End of chapter exercises

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1 End of chapter exercises Problem 1: Write only the word/term for each of the following descriptions: 1. the mass of one mole of a substance 2. the number of particles in one mole of a substance Answer 1: a) Molecular mass b) Avogadro's number Problem 2: 5 g of magnesium chloride is formed as the product of a chemical reaction. Select the true statement from the answers below: 1. 0,08 moles of magnesium chloride are formed in the reaction 2. the number of atoms of Cl in the product is 0, the number of atoms of Mg is 0,05 4. the atomic ratio of Mg atoms to Cl atoms in the product is 1:1 Answer 2: d. 0,05 moles of magnesium chloride are formed in the reaction Problem 3: 2 moles of oxygen gas react with hydrogen. What is the mass of oxygen in the reactants?

2 1. 32 g 2. 0,125 g g 4. 0,063 g Answer 3: d. 32 g Problem 4: In the compound potassium sulphate (K2SO4), oxygen makes up x% of the mass of the compound. x =? 1. 36,8 2. 9, ,3 Answer 4: d. 36,8 Problem 5: The concentration of a 150 cm 3 solution, containing 5 g of NaCl is: 1. 0,09 mol dm , mol dm ,57 mol dm ,03 mol dm 3

3 Answer 5: d. 0,57 M Problem 6: Calculate the number of moles in: 1. 5 g of methane (CH4) 2. 3,4 g of hydrochloric acid 3. 6,2 g of potassium permanganate (KMnO4) 4. 4 g of neon 5. 9,6 kg of titanium tetrachloride (TiCl4) Answer 6: a) The molar mass of methane is 12,011+4(1,008)=16,043g mol 1 So the number of moles is: n=mm n=516,043 n=0,31mols b) The molar mass of hydrochloric acid (HCl ) is 35,45+1,008=36,46g mol 1

4 So the number of moles is: n=mm n=3,436,46 n=0,09mols c) The molar mass of potassium permanganate is: 39,09+54,94+4(16)=158,03g mol 1 So the number of moles is : n=mm n=6,2158,03 n=0,04mols d) The molar mass of neon is 20,18g mol 1 So the number of moles is: n=mm n=420,18

5 n=0,2mols e) The molar mass of titanium tetrachloride is 47,88+4(35,45)=189,68g mol 1 So the number of moles is: n=mm n=960189,68 n=5,06mols Problem 7: Calculate the mass of: 1. 0,2 mol of potassium hydroxide (KOH) 2. 0,47 mol of nitrogen dioxide 3. 5,2 mol of helium 4. 0,05 mol of copper (II) chloride (CuCl2) 5. 31, molecules of carbon monoxide (CO) Answer 7: a) The molar mass of potassium hydroxide is: 39,09+1,008+16=56,1g mol 1 So the mass is:

6 m=nm m=(0,2)(56,1) m=11,2g b) The molar mass of nitrogen dioxide (NO2 ) is: 14+2(16)=46g mol 1 So the mass is: m=nm m=(0,47)(46) m=21,62g c) The molar mass of helium is 4g mol 1 So the mass is : m=nm m=(5,2)(4) m=20,8g

7 d) The molar mass of copper (II) chloride is 63,55+2(35,45)=134,45g mol 1 So the mass is: m=nm m=(0,05)(134,45) m=6,72g e) The molar mass of carbon monoxide is 12,011+16=28,011g mol 1 The number of moles of carbon dioxide is 31, , =5,2mols So the mass is : m=nm m=(5,2)(28,011) m=145,66g Problem 8: Calculate the percentage that each element contributes to the overall mass of: 1. Chloro-benzene (C6H5Cl)

8 2. Lithium hydroxide (LiOH) Answer 8: a) The molar mass is: 112,56g mol 1 For each element we get: Cl: 35,45112,56 100=31,49% C: 72,07112,56 100=64,02% H: 5,04112,56 100=4,48% b) The formula mass of LiOH is 24g mol 1 For each element we get: Li: =29,17% O: =66,67%

