Chapter 12. Reacting masses

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1 Chapter 12 Reacting masses 12.1 The mole, Avogadro constant and molar mass 12.2 Percentage by mass of an element in a compound 12.3 Chemical formulae of compounds 12.4 Empirical formulae and molecular formulae derived from experimental data 12.5 Reacting masses from chemical equations P. 1 / 52

2 Key terms Progress check Summary Concept map P. 2 / 52

3 12.1 The mole, Avogadro constant and molar mass Mole and Avogadro constant In daily life, some special units are used to describe the quantity of items. Figure 12.1 Socks are in pairs, eggs are often packed in dozens, and papers are often packed in reams. P. 3 / 52

4 Learning tip 1 ream of paper refers to 500 identical sheets of paper. Unlike socks, eggs and papers, particles (i.e. atoms, ions or molecules) are too small to be seen. A special unit called mole (abbreviation: mol) is used to describe the quantity of particles in a substance The mole, Avogadro constant and molar mass P. 4 / 52

5 The number of atoms in exactly 12.0 g of carbon-12 the reference unit for the mole The number of atoms in exactly 12.0 g of carbon-12 is Avogadro constant (symbol: L; unit: mol 1 ) One mole of any substance contains formula units The mole, Avogadro constant and molar mass P. 5 / 52

6 Learning tip A pure substance has a formula. The simplest unit of a substance is its formula unit. For example, Substance Water Copper Carbon Sodium chloride Formula unit H 2 O Cu C NaCl 12.1 The mole, Avogadro constant and molar mass P. 6 / 52

1 mole of water contains 6.02 10 23 H 2 O molecules 1 mole of copper metal contains 6.

02 10 23 NaCl formula units (or contains 1 mole of Na + ions and 1 mole of Cl ions) Figure 12.

7 1 mole of water contains H 2 O molecules 1 mole of copper metal contains Cu atoms 1 mole of carbon contains C atoms 1 mole of sodium chloride contains NaCl formula units (or contains 1 mole of Na + ions and 1 mole of Cl ions) Figure 12.2 One mole of each of the four substances. They all contain formula units The mole, Avogadro constant and molar mass P. 7 / 52

8 Key point One mole is the amount of a substance that contains the same number of formula units as the number of atoms in exactly 12.0 g of carbon-12. Class practice The mole, Avogadro constant and molar mass P. 8 / 52

9 Mole and molar mass One mole of a substance has a mass equal to its formula mass expressed in gram unit. The mass of one mole of a substance is called its molar mass. Unit of molar mass: gram per mol (g mol 1 ) 12.1 The mole, Avogadro constant and molar mass P. 9 / 52

10 Substances consisting of atoms One mole of carbon and one mole of copper both contain atoms. One mole of carbon weighs 12.0 g while one mole of copper weighs 63.5 g. Relative atomic mass of carbon = 12.0 Relative atomic mass of copper = 63.5 Molar mass of carbon = 12.0 g mol 1 Molar mass of copper = 63.5 g mol The mole, Avogadro constant and molar mass P. 10 / 52

11 Substances consisting of molecules One mole of water weighs 18.0 g. Relative molecular mass of water = = 18.0 Molar mass of water = 18.0 g mol The mole, Avogadro constant and molar mass Think about P. 11 / 52

12 Substances consisting of ions One mole of sodium chloride weighs 58.5 g. Formula mass of sodium chloride = = 58.5 Molar mass of sodium chloride = 58.5 g mol 1 Key point The mass of one mole of a substance is called its molar mass. (Unit of molar mass: g mol 1 ) 12.1 The mole, Avogadro constant and molar mass Class practice 12.2 P. 12 / 52

13 Relationship between number of formula units, number of moles and mass Avogadro constant molar mass number of formula units number of moles mass Avogadro constant molar mass 12.1 The mole, Avogadro constant and molar mass P. 13 / 52

Key point Example 12.1 Class practice 12.3 Example 12.

14 Key point Example 12.1 Class practice 12.3 Example The mole, Avogadro constant and molar mass P. 14 / 52

15 12.2 Percentage by mass of an element in a compound Calculating percentage by mass of an element in the compound From the formula of a compound percentage by mass of each element in the compound. Key point Percentage by mass of element A in a compound Example 12.3 P. 15 / 52

16 Calculating the mass of an element in the compound The mass of an element in a compound can be calculated from: formula of the compound percentage by mass of that element in the compound Example Percentage by mass of an element in a compound P. 16 / 52

17 Calculating the relative atomic mass of an element The relative atomic mass of an element may be calculated from: formula of the compound percentage by mass of that element in the compound Example 12.5 Class practice Percentage by mass of an element in a compound P. 17 / 52

18 12.3 Chemical formulae of compounds Empirical formula Empirical formula: formula which shows the simplest whole number ratio of the atoms or ions present. Applicable to all compounds. P. 18 / 52

19 Molecular formula Molecular formula: formula which shows the actual number of each kind of atoms in one molecule of the substance. Only applicable to molecular compounds and elements consisting of molecules. Structural formula Structural formula: formula which shows how the constituent atoms are joined up in one molecule of the substance Chemical formulae of compounds P. 19 / 52

20 Substance Empirical formula Molecular formula Structural formula Nitrogen N 2 N N Carbon dioxide CO 2 CO 2 O=C=O Ethene CH 2 C 2 H 4 Propene CH 2 C 3 H 6 Ethanol C 2 H 6 O C 2 H 6 O Methoxymethane C 2 H 6 O C 2 H 6 O Quartz SiO 2 Table 12.1 The different formulae of some substances Chemical formulae of compounds P. 20 / 52

21 Note: Empirical and molecular formulae of a compound may be the same or different. Molecular formula is the empirical formula multiplied by some whole number. Different compounds may have the same empirical formula and the same molecular formula but they have different structural formulae Chemical formulae of compounds Class practice 12.5 P. 21 / 52

22 12.4 Empirical formulae and molecular formulae derived from experimental data Determination of empirical formulae The empirical formula of a compound can be calculated from its composition by mass. P. 22 / 52

23 Determining the empirical formula of an oxide of copper Pass town gas into a combustion tube. Heat a known mass of oxide of copper (black) in the combustion tube. Hydrogen and carbon monoxide in the town gas reduce the oxide to reddish brown copper. Find the mass of copper Empirical formulae and molecular formulae derived from experimental data P. 23 / 52

24 town gas supply oxide of copper hole excess town gas burns here heat combustion tube Figure 12.3 To determine the empirical formula of an oxide of copper by passing town gas over the heated oxide. Learning tip Hydrogen and carbon monoxide reduce the oxide by removing oxygen from it Empirical formulae and molecular formulae derived from experimental data P. 24 / 52

SBA note At the beginning of the experiment, town gas is passed into the combustion tube. This is to expel the air inside the tube. The hot copper formed may react with the oxygen in air again.

25 SBA note At the beginning of the experiment, town gas is passed into the combustion tube. This is to expel the air inside the tube. The hot copper formed may react with the oxygen in air again. Therefore, it is necessary to pass the town gas through the combustion tube, even after heating has stopped Empirical formulae and molecular formulae derived from experimental data P. 25 / 52

26 Specimen results Item P. 26 / 52 Mass (g) Combustion tube Combustion tube + oxide of copper Combustion tube + copper Mass of copper in oxide = Mass of oxygen in oxide = Table 12.2 The specimen results of the experiment Empirical formula of the oxide of copper can be worked out as shown in Problem-solving strategy Empirical formulae and molecular formulae derived from experimental data Problem-solving strategy 12.1

27 Determining the empirical formula of an oxide of magnesium Heat a known mass of magnesium strongly in a crucible (also of known mass) until it catches fire. Lift the crucible lid slightly from time to time for letting in air to react with magnesium. From experimental results empirical formula of the oxide of magnesium: MgO 12.4 Empirical formulae and molecular formulae derived from experimental data P. 27 / 52

Empirical formula of a compound can also be determined if the percentage by mass of each element in the compound is

28 crucible pipe-clay triangle magnesium ribbon tripod heat very strongly rocksil Figure 12.4 To find the empirical formula of an oxide of magnesium by heating magnesium in air. Empirical formula of a compound can also be determined if the percentage by mass of each element in the compound is known. Experiment 12.1 Experiment 12.1 Example 12.6 Experiment 12.2 Experiment 12.2 Class practice Empirical formulae and molecular formulae derived from experimental data P. 28 / 52

29 Determination of molecular formulae A compound with known empirical formula and relative molecular mass determine the molecular formula of the compound Example 12.7 Example Class practice 12.7 Example 12.8 Experiment 12.3 Example 12.9 Experiment Empirical formulae and molecular formulae derived from experimental data P. 29 / 52

30 12.5 Reacting masses from chemical equations Chemical equations and reacting masses Consider the equation representing the reaction between magnesium and oxygen: 2Mg(s) + O 2 (g) 2MgO(s) P. 30 / 52

31 2Mg(s) + O 2 (g) 2MgO(s) 2 magnesium atoms 1 oxygen molecule 2 formula units of magnesium oxide magnesium atoms react with 2 moles of magnesium atoms oxygen molecules 1 moles of oxygen molecule to form formula units of magnesium oxide 2 moles of formula units of magnesium oxide 2 mol 24.3 g mol 1 = 48.6 g of magnesium atoms 1 mol 32.0 g mol 1 = 32.0 g of oxygen molecules 2 mol 40.3 g mol 1 = 80.6 g of formula units of magnesium oxide 12.5 Reacting masses from chemical equations P. 31 / 52

32 Learning tip The molar masses of Mg, O 2 and MgO are 24.3 g mol 1, ( = 32.0) g mol 1, and ( = 40.3) g mol 1 respectively. A balanced equation shows the quantitative relationship of the reactant(s) and the product(s) in a reaction. Stoichiometric coefficients in the equation indicate the relative number of moles (i.e. mole ratio) of reactants and products involved in the reaction Reacting masses from chemical equations P. 32 / 52

33 The total mass of reactants is equal to the total mass of products (i.e. conservation of mass). Learning tip The quantitative study of reactants and products in a reaction is called stoichiometry Reacting masses from chemical equations P. 33 / 52

34 Calculations from chemical equations reacting masses The mass of one of the substances in the reaction is known. Masses of other substances reacted or formed can be calculated based on the balanced chemical equation. Problem-solving strategy Reacting masses from chemical equations P. 34 / 52

35 Steps for determining the reacting masses from a chemical equation: divided by Known mass of A molar mass of A Number of moles of A by mole ratio (shown in the equation) Mass of the substance asked in the question multiplied by molar mass of that substance Number of moles of the substance asked in the question (Note: A represents the chemical formula of a particular substance.) 12.5 Reacting masses from chemical equations Example Example P. 35 / 52

36 Limiting reactant Consider the reaction between hydrogen and oxygen to form water: 2H 2 (g) + O 2 (g) 2H 2 O(l) O 2 molecules H 2 molecule H 2 O molecule 2 H 2 molecules + 2 O 2 molecules 2 H 2 O molecules + 1 O 2 molecule Figure 12.5 Two hydrogen molecules require one oxygen molecule for complete reaction. Therefore, oxygen is in excess Reacting masses from chemical equations P. 36 / 52

37 Only 1 molecule of oxygen is required to react with 2 molecules of hydrogen for complete reaction. In this case, oxygen is in excess. Amount of water produced is limited by the amount of hydrogen used. hydrogen is called the limiting reactant. Limiting reactant limits the amount of the product formed in a reaction. Example Reacting masses from chemical equations P. 37 / 52

38 Theoretical yield, actual yield and percentage yield Theoretical yield: amount of product expected if the reaction proceeds exactly as shown in the chemical equation. Actual yield: amount of product actually obtained from a reaction Reacting masses from chemical equations P. 38 / 52

