Chapter 10
“Chemical Quantities”
Yes, you will need a
calculator for this chapter!
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Section 10.1
The Mole: A Measurement of Matter

2
OBJECTIVES:
– Describe methods of measuring the
amount of something.
– Define Avogadro’s number as it relates
to a mole of a substance.
– Distinguish between the atomic mass of
an element and its molar mass.
– Describe how the mass of a mole of a
compound is calculated.
How do we measure items?
3

We measure mass in grams.

We measure volume in liters.

We count pieces in MOLES.
What is the mole?
4
We’re not talking about this
kind of mole!
Moles (is abbreviated: mol)
 It is an amount, defined as the
number of carbon atoms in exactly
12 grams of carbon-12.
1
5
mole = 6.02 x
23
10
of the
representative particles.
 Treat it like a very large dozen
 6.02 x 1023 is called:
Avogadro’s number.
Similar Words for an amount




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Pair: 1 pair of shoelaces
= 2 shoelaces
Dozen: 1 dozen oranges
= 12 oranges
Gross: 1 gross of pencils
= 144 pencils
Ream: 1 ream of paper
= 500 sheets of paper
What are Representative Particles?

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The smallest pieces of a
substance:
1) For a molecular compound: it is
the molecule.
2) For an ionic compound: it is the
formula unit (made of ions).
3) For an element: it is the atom.
» Remember the 7 diatomic
elements? (made of molecules)
Types of questions
How many oxygen atoms in the
following?
3 atoms of oxygen
CaCO3
Al2(SO4)3 12 (3 x 4) atoms of oxygen
 How many ions in the following?
3 total ions (1 Ca ion and 2 Cl ions)
CaCl2
2 total ions (1 Na ion and 1 OH ion)
NaOH
Al2(SO4)3 5 total ions (2 Al + 3 SO ions)

2+
1-
1+
1-
3+
8
4
2-
Measuring Moles
 Remember relative atomic mass?
- The amu was one twelfth the
mass of a carbon-12 atom.
 Since the mole is the number of
atoms in 12 grams of carbon-12,
 the decimal number on the
periodic table is also the mass of 1
mole of those atoms in grams.
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Gram Atomic Mass (gam)




10
Equals the mass of 1 mole of an
element in grams (from periodic table)
12.01 grams of C has the same
number of pieces as 1.01 grams of H
and 55.85 grams of Fe.
We can write this as:
12.01 g C = 1 mole C
(this is also the molar mass)
(**Remember we can count things by weighing them.)
Find the Gram Atomic Mass (gam) of
the following:

nitrogen
14.01 g

aluminum
26.98 g

zinc
65.39 g
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What about compounds?
12
In 1 mole of H2O molecules there are
2 moles of H atoms and 1 mole of O
atoms (think of a compound as a molar ratio)
 To find the mass of one mole of a
compound:
–Determine the number of moles of
the elements present
–Multiply the number times their
mass (from the periodic table)
–Add them up for the total mass
Calculating Gram Formula Mass (gfm)
Calculate the formula mass of
magnesium carbonate, MgCO3.
24.31 g
13
+
12.01 g
+ 3 x (16.00 g)= 84.32 g
Thus, 84.32 grams is the formula
mass for MgCO3.
GFM Practice Problem:

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What is the mass of one mole of CH4?
1 mole of C = 12.01 g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = 12.01 + 4.04 = 16.05g
Section 10.2
Mole-Mass and Mole-Volume Relationships

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OBJECTIVES:
– Describe how to convert the mass of
a substance to the number of moles
of a substance, and moles to mass.
– Identify the volume of a quantity of
gas at STP.
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Since Molar Mass is…
 The number of grams in 1
mole of atoms, ions, or
molecules
 We can make conversion
factors from these.
- To change between grams of
a compound and moles of a
compound.
For example

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How many moles is 5.69 g of NaOH?
For example

How many moles is 5.69 g of NaOH?

5.69 g

need to change grams to moles
for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g

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1 mole 
 = 0.142 mol NaOH
40.00 g 
Practice Problems:


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How much would 2.34 moles of
carbon weigh? 28.1 grams C
How many moles of magnesium is
24.31 g of Mg? 1.000 mol Mg
The Mole-Volume Relationship
 Many of the chemicals we deal with
are: gases.
 Two things effect the volume of a gas:
a) Temperature and b) Pressure
**We need to compare all gases at the
same temperature and pressure.
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Standard Temperature and Pressure
 abbreviated
STP
 0ºC (273K) and 1.0 atm pressure
 At STP 1 mole of gas occupies
22.4 L= molar volume
1 mole = 22.4 L of any gas at STP
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Practice Problems:
 What
is the volume of 4.59
mole of CO2 gas at STP? = 103 L CO
 How many moles is 5.67 L of
O2 at STP? = 0.253 mol O2
 What is the volume of 8.8 g of
CH4 gas at STP?
2
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Summary:
 These
four items are all equal:
a) 1 mole
b) molar mass (in grams/mol)
c) 6.02 x 1023 particles (atoms, molecules, or formula
units)
d) 22.4 L of a gas at STP
**Thus, we can make conversion factors
from them.
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Avog. # Practice problems:
How many molecules of CO2 are
in 4.6 moles of CO2? 2.8 x 10 m.c. CO
 How many moles of water is
5.87 x 1022 molecules?
0.0975 mol H O (or 9.75 x 10 )
 How many atoms of carbon are in
1.230 moles of Carbon? 7.405 x 10 atoms C
 How many moles is 7.78 x 1024
formula units of MgCl2? 12.9 moles MgCl

