moles - As/ gce chemistry notes

CALCULATIONS FROM EQUATIONS


Amounts of substances, the mole , molar mass, molar volume of gases (24 litre/dm3) at room temperature and pressure, the Avogadro constant and their use in calculations. (Determination of the Avogadro constant is not required.)


A major problem confronting a chemist when carrying out reactions is to try and understand what they see in the reaction vessels in terms of what is happening between individual atoms and ions, considering their small size;
• The diameter of an atom is about 10 -10 m
OR
• Two million hydrogen atoms would cover an average full stop.
• If a single drop of water could be magnified to the size of the Earth, then on the same scale its atoms would be the size of golf balls.

The masses of atoms
Atoms are very small. A single hydrogen atom for instance, weighs about 2 x 10-24 g (0.000 000 000 000 000 000 000 002g). Numbers this small are awkward to use, so instead of using the actual masses of atoms, a simpler way of expressing the mass of an atom has been found.


Relative atomic mass Ar

When you have finished this section you should be able to:
· Define relative atomic mass in terms of carbon-12
· Deduce relative atomic masses for the elements from the Periodic Table

Since hydrogen is the smallest and lightest atom it was originally used as the standard atom against which all other atoms would be compared.

Relative atomic mass = mass of one atom of an element
mass of one atom of hydrogen

Thus Ar(H) = 1
Ar(O) = 16 each oxygen atom is 16 times heavier than a hydrogen atom
Ar(S) = 32
Ar(C) = 12
Relative atomic masses are now determined by mass spectrometry (see later), and since volatile carbon compounds are used a lot in mass spectrometry, the mass of an atom of 126C is now taken as the reference standard. The modern definition of relative atomic mass is:
Relative atomic mass = mass of one atom of an element
1/12 the mass of one atom of carbon-12


On this scale, carbon-12 has a relative atomic mass of 12.000000, carbon has a relative atomic mass of 12.011011 and hydrogen has a relative atomic mass of 1.00797.
Relative atomic masses are not always whole numbers due to the existence of isotopes. e.g.. Ar(Cl) = 35.5
Note that since relative atomic masses are ratios they have no units.
Approximate relative atomic masses can be found from the mass numbers in your Periodic Table.


Relative molecular mass Mr

When you have finished this section you should be able to:
· Calculate the relative molecular mass of a substance given its formula and a table of relative atomic masses.

A molecule consists of atoms joined together. The mass of a molecule can be found by adding up the masses of the atoms it contains.
The relative molecular mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of the compound.

carbon dioxide CO2
1 atom of C, Ar(C) = 12 12
2 atoms of O, Ar(O) = 16 32
Mr(CO2) = 44
i.e. 1 molecule of CO2 weighs 44 times as much as 1 atom of hydrogen.

What is the relative molecular mass of sulphuric acid H2SO4?

Mr(H2SO4) = (2 x1) + 32 + (4 x 16) = 98

What is the relative molecular mass of magnesium nitrate Mg(NO3)2
Mr[Mg(NO3)2] = 24 + {2 x (14 + 3 x 16)}
= 24 + {2 x (14 + 48)}
= 148

What is the relative molecular mass of copper sulphate crystals CuSO4.5H2O.
Mr[CuSO4.5H2O] = 64 + 32 + (4 x 16) + 5 x {(2 x 1) + 16}
= 64 + 32 + 64 + (5 x 18)
= 250

Many compounds consist of ions, not molecules. For ionic compounds the formula represents a formula unit , rather than a molecule of the compound. A formula unit for sodium sulphate is Na2SO4. The term relative formula mass can be used for ionic compounds.

Exercise 1
Work out the relative molecular masses of these compounds:
(a) NaOH [40]
(b) KCl [ ]
(c) MgO
(d) Ca(OH)2
(e) HNO3
(f) CuCO3
(g) NH4NO3
(h) CaSO4
(i) Na2CO3.10H2O
(j) Mg(HCO3)2

THE MOLE

The mole is a convenient way of describing a large number of objects (in this case atoms, molecules or ions)
c.f. a dozen eggs, a gross of apples, a ream of paper or a bag of sugar.

It is similar in principle to the way a bank cashier 'counts' the number of coins in a bag by weighing the whole bag.

Suppose
A 2p coin is twice as heavy as a 1p coin.
A 5p coin is five times as heavy as a 1p coin.
A 10p coin is ten times as heavy as a 1p coin.
If a shopkeeper gives the cashier four bags of coins with the following weights
1p bag (100g), 2p bag (200g), 5p bag (500g) and 10p bag (1000g)
What do you know about the number of coins in each bag?


