Name : ( ) Class: 3 Date :

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1 Name : ( ) Class: 3 Date : Topic 7.2 : The From Equation Syllabus Objectives : (a) Define relative atomic mass, A r (b) Define relative molecular mass, M r, and calculate relative molecular mass (and relative formula mass) as the sum of relative atomic masses (c) Calculate the percentage mass of an element in a compound when given appropriate information. (d) Calculate number of moles of atoms and molecules and molar mass (e) Calculate molar gas volume (g) Calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24 dm 3 at room temperature and pressure); calculations involving the idea of limiting reactants may be set. (The gas laws and the calculations of gaseous volumes at different temperatures and pressures are not required.) (h) Apply the concept of solution concentration (in mol/dm 3 or g/dm 3 ) to process The results of volumetric experiments and to solve simple problems. (Appropriate guidance will be provided where unfamiliar reactions such as redox are involved. (Calculations on % yield and % purity are NOT required. ) First 40 years after Atomic Theory ( ), there was chaos in Chemistry. Chemists were groping about trying to solve the problem of comparing the masses of atoms, they were making little progress. It took 100 years to find the mass of a hydrogen atom. Mass of one hydrogen atom = g Atoms are so small that it seems inconvenient to express their absolute masses in grams. Therefore, it is preferable to express their atomic masses by comparing theirs with that of a standard atom. It was then agreed to take Carbon-12, the commonest isotope of carbon, as a standard, and compare the masses of other atoms with it. The carbon-12 atom is taken to have a mass of exactly 12 units. 1. Relative Atomic Mass Definition: The relative atomic mass, A r, of an element is defined as the average mass of an atom of the element compared with 1 of the mass of an atom of carbon Note: * Symbol for the relative atomic mass is A r and A r has no unit. The values of the Ar of atoms of the elements are given in the Periodic Table. Secondary 3 Science Chemistry 1

2 Relationship between Nucleon number and Relative atomic mass, A r The mass of an atom is due mainly to the total number of protons and neutrons in its nucleus (Nucleon number). Thus, we would expect the A r of the element to be the same as its nucleon number. This is true for elements which have no isotopes. For most elements, the nucleon number and the A r are not the same due to the existence of isotopes. The nucleon number is always a whole number (since we can t have fractions of protons or neutrons in the nucleus), but the A r values may not be whole numbers. If the Ar is not a whole number and is different from nucleon number, we can conclude that the element has naturally occurring isotopes. Question 1 Chlorine has a relative atomic mass of How can we get a mass number that is not a whole number? Reason: Most naturally occurring samples of an element contain a mixture of isotopes. A normal sample of chlorine contains 75% of chlorine-35 and 25% of chlorine- 37. Thus, the average mass of a chlorine atom = [(75/100 x 35) + (25/100 x 37) ] = 35.5, which is not a whole number. Question 2 [ O Level, Nov 2001 Sect. A 3b(ii) ] 10 Naturally occurring boron contains atoms represented by the symbols B 11 and B 5 5. Suggest why the relative mass of naturally occurring boron is not a whole number. Reason : This is because there are different isotopes of boron. The relative atomic mass is the average of the atomic mass of all the different isotopes of boron. Relative Molecular Mass Many elements and compounds exist as molecules. The mass of a molecule is measured in terms of its relative molecular mass, M r Definition: The relative molecular mass, M r, of a substance is defined as the average mass of a molecule of the substance compared with 1 of the mass of a carbon-12 atom. 12 Relative molecular mass is often abbreviated to molecular mass. Symbol is M r and M r has no units. It is calculated as the sum of the atomic masses of all the atoms in the formula. Example 1 : A r of Oxygen (O) is 16, thus M r of Oxygen (O 2 ) = 16 x 2 = 32 Example 2 : A r of Hydrogen(H) is 1; A r of Oxygen(O) is16, Thus, M r of H 2 O = (2x1) + 16 = 18 Secondary 3 Science Chemistry 2

