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Mole ConceptChapter 5
LEARNING OUTCOMES
Define the mole Derive empirical and molecular formulae State Avogadro’s Law Apply the mole concept to ionic and molecular
equations
A mole is the number of atoms or molecules in 1 g of hydrogen or 12 g of carbon.
The number 6 x 1023 is called one mole or Avogadro’s constant in honour of Amedeo Avogadro.
1 mole of atoms of any element will have a mass equal to its relative atomic mass, expressed in grams.
Introducing the Mole
Chapter 5Mole Concept
Chapter 5
Formulas
1. Mass of 1 mole of atoms = Ar in grams
2. Number of moles of atoms =
3. Mass of 1 mole of molecules = Mr in grams
4. Number of moles of molecules =
5. Mass of substance containing 1 mole of particles = Molar mass
6. Percentage yield =
Mass of the element in grams Relative atomic mass, Ar
Mass of the substance in grams Relative molecular mass, Mr
Actual mass of product obtained Theoretical mass of product obtainable
Mole Concept
The empirical formula is the simplest formula. It shows the simplest ratio of the elements present in a
compound. Examples of empirical formulae are:
H2O, CO2, H2SO4, CH2, CH3
The following are not empirical formulae: (a) C2H4, (b) C2H6, (c) C2H4O2
because they can be reduced to:(a) CH2, (b) CH3, (c) CH2O
Chapter 5
Empirical Formula
Mole Concept
The molecular formula is the true formula. It shows all the atoms present in the molecule. Examples of molecular formulae are:
H2O, H2O2, CO2, H2SO4, Cu(NO3)2
Note that H2O is water, and the molecular formula is the same as the empirical formula.
H2O2 is hydrogen peroxide. Its empirical formula is HO.
Chapter 5
Molecular Formula
Mole Concept
Let the molecular formula of propene be (CH2)n. Since the Mr is 42, (12 + 1x2)n = 42
14n = 42 n = 42 = 3 14
Hence the molecular formula is 3x(CH2) = C3H6
Chapter 5
Finding the Molecular Formula
Propene has the empirical formula CH2. The relative molecular mass of propene is 42. Find the molecular formula of propene.
Worked example 1
Solution
Mole Concept
C : H (a) 85.7 : 14.3 85.7 : 14.3
12 1 7.14 : 14.3 7.14 7.14 1 : 2 The empirical formula is CH2. (b) Let the molecular formula be (CH2)n. Since the Mr is 56, (12 + 1x2)n = 56
14n = 56 n = 56 = 4 14
Hence the molecular formula is 4 x (CH2) = C4H8
Chapter 5
Finding the Molecular FormulaWorked example 1
Solution
A hydrocarbon consists of 85.7% of carbon and 14.3% of hydrogen by mass. (a) Find the empirical formula of the compound. (b) If the molecular mass is 56, find the molecular formula.
Mole Concept
Formula of a compound A pure compound has a fixed chemical composition;
hence it can be represented by a chemical formula. For example, a molecule of water is made up of 2
atoms of hydrogen and 1 atom of oxygen, and its molecular formula is H2O.
We can find the formula of a compound from its percentage composition.
Chapter 5
Mole Concept
Step 4 : Write down the formula: Na2SO4
A compound of sodium contains the following percentage composition by mass: 32.4% of sodium, 22.6% of sulphur and 45.0% of oxygen. Find the formula of the compound.
Worked example 1
Solution Na : S : O
Step 1: Write down the percentage: 32.4 : 22.6 : 45.0Step 2: Divide each by the Ar: 32.4 : 22.6 : 45.0 (to get number of moles) 23 32 16 1.41 : 0.706 : 2.81Step 3: Divide by the smallest number: 1.41 : 0.706 : 2.81 0.706 0.706 0.706 2 : 1 : 4
Chapter 5
Finding the formula of a compound
Mole Concept
Step 4 : Write down the formula: CO2H2 (or HCOOH)
A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g ofhydrogen. Find the formula of the compound.
C : O : HStep 1: Write down the mass ratio: 1.2 : 3.2 : 0.2Step 2: Divide each mass by the Ar: 1.2 : 3.2 : 0.2 12 16 1 0.1 : 0.2 : 0.2Step 3: Divide by the smallest number: 0.1 : 0.2 : 0.2 0.1 0.1 0.1 1 : 2 : 2
Chapter 5
Finding the formula of a compoundWorked example 2
Solution
Mole Concept
Step 1: Find the % of hydrogen: 100 – 48.6 – 43.2 = 8.2 %
Step 2: Write down the % ratio: C : O : H 48.6 : 43.2 : 8.2Step 3: Divide each mass by the Ar: 48.6 : 43.2 : 8.2 12 16 1 4.05 : 2.7 : 8.2Step 4: Divide by the smallest number: 4.05 : 2.7 : 8.2 2.7 2.7 2.7 1.5 : 1 : 3Step 5: Multiply each number by 2: 3 : 2 : 6Step 6 : Write down the formula: C3O2H6 (or C2H5COOH)
A compound contains 48.6% of carbon, 43.2% of oxygen, with the remainder being hydrogen. Find the formula of the compound.
Chapter 5
Finding the formula of a compoundWorked example 3
Solution
Mole Concept
Chapter 5
Molar Volume of GasesAvogadro’s Law states that equal volume of gases under the same temperature and pressure contain the same number of molecules.
Volume of 1 mole of gas = 24 dm3
Volume of gas = Number of moles x 24 dm3
Number of moles = Volume of gas in dm3
24 dm3
Mole Concept
Limiting Reactants 2H2(g) + O2(g) 2H2O(g)
2 moles of hydrogen gas react with one or more moles of oxygen to form 2 moles of steam or water vapour.
Therefore we say that oxygen is in excess and hydrogen is called the limiting reactant because the reaction stops when hydrogen is used up.
Chapter 5
Mole Concept
Find the formula of each of the following:1. A compound containing 75% carbon and 25% hydrogen by mass.2. A compound containing 46.7% silicon and 53.3% oxygen by mass.3. A compound consisting of 43.4% sodium, 11.3% carbon and 45.3%
oxygen by mass.4. A compound consisting of 2.8 g of iron combined with 1.2 g of oxygen.5. (a) A compound containing 18.9% lithium,
64.9% oxygen and the rest carbon by mass.(b) Give the name of this compound.
Solution
Chapter 5
Quick check 1
Mole Concept
1. A hydrocarbon consists of 80% carbon and 20% hydrogen by mass. (a) Find the empirical formula of the compound.(b) If the molecular mass is 30, find the molecular formula.
2. An acid contains 40% carbon, 6.67% hydrogen and 53.33% oxygen by mass. (a) Find the empirical formula of the acid. (b) If the molecular mass of the acid is 60, what is its molecular formula?
3. A compound called borazine has the following percentage composition by mass: 40.74% boron, 51.85% nitrogen and the rest hydrogen. Find the molecular formula of borazine, given it has a relative molecular mass of 81.
Solution
Chapter 5
Quick check 2
Mole Concept
1. CH4
2. SiO2
3. Na2CO3
4. Fe2O3
5. (a) Li2CO3
(b) lithium carbonate
Return
Chapter 5
Solution to Quick check 1
Mole Concept
1. (a) Empirical formula: CH3 (b) Molecular formula: C2H6
2. (a) Empirical formula: CH2O (b) Molecular formula: C2H4O2 (CH3COOH)
3. Molecular formula: B3N3H6
Chapter 5
Solution to Quick check 2
Return
Mole Concept