Chemical
Proportions in
Compounds
Chapter 3
Chemical Composition
• The chemical make-up or composition
of a compound can be described in 2
ways:
– Formula – which will tell the ratio of each
type of atom in the compound, ex. H2O
– Percent composition by mass - which gives
the percent by mass of each element in the
compound.
Law of Definite Proportions
• The Law of Definite Proportions
states that the elements in a given
chemical compound are always present
in the same proportion by mass.
• Water, no matter the source, is always
H2O, which as a mass ratio of 2.02 g :
16.00 g.
Law of Multiple Proportions
• The Law of Multiple Proportions states that
when 2 elements combine, 2 or more
different compounds may form
• Examples:
– C + O2  CO
– C + O2  CO2
– H2 + O2  H2O
– H2 + O2  H2O2
N2 + O2  NO
N2 + O2  N2O4
• Each compound has its own unique
properties.
• What compound will form, would
depend on the experimental conditions
(temperature, pressure, catalysts,
activation energy, etc.).
Mass Percent
• To find the mass percent of an element
in a compound divide that mass by the
mass of the compound and multiply by
100%. (You do the same thing when
finding your percentage on a test.)
• Example:
– If a sample compound that has 2 different
element, A and B, has a mass of 100.0 g, and
has 20.0 g of element A in the sample. Then
the percent by mass of element A in the
compound would be calculated using the
formula:
– Mass of element A =
• By subtraction you can calculate the
percent of element B.
– Element B = 100% - 20% = 80%
Or just use the formula again solving for the
mass of element B.
Percentage Composition
• We have used the mole to describe the
composition of compounds.
• One mole of carbon dioxide contains 1
mole of carbon to 2 moles of oxygen.
• The percentage composition of a
compound refers to the relative mass of
each element in the compound.
• You need to find the mass percent of
every element in the compound.
• Example:
– Vanillin – C8H8O3 contains 63.1% carbon,
5.3% hydrogen, and 31.6% oxygen
– Why don’t the subscripts and percentages
seem to agree?
– Remember it is based on mass. Hydrogen
has by far the smallest mass.
• Percentage composition can be found
experimentally and used to help identify
unknown compounds.
• Do Sample Problem on Page 81
• Do Practice Problems on Page 82 #s 14
• Do Thought Lab on Page 83
Calculating Percentage
Composition from a Chemical
Formula
• In the previous practice problems you
used mass data to find the percentage
composition.
• This method is used to identify unknown
substances in a lab.
• Another use of percentage composition is
finding the exact amount of an element in
a known substance for extracting
purposes.
• Example – mercury is often found in
nature as mercury (II) sulfide.
• Knowing the percentage composition of
HgS, enables a metallurgist to predict
the amount of mercury that can be
extracted.
• The sample size is not important to
determine the percentage composition
by mass of a homogeneous sample.
• According to the law of definite
proportions, there is a fixed proportion
of each element in the compound, no
matter the size.
• This means you can chose a
convenient sample size.
• Example:
• Using a sample of mercury (II) sulfide,
calculate the percentage composition of each
element. Assume a sample size of 1 mole of
HgS.
• Find the mass percent of each element in the
compound.
• 86.2% Hg and 13.8% S
• Problems with more than 2 elements can get
a little more complex.
• Do Sample Problem on Page 84
• Do Practice Problems on Page 85 #s 58
• Do Section Review on Page 86 #s 1-6
The Empirical Formula of a
Compound
Section 3.2
• Atoms combine with each other in
simple whole number ratios.
• The empirical formula is the simplest
formula of the compound. It is the
lowest whole number ratio.
• The molecular formula is the actual
formula. It says exactly how many
atoms are in the compound.
• See Table 3.1 on Page 87
• The molecular formula is a multiple of
the empirical formula.
• You can determine a compound’s
empirical formula from its percentage
composition.
• If we know the mass percentages of the
elements in the compound we can
determine the simplest formula.
• Example: Calcium and fluorine react to
form a compound called fluorite.
1. Using the molar mass you can
determine the percentage composition
of fluorite.
51.3% Ca and 48.7% F
2. Always assume a 100 g sample (it
makes the calculation to grams
simpler).
3. Convert the grams to moles (use
molar mass).
4. Look at the ratio of moles of each
element that reacted to determine the
mole ratio. (Divide the largest by the
smallest.)
• 100 g × 0.531 = 53.1 g Ca
• 100 g × 0.487 = 48.7 g F
• 53.1 g × 1 mol Ca = 1.28 mol Ca
40.08 g
• 48.7 g × 1 mol F = 2.56 mol F
19.00 g
• 2.56 mol F = 2
1.28 mol Ca
1
• Or 2 F:1 Ca or CaF2
• The simplest ratio of atoms or moles is
1 Ca for 2 F. Therefore the empirical
formula for fluorite is CaF2.
