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.

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.

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.

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%.

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.

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)

(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?

(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?

(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?

(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.

 

1. Definitions

The relative atomic mass (Ar) of an element is the mass of one atom of the element, relative to the mass of 1/12 the mass of one atom of carbon- 12.

The relative molecular mass (Mr) is the mass of one molecule, relative to the mass of 1/12 the mass of one atom of carbon- 12.

2. Mole

The mole is used to represent the amount of particles. One mole refers to 6.02 x 1023 particles.

The Avogadro constant (L) has a value of 6.02 x 1023.

Hence,  number of particles = number of moles x Avogadro constant

Another formula for finding number of  moles = mass in grams / molar mass (or mass you get from periodic table)

Example. Find the mass of 5 mol of chlorine atom.

Step 1: Since questions ask for chlorine atom, we make use of Ar. From the periodic table, Ar of chlorine is 35.5.

Step 2: moles = mass/ molar mass ==> mass = molar mass x moles ===> mass = 5 x 35.5 = 177.5g

Example. Find the mass fo 5 mol of chlorine gas.

Step 1: Chlorine gas exists as Cl2. Since the Ar of one chlorine is 35.5, the Mr of Cl2 is 35.5 x 2 = 71.

Step 2: moles = mass/ molar mass ==> mass = molar mass x moles ===> mass = 5 x 71 = 355g

3. Calculating relative molecular mass

You will need to use the mass number of the elements, to determine relative molecular mass.

Example: Find the relative molecular mass of carbon dioxide.

Carbon dioxide has a formula of CO2. This means that it contains 1 carbon atom, and 2 oxygen atoms per molecule. From the periodic table, the mass number of carbon is 12, and the mass number of oxygen is 16. Hence, the Mr of carbon dioxide = 12 + 16 x 2 = 44.0

4. Calculating percentage mass (or percentage composition)

Example: Calculate the percentage composition of oxygen in carbon dioxide.

Step 1- Find the Mr of carbon dioxide.

Mr of CO2 = 12 + 16 x 2 = 44.0

Step 2- Determine relative mass of oxygen per molecule.

Since there are 2 oxygen per molecule of carbon dioxide, and the mass number of each oxygen is 16, hence the relative mass of oxygen per molecule is 16 x 2 = 32.

Step 3- Calculate percentage composition, by taking answer from step 2 divided by answer from step 1, and multiply by 100%.

Percentage composition of oxygen = 32/44 x 100% = 72.7% (3 significant figures)

Example: Calculate the mass of aluminium, in 25g of aluminium oxide.

Step 1- Write out the formula of aluminium oxide, since it is not given.

Al2O3.

Step 2- Calculate the Mr of Al2O3.

Mr of Al2O3 = 2 x 27 + 3 x 16 = 102

Step 3 – Determine the relative mass of aluminium in one compound of aluminium oxide.

There are 2 aluminium atoms per Al2O3. Hence, relative mass of Al in Al2O3 = 2 x 27 = 54.

Step 4- Step 2 and 3 tell you that for every 102 g of Al2O3, there is 54g of Al. Hence, in 25g of Al2O3, mass of Al = 54/102 x 25 = 13.2 g (3 significant figures).

5. Calculating empirical and molecular formulae

Empirical formula gives the simplest whole number ratio of each element in a compound.

Molecular formula shows the exact number of each element in a compound.

5.1. Calculating empirical and molecular formulae, given mass of each element.

Example. 50 g of a hydrocarbon is make up of 42.9g carbon. Find the empirical formula of this compound. Given that the molecular mass of this hydrocarbon is 42.0, determine its molecular formula.

Step 1. Since this is a hydrocarbon, it means that the compound consists of hydrogen and carbon only. Since it contains 42.9g of carbon, mass of hydrogen = 50 – 42.9 = 7.10

Step 2. Set up the table.