Consider
1p 2p 5p 10p
1 coin 1 2 5 10
2 coins 2 4 10 20
3 coins 3 6 15 30
4 coins 4 8 20 40

Whatever the actual weight of the coins the ratio of weights is always constant 1 : 2 : 5 : 10

As long as equal numbers of coins are taken their weights will always be in the same ratio.

Moles of Atoms
Now consider the following atoms

H C S Mg Ca
Ar 1 12 32 24 40
1 atom
2 atoms
10 atoms Ratio of their weights will always be the same
100 atoms as the ratio of their Ar.
1000 atoms
1 million atoms

??? atoms 1g 12g 32g 24g 40g

Atoms, like coins, can be counted by weighing. The only difference is that millions and millions of atoms are needed before we get a weight that registers on a normal balance.
The number of atoms in 1g of hydrogen atoms is 602 000 000 000 000 000 000 000 (6 x 1023 ) and this amount is called one mole.
A mole of any substance always contains this number (called the Avogadro number) of particles.
Therefore:
Mass of one mole of carbon atoms is 12g.
Mass of one mole of sulphur atoms is 32g.
Mass of one mole of magnesium atoms is 24g.
Mass of one mole of calcium atoms is 40g.


What is the mass of 5 moles of fluorine atoms?
Ar(F) = 19
1 mole of fluorine atoms weighs 19g
5 moles of fluorine atoms weigh 5 x 19g = 95g

How many moles of atoms are there in 1.6g of copper.
Ar(Cu) = 64
1 mole of copper weighs 64g
1.6g copper must be less than 1 mole
Amount of copper = 1.6g
64g
= 0.025 mol



Mass of 1 mole of an element = relative atomic mass in grams
Mass of 1 mole of a compound = relative molecular mass in grams
Number of moles = mass of substance
mass of 1 mole of substance

Exercise 1

1. What is the mass of 1 mole of (a) sodium atoms
(b) cobalt atoms
(c) lead atoms

2. What is the mass of 0.1 moles of (a) barium atoms
(b) copper atoms
(c) tin atoms

Moles of compounds

What is the mass of 2 mol of sulphuric acid?
Mr(H2SO4) = 98
1 mol of sulphuric acid weighs 98g
2 mol sulphuric acid weighs 2 x98g = 196g



What is the mass of 0.1 mol of water?
Mr(H2O) = 18
1 mol water weighs 18g
0.1 mol water weighs 0.1 x 18g = 1.8g

How many moles of sodium hydroxide is 8.0g?
Mr(NaOH) = 40
40g of NaOH is 1 mol
8.0g is less than 1 mol
Amount of NaOH = 8.0g
40g
= 0.2 mol


Exercise 2
1. Calculate the mass of 2 moles of
(a) sodium carbonate Na2CO3
(b) potassium hydroxide KOH
2. Calculate the mass of
(a) 1 mol of sodium chloride NaCl
(b) 0.5 mol of magnesium hydroxide Mg(OH)2
(c) 4 mol of iron (II) chloride
(d) 2.5 mol of sodium carbonate
(e) 0.1 mol of zinc (II) chloride

Exercise 3
Calculate the amount of each of the following:
(a) 30.0 g of oxygen molecules, O2.
(b) 31.0 g of phosphorus molecules, P4.
(c) 50.0 g of calcium carbonate, CaCO3.

Exercise 4
Calculate the mass of each of the following:
(a) 2.50 mol of hydrogen, H2.
(b) 0.500 mol of sodium chloride, NaCl.
(c) 0.250 mol of carbon dioxide, CO2.

Exercise 5
A sample of ammonia, NH3, weighs 1.00 g.
(a) What amount of ammonia is contained in this sample?
(b) What mass of sulphur dioxide, SO2, contains the same number of molecules as are in 1.00 g of ammonia?

Calculating reacting masses
A chemical equation such as
N2 (g) + 3H2 (g) 2NH3 (g)
is a kind of chemical balance sheet; it tells us that one mole of nitrogen reacts with three moles moles of hydrogen to yield two moles of ammonia. (it does not tell us anything about the rate of the reaction or the conditions necessary to bring it about).
Such an equation is an essential starting point for many experiments and calculations; it tells us the proportions in which the substances react and the products are formed.

When you have finished this section you should be able to:
· Do simple reacting mass calculations based on a given chemical equation.

Worked Example
What mass of iodine will react completely with 10.0 g of aluminium?

