Programme in mole calculation
Step 1
Start at the very beginning
Atoms are very small indeed! If we draw a line 1 metre long, 6,000,000,000 (6 billion) atoms
could be lined end to end.
So a scientist cannot count atoms, ions or molecules directly. They are far too small and
numerous. Instead a scientist counts particles by weighing. This is rather like in a bank. A bank
cashier has not got the time to count every coin. She weighs the bags of coins on special scales
which tell her how much they are worth.
Hydrogen is the lightest atom, we say that hydrogen has a relative atomic mass of 1. In 1g of
hydrogen atoms we say that we have 1 mole of hydrogen atoms.
Experiments have shown that an atom of carbon weighs 12 times as much as an atom of hydrogen,
so the relative atomic mass of carbon is 12. In 12g of carbon atoms we say that we have 1 mole
of carbon atoms.
Let's look at the following table which shows the relation between the mass of one mole element
and the relative atomic mass.
Element
hydrogen
carbon
relative atomic mass
mass of one mole element
1
12
1g
12g
All answers throughout the programme should be collected to 3 sig. figures
Class work 1:
Use your Periodic Table to find out the mass of one mole of the following atoms.
a.
Helium
b.
Iron
c.
Potassium
d.
Sulphur
e.
Platinum
f.
Tin
h.
Phosphorus.
g.
Silver
Answer:
a.
b.
c.
d.
e.
f.
g.
h.
Actually one mole of atoms has 6.00x1023 atoms. The scientific "magic number" is called
Avogadro's number or Avogadro's constant (L).
P.2
Example 1:
How many atoms are there in 6g of carbon?
Answer:
The relative atomic mass of carbon is 12. In 12g of carbon there are 6.00x1023 atoms which equal
one mole of carbon atoms. In 6g of carbon there is only half mole,
no. of moles of carbon =
6
= 0.5
12
Class work 2a:
Use your periodic table to help you. How many atoms are there in
a.
1.0g of helium
b.
28.0g iron
c.
13.0g potassium
d.
6.4g sulphur
e.
32.5g platinum
f.
119g tin
g.
27.0g silver
h.
10.3g phosphorus?
Answer:
a.
b.
d.
e.
g.
h.
Class work 2b:
c.
f.
How many moles of the following atoms?
a.
3.0x1023 hydrogen atoms
a.
b.
1.5x1023 oxygen atoms
b.
c.
1.5x1023 nitrogen atoms
c.
d.
3.0x1023 carbon atoms.
d.
Put your answers in table form:
Hydrogen
number of atoms
number of mole
What do you discover?
3.0x1023
oxygen
1.5x1023
nitrogen
1.5x1023
carbon
3.0x1023
P.3
Step 2
Masses into moles - moles into masses
The following calculation will make use of the equation:
no. of moles of atoms =
actual mass of the atoms
molar mass of the atoms
Example 2:
How many moles of atoms are there in 4g of calcium? [Ar (Ca) = 40]
Answer:
By rearranging equation above, it is just as easy to find out the mass of a fraction of a mole.
no. of moles of calcium =
4
= 0.1
40
Example 3:
What is the mass of 1/8 mole of copper? [Ar (Cu) = 64]
Answer:
First equation above must be rearranged:
mass = no. of moles x molar mass = 1/8 x 64 = 8 (g) (don't forget the unit)
Classwork 3:
1.
How many moles of atoms are there in
a.
d.
20g calcium
78g potassium
b.
e.
54g aluminium
8g sulphur
c.
f.
11.2g iron
2.4g magnesium?
Answer:
a.
b.
c.
d.
e.
f.
2.
What is the mass of
a.
1/10 mole of sodium atoms
c.
1/3 mole of carbon atoms
e.
1/16 mole of magnesium atoms
Answer:
a.
e.
b.
f.
b.
d.
f.
c.
2 moles of silver atoms
8 moles of iron atoms
1/4 mole of copper atoms?
d.
P.4
Classwork 4:
Here are the creep point questions. Avogadro's number = 6.00x1023.
How many atoms are there in
a.
1 mole of carbon atoms
b.
2 moles of oxygen atoms
c.
1/2 mole of sulphur atoms
d.
1/3 mole of lead atoms?
e.
What mass of magnesium contains 2.00x1023 atoms?
f.
What mass of carbon contains 2.00x1023 atoms?
g.
What mass of magnesium has five times as many atoms as 2g of carbon?
h.
What mass of potassium has the same number of atoms as 8g of magnesium?
Answer:
a.
b.
c.
d.
e.
f.
g.
h.
Step 3
Diatomic molecules
Sometimes an element exists as molecules. Molecules are groups of atoms held together by
chemical bonds.
A hydrogen atom has a relative atomic mass of 1. A hydrogen gas molecule (H2) has two
hydrogen atoms, then the relative molecular mass (Mr) of the molecule is 2.
To have one mole of molecules we should need to weigh out 2g of hydrogen gas.
