Unit 4
COUNTING PARTICLES
Counting By Weighing
We can weigh a large number of the
objects and find the average mass.
Once we know the average mass we can
equate that to any number of the
objects.
EXAMPLE:
◦ The average mass of a book is 40.0 grams.
◦ How many books are present in a sample with a mass of 2000.0 grams?
◦ 2000.0g/40.0g = 50.0 books
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Counting By Weighing
When we know the average mass of the
atoms of an element, as that element
occurs in nature, we can calculate the
number of atoms in any given sample of
that element by weighing the sample.
The atomic mass of an element, as
found on the periodic table, allows us
to count by weighing.
6-3
Avogadro’s Number
Avogadro’s Number is the number of atoms in 12.01
grams of carbon.
Its numerical value is 6.022 × 1023.
Therefore, a 12.01 g sample of carbon contains
6.022 × 1023 carbon atoms.
The Mole
The mole (mol) is a unit of measure for an amount of
a chemical substance.
A mole is Avogadro’s number of particles, that is
6.022 × 1023 particles.
1 mol = Avogadro’s Number = 6.022 × 1023 units
We can use the mole relationship to convert between
the number of particles and the mass of a substance.
Section 6.3: The Mole
The mole (mol) is known as the
“chemists dozen” and
represents
Not this type of mole!
6.022 x 1023 things (atoms,
particles, molecules, etc).
•A mole (from the Latin - “lump
of stuff”), was originally defined
as 1.0g of the lightest element
known – hydrogen.
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How Big Is a Mole?
The volume occupied by one mole of softballs would
be about the size of the Earth.
One mole of Olympic shot put balls has about the
same mass as the Earth.
Molar Mass
The atomic mass of any substance expressed in
grams is the molar mass of that substance.
The atomic mass of iron is 55.85 amu.
Therefore, the molar mass of iron is 55.85 g/mol
(grams in a mole)
Since oxygen occurs naturally as a diatomic, O2, the
molar mass of oxygen gas is 2 times 16.00 g or 32.00
g/mol.
Calculating Molar Mass
The molar mass of a substance is the sum of the
molar masses of each element.
What is the molar mass of magnesium nitrate,
Mg(NO3)2?
The sum of the atomic masses is:
24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) =
24.31 + 2(62.01) = 148.33 amu
The molar mass for Mg(NO3)2 is 148.33 g/mol.
Molar Mass (g/mol)
•The molar mass of any substance is the
mass (in grams) of one mole of the
substance.
•The molar mass of a compound is
obtained by summing the masses of ALL
component atoms.
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Molar Mass: Practice
1) Calculate the following molar masses:
a) Water – H2O
b) Ammonia – NH3
c) Propane – C3H8
d) Glucose – C6H12O6
6-11
Mole Calculations
We will be using Dimensional analysis again.
Recall:
◦First we write down the unit asked for
◦Second we write down the given value
◦Third we apply unit factor(s) to convert the given
units to the desired units
Mass-Mole Calculations
What is the mass of 1.33 moles of titanium, Ti?
We want grams, we have 1.33 moles of titanium.
Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti
47.88 g Ti
1.33 mole Ti ×
= 63.7 g Ti
1 mole Ti
Mole Calculations
Now we will use the molar mass of a compound to
convert between grams of a substance and moles or
particles of a substance.
6.02 × 1023 particles = 1 mol = molar mass
If we want to convert particles to mass, we must first
convert particles to moles and than we can convert
moles to mass.
Mole Calculations
How many sodium atoms are in 0.120 mol Na?
◦ Step 1: we want atoms of Na
◦ Step 2: we have 0.120 mol Na
◦ Step 3: 1 mole Na = 6.02 × 1023 atoms Na
0.120 mol Na ×
6.02 × 1023 atoms Na
1 mol Na
= 7.22 × 1022 atoms Na
Mole Calculations I
How many moles of potassium are in 1.25 × 1021
atoms K?
◦ Step 1: we want moles K
◦ Step 2: we have 1.25 × 1021 atoms K
◦ Step 3: 1 mole K = 6.02 × 1023 atoms K
1.25 ×
1021
atoms K ×
1 mol K
6.02 × 1023 atoms K
= 2.08 × 10-3 mol K
Mole Calculations
What is the mass of 2.55 × 1023 atoms of lead?
We want grams, we have atoms of lead.
