1. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
LEARNING OUTCOMES
Define relative atomic mass, Ar
Define relative molecular mass, Mr

2. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass
The masses of atoms and molecules are very small and it very hard for us to compare them or use them in calculations.
For e.g. the mass of a hydrogen atom is or (1.4 x 10-24)g and the mass of a carbon atom is 1.68 x g.

3. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass
Instead of using the true masses of atoms, scientists compare the masses of atoms with the mass of a hydrogen atom, which is assigned a mass of one unit.
If we take the mass of a hydrogen atom to be 1, then the mass of a carbon atom will be 12, since a carbon atom is 12 times as heavy as a hydrogen atom.

4. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
12 H atoms
Relative Atomic Mass
The mass of a carbon atom is 12 times as heavy as a hydrogen atom, so we say that the relative atomic mass of carbon is 12. C
Scientists prefer to use the carbon atom instead of the hydrogen atom as a standard unit, so if we take 1/12 of the mass of a carbon atom, we will still get 1 unit.
Hence we define:
The relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12 of the mass of a carbon-12 atom.

5. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Relative Atomic Mass
Relative atomic mass Mass of one atom of the element (Ar) Mass of of an atom of carbon-12 = 1 12
The symbol for Relative Atomic Mass is Ar
Relative atomic mass has no units.
The relative atomic mass of an element can be obtained from the mass number (nucleon number) of the element in the Periodic Table.
Examples : Na, 5626Fe Hence, Ar of Na = 23, Ar of Fe = 56

6. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Quick check 1
Define the relative atomic mass of an element.
Using the Periodic Table, find the relative atomic mass of the following: Mg, Ca, Cl, O, S, Ne, Br.
What is the mass of a calcium atom compared to the mass of a helium atom?
How many hydrogen atoms have the same mass as one potassium atom?
How many oxygen atoms have the same mass as one bromine atom?
Solution

7. Solution to Quick check 1
Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 1
The relative atomic mass of an element is the average mass of an atom of the element compared to the mass of 1/12 of a carbon-12 atom.
Ar: Mg=24, Ca=40, Cl=35.5, O=16, S=32, Ne=20, Br=80.
Mass of a Ca atom =10 x Mass of a He atom
39 hydrogen atoms
5 oxygen atoms
Return

8. Relative Molecular Mass
Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass
The picture shows the mass of a molecule of water compared to the mass of hydrogen atoms.
It shows that one molecule of water is 18 times as heavy as one hydrogen atom.
Therefore, the relative molecular mass of water is 18.

9. Relative Molecular Mass
Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass
If we use the mass of a carbon-12 atom as the standard unit of comparison, then the relative molecular mass of a substance is defined as:
The relative molecular mass of a substance is the average mass of one molecule of the substance compared to the mass of 1/12 of a carbon-12 atom.
Relative molecular mass Mass of one molecule of a substance
(Mr) Mass of of a carbon-12 atom = 1 12

10. Relative Molecular Mass
Chapter 3
Relative Masses of Atoms and Molecules
Relative Molecular Mass
The symbol for Relative Molecular Mass is Mr
The relative molecular mass of a molecule can be found by adding up the relative atomic masses of all the atoms present in the molecule.
E.g. Mr of water, H2O = Mass of 2 H atoms mass of 1 O atom = 2 x = 18

11. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Relative Formula Mass
Ionic compounds (e.g. sodium chloride) are not made up of single molecules; instead they are made up of a crystal lattice consisting of many oppositely charged ions.
Hence, instead of calculating the relative molecular mass of an ionic compound, we calculate the mass based on its formula (formula mass).
We can take the relative formula mass as equivalent to the relative molecular mass in our calculations.

12. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Finding Relative Molecular Mass
Find the relative molecular mass of carbon dioxide, CO2.
Mr of CO2 = x = 44
Find the relative molecular mass (formula mass) of copper(II) nitrate, Cu(NO3)2.
Mr of Cu(NO3)2 = ( x3 )x = x2 = 188
Find the relative molecular mass (formula mass) of ammonium sulphate, (NH4)2SO4.
Mr of (NH4)2SO4 = (14 + 1x4)x x = = 132
Worked examples

13. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Quick check 2
Find the relative molecular mass (or formula mass) of each of the following:
Magnesium sulphate, MgSO4
Calcium nitrate, Ca(NO3)2
Ammonium carbonate, (NH4)2CO3
Benzoic acid, C7H6O2
Hydrated sodium carbonate, Na2CO3.10H2O
Solution

