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Chapter Six Chemical Reactions: Mole and Mass Relationships Fundamentals of General, Organic, and Biological Chemistry 7th Edition Chapter 6 Lecture ©

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Presentation on theme: "Chapter Six Chemical Reactions: Mole and Mass Relationships Fundamentals of General, Organic, and Biological Chemistry 7th Edition Chapter 6 Lecture ©"— Presentation transcript:

1 Chapter Six Chemical Reactions: Mole and Mass Relationships Fundamentals of General, Organic, and Biological Chemistry 7th Edition Chapter 6 Lecture © 2013 Pearson Education, Inc. Julie Klare Gwinnett Technical College McMurry, Ballantine, Hoeger, Peterson

2 6.1 The Mole and Avogadro’s Number 6.2 Gram–Mole Conversions 6.3 Mole Relationships and Chemical Equations 6.4 Mass Relationships and Chemical Equations 6.5 Limiting Reagent and Percent Yield

3 Goals  1.What is the mole, and why is it useful in chemistry?  Be able to explain the meaning and uses of the mole and Avogadro’s number.  2.How are molar quantities and mass quantities related?  Be able to convert between molar and mass quantities of an element or compound.  3.What are the limiting reagent, theoretical yield, and percent yield of a reaction?  Be able to take the amount of product actually formed in a reaction, calculate the amount that could form theoretically, and express the results as a percent yield.

4 6.1 The Mole and Avogadro’s Number

5 How to balance reactions by counting molecules

6 Counting molecules: Conceptually The reactants in a chemical reaction must be balanced One molecule of hydrogen (H 2 ) reacts with one molecule of iodine (I 2 ), creating two hydrogen iodide molecules (HI)

7 Of course, we can’t visually count molecules to achieve this one-to-one ratio In fact, the only tool available is an analytical mass balance (using grams!)

8 Molecular Weight But how do we relate a sample’s mass in grams to the number of molecules it contains? We begin with the concept of ‘molecular weight’ (molecular mass) Let’s start by utilizing the Ch 2 concept of mass by ‘amu’ – Note: We will be focusing on molecules for a while

9 Recall, atomic weight is the average mass of an element’s isotope atoms (review next slide) By analogy, molecular weight (MW) is the mass of a compound’s molecules –A molecule’s molecular weight is the sum of the atomic weights for all the atoms in the molecule –A salt’s formula weight is the sum of the atomic weights for all the atoms in the formula unit

10 fyi: Atomic Weight Calculation C-12 is 98.89% of all natural carbon C-13 is 1.11% of all natural carbon The mass of C-12 is: 12 amu (by definition) The mass of C-13 is: amu Atomic weight =  [(% isotope abundance) × (isotope mass)] [98.89% x 12 amu] + [1.11% x amu] = (0.9889)(12 amu) + (0.0111)( amu) = amu

11 fyi: Disambiguation For molecules, these terms are identical – molar weight – molar mass – molecular weight – molecular mass For atoms, these terms are identical – molar weight – molar mass – atomic weight – atomic mass

12 Molar Mass of N = g/mol Molar Mass of N 2 = g/mol Molar Mass of H 2 O = g/mol (2 × g) g Molar Mass of Ba(NO 3 ) 2 = g/mol g + (2 × g) + (6 × g) Copyright © Cengage Learning. All rights reserved 12 Molecular Weight (Mass): Mass in grams of one mole of the substance:

13 Molecular weight So if the mass (atomic weight) of one atom of hydrogen (H) is amu – then the mass (molecular weight) of one hydrogen molecule (H 2 ) is amu If the atomic weight of one iodine atom (I) is amu – then the molecular weight of one iodine molecule (I 2 ) is amu Finally, the molecular weight of each HI molecule must be: – amu amu = amu

14 Counting molecules: by mass

15 Here is where we left our reaction making two HI molecules from H 2 and I 2

16 We learned in chapter 2 that the iodine atom (I) is approximately 126 times heavier than the hydrogen atom (H) – H = amu – I = amu So mathematically, it is necessary that the iodine molecule (I 2 ) is approximately 126 times heavier than the hydrogen molecule (H 2 ) – H 2 = amu – I 2 = amu

17 Copyright © Cengage Learning. All rights reserved 17 Meaning that this reaction is balanced – in amu units H 2(g) + I 2(g) → 2HI (g) amu amu 1 : 126

18 Copyright © Cengage Learning. All rights reserved 18 So by the law of identical ratios, this reaction must also be balanced – in gram units H 2(g) + I 2(g) → 2HI (g) grams grams 1 : 126 Now notice we have gone from amu to grams

19 Copyright © Cengage Learning. All rights reserved 19 And again by The Law of Identical Ratios, this must also be balanced H 2(g) + I 2(g) → 2HI (g) 1.00 g 126 g

20 Mass / Number relationship for I 2 & H 2 So in order control the number ratio of I 2 to H 2 molecules at 1:1 we can measure out any of the following – so long we keep the mass ratio at 1:126 H 2(g) + I 2(g) → 2HI (g) amu amu g g 1.00 g 126 g g g