Use Avogadro’s number and the molar mass of Pb
2.55 ×
1023
1 mol Pb
207.2 g Pb
×
atoms Pb ×
23
1 mole Pb
6.02×10 atoms Pb
= 87.8 g Pb
Mole Calculations
How many O2 molecules are present in 0.470 g of
oxygen gas?
We want molecules O2, we have grams O2.
Use Avogadro’s number and the molar mass of O2
1 mol O2
6.02×1023 molecules O2
×
0.470 g O2 ×
1 mole O2
32.00 g O2
8.84 × 1021 molecules O2
How many Oxygen atoms in 8.84 × 1021
molecules O2?
8.84 × 1021 molecules O2 x 2 oxygen atoms =
molecule
Mole Video
6.022 X 1023 pennies: would make at least 7
stacks that would reach the moon.
6.02 X 1023 blood cells: would be more than the
total number of blood cells found in every human on
earth.
A mol is A LOT of particles:
602,200,000,000,000,000,000,000
The Mole
•It is VERY important to understand that the value listed as the mass
number on the periodic table really tells us TWO things:
– The mass of one atom in units of amu’s.
– The mass of one mole of atoms in grams.
–One mol of Al = 26.98 g & one atom = 26.98 amu
–One mol of Au = 196.97 g & one atom = 196.97 amu
–One mol of B = 10.81 g
& one atom = 10.81 amu
–Remember one mol = 6.022 X 1023
atoms (or particles or molecules or
donuts)
6-22
•Remember:
can talk about one mole of atoms or
Molar we
Mass
one mole of molecules.
•One mole of oxygen atoms (O) weighs 16.00 g.
•One mole of oxygen molecules (O2) weighs 32.00 g.
•Two moles of O atoms weigh 32.00 g.
•Two moles of O2 molecules weigh 64.00 g.
•And so on . . .
6-23
The Mole
•To summarize: a sample of any element that
weighs a number of grams equal to the molar
mass (from the periodic table) contains
6.022 x 1023 atoms (1 mol) of that element.
•Therefore, atomic weight of an element =
#g in one mol of that element
•We write that as g/mol.
6-24
The Mole: Practice
1) Calculate the number of moles in a 25.0 g
sample of calcium. How many atoms are there?
2) Calculate the number of moles and the number
of atoms in a 57.7 g sample of sulfur.
3) Calculate the number of atoms in a 23.6 g
sample of zinc.
4) Calculate the number of molecules in a 12.83 g
sample of calcium carbonate.
6-25
1. .624 mol and 3.76 x 1023 atoms
2. 1.80 mol and 1.08 x 1024 atoms
3. 3.61 x 10-4 mol and 2.17 x 1020 atoms
4. 1.189 x 10-3 mol and 7.160 x 1020 atoms
Molar Mass: Practice
1. Calculate the number of moles of H2CO3 present
in a 7.55 g sample.
2. How many water molecules are present in 10.0 g
of water? (Hint: find moles first)
3. How many molecules of sucrose (C12H22O11) are
present in a 5.0 kg bag of sugar?
(Hint: find moles first)
6-27
Representative
Particles=
atoms, molecules,
Items, etc
Percent Composition of Compounds
•Sometimes it is not enough to know a
compound’s composition in terms of numbers
of atoms; it may also be useful to know its
composition in terms of the masses of its
elements.
•We can calculate the mass fraction by
dividing the mass of a given element in one
mole of a compound by the mass of one mole
of the compound.
6-29
Percent Composition of Compounds
•Once we know the mass fraction we can
multiply by 100 to get the percent.
•Remember: percent = part/whole X 100.
6-30
In one mole of methane (CH4), there is one mole of
C and four moles of H:
1 mol C = 12.01 g
4 mol H = 4(1.01) = 4.04 g
1 mol CH4 = 12.01 g + 4.04 g = 16.05 g
%C= mass of 1mol C
x 100
mass of 1 mol CH4
%C = 12.01 X 100 = 74.83 %C
16.05
%H= mass of 4 mol H
x 100
mass of 1 mol CH4
%H = 4.04 X 100 = 25.17 %H
16.05
Percent Composition Practice
Determine the mass percent of each
element in the following:
◦ H2SO4 (sulfuric acid)
◦ C3H7OH (isopropyl alcohol)
◦ C6H12O6 (glucose)
6-32
Formulas of Compounds
•The object of this section is to do the opposite of
the previous section. Instead of getting the mass
from the formula, we will determine the formula
from the mass.
