Chapter 6
Chemical Quantities
Homework
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Assigned Problems (odd numbers only)
“Questions and Problems” 6.1 to 6.53
(begins on page 168)
“Additional Questions and Problems” 6.59 to
6.77 (page 190-192)
“Challenge Questions” 6.79, 6.81 (page 192)
Counting Particles By Weighing


If a person requests 500 quarter inch
hexagonal nuts for purchase
How would you count 500 hex nuts?
a) You could count the hex nuts individually
b) Or, count the number of hex nuts by weight

First, find the average weight by weighing out 10 hex nuts
and obtaining the total weight (10 hex nuts weigh 105 g)
105 g
10.5 g / hex nut
10 hex nuts
Average
Weight
Counting Particles By Weighing


What size (wt.) will
contain 500 hex nuts?
Calculate what weight
will contain 500 hex
nuts
10.5 g
500 hex nuts 
 5250 g
hex nut

So, weigh out 5.25 kg
of the hex nuts
Using Atomic Mass to Count Atoms

Can do the exact same thing with atoms
 Too small to conveniently count
 Atoms of the same element don’t always have
exactly the same mass (isotopes) so use an
Average Mass
 Grams are too large to use to measure an atom so
use Atomic Mass Units (amu)
1 amu  1.66  10
24
g
Atomic Mass and Formula Mass


To calculate the mass
of a sample of atoms
 Each element exists
as a mixture of
isotopes
 Use a “weighted
average” for the
atomic mass
Number on the bottom
of each square in the
periodic table is the
average weight of the
element (in amu)
Atomic Mass and Formula Mass
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Atomic masses are
determined on a
relative scale
The standard scale
references the
carbon-12 isotope =
12.000 amu
All other atomic
masses are
determined relative to
carbon-12
Atomic Mass and Formula Mass
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Using Atomic Mass to Count Atoms
Calculating the number of atoms in a specific mass
If you have a sample of an element, can calculate the
number of atoms in that sample
From the atomic mass per one atom a conversion
factor can be made
For example: One nitrogen atom has an atomic mass
of 14.01 amu
1 N atom  14.01 amu
14.01 amu
1 N atom
and
1 N atom
14.01 amu

Calculating The Number of Atoms in a
Specific Mass
You have a 1.00 g sample of lead. How many
atoms of lead are present?
1.00 g Pb
1 atom Pb = 207.2 amu
1.66 10-24 g = 1 amu
1.00 g Pb
1 amu
1.66 x 1024 g
1 atom Pb
207.2 amu

2.9 10 atoms Pb
21
Calculating Mass Example

Calculate the mass (in amu) of 1.0 х 104
carbon atoms
4
1) Given: 1.0  10 atoms C
3) CF 1atom C = 12.01 amu
2) Plan: Convert from atoms to amu
12.01 amu
1 C atom
and
1 C atom
12.01 amu
4) Set Up Problem
12.01 amu
1.0 10 C atom 
 1.2 10 5 amu
1 C atom
4
Formula Mass
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The sum of atomic masses of all atoms in
its formula
Important role in nearly all chemical
calculations
Can be calculated for compounds and
diatomic elements
Calculating Formula Mass
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Calculate the formula mass of calcium chloride
Write the formula from the name given
Ca2+ (from group II) and Cl- (from group VII)
Formula is CaCl2 due to charge balance
Formula mass: Sum of the atomic masses of
atoms in the formula (1 Ca atom + 2 Cl atoms)
1 Ca atom 
2 Cl atom 
40.08 amu
= 40.08 amu
1 Ca atom
35.45 amu
= 70.90 amu
1 Cl atom
110.98 amu
Formula mass of CaCl2
Counting Large Quantities
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Many chemical calculations require counting atoms
and molecules
It is difficult to do chemical calculations in terms of
atoms or formula units
Since atoms are so small, extremely large numbers
are needed in calculations
Need to use a special counting unit just as used for
other items
 A ream of paper
 One dozen donuts
 A pair of shoes
The Mole
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It is more convenient to use a special counting unit
for such large quantities of particles
Mole: A unit that contains 6.022 х 1023 objects
It is used due to the extremely small size of atoms,
molecules, and ions
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6.022x1023 particles in 1 mole
Called Avogadro’s Number
Periodic Table
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The average atomic mass in amu (one atom)
The weight of 1 mole of the element in grams
Avogadro’s number provides the connecting
relationship between molar masses and atomic
masses
Calculating the Number of
Molecules in a Mole