3 Questions 1. Calculate the relative molecular mass of the following: (a) Carbon dioxide (CO 2 ) (b) Sugar (C 12 H 22 O 11 ) (c) Fullarene ( C 60 ) (d) Ethanoic acid ( CH 3 COOH) 2. Define relative atomic mass. [ Nov 2001 B6a ] 3. Define relative molecular mass. [ Nov 04 A3a, Nov 06 BQ10, Nov 09 A3b, Nov 2010,B11b(ii) ] Relative Formula Mass Compounds can also exist as ions, such as ionic compounds, sodium chloride containing Na + and Cl ions and silver chloride containing Ag + and Cl ions. Since ionic compounds do not contain molecules, the sum of the A r of the atoms in the formula is called the relative formula mass (also given the symbol M r ). The calculation of relative formula mass is the same as above, since the charge does not affect the overall mass. Example 1 : A r of Na = 23 and A r of Cl = 35.5 Thus, relative formula mass/m r of NaCl = = 58.5 Example 2 : A r of Ag = 108 and A r of Cl = 35.5 Thus M r of AgCl = = Questions Calculate the relative molecular mass of the following: (a) Zinc chloride (ZnCl 2 ) (b) Aluminium sulfate ( Al 2 (SO 4 ) 3 ) (c) Lead(II) nitrate (Pb(NO 3 ) 2 ) (f) Hydrated copper(ii) sulfate(cuso 4.5H 2 O ) Secondary 3 Science Chemistry 3

4 2. Calculating percentage by mass composition of element in compound 2.1 The percentage composition of a compound can be calculated from its formula, and the relative atomic masses of its elements The percentage composition = No. of atoms of the element x A r x 100% of an element in a compound M r of the compound Example 1 Calculate the percentage composition of (a) sulfur; (b) oxygen, in sulfuric acid. M r of H 2 SO 4 = ( 2 x A r of H ) + A r of S + ( 4 x A r of O) = (2x1) + (32 x 1) + (16 x 4 ) = = 98 (a) % of S in H 2 SO 4 (b) % of O in H 2 SO 4 = 1 x A r of S x 100 % = 4 x A r of O x 100% M r of H 2 SO 4 M r of H 2 SO 4 = 32 x 100 % = 64 x 100 % = 32.7 % = 65.3% Example 2 (a) Calculate the percentage by mass of sodium in Na 2 CO 3.10H 2 O. [A r of Na = 23, A r of O = 16, A r of C = 12, A r of H = 1, M r of H 2 O = (2 + 16) ] M r of Na 2 CO 3.10H 2 O = (23x2) + (12x1) + (16x3 ) + 10(2 + 16) = = 286 % of Na = 2 x 23 x = = 16.1% (3 sig fig) (b) Calculate the percentage by mass of water of crystallization in Na 2 CO 3.10H 2 O. % of H 2 O = 10 x 18 x = 62.9 % Secondary 3 Science Chemistry 4

5 Question 1. Calculate the percentage by mass of germanium in germanium (IV) oxide,geo 2. [ O Level 2005 A5c(i)] 2. Calculate the percentage by mass of carbon in ethanol, C 2 H 5 OH. [Nov 08 B11c] 2.2 Calculating the mass of an Element in a Compound The mass of an element = No. of atoms x Ar of the element x mass of sample in a compound Mr of compound Example: Calculate the mass of oxygen in 10g of sodium carbonate crystals, Na 2 CO 3.10H 2 O. Solution : Mass of oxygen = 13 x 16 x 10 g 286 = 7.27g (3 sig. fig.) Questions 1. Calculate the mass of water in 10g of sodium carbonate crystals, Na 2 CO 3.10H 2 O. 2. What is the mass of aluminium in 204 g of aluminium oxide, Al 2 O 3? 3. An ore contains 32% iron(iii) oxide(fe 2 O 3 ). Calculate the mass of iron(fe) in 1000g of the ore. [Nov 1998 P3 B8b] Secondary 3 Science Chemistry 5

6 3. The Mole Concept 3.1 Avogadro number (Avogadro constant) The mole is the standard method in chemistry for communicating how much of a substance is present. A mole is the amount of substance which contains 6x10 23 of particles of that substance. This number, 6 x 10 23, is known as the Avogadro number. The particles may be atoms, molecules, ions or electrons. e.g 1 mole of copper contains 6x10 23 copper atoms. 1 mole of carbon dioxide contains 6x10 23 CO 2 molecules. 1 mole of sodium ions contains 6x10 23 sodium ions. 1 mole of electrons contains 6x10 23 electrons. No. of moles of substance = no. of particles of substance 6 x Molar Mass Molar mass is the mass of one mole of any substance. It is the relative atomic mass (A r ) or relative molecular mass (M r ) expressed in grams. (i) Molar Mass of Atoms e.g. One mole of carbon atoms has a mass of 12 g. (A r of C = 12) No. of atoms in 1 mole = 6 x e.g. One mole of aluminium atoms has a mass of 27 g. (A r of Al = 27) No. of atoms in 1 mole = 6 x No. of moles of atoms in = mass of the sample (g) a sample of an element molar mass of element (g/mol) or simply, No. of moles = mass A r Note: all masses measured are in grams. E.g. 1 No of moles in 46g sodium = 46 (A r of sodium = 23) 23 = 2 E.g. 2 Mass of 4 moles of calcium = 4 x 40 g = 160g (A r of calcium = 40) Secondary 3 Science Chemistry 6

7 (ii) Molar mass of molecules E.g. One mole of water has a mass of 18g. [since M r of H 2 O = (2x1)+ 16 = 18 ] No. of water molecules in one mole is 6 x E.g. One mole of carbon dioxide has a mass of 44g [ since M r of CO 2 = 12 + (16x2) = 44 ] No. of carbon dioxide molecules in one mole is 6 x No. of moles of molecules in = a sample of a substance mass of sample (g) molar mass of substance(g/mol) or simply, No. of moles = mass M r E.g. 1 No. of moles in 80g oxygen = 80 = 2.5 ( since M r of O 2 = 16x2 = 32) 32 E.g. 2 Mass of 2 moles of hydrogen = 2 x 2 g = 4 g ( since M r of H 2 = 2 x1 = 2) E.g. 3 No. of Hydrogen molecules in 2 moles = 2 x 6 x = 1.2 x Questions : 1. Calculate the mass of one mole (molar mass) of the following substances : a) Cl 2 : b) C 3 H 8 : c) Na 2 CO 3 : d) Zn(NO 3 ) 2 : 2. Calculate the mass of the following : (a) 2 moles of oxygen atoms (b) 4 moles of iron(iii) oxide 3 How many chlorine molecules are there in 35.5 g of gaseous chlorine? Secondary 3 Science Chemistry 7

8 4. Molar Volume of a Gas Avogadro s Law : Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. Hence, 1 mole of any gas always has the same volume. One mole of all gases at room temperature and pressure (r.t.p) occupies a volume of 24 dm 3, known as the molar volume. i.e. The molar volume for all gases at r.t.p = 24 dm 3 or cm 3 No. of moles of gas = Volume of gas in dm 3 = Volume Molar volume (dm 3 /mol) 24 Since 1 dm 3 is 1000 cm 3, we can also calculate no. of moles of gas as below, if volume of gas is given in cm 3 : No. of moles of gas = Volume (when volume of gas is in cm 3 ) E.g.1 Calculate the number of moles of CO 2 when 12 dm 3 of it was evolved in a reaction. Solution : No. of moles of CO 2 = 12 = E.g. 2 Calculate the mass of oxygen if 4 dm 3 of the gas was liberated at r.t.p. Solution : No. of moles of O 2 = 4 = 1 (not final answer, can be in fraction) 24 6 Mass of O 2 = mole x Mr = 1 x 32 g = 5.33g ( final ans in 3 s.f. ) 6 Questions A sample of carbon monoxide has a volume of 6 dm 3 at r.t.p. Calculate (a) the no. of moles of gas in the sample, (b) the no. of gas molecules in the sample, (c) the mass of the sample. Secondary 3 Science Chemistry 8

9 5. Concentration of Solutions The concentration of a solution tells you the amount of solute in 1 dm 3 of solution. It can be measured in mol/dm 3 or in g/dm 3. Note : 1 dm 3 = 1 litre = 1000 cm 3 Mol/dm 3 is commonly used because it is convenient to express concentrations of substances in it. It is sometimes referred as Molarity of a solution represented by the symbol M. Molarity, M, means mol/dm 3 A molar solution (1M ) or 1 mol/dm 3, contains 1 mole of the substance in 1 dm 3 of the solution. 2 M solution or 2 mol/dm 3, contains 2 moles of the substance in 1 dm 3 of solution M solution or 0.05 mol/dm 3, contains 0.05 moles of the substance in 1 dm 3 of solution. In summary : (a) Concentration in mol/dm 3 = No of moles of solute Volume of solution in dm 3 Hence, in a volume of a solution : No. of moles = volume of solution(dm 3 ) x concentration in mol/dm 3 (b) Concentration in g/dm 3 = mass of solute in gram Volume of solution in dm 3 Secondary 3 Science Chemistry 9

10 Questions 1. Complete the following table : Solution Sodium hydroxide (NaOH) Sulfuric acid (H 2 SO 4 ) No. of moles of substance in a volume of the solution Mass of substance in a volume of the solution x ( ) = 8g A volume of the solution 50 cm 3 = = 0.05 dm 3 Concentration of solution in mol/dm 3 = no. of mole vol of solution Concentration of solution in g/dm 3 = mass vol of solution 4.9g 1 dm = 4.9g/dm 3 Nitric acid (HNO 3 ) Potassium hydroxide (KOH) 2 x = cm mol/dm cm mol/dm 3 (Apply The Mole Concept : No. of mole = concentration in mol/dm 3 x vol in dm 3, No of mole = mass/m r and Mass = No. of mole x M r ) 2(a) Calculate the relative molecular mass of sodium hydroxide, NaOH. (b) Calculate the mass of sodium hydroxide in (i) 1000 cm 3 of 0.2 mol/dm 3 sodium hydroxide solution, (ii) 20 cm 3 of 2.0 mol/dm 3 sodium hydroxide solution. [Nov 2002P3A3] 3. Calculate the mass of sodium carbonate required to make 250cm 3 of a 2.0 mol/dm 3 solution of sodium carbonate. Secondary 3 Science Chemistry 10

11 6. Summary Exercise on The Mole Concept Carry out the following calculations, starting with the sample of carbon dioxide, CO 2, with a mass of 11g. Write your answers and working in the boxes. (Source of Question: Science in Focus- Chemistry for GCE O Level Theory Workbook by JGR Briggs,Longman) Secondary 3 Science Chemistry 11