The problem is a bit more complicated than those you have done previously because it involves several steps. Each step is simple but you may not immediately see where to start. One approach to solving multi-step problems is given below. You may find the approach useful to you in solving more difficult problems.
Ask yourself three questions:

1. What do I know?
In this case the answer would be:
(a) the equation for the reaction
(b) the mass of aluminium used

In some problems you will be given the equation; in this case the equation is not provided and you would be expected to write it down from your general chemical knowledge.

2. What can I get from what I know?
(a) from the equation , you can find the ratio of reacting amounts
(b) from the mass of aluminium you can calculate the amount, provided you can look up the molar mass.

3. Can I now see how to get the final answer?
Usually the answer will be ‘yes’, but you may have ask Question 2 again, now you know more than you did at the start.
(a) from the amount of aluminium and the ratio of reacting masses, you can calculate the amount of iodine
(b) from the amount of iodine, you can get the mass, using the molar mass.

Now we will go through each step, doing the necessary calculations.

1. the balanced equation for the reaction is
2Al (s) + 3I2 (s) 2AlI3 (s)
this equation tells us that 2 mol of Al reacts with 3 mol of I2; so we can write the ratio

amount of Al = 2
amount of I2 3

2. calculate the amount of aluminium using n=m/M.
n = 10.0 g = 0.370 mol
27.0 g mol-1

3. calculate the amount of iodine which reacts with this amount of aluminium by substituting in the expression based on the equation.
amount of Al = 2
amount of I2 3

amount of iodine = 3/2 x 0.370 mol = 0.555 mol

4. Calculate the mass of iodine from the amount using n = m/M in the form m = nM.
M = nM = 0.555 mol x 254 g mol-1 = 141 g


Example 2
What mass of magnesium oxide can be obtained from the complete combustion of 12g of magnesium?

2Mg (s) + O2 (g) 2MgO (s)
2 mol 1 mol 2 mol
Since Ar (Mg) = 24 Mr (MgO) = 40
2 x 24g Mg 2 x 40g MgO
48g magnesium forms 80g magnesium oxide
1g magnesium forms 80g magnesium oxide
48
12g magnesium forms 12 x 80g magnesium oxide
48
= 20 g magnesium oxide

c.f. 3 packets of crisps cost 39p. What do 7 packets of crisps cost?

Exercise 6
(a) What mass of magnesium would react completely with 16.0 g of sulphur?
Mg (s) + S (s) MgS (s)

(b) What mass of oxygen would be produced by completely decomposing 4.25 g of sodium nitrate, NaNO3?
2NaNO3 (s) 2NaNO2 (s) + O2 (g)

(c) What mass of phosphorus(V) oxide, P2O5, would be formed by complete oxidation of 4.00 g of phosphorus?
4P (s) + 5O2 (g) 2P2O5 (s)

(d) When 0.27 g of aluminium is added to excess copper(II) sulphate solution, 0.96 g of copper is precipitated. Deduce the equation for the reaction.


Molar Volumes of Gases


Calculation of empirical and molecular formulae from analytical data and molar masses

The empirical formula of a compound is the simplest form of the ration of the atoms of different elements in it. The molecular formula tells the actual number of each kind of atom in one molecule of the substance.
For example, the molecular formula of phosphorus(V) oxide is P4O10, whereas its empirical formula is P2O5.

When you have finished this section you should be able to:
· Calculate the empirical formula of a compound given either
(a) the masses of constituents in a sample, or
(b) the composition in terms of the mass percentages of the constituents.

Calculating empirical formula from the masses of constituents
To determine the empirical formula of a compound, we must first calculate the amount of each substance present in a sample and then calculate the simplest whole number ratio of the amounts.
It is convenient to set out the results in the form of a table. In the following example we will go through the procedure step by step, establishing the procedure as we go.

Worked Example
An 18.3 g sample of a hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.

Solution
1. List the mass of each component and its molar mass. Although water is a molecule, in the calculation treat it in the same way as we do atoms.


Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0


2. From this information calculate the amount of each substance present using the expression n = m/M.


Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40

This result means that in the given sample there is 0.10 mol of calcium, 0.20 mol of chlorine and 0.40 mol of water.

3. Calculate the relative amount of each substance by dividing each amount by the smallest amount.

Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40
Amount/smallest amount
= relative amount
0.10/0.10
= 1.0
0.20/0.10
= 2.0
0.40/0.10
=4.0
The relative amounts are in the simple ratio 1:2:4.
From this result you can see that the empirical formula is CaCl2.4H2O

Exercise 1
A sample of hydrated compound was analysed and found to contain 2.10 g of cobalt, 1.14 g of sulphur, 2.28 g of oxygen and 4.50 g of water.
Calculate its empirical formula.
CoSO4.7H2O

A modification of this type of problem is to determine the ratio of the amount of water to the amount of anhydrous compound.

