is the percent composition of
Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
% K
39.1 g K
158 g
= 158 g
x 100 = 24.7 %
54.9 g Mn x 100 = 34.8 %
% Mn 158 g
K=
1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0
MM = 158
64.0 g O x 100 = 40.5 %
% O
158 g
Percent Composition
Determine the percentage composition of sodium carbonate
(Na2CO3)?
Molar Mass
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106 g
Percent Composition
46.0 g
% Na =
106 g
12.0 g
% C =
106 g
48.0 g
% O = 106 g
x 100% = 43.4 %
x 100% = 11.3 %
x 100% = 45.3 %
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%
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.
3.
79.90 g
___________ = 0.6714
119.0 g
0.6714 x 50.0g = 33.6 g Br
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.
3.
28.02 g
___________
= 0.192
146.2 g
0.192 x 85.0 mg = 16.3 mg N
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.
Percent Composition
Calculate the percentage of water in sodium carbonate
decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2O
Na
C
H
O
2.
= 2(22.99) = 45.98
= 1(12.01) = 12.01
= 20(1.01) = 20.2
= 13(16.00)= 208.00
3.
MM = 286.2
Water
180.2 g
_______ x 100%= 67.97 %
286.2 g
H = 20(1.01) = 20.2
O = 10(16.00)= 160.00
MM = 180.2
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
Percent Composition
Calculate the percentage of water in Aluminum bromide
hexahydrate, AlBr3•6H2O.
1. Molar Mass of AlBr3•6H2O
Al
Br
H
O
2.
= 1(26.98) = 26.98
= 3(79.90) = 239.7
= 12(1.01) = 12.12
= 6(16.00) = 96.00
MM = 374.8
3.
Water
H = 12(1.01) = 12.1
O = 6(16.00)= 96.00
MM = 108.1
or
MM = 18.02
For 6 H2O = 6(18.02) = 108.2
108.1 g
_______
x 100%= 28.85 %
374.8 g
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 X 100 = 48.84 %
246.5 g
3. Grams anhydrous MgSO4
0.4884 x 125 = 61.1 g
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 X 100 = 63.92 %
249.7 g
3. Grams anhydrous CuSO4
0.6392 x 145 = 92.7 g
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 x 100 = 14 %
5.0 g hydrate
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 x 100 = 29 %
7.5 g hydrate
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 x 100 = 22 %
5.0 g hydrate
3. Amount of water
0.22 x 18.02 = 4.0
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 x 100 = 28 %
7.5 g hydrate
3. Amount of water
0.28 x 18.02 = 5.0
Empirical and Molecular Formulas
Empirical Formula
Empirical Formula
A formula that gives the simplest whole-number ratio of
the atoms of each element in a compound.
Molecular Formula
Empirical Formula
H2O2
HO
C6H12O6
CH2O
CH3O
CH3O
CH3OOCH = C2H4O2
CH2O
EMPIRICAL FORMULA
Mass % of
elements
Assume
Empirical
Formula
100g sample
Calculate
Grams of
each
element
mole ratio
Moles of
each
element
Use Atomic Masses
What is an empirical formula?
• A chemical formula in which the ratio of
the elements are in the lowest terms is
called an empirical formula.
Example:
• The empirical formula for a glucose
molecule (C6H12O6) is CH2O. All the
subscripts are divisible by six.
C6
6
C
H12
6
H2
O6
6
O
Exceptions:
Some formulas, such as the one for carbon
dioxide, CO2, are already empirical formulas
without being reduced.
Determine the empirical formula for a
compound containing 2.128 g Cl and
1.203 g Ca.
Steps
1. Find mole amounts.
2. Divide each mole by the smallest
mole.
1. Find mole amounts.
2.128 g Cl x 1 mol Cl = 0.0600 mol Cl
35.45 g Cl
1.203 g Ca x 1 mol Ca = 0.0300 mol Ca
40.08 g Ca
2. Divide each mole by the smallest
mole.
Cl = 0.0600 mol Cl = 2.00 mol Cl
0.0300
Ca = 0.0300 mol Ca = 1.00 mol Ca
0.0300
Ratio – 1 Ca: 2 Cl
Empirical Formula = CaCl2
A compound weighing 298.12 g consists of
72.2% magnesium and 27.8% nitrogen by
mass. What is the empirical formula?
Hint
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til
whole”
A compound weighing 298.12 g consists of 72.2%
magnesium and 27.8% nitrogen by mass. What is the
empirical formula?
Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g
N – (27.8%/100)*298.12 g = 82.88 g
Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole
24.3 g
N – 82.88 g * ( 1 mole ) = 5.92 mole
14.01 g
Divide by small: Mg - 8.86 mole/5.92 mole = 1.50
N - 5.92 mole/5.92 mole = 1.00 mole
Multiply ‘til whole: Mg – 1.50 x 2 = 3.00
N – 1.00 x 2 = 2.00
Mg3N2
Molecular Formula
The molecular formula gives the actual number of atoms of each
element in a molecular compound.
1.
2.
3.
4.
Steps
Find the empirical formula.
Calculate the Empirical Formula Mass.
Divide the molar mass by the “EFM”.
Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is ~124.06 and
empirical formula is CH2O3.
2.
“EFM” = 62.03 g
3.
124.06/62.03 = 2
4.
2(CH2O3) = C2H4O6
What is a molecular formula?
• A molecular formula is the “true formula”
of a compound. The chemical formula for
a molecular compound shows the actual
number of atoms present in a molecule.
To find the molecular formula
from the empirical formula:
 Find the empirical formula.
 Determine the empirical formula mass.
 Divide the molecular mass by the empirical
formula mass to determine the multiple.
 Multiply the empirical formula by the multiple to
find the molecular formula.
MF mass = n
EF mass
(EF)n = molecular formula
EXAMPLE:
The empirical formula for ethylene is CH2. Find the
molecular formula if the molecular mass is
28.1g/mol.
C = 1 x 12 = 12
H = 2 x 1 = +2
14g/mol = empirical formula mass
28.1 g/mol = 2
14 g/mol
(CH2)2  C2H4
Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The
molar mass of the compound is 92.0 g/mol.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula
Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.
Empirical formula.
A. Find mole amounts.
4.90 g N x 1 mol N = 0.350 mol N
14.01 g N
11.2 g O x 1 mol O = 0.700 mol O
16.00 g O
B.
Divide each mole by the smallest mole.
N = 0.350 = 1.00 mol N
0.350
O = 0.700 = 2.00 mol O
0.350
Empirical Formula = NO2
Empirical Formula Mass = 46.01 g/mol
Molecular formula
Molar Mass
=
Emp. Formula Mass
92.0 g/mol = 2.00
46.01 g/mol
Molecular Formula = 2 x Emp. Formula =
N2O4
A 528.39 g compound containing only
carbon, hydrogen, and oxygen is found to be
48.38% carbon and 8.12% hydrogen by
mass. The molar mass of this compound is
known to be ~222.25 g/mol. What is its
molecular formula?
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found
to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this
compound is known to be ~222.25 g/mol. What is its molecular formula?
g C – (48.38/100)*528.39 g = 255.64 g
g H – (8.12/100)*528.39 g = 42.91 g
g O – (43.5/100)*528.39 g = 229.85 g
mole C - 255.64 g * ( 1 mole ) = 21.29 mol
12.01 g
mole H – 42.91 g * ( 1 mole ) = 42.49 mol
1.01 g
mole O – 229.85 g * ( 1 mole ) = 14.37 mol
16.00 g
A 528.39 g compound containing only carbon, hydrogen, and oxygen is
found to be 48.38% carbon and 8.12% hydrogen by mass. The molar
mass of this compound is known to be ~222.25 g/mol. What is its
molecular formula?
From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O
C – 21.29/14.27 = 1.49
H – 42.49/14.27 = 2.98 (esentially 3)
O – 14.27/14.27 = 1.00
C – 1.49 x 2 = 3
H–3x2=6
O–1x2=2
C3H6O2
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found
to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this
compound is known to be ~222.25 g/mol. What is its molecular formula?
From last slide: Empirical formula = C3H6O2
“EFM” = 74.09
Molar mass = 222.24 = ~3
EFM
74.09
3(C3H6O2) = C9H18O6
Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
• I have 5 eggs. How many cookies can I
make?
Ratio of eggs to cookies
5 eggs
5 doz.
2 eggs
= 12.5 dozen cookies
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Proportional Relationships
• Stoichiometry
– mass relationships between substances in a chemical
reaction
– based on the mole ratio
• Mole Ratio
– indicated by coefficients in a balanced equation
2 Mg + O2  2 MgO
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
–
–
–
–
–
Mole ratio Molarratio
mass
Mole
- Molarity Molar volume -
moles  moles
moles
grams
moles

moles
moles  liters soln
moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Pressure
Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02 
MOLES
1023
particles/mol
NUMBER
OF
PARTICLES
Molarity (mol/L)
LITERS
OF
SOLUTION
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem