Download AQA-H_CALCULATIONS_WITH_MOLS

How do we know how much of each reactant to
use in a chemical reaction?
Ar Mr and percentages
The mol
Calculating formulae of compounds
Masses and equations
Volumes of gases and equations
Calculations involving half equations
in electrolysis
brings you
back here
end
RELATIVE ATOMIC MASSES
Atoms of each element have a different mass.
They have varying numbers of protons and neutrons.
We can measure masses of the atoms of each
element, comparing each with the isotope C-12.
This is the relative atomic mass, written Ar
It will always be given to you
H = 1
He = 4
C = 12
N = 14
O = 16
Na = 23
Mg = 24
Al = 27
S = 32
K = 39
Ca = 40
Fe = 56
Cu = 63
Ag =108
Cl = 35.5
(it’s a mixture of
isotopes :
3 parts Cl-35
and 1 part Cl-37)
RELATIVE FORMULA MASS OF COMPOUNDS (Mr)
The Compounds
formula of the
compound
tells
you thecombine.
numbers of
are
made when
elements
atoms
thatmay
have
together.
Bonds
becombined
ionic or covalent.
Add up the Ar of the individual atoms. Take into account
the number of atoms combined as shown in the formula.
magnesium chloride, MgCl2
Mr = 24 + (2 x 35.5) = 95
calcium nitrate,
Ca(NO
Mand
40 Cl
+ (2 x 62) = 164
you added
the A
3)r2of one Mg
r = two
102
62 164 150
(3 x12)is+doubled
(8 x 1) = 44
the alkane,
propane,
Mr =) group
3H8
Careful
! -- theCWHOLE
(NO
3
44 128 13
+ 32= +62(4 x 16)
ammonium
sulphate,
(NH4)2SO
= (18
one (NO
up
tor 14
+ (3x x2)16)
4 M
3) group adds
66 118 132
14 + 4 = 18
Click on the right answer for Mr (in the black box).
= 132
Two ammonium (NH4) groups plus one sulphur atom
Elements such as Oplus
Cloxygen
their Ar multiplied by two
2 and
2 needatoms.
four
Mr of oxygen gas = 2 x 16 = 32
CALCULATING THE % OF AN ELEMENT IN A COMPOUND
%=
mass of atoms of the element
 100
total M r of the compound
What is the % of nitrogen in the fertiliser ammonium nitrate, NH4NO3 ?
Mr = 14 + 4 + 14 + (3 x 16) = 80
Note : there are two nitrogen atoms present
2  14
 100  35 %
%=
80
What is the % of aluminium in its ore, bauxite, Al2O3 ?
54
 100  53 %
Mr = (2 x 27) + (3 x 16) = 102 % =
102
A mining engineer asks how much aluminium can we get
from 5000kg of bauxite ?
53
 5000  2650kg
100
What is a ‘mol’ ?
• it’s the amount of substance give by the relative
atomic or molecular mass, in grams.
Choose the answer
40 g
1 mol of calcium is ………..
71.0 g
1 mol of chlorine gas is ………..
12 80 40
35.5 71.0 24
16
1 mol of methane is ………..
g
16 30 13
106 g
1 mol of sodium carbonate is ………..
106 83 143
1 mol
all substances
contain the same number of particles.
It of
works
out that ……….
eg. 40g Ca has the same no. of atoms as there are
molecules in 16g of CH4
Ar : C = 12 Cl = 35.5WHY
Ca = 40 H = 1 Na = 23 O = 16
Ions : Na+ CO32-
‘Mols’ formulae.
mass
mols 
Mr
Also :
1 mol
of mass
zinc
g is : x M
formula
mols