9 H: =4,17% Problem 9: CFC's (chlorofluorocarbons) are one of the gases that contribute to the depletion of the ozone layer. A chemist analysed a CFC and found that it contained 58,64% chlorine, 31,43% fluorine and 9,93% carbon. What is the empirical formula? Answer 9: We assume that we have 100 g of compound. So the mass of each element is: Cl: 58,64 g ; F: 31,43 g and C: 9,93 g. Next we find the number of moles of each element: Cl: n=mm=58,6435,45=1,65 F: n=mm=31,4319=1,65 C: n=mm=9,9312,011=0,77 Dividing by the smallest number (0,77) gives: Cl: 2 ; F: 2 ; C: 1

10 So the empirical formula is: CF2Cl2 Problem 10: 14 g of nitrogen combines with oxygen to form 46 g of a nitrogen oxide. Use this information to work out the formula of the oxide. Answer 10: We first calculate the mass of oxygen in the reactants: 46 14=32g. Next we calculate the number of moles of nitrogen and oxygen in the reactants: Nitrogen:n=mm=1414=1 Oxygen: n=mm=3216=2 So the formula of the oxide is: NO2. Problem 11: Iodine can exist as one of three oxides (I2O4; I2O5; I4O9). A chemist has produced one of these oxides and wishes to know which one they have. If he started with 508 g of iodine and formed 652 g of the oxide, which oxide has he produced?

11 Answer 11: We first calculate the mass of oxygen in the reactants: =144g Next we calculate the number of moles of iodine and oxygen in the reactants: I: n=mm=508127=4 O: n=mm=14416=9 So he has produced I4O9 Problem 12: A fluorinated hydrocarbon (a hydrocarbon is a chemical compound containing hydrogen and carbon) was analysed and found to contain 8,57% H, 51,05% C and 40,38% F. 1. What is its empirical formula? 2. What is the molecular formula if the molar mass is 94,1 g mol 1? Answer 12: a) We assume that there is 100 g of the compound. So in 100 g, we would have 8,57 g H, 51,05 g C and 40,38 g F. Next we find the number of moles of each element:

12 H: n=mm=8,571,008=8,5mols C: n=mm=51,0512,011=4,3mols F: n=mm=40,3819=2,1mols Now we divide this by the smallest number of moles, which is 2,1: H: 4 C: 2 F: 1 Therefore the empirical formula is: C2H4F b) The molar mass of the empirical formula is 47,05g mol 1. Therefore we must double the number of moles of each element to get the molecular formula. The molecular formula is: C4H8F2 Problem 13: Copper sulphate crystals often include water. A chemist is trying to determine the number of moles of water in the copper sulphate crystals. She weighs out 3 g of copper sulphate and heats this. After heating, she finds that the mass is 1,9 g. What is the number of moles of water in the crystals? (Copper sulphate is represented by CuSO4.xH2O). Answer 13:

13 We first work out how many water molecules are lost: 3g 1,9g=1,1g The mass ratio of copper sulphate to water is: 1,9g:1,1g Now we divide this by the molar mass of each species. CuSO4 has a molar mass of 159,55g mol 1 and water has a molar mass of 18g mol 1. So the mole ratio is: 1,9159,55:1,118 0,012:0,06 Dividing both of these numbers by 0,012 we find that the number of water molecules is 5. So the formula for copper sulphate is: CuSO4 5H2O Problem 14: 300 cm 3 of a 0,1 mol dm 3 solution of sulphuric acid is added to 200 cm 3 of a 0,5 mol dm 3 solution of sodium hydroxide.

