For whoever asked for mole calculations. :)
It's detailed and has some example questions with working out, and some practice questions and tables for you to do too. Hope it helps!
Note that these are taken from my class book. They are not my materials.
It's detailed and has some example questions with working out, and some practice questions and tables for you to do too. Hope it helps!
Note that these are taken from my class book. They are not my materials.
1.15 calculate relative formula masses (Mr) from relative atomic
masses (Ar)
Relative formula
mass: the sum of masses of all the atoms in a formula
Relative atomic
mass: the average mass of all the isotopes of an element
Examples:
Species

Formula

No.
of atoms of each element in one molecule

Ar
of each element

Sum
of Ar

Relative
Formula Mass (Mr)

Carbon dioxide

CO2

1 carbon
2 oxygen

C=12
O=16

12 + (16 x 2)

44

Oxygen

O2

2 oxygen

O=16

16 x 2

32

Aluminium oxide

Al2O3

2 aluminium
3 oxygen

Al=27
O=16

(27 x 2) + (16 x
3)

102

Aluminium sulphate

Al2(SO4)3

2 aluminium
3 sulphur
12 oxygen

Al=27
S=32
O=16

(27 x 2) + (32 x
3) + (16 x 12)

342

To calculate
relative atomic mass (Ar): the sum of the relative
abundance of each isotope multiplied by its mass number, divided by 100.
E.g. About 75% of
all chlorine atoms have a mass number of 35 (18 neutrons), about 25% have a
mass number of 37 (20 neutrons).
The relative atomic
mass of chlorine is:
(75 x 35) + (25 x
37) / 100 = 35.5
1.18 carry out mole calculations using relative atomic mass (Ar)
and relative formula mass (Mr)
n=m/Mr
Where n=number of
moles, m=mass in grams and Mr=relative formula mass
e.g. how many moles
are there in 120g of NaOH(s)?
n=m/Mr = 120g/40 = 3
There are 3 moles.
Try filling this
out:
Substance

Formula

Mr

Mass of sample (g)

Number of moles in
sample (n=m/Mr)

E.g. Water

H_{2}O

18

9

0.5

Carbon dioxide

88


Ammonia

1.7


Sulphur dioxide

0.64


Sulphur trioxide

80


Hydrogen bromide

24.3


Sulphuric acid

0.098


Nitric acid

3.15


Sodium nitrate

21.25


Sodium carbonate

53

Example:
What mass of
hydrogen is produced when 192g of magnesium is reacted with hydrochloric acid?
Step 1: write a
balanced equation for the reaction:
Mg + 2HCl à MgCl2
+ H2
Step 2: under the
equation, fill in all the information you have been given, and also the Mr of
the species involved:
Equation

Mg
+ 2HCl à MgCl_{2 } + H_{2}


Mass

192g

This is the value you want to find


M_{r}

24

2


n

Step 3: work out the number of moles of the reactant, using n=m/M_{r }then use the mole ratio of the equation to fill in the number of moles of the product:
Equation

Mg
+ 2HCl à MgCl_{2 } + H_{2}


Mass

192g

This is the value you want to find


M_{r}

24

2


n

8

à
1:1
mole ratio

8

Step 4: now you have the M_{r }and number of moles of the product, rearrange n=m/M_{r }to find the mass:
m=n x M_{r}
_{}
Equation

Mg
+ 2HCl à MgCl_{2 } + H_{2}


Mass

192g

16g


M_{r}

24

2


n

8

à
1:1 mole ratio

8

_{}
Once you are familiar with the process, you don't need a calculation frame, but keep your workings well ordered and logical.
 What mass of oxygen is needed to react with 8.5g of hydrogen sulphide (H2S)?
2H2S + 3O2 à 2SO2 + 2H2O
Find
out how many moles of hydrogen sulphide there are, since you have its mass.
n=m/Mr
n=8.5/34=0.25
Mole
ratio= 2:3 (2 moles of H2S : 3 moles of O2)
(0.25/2)
x 3 = 0.375  0.375 moles of O2 react with 0.25 moles of H2S, now you want to
find the mass.
n=m/Mr
m=
n x Mr
m=
0.375 x 32 = 12g
 Railway lines are welded together by the thermite reaction, which produces molten iron. What mass of iron is formed from 1kg of iron oxide?
Fe2O3 + 2Al à 2Fe + Al2O3
n=m/Mr
n=1000g/160=6.25
Mole
ratio= 1:2
6.25
x 2=12.5
m=n
x Mr
m=
12.5 x 56 = 700g
Practice:
 What mass of sodium hydroxide is formed when 46g of sodium reacts with excess water?
 Calculate the mass of water formed when 32g of oxygen reacts with excess hydrogen.
 Calcium carbonate thermally decomposes to form calcium oxide and carbon dioxide. If 44g of carbon dioxide are collected, what mass of calcium oxide is formed?
1.26 carry out mole calculations using volumes and
molar concentrations
A
solution contains a dissolved solute in a certain amount of solvent.
The
concentration of a solution tells us how many moles of the solute are dissolved
in one litre
(1 dm3) of the solvent.
(1 dm3) of the solvent.
The
units for concentration are mol/dm3, and this is often shortened to M.
Concentration
can also be measured in grams per litre.
n=
v x c
Where
v= volume (in dm3) and c= concentration (in mol/dm3)
NB:
sometimes they give you the volume in cm3 so be careful, convert it to dm3 by
dividing by 1000
(1
dm3 = 1000 cm3)
Or,
you can just use: n= (v x c) / 1000
Remember
to convert cm3 to dm3!
Solution