1 mol
ofThe
Cl2is
is=65
71g
r
2 mols
of(zinc
? g
0.1 mol
is g)
?areg 130
in
7.1
Ar : N = 14 Ca = 40 O = 16 H = 1 S = 32
What is the mass of 3 mols of nitrogen gas, N2 ?
Choose
84g
48 84 16
Choose18.5g
What is the mass of 0.25 mol of
calcium hydroxide, Ca(OH)2 ? 296 14.25 18.5
How many mols are there in 4.9g of
sulphuric acid, H2SO4 ?
Choose
0.05
0.05 20 0.5
How many mols are there in 6.8g of
ammonia, NH3?
Choose
0.4
2.5 0.4 115.6
HOW WE CALCULATE THE EMPIRICAL FORMULA
OF A COMPOUND 1
The question will supply the masses of the elements which
combine together, or the % composition of the compound.
From this, we can calculate the RATIO of atoms that have
combined.
This gives us the empirical formula.
In the case of magnesium chloride, the
ratio Mg : Cl = 1 : 2 also gives the correct, final formula MgCl2
However, for the alkane, ethane, this calculation would
come up with a ratio C : H = 1 : 3
The empirical formula would be CH3
But the formula is not CH3 , it is actually C2H6.
A further calculation would be needed to move on from the
empirical formula to the molecular formula in this case.
HOW WE CALCULATE THE EMPIRICAL FORMULA
OF A COMPOUND 2
number of mols of atoms 
AlCl
AlCl
total mass in sample
3
Ar
When aluminium
is made, experiment
shows that
This 3
numberchloride
IS PROPORTIONAL
TO the
1.350g
of Al
combined
with 5.325g Cl
number
of atoms
present.
(relative atomic masses, Ar : Al = 27, Cl = 35.5)
1.350
 0.05
Number of mols of aluminium =
27
5.325
 0.15
Number of mols of chlorine =
35.5
AlCl3
Ratio
1:3
So theThe
ratioempirical
of atomsformula
combined
is : in aluminium chloride is 1 : 3
HOW WE CALCULATE THE EMPIRICAL FORMULA
OF A COMPOUND 3
total
massore,
in sample
Experiment
shows
that
32.0g
of
the
iron
haematite
number of mols of atoms 
contains 22.4g of iron. Ar
(relative atomic masses, Ar : Fe = 56, O = 16)
We need to subtract the mass of iron from the mass of the
get the
mass=of
oxygen combined.
Mass of compound
oxygen = to32.0
– 22.4
9.6g
22.4
 0 .4
Number of mols of iron =
56
9 .6
 0.6
Number of mols of oxygen =
16
0 .4
 1 .0
0 .4
0 .6
 1 .5
0 .4
Ratio
2:3
You need to get the ratio 0.4 : 0.6 as simple whole numbers.
A trick to help you : divide them both by the smallest of the numbers
Haematite is therefore
The iron ion here is Fe3+
and its chemical name is iron(III) oxide
Fe2O3
EMPIRICAL AND MOLECULAR FORMULAE
Experiment shows that the hydrocarbon butane is made up of
82.76% carbon and the rest is hydrogen.
We find that 14.5g of butane occupy a volume of 6 litres
(relative atomic masses, Ar : C = 12 , H = 1)
The ratio of atoms C : H = 2 : 5
Empirical formula : C2H5
% hydrogen = 100.00 – 82.76 = 17.24
Using ratios :
The relative formula mass (Mr) of any gas in grams always
82 .76
6.9conditions.
occupies
a
volume
of
24
litres
under
room