12 7. Chemical Calculations From Equations Three Basic Methods to do chemical calculations: Method 1 - Using mole ratio (when masses in g, volume of gas, concentration are given) Method 2 - Using mass ratio ( where masses are not given in grams, but, in kg or tonnes) Method 3 - Using volume ratio (if only volumes of gases are involved) Method 1 Using Mole Ratio Steps involved in working out the required calculation: 1. Write the balanced equation if not given in question. 2. Get mole ratio from equation 3. Convert information given (whether in mass, volume or concentration of substance) in the question into moles 4. Use the mole ratio and work out by proportion the no. of moles of unknown 5. Convert no of moles to mass or volume or calculate concentration of solution as required by question Eg 1. What mass of magnesium oxide can be obtained from the combustion of 2.4 g of magnesium? Chemical Equation: 2Mg + O 2 2MgO (write it if not given) No. of moles in 2.4g Mg = mass = 2.4 = 0.1 ( A r used, Mg made up of atoms) A r 24 From equation, 2 mol Mg produces 2 mol MgO mole ratio is 2 mol Mg : 2 mol MgO Hence, 0.1 mol Mg produces 0.1 x 2 2 = 0.1mol MgO (Qn requires calculation of these) (Using proportion from mole ratio)* M r of MgO = = 40 ( Always show calculation of M r value) Mass of 0.1 mol MgO produced = No. of moles x M r g = 0.1 x 40 g = 4g ================================================================ *Some students may find it easier writing it as No.of moles of Mg = 2 No. of moles of MgO 2 Substitute no. of moles of Mg, we have 0.1 = 2 No. of moles of MgO 2 Then, cross multiply to get no. of moles of MgO = 0.1 x 2 = Secondary 3 Science Chemistry 12

13 Eg 2. Copper(II) oxide reacts with carbon to form copper and carbon dioxide. Calculate the mass of copper(ii) oxide used if 12.8g of copper was obtained in the reaction. Chemical Equation: 2CuO + C 2Cu + CO 2 (write it if not given) No. of moles of 12.8g Cu = mass = 12.8 = 0.2 (mass is given in qn) Ar 64 From equation, 2 mol Cu is obtained from 2 mol CuO Mole ratio is 2 mol Cu : 2 mol CuO * Hence, 0.2 mol Cu is obtained from 0.2 x 2 2 = 0.2 mol CuO M r CuO = = 80 (essential working) mass of 0.2mol CuO used = No. of moles x Mr g = 0.2 x 80g = 16g * Alternative presentation : No of moles of Cu = 2 No of moles of CuO 2 Substitute known mole 0.2 = 2 No. of CuO 2 Cross multiply to get No of moles of CuO = 0.2 x2 = Eg 3. Calcium carbonate decomposes to form carbon dioxide and calcium oxide. Calculate the volume of carbon dioxide (at r.t.p) produced from the decomposition of 20g of calcium carbonate. Chemical Equation : CaCO 3 (s) CaO (s) + CO 2 (g) M r CaCO 3 = (16x3) = 100 (essential working) No. of moles in 20g CaCO 3 = From equation: CaCO 3 produces CO 2 gas Mole ratio is Hence, 0.2mol CaCO 3 produces volume of 0.2 mol CO 2 = Hence, 20g calcium carbonate decomposes to produce 4.8dm 3 carbon dioxide. Secondary 3 Science Chemistry 13

14 Questions [Calculation from equation using mole ratio method] (Note : All working must be clearly presented, final answers must be in 3 sig fig) 1. Copper(II) oxide reacts with hydrogen to form copper and water. What mass of copper would be obtained from 16g of copper(ii) oxide? Chemical Equation : kg of magnesium reduce copper(ii) oxide to copper. What mass of copper is obtained? Chemical Equation : 3. In a reaction, silver nitrate decomposed on heating producing silver metal, nitrogen dioxide and oxygen. Given that 42.5g of silver nitrate completely decomposed when heated in an experiment. Chemical Equation : 2AgNO 3 2Ag + 2NO 2 + O 2 Calculate (a) the mass in grams of the silver metal formed; (b) the volume of nitrogen dioxide gas formed at r.t.p. Secondary 3 Science Chemistry 14

15 4. 50cm 3 of a 2M hydrochloric acid was added to excess zinc, producing zinc chloride and hydrogen gas. Calculate the volume of hydrogen gas liberated at r.t.p. 5. A solution of sulfuric acid contained 4.9g of H 2 SO 4 per dm cm 3 of the acid reacted with 24 cm 3 of a solution of NaOH. Calculate concentration of NaOH in (a) mol/dm 3 and (b) g/dm cm 3 of aqueous 0.1 mol/dm 3 hydrochloric acid exactly neutralizes 20 cm 3 of aqueous sodium hydroxide. The equation for this reaction is shown. NaOH + HCl NaCl +H 2 O What is the concentration of the sodium hydroxide solution? (GCE O Nov2009P1Q7) (Hint : This question involves use of given concentration and volume of solution to work out the no. of moles, using mole ratio to calculate unknown no. of moles and then using calculated no. of mole and given volume of solution to calculate the unknown concentration) Secondary 3 Science Chemistry 15

