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C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
C05 the mole concept
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C05 the mole concept

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  • 1. Chapter 5 Mole Concept LEARNING OUTCOMES     Define the mole Derive empirical and molecular formulae State Avogadro’s Law Apply the mole concept to ionic and molecular equations
  • 2. Chapter 5 Mole Concept Introducing the Mole 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.
  • 3. Chapter 5 Mole Concept Formulas 1. Mass of 1 mole of atoms = Ar in grams Mass of the element in grams 2. Number of moles of atoms = Relative atomic mass, Ar 3. Mass of 1 mole of molecules = Mr in grams Mass of the substance in grams 4. Number of moles of molecules = Relative molecular mass, Mr 5. Mass of substance containing 1 mole of particles = Molar mass Actual mass of product obtained 6. Percentage yield = Theoretical mass of product obtainable
  • 4. Chapter 5 Mole Concept Empirical Formula 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
  • 5. Chapter 5 Mole Concept Molecular Formula      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.
  • 6. Chapter 5 Mole Concept Finding the Molecular Formula Worked example 1 Propene has the empirical formula CH2. The relative molecular mass of propene is 42. Find the molecular formula of propene. Solution   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
  • 7. Chapter 5 Mole Concept Finding the Molecular Formula Worked example 2 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. Solution 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
  • 8. Chapter 5 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.
  • 9. Chapter 5 Mole Concept Finding the formula of a compound Worked example 1 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. Solution Na : S : O Step 1: Write down the percentage: 32.4 : 22.6 : 45.0 Step 2: Divide each by the Ar: 32.4 : 22.6 : 45.0 (to get number of moles) 23 32 16 1.41 : 0.706 : 2.81 Step 3: Divide by the smallest number: 1.41 : 0.706 : 2.81 0.706 0.706 0.706 2 : 1 : 4 Step 4 : Write down the formula: Na2SO4
  • 10. Chapter 5 Mole Concept Finding the formula of a compound Worked example 2 A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound. Solution Step 1: Write down the mass ratio: Step 2: Divide each mass by the Ar: Step 3: Divide by the smallest number: Step 4 : Write down the formula: C 1.2 1.2 12 0.1 0.1 0.1 1 : O : H : 3.2 : 0.2 : 3.2 : 0.2 16 1 : 0.2 : 0.2 : 0.2 : 0.2 0.1 0.1 : 2 : 2 CO 2H2 (or HCOOH)
  • 11. Chapter 5 Mole Concept Finding the formula of a compound Worked example 3 A compound contains 48.6% of carbon, 43.2% of oxygen, with the remainder being hydrogen. Find the formula of the compound. Solution Step 1: Find the % of hydrogen: 100 – 48.6 – 43.2 = 8.2 % Step 2: Write down the % ratio: Step 3: Divide each mass by the Ar: Step 4: Divide by the smallest number: Step 5: Multiply each number by 2: Step 6 : Write down the formula: C : O : H 48.6 : 43.2 : 8.2 48.6 : 43.2 : 8.2 12 16 1 4.05 : 2.7 : 8.2 4.05 : 2.7 : 8.2 2.7 2.7 2.7 1.5 : 1 : 3 3 : 2 : 6 C3O2H6 (or C2H5COOH)
  • 12. Chapter 5 Mole Concept Molar Volume of Gases Avogadro’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 dm 3 Number of moles = Volume of gas in dm3 24 dm3
  • 13. Chapter 5 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.
  • 14. Chapter 5 Mole Concept Quick check 1 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
  • 15. Chapter 5 Mole Concept Quick check 2 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
  • 16. Chapter 5 Mole Concept Solution to Quick check 1 1. CH4 2. SiO2 3. Na2CO3 4. Fe2O3 5. (a) Li2CO3 (b) lithium carbonate Return
  • 17. Chapter 5 Mole Concept Solution to Quick check 2 1. (a) Empirical formula: CH3 (b) Molecular formula: C2H6 2. (a) Empirical formula: CH2O (b) Molecular formula: C2H4O2 (CH3COOH) 3. Molecular formula: B3N3H6 Return

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