The Purpose of Moles
Dalton Ma
May 2024
The Purpose of Moles
Dalton Ma
Contents
1
The Mole
1
2
The Mole Ratio
1
3
Further Calculations
2
4
Empirical Formulae
4
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The Purpose of Moles
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Dalton Ma
The Mole
A mole is a quantity, like "a dozen" (or 12). We can have a mole of anything, even a mole of moles!
Definition 1.1
A mole is defined as the number of carbon-12 atoms in 12 g of carbon-12.
Numerically, 1mol = 6.022 × 1023 of anything.
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&
We can use the mole to relate masses. For example, 1 mol H has a mass of 1 g.
Relative molecular mass, written as Mr is the sum of the relative atomic masses of its component
elements.
Example 1.2
Find the relative molecular mass of CuSO4 .
Solution. The molecular masses of Cu is 63.5 u, S is 32 u, and O is 16 u. Thus, the calculation
would be 63.5 + 32 + 4(16) or 159.5u.
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m
We can use the formula n = M
to find the number of moles in a substance. Here, n is the number of
moles, m is the mass given, and m is the unit mass for the substance given.
Example 1.3
How many moles are in 27g of O2 ?
g
Solution. The molar mass of O2 is 32 mol
. So, n = 3227gg = 0.84molO2 .
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mol
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We can also rearrange this equation using algebra to solve for different unknowns.
m
M
m = nM
m
M=
n
n=
IMPORTANT Make sure you know what you should be solving for. If you are asked to find m, if
you find n you are WRONG!
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The Mole Ratio
We can use the coefficients in a balanced equation to find amounts.
1
(1)
(2)
(3)
.
The Purpose of Moles
Dalton Ma
Example 2.1
2 Mg + O2 −−→ 2 MgO
This means that 2 moles of Mg reacts with 1 mol O2 to produce 2 mol of MgO.
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We can use this mole ratio to find quantities that are unknown.
Example 2.2
2 g of magnesium ribbon was reacted in a hot bunsen flame. A bright white light was observed and
a white/grey ash remained.
1. Write a balanced equation for the reaction.
2 Mg + O2 −−→ 2 MgO
2. Calculate the number of moles of magnesium reacting.
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n
m
=M
n
= 245gg n
mol
(4)
= 0.21mol
(5)
3. Calculate the mass of product formed
Since it needed 2 mol of Mg to make 2 mol MgO, we get 0.21 mol MgO, or 8.4 g.
4. If you needed to produce 5 g of the product, what mass of Mg do you need to start with?
5 g MgO is 0.125 mol MgO, which means also 0.125 mol Mg. 0.125 mol Mg is 3 g Mg.
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Further Calculations
Sometimes, in a reaction, we don’t get the output due to impurities or losses. We call the amount we should
get theoretical yield, and the amount we actually get actual yield. The amount of product we can generate
is limited by the limiting reagent, which is the reactant that is completely consumed in a chemical reaction.
Once a reactant is completely used up, the reaction stops. In this way, the limiting reagent limits the
reaction and the amount of product generated.
In EXAMPLE 2.2, we have only 2 g of Mg, but excess O, so the Mg would be the limiting reagent.
Definition 3.1
Percent Yield is a measure of how much product is produced relative to how much should be produced
in theory.
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The Purpose of Moles
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Dalton Ma
Percent yield =
Actual Yield
× 100
Theoretical Yield
(6)
&
In order to find % yield, we need to first find ACTUAL and THEORETICAL yield.
We can use DEFINITION 3.1 to perform calculations around percent yield.
Example 3.2
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16g of CaCO3 is decomposed according to the equation below to give 7.54g of CaO. Explain how to
calculate the % yield of CaO.
We first need to write a balanced equation.
CaCO3 −−→ CaO + CO2
We also need to find the theoretical yield using mole ratios. We can use the process in EXAMPLE 1.2
g
g
to find the relative molar masses of CaCO3 and CaO are 100 mol
and 56 mol
respectively. Now, we
can use the equation in section 1 to find the number of moles of each. There are 0.16 mol of CaCO3 .
Since our mole ratio is 1:1, we get also 0.16 mol CaO, which is 8.96 g theoretical yield.
7.54
Now, we need to find percent yield, which is 8.96
= 84.2%
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Sometimes, we also need to know how much of a substance is some particular element. This is where
percent composition comes in.
Definition 3.3
Percent Composition is the percent by mass of each element in a compound.
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%by mass =
mass of element in 1 mol
× 100%
molar mass of compound
(7)
&
Example 3.4
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Calculate the % of each element in CaCO3 .
g
g
. The molar mass of Ca is 40 mol
. So the percent
We can find the molar mass of CaCO3 to be 100 mol
composition of Ca is 40 %. Similarly, we can find the percent composition of C to be 12 % and O to
be 48 %.
We should check to make sure our percents add up to 100 %. Otherwise, we have done something
wrong.
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Sometimes it is useful to know how pure a substance is. For example, if a sample of sugar is found to
contain 1 % sugar, 9 % lead powder, and 90 % asbestos, we have a problem. Here, we can say the ‘sugar’ is
only 1% pure.
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The Purpose of Moles
Dalton Ma
Definition 3.5
Percent Purity the ratio of pure substance to impure substance.
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Percent Purity =
mass of pure substance
mass of impure substance
(8)
&
Example 3.6
A 150g sample of copper ore was analysed and found to contain 87.3g of pure copper. Calculate the
% purity.
We can use the calculation 87.3
150 = 87.3%.
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Empirical Formulae
Definition 4.1
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Empirical Formulae tells us relative ratios of elements in a compound.
&
We can find mole ratios to find empirical formulae. If given % composition, we must first assume 100g
of substance to find mole ratios.
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