Unit 5
Counting Atoms
1, 2, 3, 4, …..
10, 11, ………
1000, 1001, 1002…………
Counting
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Suppose I asked you
to count all the
peanuts in this bag?
But what if I knew
that 100 peanuts had
a mass of 5 grams –
can you think of
another way to
estimate the number
of peanuts in the
sample?
Connecting Mass to Number of
Particles
One of the greatest challenges early
chemists faced was trying to find a way to
connect the mass of a substance to the
number of particles in the sample.
 Recall it was determined that particles
combined in fixed ratios by weight.

Connecting Mass to Number of
Particles

This led Dalton to the “atomic model” of
matter

Example: The mass ratio of oxygen to
hydrogen in water is 8:1
 This
does not tell us how many atoms of each
element are involved
 It could tell us this if we knew the relative mass of
each kind of atom
Relative Mass

To assign relative masses to elements it
is necessary to know that the samples
being compared have the same number
of particles
Relative Mass
Which has more balls in it: a bucket of
golf balls or an identical bucket of
baseballs?
If this is true in the macroscopic world,
why wouldn’t it be true in the submicroscopic one?
Relative Mass
Was iron more dense than aluminum
because iron had more particles per given
volume than aluminum or because iron’s
individual particles were more massive
than aluminum’s?
 Could it be some combination of both?

Relative Mass
The truth is, based on the experiments we
conducted earlier in the year, we couldn’t
say which was true.
 Dalton did not know what was true during
his time either.
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Since the mass of individual atoms could not
be determined, a system of atomic masses
had to be determined by comparison.
Relative Mass
To determine a system of masses by
comparison, one element would have to
be chosen as the basis of comparison for
all others
 Dalton chose hydrogen and assigned it a
mass of 1.

Relative Mass

To find the mass of another element like
oxygen:
Compare the masses of equal number of
oxygen and hydrogen atoms OR
 Find the combining masses of oxygen and
hydrogen in water

Avogadro’s Hypothesis

Remember..Avogadro assumed:
Equal volumes of gases have equal numbers
of molecules
 These molecules can be split during chemical
reactions
 That molecules of elemental gases could
contain more than a single atom

Avogadro’s Hypothesis
If we accept Avogadro’s Hypothesis, we
can compare the mass of various gases
and deduce the relative mass of the
molecules
 To do this, we pick a weighable amount of
the lightest element (how about 1.0 g?)
then use mass ratios to assign atomic
masses to the other elements

Implications

If two volumes of hydrogen combine with
one volume of oxygen gas, it is reasonable
to assume that two molecules of hydrogen
are reacting with each molecule of oxygen
Relative mass activity
vials containing several different items
 pretend that the items in the vials are
enlarged particles of various gases.
 Since the vials are the same size, we will
assume that each contains the same
number of particles.
 Your job is to determine the relative mass
of each kind of item in the vial, without
opening the vial to examine single items.

Comparison of Counting Methods
P vs. n Lab to Avogadro’s Hypothesis
Assumptions
Unit
Definition
Our Previous Way of
Counting Particles
Avogadro’s Way of
Counting Particles
Equal volumes of air in an
open syringe contain equal
numbers of particles (based on
density – equal vol – equal
mass, then equal # of particles)
Equal volumes of any gas
contain equal numbers of gas
molecules.
Comparing masses of gas with
the same volume (therefore,
count) gives us relative masses
of substances.
Volume of 1 syringe-ful at
constant density = 1 ‘puff’’
Actual count unknown;
must be measured
indirectly using volume
2 grams of hydrogen gas = 1
‘mole’ of gas molecules (mole =
‘lump’)
Actual count unknown;
must be measured
indirectly using volume and
mass.
The Mole

The word chosen to represent
the standard weighable amount
of stuff, the mole, comes from
the Latin “mole cula” or little
lump
Mole
A mole of any element is defined as:
the amount of the element that contains as
many atoms as there are in exactly 12 g of
the carbon-12 isotope.