6
.
90
 1 .0
Number
of
mols
of
carbon
=
14.5 g butane occupies
6
litres
(you will always12
be told this) 6.9
Ratio
2:5
.24
17.24
Mr g butane occupies 2417litres
 2 .5
Number of mols hydrogen =
 17
.24
If this was the correct,
24
Mr 
 14.5  58
6
1
Empirical formula : C2H5
So the molecular formula is
This has the ratio 2 : 5 but Mr = 58
6 .9
molecular formula,
Mr = 24 + 5 = 29
C4H10
PRODUCING BALANCED EQUATIONS .
The products of a chemical reaction are made up from exactly
the same numbers of atoms as the reactants.
The atoms have just been redistributed into new combinations

So, balancing an equation means having the same
numbers of atoms in reactants & products.
balanced
Let’s have a look at this displacement reaction :
Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
Copper(s)
+ silver
nitrate(aq)
= groups
copper
nitrate(aq)
+ silver(s)
We
must
start
with
two
nitrate
in
the
reactants.
start
with
two
silver
ions
in
the
reactants.
1 mol of copper has reacted with 2 mols of silver nitrate. 1 mol
- produced
There
are
already
two
insilver
the2+copper
nitrate
made.
must
also
be
two
silver
atoms
produced.
Ag+ and
NO
Cu
and
NO
of copper
nitrate
and
2present
mols
of
have
been
3
3
These are the ions present in the compounds.
You use them to work out the formula of the compound which is neutral
The total masses of the two products of this reaction must
overall.
the total
masses
oflater
the during
two reactants.
You must equal
never change
these
formulae
balancing.
(this is shown in the next slide)
USING BALANCED EQUATIONS .
Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
1 mol is
64g
2 mols is
(2 x 170)
= 340g
1 mol is
188g
2 mols is
(2 x 108)
= 216g
Using the number of mols in the balanced equation, these are the
How muchmasses
silverofcan
bechemical
made ifinvolved
we added
of copper to
each
in the0.64g
reaction.
an excess of silver nitrate(aq)
?
404g
404g
64g
of
Cu
produce
216g
ofofsilver.
Total
mass
of will
reactants
= total
mass
products.
First
write
down
the
relevant
masses
from
the equation.
0.64g of Cu will produce ? g of silver.
Thennitrate)
write down
a (2
statement
of the question.
Mr (copper
=064
+
x
62)
=
188
.64
 216  2.16 g of silver
? == 108 + 14 + (3 x 16) = 170
Mr (silver nitrate)
64
(NO
= 14to
+ compare
(3 x 16) the
= 62
You
information from the equation with the
3) have
masses asked about in the question.
ATOMIC
MASSESin
Arthe
: Cu
= 64, N =
14, O
= 16,
Ag = 108of
InRELATIVE
this example,
the amounts
question
were
one
hundredth
(check all the figures on this slide)
those in the equation.
Using balanced equations to calculate masses of
substances in a reaction.
Magnesium2gaining
Mg + mass
O2 as
it2burns
MgO
2 mols are 1 mol is
2 x 24 =48g
32g
What mass of oxygen does 1.2g Mg need for burning ?
48g Mg needs 32g O2
1.2 48
1 .2
1 .2
1.2g Mg needs ? g O2
Choose
?=
 32
 32 0.8 g oxygen
32
 48
48 1.2
48
32
How much magnesium oxide is made if we burn 3.0g Mg ?
48g Mg produce 80
?? g magnesium oxide.
80
3.0g Mg produce ? g magnesium
oxide.
Use the equation
and Ar to choose :
40
Choose :
3.0
 80
48
48
 80
3.0
Ar : Mg = 24 O = 16 : mass = mols x Mr
48
 3 .0
80
?=
64
3.0
 80  5.0 g
48
Using balanced equations to calculate
masses of substances in a reaction.
Neutralisation.
2 NaOH +

H2SO4  Na2SO4 + 2 H2O
balanced
2this
mols
NaOH
1 mol Unbalanced
of H2SO4
is of
1 mol,
of neutralise
course
What mass of sulphuric acid is needed to neutralise
2.0g sodium hydroxide ?
98g H2SO4 neutralises 80
?? g NaOH alkali
Choose 40 , 80 or 20 g of NaOH
?? g sulphuric acid neutralises 2.0g NaOH ?
2.0
 98
80
80
 98
2.0
2.0
 80
98
2.0
 98  2.45 g H 2 SO4
?=
80
Mr (NaOH) = 23 + 16 + 1 = 40 Mr (H2SO4) = 2 + 32 + (4 x 16) = 98
mass = mols x Mr
Calculating masses and volumes of
substances in a reaction 1.
The volume of 1 mol of ANY gas under room conditions is
always 24 litres. (Ar : Ca = 40, C = 12, O = 16)
150g of limestone are heated. Calculate
(a) the mass of quicklime, CaO produced
(b) the volume of carbon dioxide gas produced.
CaCO3  CaO + CO2

balanced
mass
of 1 mol
volume
of 1 moloxide
mass ofcarbonate
1 mol
1 mol of calcium
produces
1 mol
of calcium
= 56g
= 24 litres
= 100g
and 1 mol of CO2 gas
150
 56  84 g
?=
100
150
(b) 100g limestone make 24 litres of CO2
 24  36litres
?=
100
150g limestone make ? litres of CO2
(a) 100g limestone produce 56g CaO
150g limestone produce ? g CaO
Mr : CaCO3 = 100 CaO = 56
mass in grams = mols x Mr
Calculating masses and volumes of
substances in a reaction 2.
The volume of 1 mol of any gas under room conditions is
always 24 litres. (Ar : Zn = 65, S = 32, O = 16, H = 1)
13g of zinc are added to excess sulphuric acid. Calculate
(a) the mass of zinc sulphate produced
(b) the volume of hydrogen gas produced.

