Mole Concept - Learning from examples

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 Al2O= 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 Al2O= 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 g
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/dmx 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 cmof 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.
no. 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  H2SO =  0.00250/ 0.015 = 0.167 mol/dm3
Mr of  = 2 x 1 + 32 + 16 x 4= 98.0
Concentration (g/dm3) of  H2SO= 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 cmof 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 H2SOis added. This means that H2SOis in excess, and NaOH is the limiting reagent.

Alternatively
From equation,  1 mol of H2SOwill 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 H2SOis 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  Na2SO= 0.005 / 2 = 0.00250
Mr of Na2SO= 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 CaCO (what is given) and CaO (what you need to caculate).
Mr of CaCO= 40 + 12 + 16 x 3 = 100
Mr of CaO = 40 + 16 = 56.0
Step 3. Find mass of CaO produced from 100g of CaCO.
No. of mol of CaCO= 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:
CaCO---> CO2 + CaO
Step 2. Find the Mr of CaCO and CaO.
Mr of CaCO= 40 + 12 + 16 x 3 = 100
Mr of CaO = 40 + 16 = 56.0
Step 3. Find mass of CaCOrequired 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 CaCOreacted = 0.571 mol
Step 4. Calculate mass of CaCOpresent 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%

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