Let's look at the following table which shows the relation between the mass of one mole
molecules and the relative molecular mass.
molecule
relative molecular mass
mass of one mole molecules
hydrogen (H2)
2
2g
nitrogen (N2)
28
28g
Referring to the equation on page 3, now we come to have a similar equation:
no. of moles of molecules =
actual mass of the molecules
molar mass of the molecules
P.5
Example 4:
How many moles of nitrogen molecules are present in 7g of nitrogen gas?
Answer:
First note that the question did not ask about nitrogen atoms. It concerns nitrogen molecules, N2.
Each molecule contains two atoms of nitrogen. Since each atom has a relative atomic mass of 14,
then the relative molecular mass of nitrogen molecule is 2 x 14 = 28.
Example 5:
What is the mass of
1
mole of oxygen gas? [Ar (O) = 16]
10
Answer:
Oxygen is a diatomic gas also. It has the relative molecular mass of 2 x 16 = 32. Now rearrange
the basic equation on page 4:
mass = number of mole x relative molecular mass =
1
x 32 = 3.2 (g)
10
Classwork 5:
1.
How many moles of
a.
chlorine molecules are present in 7.1g of chlorine gas,
b.
chlorine atoms are present in 7.1g of chlorine gas,
c.
oxygen molecules are present in 64g of oxygen gas?
d.
oxygen atoms are present in 64g of oxygen gas?
Answer:
a.
b.
c.
d.
2.
What is the mass of
a.
1/8 mole of oxygen gas
b.
1/4 mole of bromine gas
c.
2 moles of chlorine gas
d.
0.25 mole of iodine?
Answer:
a.
d.
b.
c.
P.6
Step 4 Compounds
Some compounds are also made up of molecules. But a compound contains atoms of different
elements joined together by chemical bonds.
The formula for water is H2O. This means that a molecule of water contains two atoms of
hydrogen and one atom of oxygen.
To work out the mass of one mole of a compound we use the same idea as in Step 3. The relative
atomic masses of all the atoms in the compound are added together.
Let's look at the following table which shows the relation between the mass of one mole
compound and the relative formula (molecular) mass.
molecule
relative (formula)
molecular mass
mass of one mole
compound
water (H2O)
1x2+16 = 18
18g
ammonia (NH3)
14+1x3 = 17
17g
Example 6:
What is the mass of one mole of carbon dioxide? [Ar(C) = 12, Ar(O) = 16]
Answer:
The carbon dioxide molecule contains 1 carbon atom and 2 oxygen atoms. The relative molecular
mass is equal to 12 + 16x2 = 44. The mass of one mole of carbon dioxide is 44g.
Example 7:
What is the relative formula mass of magnesium nitrate Mg(NO3)2? [Ar(Mg) = 24 Ar(N) = 14,
Ar(O) = 16]
Answer:
Each mole of magnesium nitrate contains 1 mole magnesium ions, 2 moles of nitrogen and 6
moles of oxygen. So the relative formula mass is 24 + 2x14 + 6x16 = 148.
Classwork 6:
Find the relative formula mass of
a.
copper(II) oxide
b.
d.
copper(II) carbonate e.
Answer:
sulphur trioxide
c.
copper(II) sulphide
zinc nitrate
f.
ammonium carbonate.
a.
b.
c.
d.
e.
f.
P.7
Referring to the equation on pages 3 and 5, now we have another similar equation:
no. of moles of any substance =
actual mass of the substance
molar mass of the formula
Example 8:
How many moles are contained in 72g of water?
Answer:
The relative molecular mass of water is 18, so
no. of moles of water =
72
=4
18
Example 9:
How many moles of hydrogen atoms and oxygen atoms are contained in 72g of water?
Answer:
Since each mole of water contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms, 4
moles of water contain 8 moles of hydrogen atoms and 4 moles of oxygen atoms.
Classwork 7:
Find the number of moles contained in
a.
40g copper(II) sulphate
b.
282g of zinc nitrate
c.
60g sulphur trioxide
d.
32g ammonium carbonate.
Answer:
b.
a.
c.
d.
Classwork 8:
Find the mass of
Answer:
a.
0.5 mole copper(II) carbonate
a.
b.
1/4 mole copper(II) oxide
b.
c.
1/10 mole of sulphur trioxide
c.
d.
0.25 mole ammonium carbonate.
d.
P.8
Revision exercise 1
Take Avogadro's constant = 6.0x1023
1.
1.5g hydrogen gas contain
mole(s) of hydrogen molecules,
mole(s) of hydrogen atoms,
hydrogen molecules and
hydrogen
atoms.
2.
3.0x1023 particles of CO2 is equivalent to
mole(s) of CO2,
3.
4.
5.
6.
7.
g of CO2.
Find the relative molecular mass of
a.
sulphuric acid, H2SO4
c.
hydrated copper(II) sulphate, CuSO4 5H2O.
b.
ammonium sulphate, (NH4)2SO4
What is the mass of
a.