• Do Sample Problem on Page 88
• Do Practice Problems on Page 89 #s 912
Helpful Hints…
• Do not round until the end
• Use the maximum number of digits you
can.
• Round close numbers to the nearest
whole number. 0.95 to 0.99 round up.
0.01 to 0.04 round down. 0.45 to 0.55
round to 0.5.
• By dividing by the smallest mole
amount, one atom will always have the
subscript 1.
• Not all compounds contain ratios that
include 1 atom of something, e.x Fe2O3.
• You may need to multiply the ratios to
get the numbers to turn into whole
numbers.
• See Table 3.2 on Page 90
• Example:
• If you get to the end of your calculations
and end up with something like this…
2.94 mol Y = 1.5 mol Y × 2 = 3 mol Y
1.96 mol X
1 mol X
2
2 mol X
So the empirical formula would be X2Y3
• Do Sample Problem on Page 90
• Do Practice Problems on Page 91 #s
13-16
• Do Section Review on Page 94
The Molecular Formula of a
Compound
Section 3.3
• Read Page 95
Determining a Molecular
Formula
• To take an empirical formula and
calculate it into the correct molecular
formula, you need to know the correct
multiplier (n).
• Calculate the “molar mass” of the
empirical formula. Take the actual
molar mass of the molecule and divide
by the “molar mass”.
• This whole number ratio is the multiplier
(n).
• Multiply the EF by this number to obtain
the MF.
• The actual molar mass must be given.
• Do Sample Problem on Page 97
• Do Practice Problems on Page 97 #s
17-19
• Do Section Review on Page 98
Finding Empirical and Molecular
Formulas by Experiment
• One method of finding the percentage
composition and empirical formula is by
taking a known mass of an unknown
substance and reacting it with oxygen to
find the mass of the product.
• This works for simple compounds that
react in predictable ways.
The Carbon-Hydrogen
Combustion Analyzer
• This process is used for chemicals that
are mainly hydrogen, carbon, and
oxygen.
• A combustion reaction is the burning of
hydrocarbons (CH) chains in the
presence of oxygen to produce carbon
dioxide and water.
• The carbon-hydrogen combustion
analyzer burns compounds in a stream
of pure oxygen to yield carbon dioxide
and water.
• If you find the mass of the carbon
dioxide and water separately then you
can determine the mass percent of
carbon and hydrogen in the compound.
• How it works – see Figure 3.7 on Page
99.
• A sample is placed in a furnace and it is
heated as pure oxygen is streamed in.
• The products of the reaction leave the
furnace and pass through filters.
• The water vapor is collected by passing
through a tube of magnesium
perchlorate (Mg(ClO4)2)..
• The Mg(ClO4)2 absorbs all the water.
• The difference in the mass of the tube,
before and after the reaction, is the
mass of the water.
• All the hydrogen in the sample was
converted to water.
• Because hydrogen is always 11.2% of
water, you can find the amount of
hydrogen from the sample.
• Take the percent composition and
multiply by the mass of the water and
you get the mass of the hydrogen that
was reacted.
• A second filter captures the carbon
dioxide. The same process of
multiplying the mass of CO2 by the
mass percent of carbon in CO2 tells you
how much carbon was in the sample.
• The carbon-hydrogen combustion
analyzer can be used if the sample
contains a third element.
• Anything not accounted for by the
hydrogen or carbon must be another
element.
• Finally, once you have the mass of each
element you can find the empirical
formula.
• Sample Problem on Page 100
• Practice Problems on Page 101 #s 21
and 22
Hydrated Ionic Compounds
• Hydrates are compounds that have
water molecules bonded to the formula
unit.
• The water molecules are actually part of
the crystal lattice structure.
• Hydrates have a specific number of
water molecules.
• Chemists usually know the ionic part but
may have to determine the number of
water molecules.
• Hydrates occur when ionic compounds
crystallize from a water solution with
water molecules incorporated into their
crystal structure.
• Some hydrates will liquefy when in
contact with water vapor.
• Hydrates are used as desiccants –
absorb moisture from the air.
• Hydrates are written as an ionic
compound weakly bonded to a number
of water molecules:
• Magnesium sulfate heptahydrate
MgSO4●7H2O
• Every formula unit of MgSO4 has seven
water molecules weakly bonded to it.
• The dot is used to write hydrate
formulas and means weakly bonded to
it and not as a multiplication sign.
• Several ionic compounds can also
occur as hydrates.
• The term anhydrous means not in
hydrate form.
• The molar mass of a hydrate includes
all water molecules.
• Therefore the mass of MgSO4●7H2O is
higher than that of the anhydrous
MgSO4.
• Calculations for percent mass and
empirical formula for hydrates are
similar to those already done.
• Do Sample Problem on Page 102
• Do Practice Problems on Page 103 #s
23 and 24.
• Do Section Review on Page 106 #s 1-3,
6, & 7
• Do Chapter 3 Review on Pages 107 109 #s 1 - 22