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Percent Composition• Percent Composition – the
percentage by mass of each element in a compound
Percent =_______Part
Wholex 100%
Percent compositionof a compound or =molecule
Mass of element in 1 mol____________________Mass of 1 mol
x 100%
Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
K = 1(39.1) = 39.1Mn = 1(54.9) = 54.9O = 4(16.0) = 64.0
MM = 158 g
Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)?
= 158 g
% K
Molar Mass of KMnO4
39.1 g K158 g
x 100 = 24.7 %
% Mn54.9 g Mn158 g
x 100 = 34.8 %
% O64.0 g O158 g
x 100 = 40.5 %K = 1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0MM = 158
Percent Composition
Determine the percentage composition of sodium carbonate (Na2CO3)?
Molar Mass Percent Composition
% Na =46.0 g106 g
x 100% = 43.4 %
% C =12.0 g106 g
x 100% = 11.3 %
% O =48.0 g106 g
x 100% = 45.3 %
Na = 2(23.00) = 46.0C = 1(12.01) = 12.0O = 3(16.00) = 48.0
MM= 106 g
Percent CompositionDetermine 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 CompositionCalculate 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.
Percent CompositionCalculate 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) = 28.02O = 2(16.00) = 32.00
HydratesHydrated 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 CompositionCalculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2ONa = 2(22.99) = 45.98 C = 1(12.01) = 12.01
MM = 286.2
H = 20(1.01) = 20.2O = 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) = 2.02O = 1(16.00) = 16.00
MM H2O = 18.02
So… 10 H2O = 10(18.02) = 180.2
Percent CompositionCalculate the percentage of water in Aluminum bromide hexahydrate, AlBr3•6H2O.
1. Molar Mass of AlBr3•6H2OAl = 1(26.98) = 26.98 Br = 3(79.90) = 239.7
MM = 374.8
H = 12(1.01) = 12.12O = 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%=
orMM = 18.02For 6 H2O = 6(18.02) = 108.2
Percent CompositionIf 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain? MgSO4
. 7 H2O1. Molar Mass
Mg = 1 x 24.31 = 24.31 gS = 1 x 32.06 = 32.06 gO = 4 x 16.00 = 64.00 g
MM = 120.37 g
H = 2 x 1.01 = 2.02 gO = 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 g246.5 g
X 100 = 48.84 %
3. Grams anhydrous MgSO4
0.4884 x 125 = 61.1 g
Percent CompositionIf 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain? CuSO4
. 5 H2O1. Molar Mass
Cu = 1 x 63.55 = 63.55 gS = 1 x 32.06 = 32.06 gO = 4 x 16.00 = 64.00 g
MM = 159.61 g
H = 2 x 1.01 = 2.02 gO = 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 g249.7 g
X 100 = 63.92 %
3. Grams anhydrous CuSO4
0.6392 x 145 = 92.7 g
Percent CompositionA 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 hydratex 100 = 14 %
Percent CompositionA 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 hydratex 100 = 29 %
Percent CompositionA 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 hydratex 100 = 22 %
3. Amount of water
0.22 x 18.02 = 4.0
Percent CompositionA 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 hydratex 100 = 28 %
3. Amount of water
0.28 x 18.02 = 5.0
Empirical and Molecular Formulas
CH2OCH3OOCH = C2H4O2
CH3OCH3O
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
EMPIRICAL FORMULA
Assume 100g sample Calculate mole ratio
Use Atomic Masses
Mass % of
elements
Grams of
each
element
Moles of
each
element
Empirical
Formula
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 H12 O66 6 6
C H2 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 mole24.3 g
N – 82.88 g * ( 1 mole ) = 5.92 mole14.01 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 = 3.00N – 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.
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.
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 = 92.0 g/mol = 2.00
Emp. Formula Mass 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 mol1.01 g
mole O – 229.85 g * ( 1 mole ) = 14.37 mol16.00 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 – 3 x 2 = 6
O – 1 x 2 = 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
• I have 5 eggs. How many cookies can I make?
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
5 eggs 5 doz.
2 eggs= 12.5 dozen cookies
Ratio of eggs to cookies
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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 MgOCourtesy 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 - moles moles
– Molar mass - moles grams
– Molarity - moles liters soln
– Molar volume - moles liters gas
Core step in all stoichiometry problems!!
– Mole ratio - moles moles
4. Check answer.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
1 mol of a gas=22.4 Lat STP
Molar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Molar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASS
IN
GRAMS
MOLES
NUMBER
OF
PARTICLES
LITERS
OF
SOLUTION
Molar Volume(22.4 L/mol)
LITERS
OF GAS
AT STP
Molarity (mol/L)
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