21 But how many molecules is that? We know that: – amu of H 2 contains one molecule, and – amu of I 2 contains one molecule Our final question is, how many molecules are contained in: – g of H 2 – g of I 2

22 That amount we call The Mole

23 A mole is the amount of substance whose mass in grams is numerically equal to its molecular weight – g of H 2 contains one mole of H 2 molecules – g of I 2 contains one mole of I 2 molecules The mole (N A ) has now been measured – One mole of any molecule contains × molecules – One mole of any salt contains × formula units

24 Getting from amu to grams The number ‘ ,’ typically placed as shown, functions BOTH: – as the mass in amu per atom – and the mass in grams per mole of atoms So I has a mass of grams per mole of I atoms –I = g/mol

25 The technical definition of a mole

26 © 2013 Pearson Education, Inc. How many silicon atoms are in moles of silicon? a.8.83 × 10 −25 atoms b.3.81 × atoms c.2.6 × atoms d.3.8 × atoms

27 6.2 Gram–Mole Conversions

28 fyi: Synonyms The terms Molecular Weight (section 6.1) and Molar Mass (this section – 6.2) are identical – Also, the term Molecular Mass is allowed – Also, the term Molar Weight is allowed When we defined Molecular Weight in Section 6.1, we had not yet define the Mole – so you couldn’t have understood the more common term “Molar” Mass

29 The molar mass of water is g/mol So how many moles of water are there in 27 grams? Most importantly, molar mass serves as a conversion factor between numbers of moles and mass in grams for use in dimensional analysis

30 Molar Mass: Mole to Gram Conversion Ibuprofen is a pain reliever used in Advil. Its molecular weight is g/mol. If a bottle of Advil contain mole of ibuprofen, how many grams of ibuprofen does it contain? Worked Example 6.3

31 WORKED EXAMPLE 6.3 Molar Mass: Mole to Gram Conversion (Continued)

32 We now have the tools to convert: grams → moles → number of atoms grams → moles → number of molecules grams → moles → number of formula units And we can go backwards

33 33

34 © 2013 Pearson Education, Inc. What is the mass in grams of 3.2 × molecules of water? a g b.339 g c.0.90 g d.0.96 g

35 6.3 Mole Relationships and Chemical Equations Molar coefficients: using Mole ratios as conversion factors

36 Coefficients in a balanced chemical equation tell us the necessary ratio of moles of reactants –and how many moles of each product are formed

37 Especially useful, coefficients can be put in the form of mole ratios –which act as conversion factors when setting up dimensional analysis calculations

38 This reaction represents the reaction of A 2 (red) with B 2 (blue) a) Write a balanced equation for the reaction b) How many moles of product can be made from 1.0 mole of A 2 ? From 1.0 mole of B 2 ? Mole to mole

39 © 2013 Pearson Education, Inc. a.0.67 mol b.2.00 mol c.3.33 mol d.7.50 mol How many moles of NH 3 can be produced from 5.00 moles of H 2 according to the following equation? N H 2  2 NH 3

40 6.4 Mass Relationships and Chemical Equations Putting it all together

41 The actual amounts of substances used in the laboratory must be weighed out in grams Furthermore, customers generally want the quantity of product reported in terms of grams

42 We already have the necessary tools Mass to mole conversions Mole to mole conversions Mole to mass conversions

43 Mole to mole conversions (section 6.3) are carried out using mole ratios as conversion factors If you have 9.0 moles of H 2, how many moles of NH 3 can you make?

44 2) Mole-to-mass and mass-to-mole conversions (from section 6.2) are carried out using molar mass as a conversion factor

45 3) Mass to mass conversions cannot be carried out directly If you know the mass of A and need to find the mass of B first convert the mass of A into moles of A then carry out a mole to mole conversion to find moles of B then convert moles of B into the mass of B

46 So after collecting the fundamental data, always follow this three step process for mass to mass 1) convert grams of A to moles via molar mass 2) convert moles of A to B via mole ratio 3) convert mole of B to grams via molar mass

47 47 Let’s go back to our hydrogen / iodine reaction giving hydrogen iodide

48 48 I 2 = g/mol H 2 = g/mol HI = g/mol 10.0 grams of HI requires how many grams of H 2

49 © 2013 Pearson Education, Inc. How many grams of oxygen are needed to react with 25.0 g of K according to the following equation? 4 K(s) + O 2 (g)  2 K 2 O(s) a.10.2 g O 2 b.2.56 g O 2 c.8.66 g O 2 d.5.12 g O 2

50 6.5 Limiting Reagent and Percent Yield Vocabulary: Reactant = Reagent

51 limiting reactant (reagent) Reactants are not always perfectly balanced When running a chemical reaction, we generally ‘overcharge’ one of the reactants As our mechanic has ‘overcharged’ tires

52 limiting reactant (reagent) The limiting reactant is the reactant that runs out first The reactant that never runs out is called the excess reactant So which is ‘limiting here?’ Tires or car bodies.

