Calculations in Standard Grade Chemistry
1
Formula Mass and the Mole
Each element has a relative atomic mass listed in the data book.
The formula mass (measured in amu) is the sum of the relative atomic masses.
The gram formula mass is the formula mass expressed in grammes.
One mole of a substance is the same as the gram formula mass
Examples
a)
Find 1 mole of sodium hydroxide (NaOH)
The relative atomic masses are Na = 23, O = 16, H =1 (data book)
Formula mass = 23 + 16 + 1 = 40 amu
1 mole = gram formula mass = 40 g
a)
Find one mole of ammonium carbonate (NH4)2CO3
Hint: Just like in maths calculate the contents of the brackets first:
Relative atomic masses are N =14, H =1, C =12, O =16
NH4 =
(NH4)2 =
CO3 =
14 + (4x 1)
= 18
18 x 2
= 36
12 + (16 x 3) = 60
Formula mass = 36 + 60
= 96 amu
1 mole = gram formula mass = 96g
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Calculations in Standard Grade Chemistry
2
Calculations Using the no. of moles = mass / gram formula mass triangle
Mass of
substance in
question (g)
m
n
gfm
Gram formula
mass - mass of 1
mole (g)
number of moles
of substance
(mol)
Examples
a)
How many moles are in 10.6 g of sodium carbonate (Na2CO3)
The question gives us m = 10.6g. We need to find n. First calculate gfm.
Formula = Na2CO3
Formula mass = (23 x2) + 12 + (16 x 3) = 106 amu
gfm = 1 mole = 106g
Now use triangle to calculate n.
n = m / gfm = 10.6 /106 = 0.1 mol
b)
(Note mol is the abbreviation for moles)
What is the mass of 2.5 moles of oxygen gas.
The question gives us n= 2.5. We must find m. First calculate gfm.
Formula = O2
(You must know your diatomics H2, N2, O2, Cl2, F2, Br2)
Formula mass = (16 x 2) = 32 amu
gfm = 1 mole = 32g
Now use triangle to calculate m
m = n x gfm = 2.5 x 32 = 80g
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Calculations in Standard Grade Chemistry
3
Calculations involving solutions and concentrations
number of moles
(mol)
volume of solution in
litres (l)
n
c
v
Concentration of
solution mol l-1
or mol / l
In these questions you have to remember to change volumes in cm3 (=mls) into litres.
Remember 1000 cm3 = 1 litre
To change cm3 into litres divide by one thousand.
Examples
a)
Find the concentration of a sodium hydroxide solution if 0.25 mol are
dissolved in 500cm3 of solution.
The question gives us n = 0.25 mol and v = 0.5 l (since 500 cm3 = 0.5 l)
c = n /v
b)
=0.25 / 0.5
=
0.5 mol l-1
How many moles of hydrochloric acid are present in 250 cm3 of a 0.1 mol l-1
solution?
The question gives us c = 0.1 mol l-1 and v = 0.25 l (250 cm3 = 0.25 l)
n=cxv
c)
=
0.1 x 0.25 =
0.025 mol l-1
What volume of 0.2 mol l-1 copper sulphate solution would contain 1 mole of
dissolved substance.
The question gives us c = 0.2 mol l-1 and n = 1 mol
v = n/c
=
1 / 0.2
=
5l
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Calculations in Standard Grade Chemistry
4
Calculations involving both triangles
In questions of this type you will have to calculate the formula mass and then use
n=m/gfm to calculate the number of moles. This number of moles will then be used in
the n= c x v triangle.
Example
What is the concentration of a solution which contains 4 g of sodium hydroxide in
100 cm3 of solution?
We want to find c and are given v = 0.1 l (100cm3) but need to calculate n.
Formula mass NaOH = 23 + 16 + 1 = 40 amu
1 mole = gfm
= 40g
Since question gives us m = 4g, then
n = m / gfm = 4 / 40 = 0.1 mol
We have 0.1 mol in 0.1 l so,
c = n/v
=
1/1
=
1 mol l-1
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Calculations in Standard Grade Chemistry
5
Acid / Alkali Calculations
In these questions you have to know that the equation for the reaction which is usually
given tells you in what proportions the acid and alkali neutralise each other:
HCl
+
NaOH
1 mol
H2SO4
1 mol
-->
NaCl
+
H2 O
-->
Na2SO4
+
2H2O
1 mol
+
2NaOH
2 mol
You have to use n = c x v to calculate the number of moles of acid or alkali and use the
equation to find out how many moles of the other substance is needed to neutralise it.
You may then have to use n= c x v again to find out any missing information.
Example
What volume of 0.5 mol l-1 sodium hydroxide is need to neutralise 500 cm3 of 0.25
mol l-1 hydrochloric acid.
For the acid v = 0.5 l and c = 0.25 mol l-1, so
n = c x v = 0.5 x 0.25 = 0.125 mol
Now the first equation above tells us that number of moles acid = no of moles alkali so
for the alkali NaOH, n = 0.125 mol and the question tells us v = 0.5 l, so
c=n/v
= 0.125 / 0.5 = 0.25 l = 250 cm3
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Calculations in Standard Grade Chemistry
Method 2
What volume of 0.5 mol l-1 sodium hydroxide is need to neutralise 500 cm3of 0.25
mol l-1 hydrochloric acid.
Use the equation p x v x c (acid)
=
p x v x c (alkali)
v and c are read from the question as normal but volumes can be in cm3. p stands for the
power of the acid or alkali and it means the number of (H+) in the acid formula or (OH-)
in the alkali formula.
