Up until this point we have been talking about
atoms and molecules. The problem with this
approach is that atoms and molecules are very small
things. In a single drop of water for example, there
are trillions and trillions of water molecules. A
reaction between a single molecule of hydrogen and
a single molecule of oxygen, as we discussed above,
would be undetectable. Instead of talking about
single molecules in science, we talk about groups of
molecules. You can think of it like buying
eggs. You don't go to the store and buy an egg - you
buy a dozen. Contained within that dozen are the
individual eggs. Its the same thing when we talk
about molecules. We don't talk about single units,
we talk about groups.
But even a dozen molecules is a tiny
amount. What we need is a big number - a huge
number! That number is the mole.
One mole equals 6.02 x 1023 (also known as Avogadro's
number). A 6 followed by 23 zeros. Now that's a pretty big
number. But that's all it is, a number. You can't just have a mole,
you have to have a mole of something. A mole of atoms. A mole
of water molecules. A mole of pennies (which would make you
richer than you can imagine).
A penny has
approximately 6 x
1023 molecules.
600,000,000,000,000,000,000,000!
Amadeo Avagadro
•An Avogadro's number of standard soft drink cans would
cover the surface of the earth to a depth of over 200 miles.
•If you had Avogadro's number of unpopped popcorn
kernels, and spread them across the United States of
America, the country would be covered in popcorn to a
depth of over 9 miles.
•If we were able to count atoms at the
rate of 10 million per second, it would
take about 2 billion years to count the
atoms in one mole.
•If marbles that have a diameter of
one centimeter were to be lined up
end-to-end in a straight line, the
distance covered by this string of
marbles can hold in about
500,000,000 of our Solar Systems
placed end-to-end.
Why the mole? As it turns out, the mole has some interesting properties.
One mole of hydrogen atoms (6.02 x 1023 H atoms) weighs 1 g. In fact,
one mole of any element is equal to the atomic mass of that element (in
Let's think about that for a second. If we
grams).
know the molar mass of an element, and we
know how many elements make up a specific
molecule, then you can calculate the molar
mass of a compound by adding up the atomic
weights. Huh? Take water for example. How
much does a mole of water weigh? Well, one
mole of water contains one mole of oxygen
atoms and two moles of hydrogen atoms. A
mole of hydrogen weighs 1 g and a mole of
oxygen weighs 16 g (look at the atomic mass
in the periodic table). So to calculate the
weight of one mole of water:
(2 moles H * 1 g per mole) + (1 mole O * 16 g
per mole) = 18 g
One mole of water weighs 18 grams!
We often need to figure out the molecular mass of a compound,
particularly when making solutions for the lab.
To find the molecular weight of a compound, simply add the atomic
weights of each atom in the compound.
Example: CaCl2 has one Ca and 2 Cls.
According to the Periodic Table:
atomic weight of Ca = 40 g/mol
atomic weight of Cl = 35 g/mol 35 x 2 = 70
Therefore 1 mole (or 6.02 x 1023 molecules) of CaCl2
weighs 40 g/mol + 70 g/mol = 110 g/mol.
Another example: Find the molecular weight of C3H8.
C3H8 has 3 Cs and 8 Hs. According to the Periodic Table:
The atomic weight of C is 12 g/mol 12 x 3 = 36 g/mol
The atomic weight of H is 1 g/mol 1 x 8 = 8 g/mol
Therefore, 1 mole (or 6.02 x 1023 molecules) of C3H8 weighs
36 g/mol + 8 g/mol = 44 g/mol.
Try some: Calculate the molecular weight for 1 mole of:
1. HCl
2. NaH2PO4
3. CaBr2
4. RbS2
5. KMnO4
6. NH4Cl
7. C5H12
8. (NH4)2SO3
9. Al2SO3
10. C6H12O6
Moles
(mol)
Mass
(g)
Volume
(L)
Atoms
or
Molecules
Challenge: What if you wanted to know the molecular weight of a compound with
more or less than 1 mole? Think about it and try:
1. 10 moles of CH4
2. 0.1 moles of NH3
3. 0.002 moles of H2O
4. 86 moles of AlF3
5. 1.15 moles of NaCl
6. 15 moles of KMnO4
Challenge: What if you know the weight, but need to find the number of mole. Create
a way to figure out how many moles are in 24 grams of NaCl.