How many molecules of bromine are present in 0.045
mole of bromine gas?
Given: 0.045 mol Br2
Equality:
Avogadro’s number
Need: molecules of Br2
1 mol Br2  6.022  10 23 molecules Br2
6.022 10 23 molecules Br2
mol Br2
and
Conversion factors:
mol Br2
6.022 1023 molecules Br2
Set Up Problem:
0.045 mol Br2
6.022 10 23 molecules Br2


mol Br2
2.7  10 22 molecules Br2
Subscripts State Moles of Elements
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The subscripts in a chemical formula indicate the
number of atoms of each element present in a
compound
The subscripts in a chemical formula can also
indicate the number of moles of atoms of each
element present in one mole of a compound
i.e. In one molecule of glucose (C6H12O6) there
are 6 atoms of carbon, 12 atoms of hydrogen,
and 6 atoms of oxygen
Calculating the Moles of an Element
in a Compound

How many moles of carbon atoms are present in
1.85 moles of glucose?
Plan: moles of glucose
subscript
moles of C atoms
Equality: (One) mol C6H12O6 = 6 mols C atoms
Conversion Factors:
Set Up Problem:
mol C6H12 O 6
6 moles C atoms
and
mol C6H12 O 6
6 moles C atoms
1.85 mol C 6H12 O 6

6 mols C atoms
 11.1 mol C atoms
mol C 6H12 O 6
Molar Mass
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The atomic mass of a carbon-12 atom is 12.00
amu
The atomic mass of one mole of carbon-12
atoms 12.00 g
One mole of any element is the amount of
atoms (molecules or ions) that is equal to its
atomic mass (in grams)
This mass contains 6.022 х 1023 particles of
that element
Use the periodic table to obtain the molar
mass of any element
Molar Mass

When the number of grams (weighed out) of a
substance equals the formula mass of that
substance, Avogadro’s number of molecules of
that substance are present
Molar Mass of a Compound
Calculate the molar mass of iron (II) sulfate

Formula is FeSO4
1) Calculate the molar mass of each element
2) Each element is multiplied by its respective
subscript: (number of moles of each element)

Moles of
Compound
Formula
Subscript
Moles of Element
in Compound
3) The molar mass is calculated by the sum of
the molar masses of each element
Molar Mass of a Compound
1) Formula is FeSO4: The molar masses of iron, sulfur, and oxygen are
55.85 g Fe
mol Fe
32.00 g S
mol S
16.00 g O
mol O
2) Multiply each molar mass by its subscript
1 mol Fe 
55.85 g Fe
 55.85 g Fe
mol Fe
4 mol O 
16.00 g O
 64.00 g O
mol O
1 mol S 
32.00 g S
 32.00 g S
mol S
3) Find the molar mass of the compound by adding the mass of each element
55.85 g  32.00 g  64.00 g  151.85 g
Calculations Using Molar Mass
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The three quantities most often calculated
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Number of particles
Number of moles
Number of grams
Using molar mass as a conversion factor is
one of the most useful in chemistry

Can be used for g to mole and mole to g
conversions
Relationship between Moles,
Molar Mass and Avogadro’s number
Moles of
substance
Avogadro’s
Number
Particles of
substance
Molar Mass
Moles of
substance
Grams of
substance
Converting Mass of a Compound to Moles

International Foods Coffee contains 3 mg of sodium
chloride per cup of coffee. How many moles of sodium
chloride are in each cup of coffee?
3 mg NaCl
3 mg NaCl
moles of NaCl
1g NaCl
= 0.003 g NaCl
1000 mg NaCl
MM NaCl  1(Na)  1(Cl)  1(22.99)  1(35.45)  58.44
Equality: 1 mol NaCl = 58.44 g
g
NaCl
mol
58.44 g NaCl
mol NaCl
and
mol NaCl
58.44 g NaCl
0.003 g NaCl 1mol NaCl
 5.13 10 -5 mol NaCl
58.44 g NaCl
Converting Grams to Particles

Ethylene glycol (antifreeze) has the formula C2H6O2. How
many molecules are present in a 3.86 × 10-20 g sample?
Plan: convert g
Molar
mass
moles
Avog
Number
Equality 1:1mol C2H6O2 62.05 g C2H6O2
molecules of ethylene glycol
Conversion 62.05 g C2H6O2
mol C2H6O2
and
mol C2H6O 2
62.05 g C2H6O 2
Factor 1
Equality 2: 1mol C2H6O2 6.022 1023 molecules
Conversion 6.022 10 23 molecules
1mol C 2H6 O 2
or
Factor 2
1mol C 2H6 O 2
6.022  10 23 molecules
3.86 10
 20
mol C 2H6 O 2
6.022  10 23 molecules
g C 2H 6 O 2 