In the question p (acid) = 1 since HCl has one H and p (alkali) = 1 since NaOH has one
OH.
v for alkali = unknown, c = 0.5 mol l-1
v for acid = 500 cm3 and c = 0.25 mol l-1
p x v x c (acid)
= p x v x c (alkali)
1 x 500 x 0.25
125
v
= 1 x v x 0.5
= 0.5 x v
= 125 / 0.5
= 250 cm3
Method 3
Since the equation shows that one mole of acid neutralises one mole of alkali, we know
that equal amounts and concentrations of acid and alkali neutralise each other. Since we
are given all the information about the acid we can write:
500 cm3 of 0.25 mol l-1 hydrochloric acid = 500 cm3 of 0.25 mol l-1 sodium hydroxide
But the alkali we are using has a concentration of 0.5 mol l-1 which is double the
concentration of the acid.
So we only need to use half the volume of alkali, ie 250cm3
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Calculations in Standard Grade Chemistry
6
Calculations from equations.
In these questions you will be given an equation and some information about one of the
components. You will be asked to find out a quantity of one of the other components.
The equation is used to get a relationship in moles. Formula masses then proportion is
used to find the final answer in g (or Kg)
Example 1
What mass of carbon dioxide is produced by roasting 10g of calcium carbonate as
shown?
CaCO3
-->
CaO
+
CO2
We are given the mass of calcium carbonate and we are to find the mass of carbon
dioxide. We are not interested in the calcium oxide
Need to calculate gfm of calcium carbonate = 40 + 12 + (3 x 16) =100g
. . . and of carbon dioxide = 12 + (16 x 2) = 44g
Rewrite the equation putting a box round what we're interested in:
CaCO3
-->
CaO
+
CO2
1 mole
------------------------------------->
1 mole (1 formula of each in eqtn)
100g
------------------------------------->
44g
1g
------------------------------------->
44 /100 = 0.44g
10g
------------------------------------->
0.44g x 10 = 4.4g
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Calculations in Standard Grade Chemistry
Example 2
What mass of ammonia (NH3) is formed when 5.6 g of nitrogen reacts with hydrogen
according to the equation below.
N2
+
3H2
-->
2NH3
We are given the mass of nitrogen and are to find the mass of ammonia. We need the
gfm of each
Nitrogen - gfm = 14 x 2 = 28g
Ammonia - gfm = 14 + (1 x 3) = 17g
Rewrite the equation putting a box round what we're interested in:
N2
+
3H2
-->
2NH3
1mole
--------------------------------------->
2moles (2 formulas of NH3)
28 g
--------------------------------------->
2 x 17 = 34 g
1g
--------------------------------------->
34 / 28 = 1.21g
5.6 g
--------------------------------------->
1.21 x 5.6 = 6.8g
Note if you had to calculate the number of moles of ammonia formed you do not
need to calculate the gfm of ammonia you simply proceed as shown below:
N2
+
3H2
-->
2NH3
1mole
--------------------------------------->
2moles (2 formulas of NH3)
28 g
--------------------------------------->
2 moles
1g
--------------------------------------->
2 / 34 = 0.059 mol
5.6 g
--------------------------------------->
0.059 x 5.6 = 0.33 mol
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Calculations in Standard Grade Chemistry
7
Empirical Formula
In these questions you are given the composition of a compound by mass (or %) and are
asked to find the empirical (simplest) formula. This is done by converting the masses
into moles, then finding the simplest ratio. Percentages just get treated in the same way
as masses.
Working can be set out in a neat table.
Example 1
Find the simplest formula of a compound containing 4g of Cu and 0.5g of oxygen.
1) Element
Cu
O
2) Mass (%)
4
0.5
3) n = m /gfm
4/64 = 0.0625
0.5 /16 = 0.03125
4) Divide by smallest
0.0625 / 0.03125 =2
0.03125 / 0.03125 =1
Note here the smallest refers to the smallest answer in column 3
The simplest ratio is Cu:O = 2:1 so the formula is Cu2O
Example 2
(i) a 0.75 g sample of a hydrocarbon is found to contain 0.6 g of carbon. Calculate its
empirical formula
(ii)
if the gram formula mass of the hydrocarbon is known to be 30, calculate the
molecular formula of the compound.
(i) hydrogen is the only other element so its mass = 0.75 - 0.6 = 0.15 g
1) Element
C
H
2) Mass (%)
0.6
0.15
3) n = m /gfm
0.6 /12 = 0.05
0.15 / 1 = 0/15
4) Divide by smallest
0.05 /0.05 = 1
0.15 / 0.15 = 3
Empirical formula = CH3
(ii)
The gfm = 30, but the mass of CH3 is 12 + (3 x 1) = 15
No of CH3 units in the formula = 30 / 15 = 2
Thus the actual formula is CH3CH3 or C2H6 (ethane)
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Calculations in Standard Grade Chemistry
8
Percentage by mass
These questions are often related to the percentage of an element in a fertiliser, but may
not be. The calculation can be summed up as:-
% by mass = (Total mass of element in compound)/(gram formula mass) x 100
Example 1
Find the percentage by mass of potassium in potassium oxide (K2O)
Gfm = (39 x 2) + 16 = 78 + 16 = 94
The compound has 2 K’s, so mass of element = (2 x 39) = 78
% potassium = 78/94 x 100 = 83%
Example 2
Find the percentage by mass of nitrogen in ammonium nitrate (NH4NO3)
Example 3
Find the percentage by mass of phosphorus in ammonium phosphate (NH4)3PO4
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