14. Solution to Quick check 2
Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 2
Mr of MgSO4 = x4 = 120
Mr of Ca(NO3)2 = ( x3) = 164
Mr of (NH4)2CO3 = 2(14+4) x3 = 96
Mr of C7H6O2 = 12x x2 = 122
Mr of Na2CO3.10H2O = 23x x3 + 10(2x1 + 16) = 286
Return

15. Percentage composition
Chapter 3
Relative Masses of Atoms and Molecules
Percentage composition
Worked example 1
What is the percentage composition of calcium carbonate, CaCO3?
Solution % of Ca = [Ca] x 100% = x 100% [CaCO3] [ x3] = 40%
% of C = [ C ] x 100% = 12 x 100% [CaCO3] = 12%
% of O = [ O3 ] x 100% = 16x3 x 100% [CaCO3]
= 48%
Check
%Ca + %C + %O = = 100%

16. Percentage composition
Chapter 3
Relative Masses of Atoms and Molecules
Percentage composition
Worked example 2
What is the percentage of sodium in sodium carbonate, Na2CO3?
Solution % of Na = [Na2] x 100% = x x 100% [Na2CO3] [23x x3] = 43.4%
Worked example 3
Find the mass of oxygen in 90 g of water.
Solution Mr of H2O = 1x = 18 Mass of oxygen = 16 x 90 g = 80 g
Worked example 4
What mass of magnesium oxide can be made from 6 g of magnesium?
Solution Mr of MgO = = 40 Mass of MgO = 6 x 40 g = 10 g

17. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Percentage yield
The percentage yield of a product is given by the formula:
Percentage yield = Actual mass of product obtained x 100% Theoretical mass of product obtained
Worked example 1
In an experiment to prepare magnesium sulphate, 2.4 g of magnesium was dissolved completely in dilute sulphuric acid. On crystallisation, 22.0 g of magnesium sulphate crystals, MgSO4.7H2O, were obtained. What is the percentage yield?

18. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Percentage yield
Solution
Write the chemical equation: Mg + H2SO4  MgSO4 + H2
Number of mole of Mg reacted = = 0.1 mol
From equation, 1 mol of Mg  1 mol of MgSO4
Therefore, 0.1 mol of Mg  0.1 mol of MgSO4.7H2O
Mass of MgSO4.7H2O produced = 0.1 mol x 246 g/mol = g
Percentage yield = Actual mass of product obtained x 100% Theoretical mass of product obtained
= x 100 %
= %

19. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Percentage purity
The percentage purity tells us how pure a prepared product is compared to the pure form of the substance.
For example, the percentage purity of a gold bar may be given as %.

20. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Percentage purity
Worked example 2
On analysis, 5.00 g of a sample of marble (calcium carbonate) was found to contain only 4.26 g of pure calcium carbonate. What is the percentage purity of the marble?
Solution
Percentage purity = x 100 %
= %

21. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
Quick check 3
Find the percentage composition of each element in sulphuric acid, H2SO4.
Find the percentage of nitrogen in calcium nitrate, Ca(NO3)2.
Find the mass of calcium in 250 g of calcium carbonate, CaCO3.
What mass of iron can be obtained from 320 g of iron(III) oxide, Fe2O3?
24 g of hydrogen combines with 192 g of oxygen to form water. What mass of hydrogen will combine with 24 g of oxygen?
In the Haber process to manufacture ammonia, it was reported that in a certain factory, 2.8 tonnes of nitrogen gas produced 0.80 tonne of ammonia. What is the percentage yield of ammonia? The equation for the Haber process is: N2(g) + 3H2(g)  2NH3(g)
Solution

22. Solution to Quick check 3
Chapter 3
Relative Masses of Atoms and Molecules
Solution to Quick check 3
% composition of H2SO4 : H =2.04 %, S =32.7 %, O = 65.3 %
% of nitrogen in Ca(NO3)2 = 17.1 %
Mass of calcium = 40 x 250 = 100 g
Mass of iron = 112 x 320= 224 g
Mass of hydrogen = 24 x 24 = 3 g
Percentage yield = 0.8 x 100 %
= 23.5 %
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23. Relative Masses of Atoms and Molecules
Chapter 3
Relative Masses of Atoms and Molecules
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