GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME. PHYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES)

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1 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) MOLE CONCEPT, STOICIOMETRIC CALCULATIONS Learner Note: The ole concept is carried forward to calculations in the acid and base section, as well as in the cheical equilibriu. It is iportant to know this section well. Question 1: 5 inutes (Taken fro DoE Nov 2007) (Revise calculations of olecular ass. Reeber cheical calculations need forulae, substitution and an answer with the correct unit.) The copound NaCO 3 is coonly known as baking soda. A recipe requires 1,6 g of baking soda, ixed with other ingredients, to bake a cake. 1.1 Calculate the nuber of oles of NaCO 3 used to bake the cake. (3) 1.2 ow any atos of oxygen are there in1,6 g baking soda? (4) (7) Question 2: 10 inutes (Taken fro MED Nov 2009) (The su of all the percentages equals yes, 100. Therefore, in a 100 g of the substance the ratio will be the sae. Learn the ethod the steps are always repeated. The eleents are ostly given in the sae order as they appear in the forula) One of the active ingredients in vinegar is Ethanoic acid. Ethanoic acid has a olecular ass of 60 g.ol -1 and the following percentage coposition 39,9 % carbon 6,7 % hydrogen 53,4 % oxygen 2.1 Define the concept epirical forula (2) 2.2 Deterine the epirical forula of Ethanoic acid (5) 2.3 What is the olecular forula of Ethanoic acid? (3) (10) Question 3: 10 inutes (Adapted fro MED Nov 2009) The contact process is given by the equation below. SO 2 (g) + O 2 (g) SO 3 (g) 3.1 Balance the cheical equation (2) In an investigation 256 g SO 2 reacts with 80 g O 2 in a reaction vessel. 3.2 Calculate the nuber of oles of each reactant present at the start of the reaction(5) 3.3 Identify the liiting reagent in the reaction and justify your answer. (2) (Liiting reagents are frequently asked the one that liits the reaction is the one that will be used up first. You ust first work out the nuber of oles represented by the given asses of the reactants, then deterine the liiting reagent by using the oll ratio) 3.4 Calculate the ass of SO 3 produced in the reaction (4) (13) Page 1 of 8

2 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) Question 4: 5 inutes (It is a coon istake to interpret the dot as a ultiply function. This is the water that is trapped in the crystal during crystallisation and ust be ADDED to the ass of the ionic copound.) 4.1 Calculate the percentage water of crystallisation in CuSO O (4) 4.2 Calculate the concentration of a 250 l solution of sodiu hydroxide if 10 g of the solute is dissolved. (4) (8) SECTION B: SOLUTIONS AND INTS Question 1 Learner Note: Ephasize the correct use of forulae and layout. Use the correct unit 1.1 M (NaCO 3 ) = (16) = 84 g.ol -1 n = M 1,6 n = 84 = 0,02 ol (rounded to 2 decial places) 1.2 Each ato has three oxygen atos, there is 0,02 ol of atos 1 ol = 6,023 x particles therefore 0,02 ol x 3 atos x 6,023 x particles = 3,44 x atos oxygen Page 2 of 8

3 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) Question 2 Learner Note: If the steps are followed, the questions becoe relatively siple. Reeber we are working out what the ratio of the eleents are in the olecule therefore when you deterine the nuber of ol the eleent ass is usedregardless of whether the eleent is diatoic! 2.1 Epirical forula the siplest ratio of atos in a olecule Eleents C O In 100g 39,9 g 6,7 g 53,4 g Convert ass to ol n = M 39,9 12 = 3,325 ol Divide by sallest answer 325 = 1 Ratio of eleents in the epirical forula Epirical forula C 2 O 2.3 M (C 2 O) = (1) + 16 = 30 g.ol -1 Question 3 6,7 1 = 6,7 ol 6, = 2,01 53,4 16 = 3,3375 ol = Molecular ass is double epirical forula ass therefore olecular forula is C 2 4 O 2 (It is iportant to balance cheical equations before doing the calculations. The only tie where the balancing coefficients are used will be in the ratio step.) SO 2 (g) + O 2 (g) 2 SO 3 (g) 3.2 n = M n = n = 4 ol SO 2 n = M 80 n = 32 n = 2,5 ol O 2 Page 3 of 8

4 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) 3.3 A ratio of 2 ol SO 2 is needed for 1 ol O 2 according to reaction. Therefore 2,5 ol O 2 needs 5 ol SO 2 to react copletely, the SO 2 is therefore the liiting reagent ol SO 2 reacts and 4 ol SO 3 is produced n = M 4 = 80 Question 4 = 320 g SO 3 ade 4.1 M (CuSO O) = 63, (16) + 5 ( ) = 159, = 249,5 g.ol % water = x ,5 = 36 % water 4.2 c = MV 10 c = 40x0,25 c = 1 ol.d -3 Page 4 of 8