•To do this, the mass must be converted to moles
using each element’s mass number. How can we
convert mass to moles?
6-33
Formulas of Compounds
ex) An unknown compound with a mass of 0.2015 g
is is found to contain:
◦ 0.0806 g C
◦ 0.01353 g H
◦ 0.1074 g O
It must contain:
0.0806g (1 mol C/12.01 g C) = 0.00671 mol C
0.01353g(1 mol H/1.008 g H) = 0.01342 mol H
0.1074g(1 mol/16.00 g O) = 0.00671 mol O
6-34
The numbers from the previous slide allow us to
determine the C:H:O ratio.
0.00671 (C) : 0.01342 (H) : 0.00671 (O)
If we divide each number by the smallest number
we get 1:2:1 for the C:H:O ratio.
This leads us to a formula of C1H2O1 or CH2O.
This is not necessarily the TRUE formula of the
compound, but represents the RELATIVE numbers
of atoms.
6-35
This represents the lowest whole number ratio of
the compound.
The actual formula could be CH2O, C2H4O2,
C3H6O3, C4H8O4, C5H10O5, C6H12O6, . . .
Any formula with a C:H:O ratio of 1:2:1 is
possible (in theory, an infinite number).
C1H2O1 represents the simplest possible
formula or the EMPIRICAL FORMULA.
The multiples represent possible
MOLECULAR FORMULAS.
6-36
Formulas of Compounds: Practice
Determine the empirical formula from each of the
following molecular formulas:
◦ H2O2 (hydrogen peroxide)
◦ C4H10 (butane)
◦ CCl4 (name?)
◦ HC2H3O2 (acetic acid)
◦ C6H12O6 (glucose)
6-37
Calculation of Empirical Formulas
There are four steps to determine the
empirical formula of a compound:
1. Obtain the mass of each element present (in
grams).
2. Determine the number of moles of each type
of atom present (use atomic mass).
3. Divide each number by the smallest number.
4. Multiply all numbers by the smallest integer
that will make them all integers
6-38
Empirical Formulas: Practice
1) A 1.500 g sample of a compound containing only
carbon and hydrogen is found to contain 1.198 g
of carbon. Determine the empirical formula for
this compound.
2) A 3.450 g sample of nitrogen reacts with 1.970
g of oxygen. Determine the empirical formula
for this compound.
3) When a 2.000 g sample of iron metal is heated
in air, it reacts with oxygen to achieve a final
mass of 2.573g. Determine the empirical
formula for this compound.
6-39
If the relative amounts of elements are
presented as percentages, assume we are
starting with a 100 g sample (100%). Then
each percentage simply becomes a mass (in
grams).
For example if 15% of a compound is
carbon, we just assume it is 15 g of a 100 g
sample; from there we convert to moles.
6-40
Empirical Formulas: More Practice
1) The simplest amino acid, glycine, has the
following mass percents: 32.00% carbon,
6.714% hydrogen, 42.63% oxygen, and 18.66%
nitrogen. Determine the empirical formula for
glycine.
2) A compound has been analyzed and found to
have the following mass percent composition:
66.75% copper, 10.84% phosphorous, and
22.41% oxygen. Determine the empirical
formula for this compound.
6-41
Calculation of Molecular Formulas
If we know the empirical formula AND the molar
mass, we can calculate a compound’s molecular
formula.
Note: without the molar mass, the best you can
find is the empirical formula.
Once the molar mass is known, one must ALWAYS
find the empirical formula before one can calculate
the molecular formula. It is impossible to do the
reverse.
6-42
Calculation of Molecular Formulas
It is also important to note that the molecular
formula is ALWAYS an integer multiple of the
empirical formula. We can represent the molecular
formula in terms of the empirical formula:
(Empirical Formula)n = molecular formula
It should also be noted when n = 1, the empirical
and molecular formulas are identical to each other.
6-43
Molecular Formulas: Practice
1) A compound containing carbon, hydrogen, and
oxygen is found to be 40.00% carbon and
6.700% hydrogen by mass. The molar mass of
this compound is between 115 and 125 g/mol.
Determine the molecular formula for this
compound.
2) Caffeine is composed of 49.47% C, 5.191% H,
28.86% N, and 16.48% O. The molar mass is
about 194 g/mol. Determine the molecular
formula for caffeine.
6-44