39 The actual yield of a reaction is often less than the theoretical yield because: the reaction is incomplete. impurities are present in the reactants. side reactions occur in which unwanted side products are produced. some product is lost during different experimental processes, such as purification Reacting masses from chemical equations P. 39 / 52

40 The efficiency of a chemical reaction can be expressed by the percentage yield. Key point Example Reacting masses from chemical equations Class practice 12.8 P. 40 / 52

41 Key terms 1. actual yield 實際產量 2. Avogadro constant 亞佛加德羅常數 3. composition by mass 質量組成 4. empirical formula 實驗式 5. limiting reactant 限量反應物 6. molar mass 摩爾質量 7. mole 摩爾 8. molecular formula 分子式 9. percentage by mass 質量百分比 P. 41 / 52

42 10. percentage yield 百分產率 11. structural formula 結構式 12. theoretical yield 理論產量 Key terms P. 42 / 52

43 Progress check 1. What is the meaning of mole? 2. What is the Avogadro constant? 3. What is the meaning of molar mass? 4. How is the mole of a substance related to its mass and number of formula units? 5. How can we calculate the percentage by mass of an element in a compound? 6. What are empirical formula, molecular formula and structural formula? 7. How can we determine the empirical formula of a compound? P. 43 / 52

44 8. How can we determine the molecular formula of a compound? 9. What are the interrelationship between masses of reactants and products in a reaction? 10. How can we calculate masses of reactants and products in a reaction from the relevant equation? 11. What is the meaning of a limiting reactant? 12. What are the meanings of actual yield and theoretical yield? 13. How can we calculate the percentage yield of a chemical reaction? Progress check P. 44 / 52

45 Summary 12.1 The mole, Avogadro constant and molar mass 1. Chemists use mole (abbreviation: mol) to describe the quantity of particles in a substance. 2. The Avogadro constant (L) is the number of atoms in exactly 12.0 g of carbon-12. It is equal to mol The molar mass of a substance is its formula mass expressed in gram unit. The unit of molar mass is g mol 1. P. 45 / 52

46 4. Important relationships involving moles: 12.2 Percentage by mass of an element in a compound 5. The percentage by mass of an element in a compound can be found by the equation: Percentage by mass of element A in a compound Summary P. 46 / 52

47 12.3 Chemical formulae of compounds 6. Chemical formulae are part of the language of chemistry. Some common chemical formulae include empirical formula, molecular formula and structural formula Empirical formulae and molecular formulae derived from experimental data 7. Empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms or ions present. Summary P. 47 / 52

48 8. The empirical formula of a compound can be calculated from its composition by mass. The composition of a compound has to be determined by experiment. 9. Molecular formula may be determined from empirical formula and relative molecular mass. This is because molecular formula is a whole number multiple of empirical formula Reacting masses from chemical equations 10. The theoretical amounts of substances used up or produced in a reaction can be calculated from its balanced equation. Summary P. 48 / 52

49 11. Limiting reactant is the reactant that is completely used up in a reaction. It limits the amount of product(s) formed in the reaction. 12. The theoretical amounts of product predicted by calculation from its balanced equation is called theoretical yield. The actual yield of a reaction is often less than the theoretical yield. 13. Percentage yield is the ratio of actual yield and theoretical yield. It is a measure of the efficiency of a chemical reaction. Summary P. 49 / 52

50 Concept map Mass Molar mass equals Number of moles equals Number of formula units Avogadro constant equals formula units P. 50 / 52

51 Molar mass without unit, equals Formula mass equals Sum of relative atomic masses of all atoms/ions in a formula unit of a substance Relative molecular mass equals Sum of relative atomic masses of all atoms in a molecule Concept map P. 51 / 52

52 Relative molecular mass determines Empirical formula Molecular formula Chemical formulae Structural formula Concept map P. 52 / 52

Chapter 3. Mass Relationships in Chemical Reactions

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