	C	H
mass/g (1)	42.9	7.10
Ar (2)	12	1
Number of moles per 100 g (3) [Take (1) divided by (2)]	3.58	7.1
Ratio (round off to nearest whole number).   From (3), notice that 3.58 is the smallest number. Hence, divide all numbers in (3) by 3.58.	1	2
Empirical Formula	CH2

Step 3: Determine molecular formula.

Let the molecular formula be (CH2)n, where n is an integer.

Molecular mass  = (12 + 2 x 1) x n = 42

14 n = 42

n = 3

Hence, molecular formula = C3H6.

5.2 Calculating empirical and molecular formulae, given percentage composition of each element.

Example. A hydrocarbon consists of 86% by mass of carbon. Find the empirical formula of this compound. Given that the molecular mass of this hydrocarbon is 42.0, determine its molecular formula.

Step 1. Since this is a hydrocarbon, it means that the compound consists of hydrogen and carbon only. Since it contains 86% carbon, percentage composition of hydrogen = 100% – 86% = 14%

Step 2: Set up the table

	C	H
% composition (1)	86	14
Ar (2)	12	1
Number of moles per 100 g (3) [Take (1) divided by (2)]	7.17	14
Ratio (round off to nearest whole number).   From (3), notice that 7.17 is the smallest number. Hence, divide all number in (3) by 7.17.	1	2
Empirical Formula	CH2

Step 3: Determine molecular formula.

Let the molecular formula be (CH2)n,where n is an integer.

Molecular mass  = (12 + 2 x 1) x n = 42

14 n = 42

n = 3

Hence, molecular formula = C3H6.

 

6. Gases

number of moles of gases = (volume of gas in dm3)/ (molar volume)

molar volume at room temperature and pressure = 24 dm3

molar volume at standard temperature and pressure = 22.4 dm3.

 

Note: 1dm3 = 1000 cm3

Example.Find the number of moles of carbon dioxide in 250 cm3of carbon dioxide gas.

Step 1. 250 cm3= 250/1000 dm3 = 0.250 cm3

Step 2. moles = volume/ molar volume = 0.25/24 =0.0104 mol

 

7. Solutions

Concentration of solutions can be represented by mol/dm3 or g/dm3.

 

Concentration in mol/dm3= (number of moles) / (volume in dm3)

Concentration in g/dm3= (mass in grams) / (volume in dm3)

Concentration in g/dm3= Concentration in mol/dm3 x Mr

 

8. Chemical equations and mole ratios

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

From the above equation, it means that 2 moles of NaOH react with 1 mole of H2SO4 to form 1 mole of Na2SO4 and 2 moles of H2O.

 

Example. 20.0 cm3 of 0.250 mol/dm3 sodium hydroxide reacts completely with 15 cm3 of sulfuric acid.  Find the concentration of the sulfuric acid in (i) mol/dm3 (ii) g/dm3.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

Step 2. Find the number of moles of NaOH.

1. of mol of NaOH = 20/1000 x 0.250 = 0.005 mol

Step 3. Use equation to find number of moles of H2SO4.

From equation, 2 mol of NaOH reacts with 1 mol of H2SO4.

Hence, number of moles of H2SO4= 0.005/2 = 0.00250 mol

Step 4. Find concentration of H2SO4.

15 cm3= 15/1000  dm3=  0.015 dm3.

Concentration (mol/dm3) of  H2SO4  =  0.00250/ 0.015 = 0.167 mol/dm3

Mr of  = 2 x 1 + 32 + 16 x 4= 98.0

Concentration (g/dm3) of  H2SO4 = 0.167 x 98 = 16.4 g/dm3.

 

9. Limiting and excess reactant

 

The limiting reactant is one that will be used up in the reaction. The excess reactant is one that will not be used up in the reaction.

Example. 20.0 cm3 of 0.250 mol/dm3 sodium hydroxide is added to 20.0 cm3 of 0.2 mol/dm3  sulfuric acid. Find the mass of sodium sulfate formed.