24
2
-2
2
23
2
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Mixed Practice Problems:

How many atoms of lithium is 1.0 g
of Li? 8.7 x 10 atoms Li
How much would 3.45 x 1022
atoms of U weigh? 13.6 g U
What is the volume of 10.0 g of
CH4 gas at STP? 14.0 L CH4
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

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Section 10.3
Percent Composition and Chemical Formulas

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OBJECTIVES:
– Describe how to calculate the percent by
mass of an element in a compound.
– Interpret an empirical formula.
– Distinguish between empirical and
molecular formulas.
Percentage Composition
 the
percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass compound
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Calculating Percent Composition
of a Compound
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 Like all percent problems:
part
whole x 100 % = percent
1) Find the mass of each of the
components (the elements
masses from the periodic table ),
2) Next, divide by the total mass of
the compound; then x 100
Percentage Composition
 Find
the % composition of Cu2S.
%Cu =
%S =
29
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 =
79.85% Cu
 100 =
20.15% S
Percentage Composition
 Find
the mass percentage of
water in calcium chloride
dihydrate, CaCl2•2H2O?
%H2O =
30
36.04 g
147.02 g
 100 = 24.51%
H2O
Empirical Formula (EF)
 Lowest
whole number ratio of
atoms in a compound
C2H6
reduce subscripts
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CH3
Empirical Formula (EF)
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 Just find the lowest whole number ratio
C6H12O6 = CH2O
CH4N = this is already the lowest ratio.
 A formula is not just the ratio of atoms,
it is also the ratio of moles.
 In 1 mole of CO2 there is 1 mole of
carbon and 2 moles of oxygen.
 In one molecule of CO2 there is 1 atom
of C and 2 atoms of O.
Formulas (continued)
Formulas for ionic compounds are
ALWAYS empirical (the lowest whole
number ratio = cannot be reduced).
Examples:
NaCl
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MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds
MIGHT be empirical (lowest whole
number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
H2O
CH2O
C12H22O11
(Correct formula)
Empirical:
(Lowest whole
number ratio)
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Calculating EF
We can get a ratio from the percent composition.
1.) Assume you have 100 g.

-the percentage becomes grams (75.1% = 75.1 grams)
2.) Convert grams to moles.
3.) Find lowest whole number ratio by dividing by the
smallest # of moles.
**4.) If not a whole #, use a multiplier.
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Example
 Calculate
the empirical formula of a
compound composed of 38.67 % C,
16.22 % H, and 45.11 %N.
 Assume 100 g so:
 38.67 g C x 1mol C
= 3.220 mole C
12.01 gC
 16.22 g H x 1mol H
= 16.06 mole H
1.01 gH
 45.11 g N x 1mol N = 3.220 mole N
14.01 gN
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Now divide each value by the smallest value
Example
 3.220
mol C = 1 mol C
 16.06 mol H = 5 mol H
 3.220 mol N = 1 mol N
3.220
EF = C1H5N1
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= CH5N
3.220
3.220
Empirical Formula (EF)
 Find
the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g
74.1 g 1 mol
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= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts
whole numbers  multiply by 2
N2O5
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EF Practice
Find the empirical formula for a sample
of 43.64% P and 56.36% O.
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Molecular Formula (MF)
 “True
Formula” - the actual
number of atoms in a compound
empirical
formula
CH3
?
molecular
formula
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C2H6
Empirical to Molecular
 Since
the empirical formula is the
lowest ratio, the actual molecule
would weigh more.
By a whole number multiple.
 **Divide the actual molar mass by
the empirical formula mass.
MF mass
n
EF mass
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EF n  MF
Molecular Formula
 The
empirical formula for
ethylene is CH2. Find the
molecular formula if the
molecular mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
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=2
(CH2)2  C2H4
Practice Problem:
 Caffeine
(EF= C4H5N2O) has a molecular mass
of 194 g. What is its molecular formula?
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Formulas
 EF
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= lowest whole number ratio of
elements in a compound.
 The molecular formula (MF) = the actual
ratio of elements in a compound.
 The two can be the same.
 CH2 is an empirical formula
 C2H4 is a molecular formula
 C3H6 is a molecular formula
 H2O is both empirical & molecular
Density of a gas (10.2 revisited)
 D = m / V (density = mass/volume)
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- for a gas the units will be: g / L
 We can determine the density of any
gas at STP if we know its formula.
 To find the density we need: 1) mass
and 2) volume.
 If you assume you have 1 mole, then
the mass is the molar mass (from periodic table)
 And, at STP the volume is 22.4 L.
Practice Problems (D=m/V)
 Find
the density of CO2 at STP.
D = 44.01g/22.4L = 1.96 g/L
 Find
the density of CH4 at
STP.
D = 16.05g/22.4L = 0.717g/L
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Note page 313 – Gas Chromatography
used for chemical analysis
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