Avogadro’s Number

A mole was found experimentally to be
equal to 6.022 X 1023 atoms of C-12,
which is called Avogadro’s Number (NA).
12 g Carbon-12 =
 1 mol Carbon-12 atoms =
 6.022 X 1023 Carbon-12 atoms

The MOLE (mol)
The molar mass of any substance is the
mass of one mole of that substance.
Molar mass is numerically equal to the atomic
mass of the substance.
 A mole of any substance contains Avogadro’s
number of units of that substance
 (6.022 X 1023 units).

Mass
2 ways to measure mass:
 Atomic Mass Units (amu)
 The atomic mass of an element is the
mass (in amus) as shown on the periodic
table for an atom.
 It is a average relative mass.
C = 12.01 amu
Average Atomic Mass of
Elements
ELEMENT
AVERAGE ATOMIC
MASS (amu)
AVERAGE ATOMIC
MASS, rounded
(amu)
Hydrogen
1.00794
1.01
Carbon
12.0111
12.01
Oxygen
15.9994
16.00
Chlorine
35.453
35.45
Iron
55.847
55.85
Formula Mass
Formula Mass: the sum of the atomic masses
of all atoms in one compound.
Water (H2O) has 2 Hydrogen atoms and 1
Oxygen atom.
 Formula Mass of H2O is
H 2 X 1.01 amu = 2.02 amu
O 1 X 16.00 amu = 16.00 amu
18.02 amu

Your turn…..Formula Mass
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What is the formula mass of methane, CH4?
Of NaCl?
Of ammonia, NH3?
Of glucose, C6H12O6?
Molar Mass
Molar Mass (grams per mole)
 The mass of 1 mol of a compound.



SO3 = 1S + 3O =
32.10 g + 3(16.00 g) = 80.10 g
CaI2 = 1 Ca + 2 I =
40.10 g + 2(126.90 g) = 293.00 g
Molar mass and formula mass
1 atom of Carbon weighs 12.01 amus
(formula mass)
 1 mole of Carbon weights 12.01 grams
(molar mass)
 1 molecule of H20 weighs 18.02 amus
(formula mass)
 1 mole of H20 weighs 18.02 grams (molar
mass)
 Same numerically but different units!!!!!

What’s in a Mole?




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Atoms – one mole of an element contains 6.022 X 1023
atoms of the element.
Molecules – one mole of a molecular (covalent) compound
contains 6.022 X 1023 molecules.
Gizmos – one mole of gizmos contains 6.022 X 1023
gizmos.
Anything – one mole of anything contains 6.022 X 1023
anythings!
NA (6.022 X 1023) is very practical for counting small
particles, especially things like atoms, ions and molecules.
EXAMPLES

How many pens in 1 mole of pens?
How many atoms in 63.546g of Cu?
 How many atoms in 6.3546g of Cu?

EXAMPLES
How many molecules in 1 mole of sugar
(C6H12O6)?
 How many molecules in 10 moles of
sugar?
 How many carbon atoms in 1 mole of
sugar? How many oxygen atoms?

Solving ‘Mole Problems”
MASS
Molar
mass
MOLES
Mole Conversions by Factor Label
Method

Mass and Moles


Use the molar mass of the substance.
1 Mole = n grams of the substance,

so have 2 conversion factors:
1 mole
n grams
or
n grams
1 mole
Mole Conversions by Factor Label
Example A: How many moles in 75.0 g of iron?
75.0g Fe
X
1 mol Fe
55.85g Fe
= 1.34 mol Fe
Example B: How many grams in 0.250 mol Na?
0.250 mol Na X
22.99g Na
1 mol Na
= 5.75g Na
Solving ‘Mole Problems”
PARTICLES
MOLES
Number of
Particles in
1 mole
(6.02 X 1023)
Mole Conversions by Factor Label

Particles and Moles

Use Avogadro’s Number (6.022 X 1023) of
particles.