Step 1: Write the equation for the reaction, if it is not given. In this case, the equation is:

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

Step 2: The question gives sufficient information to find number of moles of both reactants — H2SO4 and NaOH. This means that you need to find limiting and excess reactant.

Step 3: Find number of moles of both reactant.

No. of mol of NaOH =  20/1000 x 0.250 = 0.005 mol

No. of mole of H2SO4 = 20/1000 x 0.2 = 0.004 mol

Step 4. Use values from step 3, and equation to determine limiting and excess reactant.

From equation, 2mol of NaOH will react with 1 mol of H2SO4.

Hence, 0.005 mol of NaOH will react with 0.005/2 = 0.0025 mol of H2SO4. However, 0.004 mol of H2SO4 is added. This means that H2SO4 is in excess, and NaOH is the limiting reagent.

 

Alternatively

From equation,  1 mol of H2SO4 will react with 2 mol of NaOH

Hence, 0.004 mol of H2SO4 will react with 0.004 x 2 = 0.008 mol of NaOH. However, only 0.005 mol of NaOH is added. This means that H2SO4 is in excess, and NaOH is the limiting reagent.

 

Step 5. Once you found the limiting reagent, use its number of moles for calculating amount or mass of other reactant/ products, because it is the one that reacts completely.

Since NaOH is the limiting reactant, we use it to calculate number of moles of Na2SO4.

No. of mol of NaOH =  20/1000 x 0.250 = 0.005 mol

From the equation, 2 moles of NaOH form 1 mole of Na2SO4.

No. of mol of  Na2SO4 = 0.005 / 2 = 0.00250

Mr of Na2SO4 = 23 x 2 + 32+ 16 x 4 =  142

Mass of Na2SO4= 142 x 0.00250 = 0.355 g

 

10. Percentage yield and purity

% yield = (actual mass)/ (theoretical mass) x 100%

% purity = (calculated mass)/ (mass of impure substance) x 100%

 

Example. When 100 g of calcium carbonate is heated, 32 g of calcium oxide is obtained. Calculate the percentage yield.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

CaCO3 —> CO2 + CaO

Step 2. Find the Mr of CaCO3  (what is given) and CaO (what you need to caculate).

Mr of CaCO3 = 40 + 12 + 16 x 3 = 100

Mr of CaO = 40 + 16 = 56.0

Step 3. Find mass of CaO produced from 100g of CaCO3 .

No. of mol of CaCO3 = 100/ 100 = 1

From equation, 1 mol of CaCO3  forms 1 mol of CaO.

No. of mol of CaO = 1

Theoretical mass of CaO = 1 x 56 = 56.0 g

Step 4. Calculate percentage yield.

percentage yield = actual mass/ theoretical mass x 100% = 32/56 x 100% = 57.1%

Example. Limestone is consist of calcium carbonate and other impurities. When 100 g of calcium carbonate is heated, 32 g of calcium oxide is obtained. Calculate the percentage purity.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

CaCO3 —> CO2 + CaO

Step 2. Find the Mr of CaCO3  and CaO.

Mr of CaCO3 = 40 + 12 + 16 x 3 = 100

Mr of CaO = 40 + 16 = 56.0

Step 3. Find mass of CaCO3 required to produce 32 g of CaO.

No. of mol of CaO = 32/ 56 = 0.571 mol

From equation, 1 mol of CaCO3  forms 1 mol of CaO.

No. of mol of CaCO3 reacted = 0.571 mol

Step 4. Calculate mass of CaCO3 present in limestone.

Mass of CaCO3 = 0.571 x 100 = 57.1 g

Step 4. Calculate percentage yield.

percentage yield = theoretical mass/ mass of impure substance x 100% = 57.1/100 x 100% = 57.1%

 

You can ask the Super AI Chemistry Alevel teacher any question of s5 and s6 chemistry and get answers