14 1. Write down a balanced equation for the reaction which takes place when these two solutions are mixed. 2. Calculate the number of moles of sulphuric acid which were added to the sodium hydroxide solution. 3. Is the number of moles of sulphuric acid enough to fully neutralise the sodium hydroxide solution? Support your answer by showing all relevant calculations. Answer 14: a) H2SO4+2NaOH 2H2O+Na2SO4 b) We first convert the volume to dm3: =0,3dm3 n=cv n=(0,3)(0,1) n=0,03mol of sulphuric acid. c) The number of moles of sodium hydroxide used is: n=cv n=(0,5)(0,2)

15 n=0,1mols The molar ratio of sulphuric acid to sodium hydroxide is 1:2 So we need twice the number of moles of sulphuric acid (i.e. 0,1 2=0,2 ) to neutralise the sodium hydroxide. The number of moles of sulphuric acid is less than twice the number of moles of sodium hydroxide and so the the number of moles added is not enough to fully neutralise the sodium hydroxide. Problem 15: A learner is asked to make 200 cm 3 of sodium hydroxide (NaOH) solution of concentration 0,5 mol dm Determine the mass of sodium hydroxide pellets he needs to use to do this. 2. Using an accurate balance the learner accurately measures the correct mass of the NaOH pellets. To the pellets he now adds exactly 200 cm 3 of pure water. Will his solution have the correct concentration? Explain your answer. 3. The learner then takes 300 cm 3 of a 0,1 mol dm 3 solution of sulphuric acid (H2SO4) and adds it to 200 cm 3 of a 0,5 mol dm 3 solution of NaOH at Write down a balanced equation for the reaction which takes place when these two solutions are mixed. 5. Calculate the number of moles of H2SO4 which were added to the NaOH solution. Answer 15: a) We first convert the volume to dm =0,2dm3

16 The number of mols is: n=cv n=(0,2)(0,5) n=0,1mols The molar mass of NaOH is: 22,99+16,00+1,01=40g mol 1 And the mass of NaOH needed is: m=nm m=(0,1)(40) m=4g b)the concentration will be incorrect. The volume will change slightly when the pellets dissolve in the solution. The learner should have first dissolved the pellets and then made the volume up to the correct amount.

17 c) H2SO4+2NaOH 2H2O+Na2SO4 d) n=cv n=(0,5)(0,2) n=0,1mols Problem 16: 96,2 g sulphur reacts with an unknown quantity of zinc according to the following equation: Zn+S ZnS 1. What mass of zinc will you need for the reaction, if all the sulphur is to be used up? 2. Calculate the theoretical yield for this reaction. 3. It is found that 275 g of zinc sulphide was produced. Calculate the % yield. Answer 16: a) We first calculate the number of moles of sulphur: Molar mass sulphur: 32,06 g mol 1 n=mm n=96,232,06 n=3,00 mol

18 The molar ratio of sulphur to zinc is 1:1 so this is also the number of mols of zinc that must be added to use up all the sulphur. The mass of zinc is therefore: m=nm m=(3)(65,38) m=196,14 g b) We can use either the moles of zinc or the moles of sulphur to determine the moles of zinc sulphide. The molar ratio of sulphur to zinc sulphide is 1:1, so the number of mols of zinc sulphide is 3 mols. The mass is: m=nm m=(3)(65,38+32,06) m=(3)(97,44) m=292,32 g Problem 17: Calcium chloride reacts with carbonic acid to produce calcium carbonate and hydrochloric acid according to the following equation:

19 CaCl2+H2CO3 CaCO3+2HCl If you want to produce 10 g of calcium carbonate through this chemical reaction, what quantity (in g) of calcium chloride will you need at the start of the reaction? Answer 17: Molar mass of CaCO3: 40,08+12,01+3(16,00)=100,09 g mol 1 Number of moles of CaCO3: n=mm n=10100,09 n=0,10 mol The molar ratio of calcium carbonate to calcium chloride is 1:1, so the number of moles of calcium chloride is 0,10 mol. The mass of calcium chloride needed is therefore: m=nm m=(0,10)(40,08+2(35,45))

20 m=(0,10)(110,98) m=11,098 g

Chapter 5 Chemical Calculations

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