(6.022 X 1023)particles  1 mol
1 mole
(6.022 X 1023)particles

Again, set up Factor Labels to cancel units!
Mole Conversions by Factor Label
Method
Example 1: How many atoms in 0.25 mol Na?
6.022 X1023 atoms Na
0.250 mol Na X
= 1.51 X 1023 atoms Na
1 mol Na
Example D: How many moles in 4.20 X 1023 molecules of CO2?
4.20 X
1023
1 mol CO2
Molecules CO2 X
6.022 X 1023 molecules CO2
= 0.697 mol CO2
Mole is the bridge…
“How much”
Mass
“How many”
MOLE
Bridge
Number of
particles
Summary: Solving ‘Mole
Problems” using the Mole Map
PARTICLES
MASS
Molar
mass
MOLES
Number of
Particles in
1 mole
(6.02 X 1023)
Summary of Mole Conversions by
Factor Label Method

Mass and Moles


Particles and Moles



Use the molar mass of the substance.
Use Avogadro’s Number (6.02 X 1023) of particles.
Set up Factor Labels to cancel units!
 DON’T GET LAZY! Include labels to ensure
that ALL units cancel correctly.
Multi-step conversions are easily done if
you are careful with the labels!
Percentage Composition
The mass of each element in a compound
compared to the entire mass of the
compound and multiplied by 100%.

Ex: CH4
C
H

1 x 12.00g =
4 x 1.01g =
12.00 g
4.04 g
16.04 g
total
%C = 12.00g/16.04g X 100 = 74.8%
%H = 4.04g/16.04g x 100 = 25.2%
Percentage Composition

Example 2

2.45 g aluminum oxide decomposes into 1.30 g
aluminum & 1.15 g oxygen. What is the percentage
composition?

%O = 1.15g O
X 100% = 46.9% O
2.45g Al Oxide note: Oxygen (O, not O2)

%Al = 1.30g Al
X 100% = 53.1% Al
2.45g Al Oxide

As a check, note that 46.9% + 53.1% = 100.0%.
Determining Empirical Formulas
Empirical Formula
 The formula that gives the simplest whole
number ratio of the atoms of the elements
in the formula.
Example 1
 What is the empirical formula of a compound
containing 2.644g of gold and 0.476g of chlorine?



0.476g Cl X 1 mol Cl
= 0.0134 mol Cl
35.45g Cl
2.644g of Au X 1 mol Au
= 0.0134 mol Au
196.97g Au
Empirical formula = Au0.0134 Cl0.0134 or AuCl
Determining Empirical Formulas
Example 2
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What is the empirical formula of a compound with 5.75 g
Na, 3.50 g N & 12.00 g O?
First, find the mole amounts.
5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na
3.50 g N X (1 mol N/14.01g N) = 0.250 mol N
12.00g O X (1 mol O/16.00g O) = 0.750 mol O
Empirical formula = Na0.250 N0.250 O0.750
Divide each mole quantity by the smallest to get whole
numbers. (0.250 in this case)
Empirical formula = NaNO3
Determining Molecular
Formulas

Molecular Formula


The formula that gives the actual number
of atoms of each element in a molecular
compound.
Example
 Hydrogen peroxide has a molar mass
of 34.00 g/mol and a chemical
composition of 5.90% H & 94.1% O.
What is its molecular formula?
Determining Molecular
Formulas
1. First, find the empirical formula, assuming the
percents are mass.
5.90gH X (1 mol H/1.01g H) = 5.84 mol H
94.1g O X (1 mol O/16.00g O) = 5.88 mol O
Empirical formula = H5.84O5.88 or HO.
2. Find the molar mass of the empirical formula =
17.01g/mol
3. Divide by molar mass of the empirical formula
34.00/17.01 = 2
4. Multiply the empirical formula subscripts by this
number: = 2(HO) or H2O2
The story so far….
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
From Avogadro’s Hypothesis we are able to
count molecules by weighing macroscopic
samples.
For gases at the same temperature and
pressure we can deduce the following:
 1. From combining volumes we can
determine the ratio in which molecules react.
 2. From masses of these gases we can
determine the relative mass of individual
molecules.
The story so far….

From these results it is possible to determine:

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the molar masses of the elements;
using these masses and formulas of compounds, one
can determine molar masses of compounds.
One can also determine empirical formulas and
molecular formulas
These tools allow one to relate
“how much stuff” to “how many particles”.
