Recap – Molar Mass and Moles
Relative Atomic Mass – weighted average mass of
all isotopes of a particular element relative to 12C.
Avogadro number – 6.022 x 1023 this is the number
of atoms of 12C in exactly 12.00 g of 12C.
The mole – name we use when we have an
Avogadro number of objects.
Amount in moles, n = mass / molar mass
1
Stoichiometry
Q: What masses of Na and Cl2 are needed to
produce 100 g NaCl?
2Na
Mass/g
Molar mass
Moles
?
23.0
+
Cl2
?
71.0

2NaCl
100
58.5
n  m
M
1.71
2
Stoichiometry
Q: What masses of Na and Cl2 are needed to
produce 100 g NaCl?
2Na
Mass/g
?
Molar mass
23.0
Moles
1.71
+
Cl2
?
71.0
0.85

2NaCl
100
58.5
1.71
Use balanced equation
3
Stoichiometry
Q: What masses of Na and Cl2 are needed to
produce 100 g NaCl?
2Na
+
Cl2
Mass/g
39.3
60.7
Molar mass
23.0
71.0
Moles
1.71
0.85

2NaCl
100
58.5
1.71
m=nM
4
Limiting Reagent
So far we have talked about reactions
where we have exactly the right amounts
of each reactant.
Most of the time, however, we will not
have enough of one reactant (reagent) to
react with all of the other.
The reagent which we are short of is
called the limiting reagent – it determines
how much product we get.
5
n  m
M
Limiting Reagent
Q: 5.00 g of H2 and 10.0 g of F2 are mixed. When
the reaction ceases, what is the mass of each
type of molecules present?
H2
Mass/g
Molar mass
Moles
5.00
2.01
2.49
+
F2
10.0
38.0
0.263

2HF
20.0
\ F2 is the limiting reagent – amount of F2
will determine amount of product.
6
n  m
M
Limiting Reagent
Q: 5.00 g of H2 and 10.0 g of F2 are mixed. When
the reaction ceases, what is the mass of each
type of molecules present?
H2
+
F2
Mass/g
Molar mass
Moles (initially)
Moles (used)
Moles (after react.)
5.00
2.01
2.49
0.263
2.23
10.0
38.0
0.263
0.263
0
Mass (after react.)
4.48
0

2HF
20.0
0.526
10.52 7
% Yield
• So far we have assumed that all of the
limiting reagent is used up – 100 %
reaction – the reaction ‘has gone to
completion’.
• This may not always be the case.
% yield = actual mass of product  100
max theoretical mass
8
n  m
M
% Yield
Q: In the reaction of 5.0 g of sodium with
excess chlorine, the mass of sodium chloride
obtained was 12.4 g. Calculate the % yield.
2Na
Mass/g
Molar mass
+
Cl2

2NaCl
5.0
23.0
70.9
58.5
Mole
9
n  m
M
% Yield
Q: In the reaction of 5.0 g of sodium with
excess chlorine, the mass of sodium chloride
obtained was 12.4 g. Calculate the % yield.
2Na
Mass/g
Molar mass
Mole
+
Cl2

5.0
23.0
2NaCl
12.7
70.9
0.217
58.5
0.217
% yield = 12.4 / 12.7  100 = 97.6 %
10
Learning Outcomes:
• By the end of this lecture, you should:
−
be able to perform stoichiometric calculations including:
−
−
−
−
−
determining the mass of product given the mass of starting materials
determine the mass of starting materials needed to give a required
mass of product
identifying the limiting reagent and using this in the calculations
determine the % yield in a reaction given the mass of starting
materials
be able to complete the worksheet (if you haven’t already
done so…)
11
Questions to complete for next lecture:
1. Write a balanced equation for the complete combustion
of carbon to form carbon dioxide.
2. How many moles of carbon are there in 100 g of carbon?
3. If the combustion is carries out in air, which is the
limiting reagent?
4. How many moles of carbon dioxide would this produce
on complete combustion of 100 g of carbon?
5. What mass of carbon dioxide is produced from the
complete combustion of 100 g of carbon?
6. If 348 g of carbon dioxide is isolated from complete
combustion of 100 g of carbon, what % yield does this
represent?
12