2 moles of magnesium atoms b.
c.
0.75 mole of calcium atoms?
1/2 mole of copper atoms
How many atoms are there in
a.
1 mole of carbon atoms
c.
1/3 mole of sulphur atoms?
b.
0.5 mole of oxygen atoms
How many atoms are there in
a.
39.0g of potassium
c.
60.0g of calcium?
b.
6.35g of copper
b.
0.5 mole of sodium thiosulphate, Na2S2O3
What is the mass of
a.
4 moles of water
c.
0.25 mole of ammonium dichromate, (NH4)2Cr2O7?
P.9
Step 5 Molarity – a unit of concentration of solution
Concentration is a measure of how much solute dissolved in a unit volume of the solution. For
example, if you dissolve 10g sugar in water so that the volume of solution is 10cm3, then the
concentration of the sugar solution is 10g in 10cm3 or 1g/cm3.
Then the unit of concentration is g/cm3, kg/cm3, kg/m3 etc.
It is very convenient to use figure to compare the concentrations of different solutions. For
example, two solutions
B
A
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10g sugar
?
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W
Molarity is one kind of units of concentration – the unit is mol/dm3.
3
m
c
0
0
0
1
=
3
m
d
1
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molarity =
no. of moles of solute
volume of solution in dm 3
no. of moles of NaCl =
mass of NaCl
formula mass of NaCl
no. of moles of NaCl =
molarity =
58.5
=1
58.5
1
(M) = 10.0M
0.1
Class work 9 :
1.
Calculate the concentration of the following solutions.
a.
14.625g sodium chloride in 250cm3 solution
no. of moles of sodium chloride =
=
n
o
i
t
u
l
o
s
e
h
t
f
o
y
t
i
r
a
l
o
m
P.10
b.
47.85g copper(II) sulphate in 0.5dm3 solution.
c. 44.8g sulphur dioxide in 700cm3 solution.
Class work 10
1.
In order to prepare the following solutions, how many grams of the solution is required?
.
a
200cm3, 0.5M copper(II) chloride solution
no. of moles of copper(II) chloride required =
mass of copper(II) chloride required =
.
b
.
c
2.
250cm3, 0.45M sodium hydroxide solution
2dm3, 1.5M potassium nitrate solution
What is the volume of the solutions in order to make
a.
0.1M solution by dissolving 13.45g copper(II) chloride?
no. of moles of copper(II) chloride dissolved =
volume of the solution (dm3) =
b. 0.025M solution by dissolving 1.825g hydrogen chloride?
no. of moles of hydrogen chloride dissolved =
volume of the solution (dm3) =
P.11
Answer of the class work
Class work 1
a.
e.
4.00g
195g
b.
f.
56.0g
119g
c.
g.
39.0g
108g
d.
h.
32.0g
31.0g
b.
f.
3.0x1023
6.0x1023
c.
g.
2.0x1023
1.5x1023
d.
h.
1.2x1023
2.0x1023
b.
0.25
c.
0.25
d.
0.5
Class work 2a
a.
e.
1.5x1023
1.0x1023
Class work 2b
a.
0.5
Hydrogen
oxygen
nitrogen
carbon
number of atoms
3.0x1023
1.5x1023
1.5x1023
3.0x1023
number of mole
0.5
0.25
0.25
0.5
Class work 3
1.
a.
e.
0.500
0.250
b.
f.
2.00
0.100
c.
0.200
d.
2.00
2.
a.
e.
2.30g
1.50g
b.
f.
216g
15.3g
c.
4.00g
d.
448g
Class work 4
a.
e.
6x1023
8.00g
b.
f.
12x1023
4.00g
c.
g.
3x1023
20.0g
d.
h.
2x1023
13.0g
Class work 5
1.
a.
0.100
b.
0.200
c.
2.00
d.
4.00
2.
a.
4.00g
b.
40.0g
c.
142g
d.
63.5g
Class work 6
a.
e.
79.5
189
b.
f.
80.1
96.0
c.
95.6
d.
124
b.
1.49
c.
0.750
d.
0.333
Class work 7
a.
0.251
P.12
Class work 8
a.
61.8g
b.
19.9g
c.
8.00g
d.
Revision exercise 1
1.
0.750, 1.50, 4x1023, 9x1023.
2.
0.500, 6.00
3.
a.
98.0
b.
132
c.
250
4.
a.
48.0g
b.
31.8g
c.
30.0g
5.
a.
6x1023
b.
3x1023
c.
2x1023
6.
a.
6x1023
b.
6x1022
c.
9x1023
7.
a.
72.0g
b.
79.0g
c.
63.0g
0.25, 1.0M
b.
0.6M
c.
1.0M
c.
303.3g
class work 9
1.
a.
class work 10
1.
a.
0.1, 13.45g
b.
4.5g
2.
a.
0.1, 1.0dm3
b.
0.05, 2.0dm3
24.0g