Chapter 4 Stoicheometry

CHEMISTRY 11 IGCSE

1. Sir . Muhammad Abdul Mageid CIE - IGCSE Chemistry Chapter 4 Stoichiometry 00965 60047622

2. Contents 4.2 The mole concept 4.1 Stoichiometry Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid

3. Saturday, July 25, 2020 4.2 The mole concept Sir. Muhammad Abdul Mageid The Mole & Avogadro’s Constant The mole This is the mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of  [ 12 C] The mole is the unit representing the amount of atoms, ions, or molecules One mole is the amount of a substance that contains  6.02 x 10 23  particles (Atoms, Molecules or Formulae) of a substance (6.02 x 10 23  is known as the  Avogadro Number)

4. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Examples 1 mole of Sodium (Na) contains 6.02 x 10 23   Atoms  of Sodium 1 mole of Hydrogen (H 2 ) contains 6.02 x 10 23   Molecules  of Hydrogen 1 mole of Sodium Chloride (NaCl) contains  6.02 x 10 23  Formula units  of Sodium Chloride Remarks Molecules represent Covalent Molecules Formula units represent Ionic compounds

5. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Linking the mole and the atomic mass One mole of any element is equal to the relative atomic mass of that element in grams For example one mole of carbon, that is if you had 6.02 x 10 23  atoms of carbon in your hand, it would have a mass of 12g So one mole of helium atoms would have a mass of 4g, lithium 7g etc For a compound we add up the relative atomic masses So one mole of water would have a mass of 2 x 1 + 16 = 18g Hydrogen which has an atomic mass of 1 is therefore equal to  1 / 12  the mass of a  12 C atom So one carbon atom has the same mass as 12 hydrogen atoms

6. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept The Mole & the Volume of Gases Molar volume This is the volume that one mole of any gas (be it molecular such as CO 2  or monoatomic such as helium) will occupy It’s value is 24dm 3  or 24,000 cm 3  at room temperature and pressure ( r.t.p .) Amount of gas (mol.) = Volume of gas (dm 3 ) ÷ 24 Calculations involving gases Amount of gas (mol.) = Volume of gas ( ) ÷ 24000  

7. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept 1. Calculating the volume of gas that a particular amount of moles occupies Equation: Volume of gas ( ) = Amount of gas ( mol. ) x 24   Volume of gas ( ) = Amount of gas ( mol. ) x 24000  

8. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept 2. Calculating the moles in a particular volume of gas Equation: Amount of gas ( ) = Volume of gas (dm 3 ) ÷ 24   Amount of gas ( ) = Volume of gas ( ) ÷ 24000  

9. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Calculating Reacting Masses, Solutions & Concentrations of Solutions in &   Calculating percentage composition, moles, mass and relative formula mass

10. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 1. Calculating Moles

11. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 2. Calculating Mass

12. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 3 . Calculating Relative Formula Mass   Example : 10 moles of Carbon Dioxide has a Mass of 440 g. What is the Relative Formula Mass of Carbon Dioxide ?  

13. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4. Calculating Percentage Composition   Example: Calculate the percentage of oxygen in CO 2    

14. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Calculations of solutions: moles, concentration and volume General Equation:      

15. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept 1. Calculating Moles         Example: Calculate the Moles of Solute Dissolved in 2 dm 3  of a 0.1 mol / dm 3  Solution Concentration of Solution :  0.1 mol / dm 3 Volume of Solution :  2 dm 3 Moles of Solute   =   0.1   x   2    =    0.2 mol

16. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept 2. Calculating Concentration          

17. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept   Remark ; In neutralization reactions 1 st Method: You can calculate the concentration of the acid or the base by using Titration method Step 1 –   Write the balanced chemical equation between the acid and the base Step 2 – Use the following equation 2 nd Method: Step 1 –   Calculate the amount, in moles, of Acid or Base from the problem (Q) Step 2 –  Calculate the amount, in moles, of Base or Acid by using mole ratio in the balanced equation   Step 3 –  Calculate the concentration, in mol / dm 3  of the Acid or the Base  

18. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid Example: 25.0 cm 3  of 0.050 mol / dm 3  sodium carbonate was completely neutralised by 20.00 cm 3  of dilute hydrochloric acid. Calculate the concentration in mol / dm 3   of the hydrochloric acid. Step 1 –  Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance ( mol ) and dividing by 1000 to convert cm 3  to dm 3 Amount of Na 2 CO 3   =  (25.0 x 0.050) ÷ 1000  =  0.00125 mol  

19. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid Step 3 –  Calculate the concentration, in mol./ dm 3  of the Hydrochloric Acid 1 dm 3  = 1000 cm 3   Volume of HCl =  20 ÷ 1000  =   0.0200 dm 3 = 0.125 mol / dm 3   Step 2 –  Calculate the amount, in moles, of hydrochloric acid reacted the amount, in moles, of hydrochloric acid reacted   4.2 The mole concept

20. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept       3. Calculating Volume   Example: Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm 3  that is required to react completely with 2.5g of calcium carbonate.

21. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Example: Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm 3  that is required to react completely with 2.5g of calcium carbonate. Step 1 –  Calculate the amount, in moles, of calcium carbonate that reacts M r  of CaCO 3  is 100 Amount of CaCO 3   =  (2.5 ÷ 100)  =  0.025 mol Step 2 –  Calculate the moles of hydrochloric acid required CaCO 3   +  2HCl  →  CaCl 2   +  H 2 O  +  CO 2 = 0.05 mol of HCl  

22. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Step 3 –  Calculate the volume of HCl Required Volume of Hydrochloric Acid =  0.05 dm 3  

23. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept The limiting reactant and reacting masses Limiting reactant The limiting reactant is the reactant which is  not present in excess  in a reaction It is always the first reactant to be used up which then causes the reaction to stop In order to determine which reactant is the limiting reagent in a reaction, we have to consider the ratios of each reactant in the balanced equation

24. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept The limiting reactant and reacting masses Example: 9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, Which reactant is in excess and which is the limiting reactant?    

25. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept The limiting reactant and reacting masses Step 2 –  Write the balanced equation and determine the molar ratio 2Na + S → Na 2 S so the molar ratios is 2 : 1 Step 3 –  Compare the moles. So to react completely 0.40 moles of Na require 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant. Step 1 –  Calculate the moles of each reactant Moles = Mass ÷ A r Moles Na = 9.2/23 = 0.40 Moles S = 8.0/32 = 0.25

26. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Chemical equations can be used to calculate the moles or masses of reactants and products Use information from the question to find the amount in moles of the substances being considered Identify the ratio between the substances using the balanced chemical equation Apply mole calculations to find answer Calculating reacting masses

27. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Example 1: Calculate the Mass of Magnesium Oxide that can be made by completely burning 6 g of Magnesium in Oxygen Magnesium (s) + Oxygen (g)  →  Magnesium Oxide (s) Symbol Equation:          Relative Formula Mass:     Magnesium : 24             Magnesium Oxide : 40 Step 1 –  Calculate the moles of Magnesium Used in reaction Moles = Mass ÷ M r       Moles = 6 ÷ 24 = 0.25

28. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Symbol Equation:          Step 2 –  Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation Step 3 –  Find the Mass of Magnesium Oxide Moles of Magnesium Oxide = 0.25 Mass = Moles x M r       Mass = 0.25 x 40 = 10 g

29. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Using the Mole to Determine Empirical & Molecular Formulae Empirical formula:  gives the  simplest whole number ratio  of atoms of each element in the compound Calculated from knowledge of the ratio of masses of each element in the compound Molecular formula:  gives the exact numbers of atoms of each element present in the formula of the compound

30. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Empirical formula Example: A compound that contains 10 g of Hydrogen and 80 g of Oxygen ,Find its Empirical Formula [   Amount of Hydrogen Atoms = Mass in grams ÷ A r  of Hydrogen = (10 ÷ 1) =  10 moles Amount of Oxygen Atoms = Mass in grams ÷ A r  of Oxygen = (80 ÷ 16) =  5 moles the  empirical formula  is H 2 O

31. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Molecular formula Relationship between Empirical and Molecular Formula:

32. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Molecular formula Calculating the Molecular Formula: Step 1 –  Calculate Relative Formula Mass of Empirical Formula Step 2 –   Divide Relative Formula Mass of X by Relative Formula Mass of Empirical Formula   Step 3–    

33. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Example: The Empirical Formula of X is C 4 H 10 S 1  and the Relative Formula Mass of X is 180. What is the Molecular Formula of X? Relative Formula Mass:        Carbon : 12      Hydrogen : 1      Sulfur : 32   Step 1 –  Calculate Relative Formula Mass of Empirical Formula (C x 4) + (H x 10) + (S x 1)    =   (12 x 4) + (1 x 10) + (32 x 1)   =   90 Step 2 –  Divide Relative Formula Mass of [X] by Relative Formula Mass of Empirical Formula 180 / 90 = 2 Step 3 –  Multiply Each Number of Elements by 2 (C 4 x 2 ) + (H 10 x 2 ) + (S 1 x 2 )     =    (C 8 ) + (H 20 ) + (S 2 ) Molecular Formula of X =  C 8 H 20 S 2

34. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Calculating Percentage Yield & Percentage Purity of the Product Percentage yield This is the calculation of the percentage yield obtained from the theoretical yield In practice, you never get 100% yield in a chemical process for several reasons These include some reactants being left behind in the equipment, the reaction may be reversible or product may also be lost during separation stages

35. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Calculating Percentage Yield & Percentage Purity of the Product Percentage yield This is the calculation of the percentage yield obtained from the theoretical yield In practice, you never get 100% yield in a chemical process for several reasons These include some reactants being left behind in the equipment, the reaction may be reversible or product may also be lost during separation stages Equation:  

36. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Example: In an experiment to displace copper from copper sulfate, 6.5 g of Zinc was added to an excess of copper (II) sulfate solution. The copper was filtered off, washed and dried. The mass of copper obtained was 4.8 g. Calculate the percentage yield of copper. Equation Of Reaction:     Step 1:  Calculate the Amount, in Moles of Zinc Reacted Moles of Zinc = 6.5 ÷ 65 = 0.10 moles Step 2:  Calculate the Maximum Amount of Copper that could be formed from the Molar ratio Maximum Moles of Copper = 0.10 moles (Molar ratio is 1:1)

37. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept Equation Of Reaction:     Step 3:  Calculate the Maximum Mass of Copper that could be Formed Maximum Mass of Copper = ( 0.10 x 64 ) = 6.4 g Step 4:  Calculate the Percentage of Yield of Copper Percentage Yield of Copper =  75%  

38. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid 4.2 The mole concept   Percentage purity

39. Saturday, July 25, 2020 Sir. Muhammad Abdul Mageid Percentage purity Example: In an experiment 7.0g of impure calcium carbonate were heated to a very high temperature and 2.5g of carbon dioxide were formed. Calculate the percentage purity of the calcium carbonate.   Calculate the theoretical mass of calcium carbonate used if pure From 2.5g CO 2 = 81.1%