Zn + H2SO4  ZnSO4 + H2
balanced
1 mol
zinc
reacts
molsulphate
of sulphuric acid to make
(a)
65g of
zinc
make
?? with
g of1zinc
161
13 gas.
1
mol
of
zinc
sulphate
plus
1
mol
of
hydrogen
 161
g
13g zinc make ?? g of zinc sulphate
? = 161
Choose from
113 32
322
Choose
(b) 65g of zinc make ??
of H2from
(g)
24 litres
6565
13
13
 65
 161
 161
161
13
65
Choose
?? from1324
48 2
13g of zinc make ?? litres
of H2(g)
? =  24  4.8litres
Choose ?? from
End of practice
13
 24
65
65 13
65
 24
 65
13
24
Calculating volumes of gases in a reaction.
1 mol of any gas under room conditions has a volume of 24 litres.
N2(g) + 3 H2(g)
2 NH3(g)
1 mol
3 mols
occupy
mols
occupy gas
1 mol
of occupies
nitrogen gas
reacts
with 3 mols2 of
hydrogen
24 litres
24)
(2 x 24)
to make(32 xmols
of ammonia gas.
= 72 litres
= 48 litres
volume
gas = no. of mols x 24 litres
We begin with
5m3 ofof
nitrogen.
(a) What volume of hydrogen is needed for complete reaction?
(b) What volume of ammonia is produced ?
In an all-gas reaction, the ratio of the numbers of mols
reacting is the same as the ratio of their volumes.
1:3:2
5m3 of N2 react with 15m3 of H2 & make 10m3 of ammonia.
Half equations for reduction at the negative electrode.
click3+
for+
solution
Cu2+ +2e  Cu
Al
3e  Al
the ion GAINS electrons
from the electrode
Ag+ +1e  Ag
Pb2+ +2e  Pb
Here are different ions that might be in a solution.
You need
to be
able
balance
the halfwould
equations.
In these
cases,
metallic
elements
appear
+ to
click
for solution

at the negative electrode.
2
2 H +2 e H
ELECTROLYSIS
AND
Half equations for oxidation at the positive electrode.
CALCULATIONS.
H+ is present in all acids and hydrogen gas
evolved
from the electrode.
Here electrons
are lostisby
the ion.
-
2Cl - 2 e  Cl
2 Br
- 2 e  Br
Cu - 2 e Cu
2 are produced, they bond together
2
When gaseous elements
in pairs to make a molecule.
2+
2

The balancing needs
 to2 include this.
2O
- 4e
O
In a special case, a positive copper electrode dissolves in a solution of
copper sulphate. Electrons are lost by the copper metal.
Quantities of elements released in electrolysis 1.
Electrolysis of copper chloride(aq)
Cu2+
+2e  Cu
Ar = 64
2Cl
- 2 e  Cl2
1 mol of any gas has a
volume of 24 litres
First, we must have the same number of electrons involved in
1 mol
of copper at
is the
produced
at the same
time
as 1time.
mol of
the reactions
two electrodes
in the
same
chlorine
This is already correct,
sogas
we need not worry.
64g of copper are produced along with 24 litres of chlorine.
We might then be asked :
What volume chlorine gas would be made in the same time as
3.2g of copper were deposited on the negative electrode ?
3.2
 24  1.2litres
64
Quantities of elements released in electrolysis 2.
Electrolysis of molten sodium bromide.
2
Na+
+1
2e 2 Na
2 Br
Ar : Na = 23
Br = 80
mass = mols x Mr
- 2 e  Br2
2 mols
ofget
sodium
are produced
time as
We
need to
two electrons
involvedin
in the
bothsame
half equations.
The same amount of electric charge must be lost/gained at the two
1electrodes
mol of in
bromine
(Br2)
the same time.
46g sodium are produced in the same time as 160g bromine.
…….. and, for example, how much bromine would be made
along with 9.2 g sodium ?
9.2
 160  32 g
46