16 Method 2 ( Using Mass Ratio ) Although the gram is the usual unit of mass used in calculations, other units such as the kilogram or the tonne can be used, especially in industry. In cases where the kg or tonne is used, it is simpler to use the Mass-Mass ratio method. Steps: 1. Write the balanced equation if not given in question 2. Get mole ratio from equation 3. Convert mole ratio to mass ratio (use Ar or Mr, applying mass =mole x Ar or mass = mole x M r ) 4. Use proportion to find unknown mass E.g kg of magnesium reduced copper(ii) oxide to copper. What mass of copper is obtained? Chemical Equation : Mg + CuO Cu + MgO Explanation: ( A r of Mg = 24; A r of Cu=64, mass of 1 mol Mg = 1 mol x 24 g = 24 g From equation, 1 mol Mg produces 1 mol Cu Mole ratio : 1 mol Mg : 1 mol Cu Mass ratio : 24 g Mg : 64 g Cu mass of 1 mol Cu = 1 mol x 64g = 64 g) 24 kg Mg : 64kg Cu Hence, 4.8 kg Mg produces 4.8 x 64 kg Cu 24 = 12.8 kg Cu E.g. 2. Ammonia is produced from the reaction between hydrogen and nitrogen gas. Calculate the mass of nitrogen needed to produce 17 tonnes of ammonia. [ 1 tonne = 1000 kg = g ] Chemical Equation : N 2 + 3H 2 2NH 3 Explanation: Mole ratio is 2 mol of NH 3 : 1 mol N 2 M r of NH 3 = 14 + (3 x 1) = 17 Mass ratio : (2 x17)g NH 3 : (1x 28) g N 2 M r of N 2 = 2 x 14 =28 34 g NH 3 : 28 g N 2 mass = no. of mole x M r g 34 tonnes : 28 tonnes N 2 Hence,17 tonnes NH 3 requires (17 x 28 ) = 14 tonnes tonnes of nitrogen is needed. Secondary 3 Science Chemistry 16

17 Questions : [Calculations form equation using the mass ratio method ( when the reacting masses are not in grams, say, in kg or in tonnes )] 1. Iron is made in the blast furnace by the following reaction: Fe 2 O 3 + 3CO 2Fe + 3CO 2 Starting from 28 tonnes of iron(iii) oxide, how much iron can be made? 2. In industry, wolframite, (FeMn)WO 4,is changed into tungsten(vi) oxide, WO 3. Metallic tungsten is formed by heating this oxide in hydrogen gas. WO 3 + 3H 2 W + 3H 2 O Calculate the mass of tungsten in tonnes that could be formed from 20 tonnes of tungsten(vi) oxide. [ A r : H,1; O,16; W,184] [Nov 2007 P3 5bii] 3. In the manufacture of metal uranium, uranium dioxide is first converted into a fluoride. Balance this equation for the reaction. [Nov 2008P3A7] UO 2 + HF UF 4 + H 2 O Uranium tetrafluoride is then reduced to metal uranium by heating with n magnesium, according to this balanced chemical equation. UF 4 + 2Mg 2MgF 2 + U Calculate the mass of magnesium that must be used to manufacture 10 tonnes of uranium. [Relative atomic masses: A r : Mg, 24; U, 238 ] Secondary 3 Science Chemistry 17