Formula

Concentration (mol m^{3})

Concentration
(g dm^{3})

Volume

Number of moles

Sodium hydroxide

NaOH

1

500 cm^{3}


Hydrochloric acid

HCl

0.5

2 dm^{3}


Sodium chloride

NaCl

58.5

4


Potassium chloride

KCl

200 cm^{3}

0.2


Ammonium chloride

NH_{4}Cl

0.25

250 cm^{3}


Silver nitrate

AgNO_{3}

2

0.5


Lithium iodide

LiI


Sulphuric acid

H_{2}SO_{4}

0.2

1.8


Potassium nitrate

KNO_{3}

150 cm^{3}

0.15

 25cm3 of a solution of 0.1M NaOH is neutralised by 50cm3 of HCl. What is the concentration of the HCl?
NaOH + HCl à NaCl + H2O
NB:
remember to convert volumes to dm3!
n=
v x c
n=
0.025 x 0.1=0.0025 mole
Mole
ratio= 1:1, so moles of HCl= 0.0025 mole
C=
n/v
C=
0.0025/ 0.05=0.05 mol/dm^{3}
^{}
Practice:
 What mass of silver chloride precipitate will be produced if 25 cm3 of 0.1M silver nitrate is added to excess sodium chloride solution?(Answer may not be exact, so you give it to 2 significant figures here because the most exact data they give you here is to 2 sig. fig. which is the volume.)
 What mass of magnesium will react with 10cm3 of 1M HCl? Extra: and what volume of hydrogen will be formed? (1 mole of any gas at RTP is 24dm3)
Answers:
Substance

Formula

Mr

Mass of sample (g)

Number of moles in sample (n=m/Mr)

E.g. Water

H_{2}O

18

9

0.5

Carbon dioxide

CO2

44

88

2

Ammonia

NH3

17

1.7

0.1

Sulphur dioxide

SO2

64

0.64

0.01

Sulphur trioxide

SO3

80

80

1

Hydrogen bromide

HBr

81

24.3

0.3

Sulphuric acid

H2SO4

98

0.098

0.001

Nitric acid

HNO3

63

3.15

0.05

Sodium nitrate

NaNO3

85

21.25

0.25

Sodium carbonate

Na2CO3

106

53

0.5

2 Na

+

2 H_{2}O

à

2 NaOH

+

H_{2}

46g

?


1:1 ratio


23g

:

40g


46g

:

80g

2. Calculate the mass of water formed when 32g of oxygen reacts with excess hydrogen.
2 H_{2}

+

O_{2}

à

2 H_{2}O

1

:

2


32g

:

36g
(m = n x M_{r}
m = 2 x 18)

3. Calcium carbonate thermally decomposes to form calcium oxide and carbon dioxide. If 44g of carbon dioxide are collected, what mass of calcium oxide is formed?
CaCO_{3}

à

CaO

+

CO_{2}

1

:

1


56g

:

44g


*Remember that one mole of a substance is equivalent to its RFM in
grams. For instance with CaO, the M_{r} of Ca is 40, and for O it is
16. Thus the RFM of CaO is 40 + 16 = 56. And since 44g of CO_{2} is 1
mole, then one mole of CaO must have been produced too (look at balanced equation),
and 1 mole of CaO = 56g.

Solution

Formula

Concentration (mol m^{3})

Concentration
(g dm^{3})

Volume

Number of moles

Sodium hydroxide

NaOH

1

40

500 cm^{3}

0.5

Hydrochloric acid

HCl

0.5

18.25

2 dm^{3}

1

Sodium chloride

NaCl

1

58.5

4 dm^{3}^{}

4

Potassium chloride

KCl

1

74.5

200 cm^{3}

0.2

Ammonium chloride

NH_{4}Cl

0.25

13.375

250 cm^{3}

0.0625

Silver nitrate

AgNO_{3}

2

340

0.25 dm^{3}^{}

0.5

Lithium iodide

LiI

1

134

1 dm^{3}

1

Sulphuric acid

H_{2}SO_{4}

0.2

19.6

9 dm^{3}

1.8

Potassium nitrate

KNO_{3}

1

101

150 cm^{3}

0.15

What mass of silver chloride precipitate will be produced if 25 cm3 of 0.1M silver nitrate is added to excess sodium chloride solution?(Answer may not be exact, so you give it to 2 significant figures here because the most exact data they give you here is to 2 sig. fig. which is the volume.)
AgNO_{3}

+

NaCl

à

AgCl

+

NaNO_{3}

1

:

1


Moles of AgNO_{3}

=

v

x

C


=

0.025

x

0.1


=

0.0025 mol


Moles of AgCl

=

0.0025 mol

Due to 1:1 ratio


m

=

n

x

RFM


=

0.0025

X

143.5


≈

0.36g

(volume is only 2 sig. fig.)