5 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) SECTION C: ADDITIONAL CONTENT NOTES The Mole The ole is one of the base units of the SI and is the base unit of the aount of atter of substance. A ole of any substance always has the sae nuber of atos. Definition: A ole of any substance is that aount of substance which contains as any eleentary particles as there are atos in 12g of carbon-12. The aount of atter or substance is not the sae as the ass of the substance. While one ole of a substance contains the sae nuber of particles, their asses are not the sae. The forula used to calculate the nuber of oles in a substance: In sybols n = M Avogadro s Nuber One ole of any substance is approxiately equal to 6,02x10 23 eleentary particles. This very large nuber is known as Avogadro s nuber or Avogadro s constant and has the sybol N A or L. Therefore: Nuber of particles = Avogadro s nuber x nuber of oles In sybols Np = N A x n Molar Volues of Substances Molar volue is the volue of one ole of a substance and can be easured in d 3 /ol. One ole of any gas occupies a volue of approxiately 22,4d 3 at Standard Teperature and Pressure (STP). Standard Teperature is 0 o C (273K) and Standard Pressure is 101,3kPa. Equal volues of all gases at STP contain the sae nuber of olecules. Therefore: olar volue = volue of substance Nuber of oles of substance V n = 3 22,4 d Epirical forula sallest ratio of atos in a olecule True Forula or olecular forula actual ratio of atos in a olecule eg. the Epirical Forula of a substance is CO 2 but its olecular forula is C 2 O 2 4 Percentage Coposition shows percentage of each eleent in a copound copared to its olecular ass Page 5 of 8

6 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) SECTION D: OMEWORK Learner Note: Revise steps for each type of calculation and follow the prescribed ethod. Attept all questions and refer to notes when in doubt. Question 1: 25 inutes 1.1 Calculate the relative forula ass of KClO 3 (3) 1.2 Calculate how any ties a olecule of ethanol (C 3 O) is heavier than a olecule of water (5) 1.3 Calculate the epirical forula of the substance with the following coposition 45,3 % O; 43 % Na; 11,3 % C (5) 1.4 ow any potassiu atos are there in 2 g K 2 SO 4 (5) 1.5 Fe + S FeS Which of the two substances will be used up if 10 g Fe and 10 g S are ixed and heated (7) (25) Page 6 of 8

7 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) SECTION E: SOLUTIONS TO SESSION 10 OMEWORK 1.1 When atos approach each other, the valence electron of the two atos attract the nucleus of the other ato, these attractive forces are stronger than the repulsive forces between the atos. The protons and electrons of respective atos attract the atos to for the bond. The atos ove closer together and the potential energy becoes negative. The atos are ost stable at the lowest value of potential energy when the orbitals overlap and bonding occurs. The two hydrogen atos each share an electron during bonding; there is a net electrostatic force of attraction between the atos. 2 is fored. When an ato approach oxygen ato, the valence electron of the atos attracts the nucleus of the other ato, these attractive forces are stronger than the repulsive forces between the atos. The protons and electrons of respective atos attract the atos to for the bond. The atos ove closer together and the potential energy becoes negative. The atos are ost stable at the lowest value of potential energy when the orbitals overlap and bonding occurs. The hydrogen and oxygen ato share an electron during bonding, there is a net electrostatic force of attraction between the atos. 2 O is fored. 1.2 Both atos require an electron to fill the orbital and obtain the noble structure which is of lower energy. Its valence energy level is not filled. The atos share an electron pair, there is a net electrostatic force of attraction, bonding occurs 1.3 The eliu ato is in the noble state, it has a filled last energy level, it is stable and requires a large aount of energy to reove an electron. No bonding occurs. 1.4 A bond is a net electrostatic force between two atos. Atos bond to obtain a filled valence orbital octet rule 8 electrons in valence orbital increases stability. Except hydrogen which follows the rule of two its valence orbital can have a axiu of 2 electrons. When two atos approach each other, the valence electrons of the two atos attract the positive nucleus of the other ato; these attractive forces are stronger than the repulsive forces between the atos. The protons and electrons of respective atos attract the atos to for the bond when orbitals overlap to get a full valence orbital. The atos ove closer together and the potential energy becoes negative. The atos are ost stable at the lowest value of potential energy when the orbitals overlap and bonding occurs. The two atos each share electrons during bonding, there is a net electrostatic force of attraction between the atos. 1.5a. Different atos, each with an unpaired valence electron can share these electrons to for a cheical bond Page 7 of 8

8 PYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES) Question 2 b. Different atos with paired valence electrons called lone pairs of electrons, cannot share these four electrons and cannot for a cheical bond c. Different atos, with unpaired valence electrons can share these electrons and for a cheical bond for each electron pair shared (ultiple bond foration) d. Atos with an incoplete copleent of electrons in their valence shell can share a lone pair of electrons fro another ato to for a co-ordinate or dative covalent bond 2.1 a. Cl b. e 2.2 a. F F b. O c. N a. N b. Lewis base c. Lewis acid The SSIP is supported by Page 8 of 8

21. Practice Problem (page 556) What is the volume of 5.65 mol of helium gas at a pressure of 98 kpa and a temperature of 18.0 C?

21. Practice Problem (page 556) What is the volume of 5.65 mol of helium gas at a pressure of 98 kpa and a temperature of 18.0 C? Section 12.2 The Ideal Gas Law Solutions for Practice Probles Student Edition page 556 21. Practice Proble (page 556) What is the volue of 5.65 ol of heliu gas at a pressure of 98 kpa and a teperature