 375 molecules
62.05 g C 2H6 O 2
mol C 2H6 O 2
Percent Composition
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Sometimes it’s useful to know the
composition of a compound in terms of
what percentage of the total is each
element
Percent
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“Parts per 100”
The number of specific items per a group of
100 items
50% of $100 is $50 (50 items/100 total items)
Percent Example
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You have 4 oranges and 5 apples. What
percent of the total is oranges?
4 oranges
 100%  44% oranges
9 total

In “parts per 100”
44 oranges
 100%  44% oranges
100 total
Percent Composition
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It is the percent by mass of each element in a
compound
Can be determined
 By its chemical formula
 Molar masses of the elements that
compose the compound
The percent of each element contributes to
the mass of the compound
mass of each element
mass percent of each 
100%
element in a compound molar mass of the compound
Calculating Percent Composition Example

What is the percent composition of each
element in NH4OH?
Determine the contribution of each
element
14.01g
N:
 100%  39.97% N
35.05 g
H:
5.04 g
 100%  14.38% H
35.05 g
16.00 g
O:
 100%  45.65% O
35.05 g
N : 1  14.01g = 14.01g
H : 5  1.0078 g = 5.04 g
O : 1  16.00 g = 16.00 g
Molar mass = 35.05 g
Empirical Formulas
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The simplest ratio of elements in a
compound
It uses the smallest possible whole
number ratio of atoms present in a
formula unit of a compound
If the percent composition is known, an
empirical formula can be calculated
Empirical Formulas
To Determine the empirical formula:
1) Calculate the moles of each element

 Use molar mass (atomic mass)
2) Calculate the ratios of the elements to
each other
3) Find the lowest whole number ratio
 Divide each number of moles by the
smallest number of moles present
Empirical Formula: Converting Decimal Numbers to
Whole Numbers
 The
subscripts in a formula are
expressed as whole numbers, not as
decimals
 The resulting numbers from a
calculation represent each element’s
subscript
 If the number(s) are NOT whole
numbers, multiply each number by
the same small integer (2, 3, 4, 5, or
6) until a whole number is obtained
Relating Empirical and
Molecular Formulas

n represents a whole number multiplier from 1
to as large as necessary
molar mass ( g / mol )
n
empirical formula mass ( g / mol )


Calculate the empirical formula and the mass of
the empirical formula
Divide the given molecular mass by the
calculated empirical mass
 Answer is a whole number multiplier
Relating Empirical and
Molecular Formulas

Multiply each subscript in the empirical formula
by the whole number multiplier to get the
molecular formula
Calculate Empirical Formula from
Percent Composition
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Lactic acid has a molar mass of 90.08 g
and has this percent composition:
40.0% C, 6.71% H, 53.3% O
What is the empirical and molecular
formula of lactic acid?
Assume a 100.0 g sample size

Convert percent numbers to grams
Calculate Empirical Formula from
Percent Composition
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Convert mass of each element to moles
Divide each mole quantity by the smallest
number of moles
40.0 g C 
6.71 g H 
mol C
 3.33 mol C
12.0 g C
mol H
 6.66 mol H
1.008 g H
53.3 g O 
mol O
 3.33 mol O
16.00 g O
The ratio of C to H to O is 1 to 2 to 1
3.33
 1.00
3.33
6.66
H:
 2.00
3.33
C:
O:
3.33
 1.00
3.33
Empirical
formula is
CH2O
Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol
Determination of the
Molecular Formula


Obtain the value of n (whole number multiplier)
Multiply the empirical formula by the multiplier
molar mass ( g / mol )
n
empirical formula mass ( g / mol )
90.08 g / mol
3
30.03 g / mol
Molecular formula = n х empirical formula
Molecular formula = 3 (CH2O)
C3H6O3
Formulas for Compounds
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Empirical Formula
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Smallest possible set of subscript numbers
Smallest whole number ratio
All ionic compounds are given as empirical formulas
Molecular Formulas
The actual formulas of molecules
It shows all of the atoms present in a molecule
It may be the same as the EF or a wholenumber multiple of its EF
Molecular formula = n х Empirical formula
end