# CHEMISTRY: HOW MUCH (CALCULATIONS)

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```CHEMISTRY: HOW MUCH (CALCULATIONS)
Relative Formula Mass (Mr)
To find the relative formula mass of a compound you add up the relative atomic
masses of all the atoms in the compound.
e.g. MgSO4 contains:
1 x Mg: 1 x 24 = 24
1 x S: 1 x 32 = 32
4 x O: 4 x 16 = 64
So the relative formula mass = 24 + 32 + 64 = 120
Moles
The relative formula mass of a substance in grams is known as ONE MOLE of that
substance.
e.g. MgSO4
Relative Formula Mass = 120
Mass of One Mole = 120grams
A mole of a substance always contains the same number of particles, just like a dozen
of anything is always 12.
Calculating the Percentage Mass of an Element in a Compound
e.g. Find the mass of hydrogen in CH4
1. Write down the formula:
CH4
2. Work out the relative formula mass:
C= 12
H= 1
H=1
H=1
H=1
relative formula mass = 12 + 1 + 1 +1 + 1 = 16
3. Work out the mass of the element you are looking for in one
formula unit (e.g. the amount of H in CH4)
CH4 : there are four hydrogens in the formula,
so the mass is 1 + 1 + 1 + 1 = 4 (1 is the atomic mass of hydrogen)
4. Calculate the percentage of the element in formula
Mass of the element you are looking for
x 100
The relative formula mass of the entire compound
4
16
x 100
=
25 %
Working Out the Formula of a Compound from Reacting
Masses (or the EMPIRICAL formula)
e.g. A sample of a compound has 16g of sulfer and 24g of oxygen. Work out the
formula of this compound.
It is easiest to do this in a table:
1. Write the name of the elements at the top of the table
Sulfur
Oxygen
2. Write the amount you have of each element
Sulfur
16g
Oxygen
24g
3. Work out how many moles of the each element you have by
dividing them by the RELATIVE ATOMIC MASS
Atomic mass of oxygen = 16
Atomic mass of Sulfer = 32
Sulfur
16g
16 ÷ 32 = 0.5
Oxygen
24g
24 ÷ 16 = 1.5
4. Divide by the smallest to find a RATIO
0.5 is smaller than 1.5, so we divide them both by 0.5
Sulfur
16g
16 ÷ 32 = 0.5
0.5 ÷ 0.5 = 1
Oxygen
24g
24 ÷ 16 = 1.5
1.5 ÷ 0.5 = 3
5. Use the RATIO to work out the formula as it tells you how many of
each element you have
Sulfur
16g
16 ÷ 32 = 0.5
0.5 ÷ 0.5 = 1
One sulfer
Oxygen
24g
24 ÷ 16 = 1.5
1.5 ÷ 0.5 = 3
Three oxygens
The formula has one sulfur and three oxygens so looks like:
SO3
Working out the Mass of a Reactant or Product Using the
Formula
You will be given the mass of one of the reactants/ products and be expected to use to
find the mass of another using the formula.
e.g. What mass of carbon will react exactly with 40g of copper oxide. Work out
using the formula:
2CuO + C  2Cu + CO2
1. Work out the RATIO of the reactants and products by looking at the
numbers in front of the compounds. If there is no number then we
assume it is one.
2CuO + C  2Cu + CO2
2
:
1 : 2 : 1
2. Covert the mass of the compound you are given (the 40g of CuO) by
dividing the mass by the relative formula mass of the compound
Relative formula mass of CuO:
1 x Cu: 1 x 64 = 64
1 x O: 1 x 16 = 16
Mass given
=
Relative formula mass
relative formula mass = 64 + 16 = 80
40
80
=
0.5 moles of CuO
3. Look at the RATIO to work out how many moles of CuO react with
C (which your trying to find)
Ratio = 2 (CuO) : 1 (C)
Looking at this we can see that there are half the number of moles of C than CuO (2
divided by 2 is 1) so we need to half the number of moles of CuO that there are:
0.5 ÷ 2 = 0.25 mole of C
4. Convert the number of moles of C to grams by multiplying the
number of moles (0.25) by the atomic (formula if a compound)
mass of C.
Atomic mass of C = 12
0.25 x 12 = 3g
Therefore there are 3g of carbon
Percentage Yield
The amount of product made in a reaction is called its yield.
The percentage yield tells us the amount of product made compared to the maximum
amount that could be made.
e.g. Using known masses if A and B it was calculated that the chemical reaction
could produce 200g of product, C. When the reaction was carried out only 140g of C
is produced.
What is the percentage yield?
Percentage = amount of product actually produced
x 100%
yield
Maximum amount of product that could be produced
140
200
x 100% =
70%
Atom Economy
Atom economy is a measure of how much of the starting material ends up as
USEFUL products.
e.g. CaCO3  CaO + CO2
This reaction is used to produce calcium oxide (CaO). What is the atom economy of
this reaction?
Atom economy = relative formula mass of useful products
x 100%
1. Work out the relative formula mass of the useful product (the useful
product is the one that the reaction is designed to produce, so in this case
CaO)
Relative formula mass of CaO:
Ca: 1 x 40
O: 1 x 16
relative formula mass: 40 + 16 = 56
2. Work out the relative formula mass of all the products
Work out the relative formula mass of the other product:
Relative formula mass of CO2
C: 1 x 12
O2: 2 x 16 = 32
relative formula mass = 32 + 12 = 44
Add up the relative formula masses of the products: CaO + CO2 = 56 + 44 = 100
3. Divide the relative formula mass of the useful products by the
relative formula mass of all the products
56
100
= 0.56
4. Multiply by 100%
0.56 x 100% = 56% atom economy
```