Exercise 2
10.00 g of hydrated barium chloride is heated until all the water is driven off. The mass of anhydrous compound is 8.53 g.
Determine the value of x in BaCl2.xH2O.
BaCl2.2H2O x=2

You should be prepared for variations to this type of problem.

Exercise 3
When 127 g of copper combine with oxygen. 143 g of an oxide are formed. What is the empirical formula of the oxide?
[notice here that the mass of oxygen is not given to you – you must obtain it by subtraction]



Calculating empirical formula from percentage composition by mass
The result of the analysis of a compound may also be given in terms of the percentage composition by mass.

Worked Example
An organic compound was analysed and was found to have the following percentage composition by mass: 48.8% carbon, 13.5% hydrogen and 37.7% nitrogen.
Calculate the empirical formula of the compound.

Solution
If we assume the mass of the sample is 100.0 g, we can write immediately the mass of each substance: 48.8 g carbon, 13.5 g hydrogen and 37.7 g nitrogen. Then we set up a table as before.


C
H
N
Mass /g
48.8
13.5
37.7
Molar mass /g mol-1
12.0
1.00
37.7
Amount /mol
48.8/12.0
= 4.07
13.5/1.00
= 13.5
37.7/14
= 2.69
Amount/smallest amount
= relative amount
4.07/2.69
= 1.51
13.5/2.69
= 5.02
2.69/2.69
= 1.00
Simplest ratio of relative amounts
3
10
2

Empirical formula = C3H10N2

Values close to whole numbers are ‘rounded off’ in order to get a simple ratio. This is justified because small differences from whole numbers are probably due to experimental error. In the above example however we cannot justify rounding off 1.51 to 1 or 2, but we can obtain a simple ratio by multiplying the relative amounts by two.

Try the following exercises where you must decide whether to round off or multiply by a factor.

Exercise 4
A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen.
Calculate its empirical formula.
CH2O

Exercise 5
Determine the formula of a mineral with the following mass composition:
Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1%, H2O = 9.48%.
Na2Al2Si3O10.2H2O

Exercise 6
A 10.0 g sample of a compound contained 3.91 g of carbon, 0.87 g of hydrogen and the remainder is oxygen.
Calculate the empirical formula of the compound.
C3H8O3

Calculation of reacting masses and volumes of substances, including examples in which some reactants are in excess.

In some chemical reactions it may be that one or more of the reactants is in excess and is not completely used up in the reaction. The amount of product is determined by the amount of the reactant which is not in excess and is therefore used up completely in the reaction. This is called the limiting reactant.
Example 1
5.00 g of iron and 5.00 g of sulphur are heated together to form iron(II) sulphide. What mass of product is formed.

Fe (s) + S (s) FeS (s)
1 mol 1 mol 1 mol
56 g 32 g 88 g
Amount of Fe = 5/56 mol = 0.0893 mol
Amount of S = 5/32 mol = 0.156 mol
There is not enough Fe to react with 0.156 mol of S so Fe is the limiting reactant.
0.0893 mol Fe forms 0.0893 mol of FeS
Mass of FeS = 0.0893 x 88 g = 7.86 g

Exercise 1
(a) In the blast furnace, the overall reaction is
2Fe2O3 (s) + 3C (s) 3CO2 (g) + 4Fe (s)
what is the maximum mass of iron that can be obtained from700 tonnes of iron(III) oxide and 70 tonnes of coke? (1 tonne = 1000 kg)

436 tonnes
(b) In the manufacture of calcium carbide
CaO (s) + 3C (s) CaC2 (s) + CO (g)
What is the maximum mass of calcium carbide that can be obtained from 40 kg of quicklime and 40 kg of coke?
46 kg

(c) In the manufacture of the fertiliser ammonium sulphate
H2SO4 (aq) + 2NH3 (g) (NH4)2SO4 (aq)
What is the maximum mass of ammonium sulphatethat can be obtained from 2.0 kg of sulphuric acid and 1.0 kg of ammonia?
2.7 kg

(d) In the Solvay process, ammonia is recovered by the reaction
2NH4Cl (s) CaO (s) CaCl2 (s) + H2O (g) + 2NH3 (g)
What is the maximum amount of ammonia that can be recovered from 2.00 x 103 kg of ammonium chloride and 500 kg of quicklime?
304 kg

(e) In the Thermit reaction
2Al (s) + Cr2O3 (s) 2Cr (s) + Al2O3 (s)
Calculate the percentage yield when 180 g of chromium are obtained from a reaction between 100 g of aluminium and 400 g of chromium(III) oxide.

93.5%