18 Method 3 Using volume ratio (For calculations involving volumes of reacting gases, it is simpler to use the volume-volume ratio method) Steps: 1. Write the balanced equation if not given in question. 2. Get mole ratio from equation 3. Write down the volume ratio which is the same as the mole ratio. 4. Use proportion to find volume of gas required E.g. 1 Under suitable conditions, nitrogen and hydrogen combine to form ammonia. Calculate the volume of ammonia formed if 100cm 3 of nitrogen is reacted with excess hydrogen. All gas volumes are measured at room temperature and pressure. Chemical Equation : N 2 + 3H 2 2NH 3 From equation, mole ratio is 1 mol of N 2 : 2 mol NH 3 volume ratio is 1 cm 3 of N 2 : 2 cm 3 of NH 3 Explanation: ( cm 3 in ratio is used as in the question) Hence,100cm 3 N 2 produces (100 x 2 ) = 200 cm 3 NH 3 1 E.g. 2 What volume of CO is produced by completely reacting 15 cm 3 CO 2 with charcoal? Chemical Equation : C + CO 2 2CO From equation, Mole ratio is 1 mol of CO 2 : 2 mol CO Volume ratio is 1 cm 3 of CO 2 : 2 cm 3 of CO 15 cm 3 CO 2 produces (15 x 2 ) = 30 cm 3 CO 1 E.g. 3 A volume of 1000 dm 3 methane is burnt completely in oxygen. Calculate the volumes of gaseous products when measured (i) above C, (ii) below C [Nov 2003 P3B11b] Chemical Equation : CH 4 + 2O 2 CO 2 + 2H 2 O Mole ratio is 1 mol CH 4 : 1 mol CO 2 : 2 mol H 2 O volume ratio is 1 dm 3 CH 4 : 1 dm 3 CO 2 : 2 dm 3 H 2 O Hence, 1000 dm 3 CH 4 : 1000 dm 3 CO 2 : 2000 dm 3 H 2 O Secondary 3 Science Chemistry 18

19 (i) Above C, gaseous products are CO 2 and H 2 O,thus total volume of gaseous product measured is 1000dm dm 3 = 3000 dm 3 (ii) Below C, carbon dioxide is the only gaseous product, thus volume volume of gaseous product measured is 1000dm 3. Questions [Calculation involving volume of gases using volume-volume ratio] 1. Phosphine, PH 3, burns in air. The change is represented by the following equation. 4PH 3 (g) + 8O 2 (g) P 4 O 10 (s) + 6H 2 O(g) (a) Calculate the relative molecular mass of phosphorus (V) oxide, P 4 O 10. (b) What volume of oxygen will be needed to burn completely 48dm 3 of phosphine? (c) What is the mass of 48dm 3 of phosphine at room temperature and pressure? cm 3 of butane (C 4 H 10 ) was mixed with an excess of oxygen and exploded. (All measurements were made at room temperature and pressure.) Equation : 2C 4 H 10 (g) + 13O 2 (g) 8CO 2 (g) + 10 H 2 O(g) Calculate (a) the volume of carbon dioxide produced and (b) the volume of oxygen required for the reaction Secondary 3 Science Chemistry 19