2. What mass of magnesium will react with 10cm3 of 1M HCl? Extra: and what volume of hydrogen will be formed? Assume RTP conditions  25°C and 1 atm (1 mole of any gas at RTP is 24dm3)
Mg_{}

+

2 HCl

à

MgCl_{2}

+

H_{2}

1

:

2

:

1


Moles of HCl_{}

=

v

x

C


=

0.01

x

1


=

0.01 moles


Moles of Mg

=

0.005

(due to 1:2 ratio, 0.01 moles of HCl will react with 0.005 moles of
Mg)


Mass of Mg

=

n

x

RFM


=

0.005

x

24


=

0.012g


1 mole of H_{2}

=

24 dm^{3}

=

24,000 cm^{3}


0.005 mole of H_{2}

=

0.005

x

24,000


=

120 cm^{3}

So that's done.. hope the answers help. And good luck with your IGCSEs! I really can't update like this anymore, apologies but this was just to complete this post. Hope it helps anyway. :)
_{}
_{}
Are there answers anywhere?
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ReplyDeletea m a z i n g.
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I was looking at the first answers table and
ReplyDeleteI'm not sure how the number of moles in the sample of sodium carbonate is 0.021?
I got m/Mr as 53/106.. so the answer was 0.5? :)
Ah yes you're right my bad! I forgot I changed some of the qs from my class exercise book to make the numbers easier, the mass in my book was 2.25g which gave 0.021..moles as an ans. (Didn't like the many decimal places so I changed the mass to 53, in which case yes the answer is 0.5 moles.)
DeleteSorry! And well spotted, will change it now. :D
Ok, just checking  thank you!! :) And thank you for this blog in general.. it is literally getting me through science revision at the moment!!! :D
DeleteYou're welcome :) Btw I thanked you at the bottom of the post haha (y) and good luck with your IGCSEs, I'm glad the blog helps! :D
Deletehow do you work out the mole ratio?! :s
ReplyDeleteYou look at the balanced chemical equation. For instance with:
DeleteMg + 2 HCl > MgCl2 + H2
The ratio of Mg to HCl would be 1:2, since 1 mole of magnesium reacts with 2 moles of hydrochloric acid.
Does that make sense? :)
Wait, so why is it 1:1 in the example?
Deletebecause that mole ratio is of Mg to H2, you're trying to find the mass of hydrogen so that is what's relevant in this example qs. :)
DeleteOhhhhh! I get it! Thank you so much for answering! I finally get it :D
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ReplyDeleteI would like to recommend your article on Sodium Hydro sulphide. For further information, you can refer Sodium Hydro sulphide
ReplyDeleteIs the Mr for HBr2 81 or 161? Because on the first table in the answers section, that is what it says for the hydrogen bromide row.
ReplyDeleteMy apologies, it's meant to be 'HBr', I've changed it and the Mr would therefore be correct at 81. Thanks for spotting it. :)
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ReplyDeletedont understand why '(due to 1:2 ratio, 0.01 moles of HCl will react with 0.005 moles of Mg)' works in the 2nd volumes question
ReplyDeletecan u put some more question for practice
ReplyDeleteThis was very useful! thank you :)
ReplyDeleteamazing blog
ReplyDeletewhy in Mg+2HCl in the end we are dividing 0.01 by 2 other then multiplying????????
ReplyDeletethank you
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ReplyDelete0.005*24 does not equal 0.012g in the magnesium question. Correct mass should be 0.12
ReplyDeleteThanks so much!!! It has helped me a lot....hope I get good marks in my exam tomorrow.. Thanks!
ReplyDeleteSorry to bother you but I was just wondering for the first answers table, the answer for Sodium Bromide is different to what I got and I was really confused.
ReplyDeleteSo Moles = Mass/Mr so therefore Moles = 24.3/81 which is = 0.3 but for your answer it was 0.22 so I was just wondering if I did something wrong...? Thank you for this Blog, it saved my life
Hi there, if you are referring to the Hydrogen Bromide answer then you are quite right! It's my bad. I'll change it, thanks for spotting that! And no problem, really glad the blog helps. :)
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ReplyDelete