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
-24
1.66 x 10 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
1) Given: 1.0  10 4 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
• 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
•
•
•
•
•
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
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:
Conversion factors:
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
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
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
=
mol C 6H12 O 6
11.1 mol C atoms
Molar Mass of a Compound
•
•
1)
2)
Calculate the molar mass of iron (II) sulfate
Formula is FeSO4
Calculate the molar mass of each element
Each element is multiplied by its respective
subscript: (number of moles of each element)
Formula
Subscript
Moles of
Compound
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
• The three quantities most often calculated
– 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
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
molecules of ethylene glycol
Equality 1: 1mol C2H6O2 =62.05 g C2H6O2
Conversion 62.05 g C2H6O2
mol C2H6O2
and
Factor 1
mol C2H6O 2
62.05 g C2H6O 2
Equality 2: 1mol C2H6O2 =6.022 1023 molecules
Conversion
Factor 2
3.86 10
- 20
1mol C 2H6 O 2
6.022 10 23 molecules
or
1mol C 2H6 O 2
6.022  10 23 molecules
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
• 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
• 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
• 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
• 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
• Empirical Formula
– 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 whole- number
multiple of its EF
Molecular formula = n х Empirical formula
Parts of Chemical Equations
• Flour + Water+ Egg+ Yeast ->
Bread



•
Reactants
yield Product
• Reactants are the starting chemicals in a chemical
reaction.
• Products are the new chemicals made in a chemical
reaction.
• H2 + Cl2 -> 2HCl
• C12H22O6 -> 12C + 11H2+ 3O2
• The chemical properties of the reactants changed when
they are combined.
What is an equation?
Think about math…what does an equation in math look like?
Something like this: a + b = c + d
Chemistry is very similar. The equations look a little different, but amount to the same
thing. For example:
2H2 + O2  2H2O
+ means to add or combine the chemicals
 is basically the same as =, but means “becomes”
so H2 and O2 combine to become H2O
The 2 is the subset and is the number of atoms in the molecule, in this case, there are 2 Hs
and 2 Os to start with, and 2 Hs in the end.
Remember, no subset number means you have one of that atom.
What about the 2 (big)?
2H2 + O2  2H2O
The big 2 is a coefficient. This is how many molecules you have. You will learn how to
figure out how many of each molecule you need in a chemical reaction later this week.
In the case of making water, you start with 2 H2s and one O2, and make 2 waters.
You will learn how to “balance” equations this week, as well.
Chemical bonds are VERY important in chemical reactions. The valence electrons
dictate the kinds of reactions that can and will happen. You must be comfortable
using them. If you aren’t, please come for extra tutoring.
Reactions
forming ionic
compounds.
Reactions
creating
single bonds.
Reactions creating
covalent double
bonds.
Balancing Chemical Equations
• 2H2 + O2 -> 2H2O (Water)
• H2 + O2 -> H2O2 (NO!)
(Hydrogen Peroxide)
• The twos in front of the Hydrogen are called Coefficients.
Coefficients tell us how much of an atom or molecule is present
without changing the formula. Never add a subscript to any
formula. You will change the formula.
How to balance equations
• H2 + O2 -> H2O
• Step 1: Write the equation out without
coefficients. Make sure the formulas are written
correctly.
• Step 2: Count the oxygen. There are two on the
left side, and one on the right. Add a coefficient of
two to the water to make the oxygens even, do
not add any subscripts.
• H2 + O2 -> 2H2O
How to balance equations
• Step 3: Count the hydrogen. There are 4 on the right.
(2H2, 2*2=4) and 2 on the left. Write a 2 in front of
the hydrogen on the right.
H2 + O2 -> 2H2O
2H2 + O2 -> 2H2O
• Step 4: Check your work. All elements should have
the same number of atoms on both sides. If not,
check your work again.
2H2 = 2H2? Yes
O2 = 2O? Yes
TIPS for balancing Equations
• To make this an easier process try doing it in this
order:
•
•
•
•
1. Balance the metals first.
2. Balance all non metals except O and H.
3. Balance the oxygen
4. Balance the hydrogen
• Also, make sure the formulas are written correctly
Let’s Practice!
1.
Cl2 +
NaBr ->
NaCl +
2.
Fe
O2 ->
FeO
3.
C3H8
4.
Al
5.
C12H22O6
+
+
+
O2
Cl2
->
->
->
CO2 +
Br2
H2O
AlCl3
C +
H2
+
O2
Write balanced chemical equations for the following reactions:
(a) Solid mercury(II) sulfide decomposes into its component elements
when heated.
(b) The surface of aluminum metal undergoes a combination reaction
with oxygen in the air.
(c) The reaction that occurs when ethanol, C2H5OH(l), is burned in air.
Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in
C12H22O11.
Without using a calculator, arrange the following samples in order of
increasing numbers of carbon atoms: 12 g 12C, 1 mol C2H2, 9  1023
molecules of CO2.
(a)How many nitric acid molecules are in 4.20 g of HNO3?
(b) How many O atoms are in this sample?
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O
by mass. What is the empirical formula of ascorbic acid?
A 5.325-g sample of methyl benzoate, a compound used in the
manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g
of hydrogen, and 1.251 g of oxygen. What is the empirical formula of
this substance?
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has
an empirical formula of C3H4 . The experimentally determined molecular
weight of this substance is 121 amu. What is the molecular formula of
mesitylene?