20 8. Chemical Calculation involving Limiting and Excess Reactants E.g.1 Nitrogen,N 2 and hydrogen, H 2 will react according to the chemical equation : N 2 + 3H 2 2NH 3 If I mole of nitrogen is reacted with 4 moles of hydrogen in the reaction vessel, (a) (b) Which reactant will be in excess? What is the maximum number of moles of ammonia produced? Solution: (a) From equation, Mole Ratio is 1 mol N 2 : 3 mol H 2 Thus, 1 mol of N 2 will react only with 3 mol of H 2 Since there is 4 mol of H 2 in the reaction vessel, H 2 is the reactant in excess and N 2 will be totally used up, it is the limiting reactant. (b) The amount of product formed depends on the amount of limiting reactant used (N 2 ) From equation, Mole Ratio is 1 mol N 2 : 2 mol NH 3 Thus, maximum no. of moles of NH 3 produced is 2 moles. E.g. 2 A mixture of 125cm 3 oxygen and 50cm 3 hydrogen at room temperature is exploded in a suitable apparatus. After reaction, the apparatus is allowed to cool to room temperature again. Give the names of the remaining gases and also their volumes. Chemical Equation : 2H 2 + O 2 2H 2 O From equation, Mole Ratio is 2 mol H 2 : 1 mol O 2 Volume Ratio: 2 cm 3 H 2 : 1 cm 3 O 2 50cm 3 H 2 will react with ( 50 x 1 ) = 25 cm 3 O 2 2 or 125 cm 3 O 2 will need (125 x 2)cm 3 = 250 cm 3 H 2 [note : only 50cm 3 H 2 is given] so, not possible to totally react 125cm 3 O 2 ] Thus, hydrogen is the limiting reactant as it will be totally used up in the reaction and oxygen is the excess reactant. So, the amount of water produced depends on the amount of limiting reactant, H 2. From equation, Mole ratio is 2 mol H 2 : 2 mol H 2 O Volume ratio is 2 cm 3 H 2 : 2 cm 3 H 2 O 50cm 3 H 2 produces 50cm 3 of H 2 O Thus, 50 cm 3 of water vapour is produced and (125 25) cm 3 oxygen remains in excess. Questions: = 100 cm 3 of Secondary 3 Science Chemistry 20

21 1. When 7g of iron reacts with 4g of sulfur, 11g of iron(ii) sulfide is produced. What will be produced if 7g of iron reacted with 7g of sulfur? Chemical equation : Fe + S FeS A. 11 g of iron(ii) sulfide and 3 g of unchanged iron B. 11 g of iron(ii) sulfide and 3 g of unchanged sulfur C. 11 g of iron(ii) sulfide only D. 14 g of iron(ii) sulfide only ( ) cm 3 of oxygen are reacted with 20 cm 3 of carbon monoxide. What are the volumes of the gases remaining, at original temperature and pressure? Chemical Equation : 2CO + O 2 2CO 2 Oxygen/cm 3 Carbon monoxide/cm 3 Carbon dioxide/cm 3 A B C D ( ) 3. Fe combines with Cl 2 according to the equation : 2Fe + 3Cl 2 2 FeCl 3 How many grams of FeCl 3 can be obtained by reacting 3 mol of Fe and 5 mol of Cl 2? cm 3 of a 0.1 mol/dm 3 solution of lead(ii) nitrate was mixed with 100 cm 3 of a 0.1 mol/dm 3 solution of sodium sulfate. The insoluble lead(ii) sulfate is precipitated according to the chemical equation : Pb(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) PbSO 4 (s) + 2NaNO 3 (aq) [precipitate is an insoluble solid formed in a solution] (i) What is the limiting reactant? (ii) Calculate the number of moles of lead (II) sulfate precipitated. (iii) Calculate the mass of lead (II) sulfate precipitated. Summary of Key Points and Formulae in Mole Concept and Chemical Calculation Secondary 3 Science Chemistry 21

22 - Define Relative Atomic Mass (A r ) and Relative Molecular Mass (M r ) - One mole of a substance has 6 x particles ( be it ions, atoms, molecules or electrons) - Percentage composition = No. of the atoms of the element x A r x 100 % by mass of element M r of compound in a compound - Mass of an element = No. of the atoms of the element x A r x mass of in a sample of a M r of compound sample compound - Molar mass = mass of one mole = A r g ( for atoms) - Molar mass = mass of one mole = M r g ( for molecules ) - Molar volume of a gas is 24 dm 3 at r.t.p No. of particles Mass (in g) No. of moles 6 x No. of moles A r Volume (in cm 3 ) No. of moles Volume (in dm 3 ) 24 dm 3 No. of moles cm 3 No. of moles Mass (in g) M r No. of moles Concentration Mol/dm 3 Volume of solution in dm 3 Secondary 3 Science Chemistry 22