# Molar Mass and Percent Composition

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```							Percent Composition
PURPOSE and PROCEDURE
I.   To find out the percentage (by mass) of a
particular element in a chemical compound.
II. Calculate the molar mass for the compound.

III. Divide the mass of the element you want to
find the percentage of by the molar mass of
the compound.
IV. Write the answer you get from step B in
percentage form.
V. Round your percentage to the tenth.
EXAMPLES
I.   What percent of Cu2O is O?

2 x Cu = 2 x 63.5 = 127.0 g
1 x O = 1 x 16.0 = 16.0 g

molar mass = 143.0 grams/mol

% O = Part ÷ Total = 16.0 ÷ 143 = .1118 x 100% =
11.2%
Example
I.   What percent of Al2O3 is Al?

2 x Al = 2 x 27.0 = 54.0 g
3 x O = 3 x 16.0 = 48.0 g

molar mass = 102.0 grams/mole

% Al = Part ÷ Total = 54 ÷ 102
= .5294 x 100% = 52.9%
Practice - Find the percent composition
   What percent of CaCl2 is Cl?
   Round to the nearest tenth of a percent.
Solution
   CaCl2
   1 x Ca = 1 x 40.1
   2 x Cl = 2 x 35.5
   Molar mass = 111.1 g
   % of Cl
   71.0/111.1 = 63.906%
   Round to the tenth.
Practice – Find the percent composition
   What percent of MgO is O?
   Round to the nearest tenth of a percent.
Solution
   MgO
   1 x Mg = 1 x 24.3
   1 x O = 1 x 16.0
   Molar mass = 40.3 g
   % of O
   16.0/40.3 = 39.702%
   Round to the tenth.
Empirical and Molecular
Formulas
Empirical Formula
   Definition
   Simplest formula
   Lowest whole-number ratio of atoms in
compound.
Empirical Formula
Ionic formulas are always in the empirical
form
Molecular formulas maybe
   Examples: H2O vs. H2O2
   H2O is empirical
   H2O2 is NOT EMPIRICAL
   The lowest ratio is HO (empirical formula)
Molecular Formula
   Definition - actual formula
   Is either the same as the empirical or a
whole number multiple of it
   H2O and H2O2 are molecular formulas
Example: Determine molecular
formulas
   To determine molecular formula, use the empirical
formula and compare to the known/given molar
mass.
   Given
   empirical formula is HO
   molar mass of the molecular formula is 34 grams/mole
   Solution: mass of HO is 17
    34/17 = 2 (multiple of 2)
   Therefore molecular formula is
             H2O2
INTERPRETING FORMULAS
   IF
    CO2 has one carbon atom and two oxygen
atoms
   THEN ASSUME
    CO2 has one mole carbon atoms and two
moles oxygen atoms
   A formula shows the mole ratios of the
elements in a compound.
    Use assumption to get molar mass in grams
The empirical formula for an ionic
compound
A)   Is the same as the simplest formula for
the ionic compound
B)   Shows the highest ratio of atoms to form
that compound
C)   Both A and B
D)   None of the above
The empirical formula for a
molecular compound
A)   Is the same as the molecular formula for
that compound
B)   Shows the lowest ratio of atoms to form
that compound
C)   Both A and B
D)   None of the above
If the molar mass of a compound is 60 g/mol and has an
empirical formula of NO, what is the molecular
formula?

A)   NO
B)   N 2O
C)   NO2
D)   N2O2
E)   None of the above
Molar Conversions
Practice problems
Mole Conversions
   The concept of the mole is useful because the
mole can be used to convert different units
   Volume
   Mass
   Representative Particles (atoms, molecules,
formula units)
Volume
   Gases will behave differently at different
temperatures and pressures
   To make comparisons, we use a Standard
Temperature and Pressure (STP) conditions
   101.3 kPa
   273 K (0 °C)
   One mole of a gas will always occupy 22.4 L
(molar volume of a gas)
Mass
   Mole to mass conversions are done using the
molar mass
Representative particles
   Gram atomic mass – convert moles to atoms
(element)
   Gram molecular mass – convert moles to a
molecules (molecular compound, type 3, nonmetal
and nonmetal)
   Gram formula mass – convert moles to formula units
(an ionic compound, type 1, 2 – metal and nonmetal)
   Atoms, molecules and formula units are all specific
examples of representative particles
Conversions
   Mass to mole (use molar mass)
   Volume to mole (use 22.4 L)
   Representative particles to mole (use
Calculate the molar mass
 Calculate the Molar mass of the
compound Potassium Chlorate.
 Round to the nearest tenth.
Solution
   Find molar mass
   KClO3
   1 x 39.1 g
   1 x 35.5 g
   3 x 16.0 g

   Total = 122.6 g/mol
Mass to moles
 How many moles are present in a
24.7 gram sample of Potassium
Chlorate?
 Remember sig figs
Solution
   Mass to mole
   Round to 3 sig figs

24.7 g KClO3     1 mole KClO3
0.20146 mole
1           122.6 g KClO3      KClO3
Molecules to mass
 What  is the mass of a sample
containing 13.20 x 10 23
molecules of Barium chloride?
 Remember sig figs.
Solution
    Molecules to mass
    BaCl2
    1 x 137.3
    2 x 35.5
    Molar mass = 208.3 g/mol
    Round to 4 sig figs

13.20 x 1023 1 mole of      208.3 g of
molecules      BaCl2         BaCl2      456.737 g
of BaCl2                                 of BaCl2
1       6.02 x 1023    1 mole of
molecules        BaCl2
of BaCl2
Mass to volume
 Calculate the number of liters in
a 3.95 gram sample of Nitrogen
dioxide.
 Remember sig figs.
Solution
    Mass to volume
    NO2
    Molar mass = 46.0 g/mol
    1 x 14.0 2 x 16.0
    Round to 3 sig figs
3.95 g NO2    1 mole of    22.4 L of
NO2          NO2
1.9234 L of
NO2
1         46.0 g of   1 mole of
NO2         NO2

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