53 limiting reactant (reagent) In order to make this ‘reaction’ balanced stoichiometrically, how many tires should our mechanic actually have on hand? 32 tires would perfectly ‘balance’ this ‘reaction’

54 Back to chemistry

55 A stoichiometric mixture of reactants contains the relative amounts (in moles) of reactants that matches the coefficients in the balanced equation N 2 (g) + 3H 2 (g)  2NH 3 (g) Copyright © Cengage Learning. All rights reserved 55

56 A stoichiometric mixture of reactants Contains the relative amounts (in moles) of reactants that matches the coefficients in the balanced equation N 2 (g) + 3H 2 (g)  2NH 3 (g) Copyright © Cengage Learning. All rights reserved 56

57 A limiting reactant mixture Copyright © Cengage Learning. All rights reserved 57 on the other hand, contains an excess of one of the reactants The other reactant is thus limiting N 2 (g) + 3H 2 (g)  2NH 3 (g)

58 A limiting reactant mixture 58 Which reactant is in excess in this reaction mixture? Which reactant is limiting? on the other hand, contains an excess of one of the reactants The other reactant is thus limiting N 2 (g) + 3H 2 (g)  2NH 3 (g)

59 Definition: Limiting Reactant N 2 (g) + 3H 2 (g)  2NH 3 (g) The limiting reactant (reagent) is the reactant that runs out first and thus constrains (limits) the amounts of product(s) that can form  H 2 in the previous example is limiting Copyright © Cengage Learning. All rights reserved 59

60 But the limiting reactant must be identified before the correct amount of product can be calculated Copyright © Cengage Learning. All rights reserved 60 2H 2 + O 2  H 2 O

61 Example 10.0 g of Chemical A are reacted with 10.0 g of B What information do you need to calculate the mass of the product (C) that will be produced? The mole ratios between A, B, and C – ie, the balanced reaction equation The molar masses of A, B, and C Copyright © Cengage Learning. All rights reserved 61

62 Example 10.0 g of Chemical A are reacted with 10.0 g of B What information do you need to calculate the mass of the product (C) that will be produced? A + 3B  2C molar masses: – A is 10.0 g/mol – B is 20.0 g/mol – C is 25.0 g/mol? 62

63 Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C 1) Convert known masses of reactants to moles 63

64 Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C 2) Convert to the number of moles of product 64 So B is limiting reactant

65 Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C Choose the least number of moles of product formed as limiting reactant: B is limiting 65

66 Example 10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C 3) Convert moles of C to grams of C using the molar mass 66

67 shortcut to limiting reactant How do we determine quickly which reactant is limiting – and by deduction, which is in excess? Three step program 1Calculate the number of moles of each reactant 2divide the number of moles of each reactant by its coefficient from the balanced equation 3the reactant with the smaller result is limiting

68 © 2013 Pearson Education, Inc. If 56 g of K is reacted with 56 g of oxygen gas according to the equation below, indicate the mass of product that can be made and identify the limiting reactant. 4 K(s) + O 2 (g)  2 K 2 O(s) a.67g K 2 O; K is the limiting reactant. b.270 g K 2 O; K is the limiting reactant. c.270 g K 2 O; O 2 is the limiting reactant. d.67g K 2 O; O 2 is the limiting reactant. K 2 O = 94.2g/mol

69 Theoretical vs actual yield Theoretical Yield – The maximum amount of a given product that can be formed once a limiting reactant has been completely consumed – it is a calculation Actual yield – The amount of product that is actually produced in a reaction – it is a measurement – It is usually less than the maximum expected (theoretical yield)

70 Theoretical yield is found by using the amount of limiting reactant calculated in a mass-to- mass calculation For the chemist in the lab, the actual yield is found by weighing the amount of product obtained FYI

71 Theoretical Yield –For our auto mechanic, the theoretical yield is 8 finished cars Actual yield –If one car body was damaged beyond repair, the actual yield would have been 7 finished cars

72 Percent yield is the percent of the theoretical yield actually obtained from a chemical reaction For our car mechanic:

73 The combustion of acetylene gas (C 2 H 2 ) produces carbon dioxide and water as indicated in the following reaction 2 C 2 H 2 (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O (g) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO 2 is 88.0 g Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO 2 Analysis—The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100 WORKED EXAMPLE 6.8 Percent Yield

74 WORKED EXAMPLE 6.8 Percent Yield (finished)

75 The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature. B 2 O 3 (l) + 3 Mg (s) → 2 B (s) + 3 MgO (s) What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield

76 WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield (Continued)

77 © 2013 Pearson Education, Inc. If g of K 2 O is produced when g K is reacted according to the following equation, what is the percent yield of the reaction? 4 K(s) + O 2 (g)  2 K 2 O(s) a.87.54% b.95% c.94.82% d.88%

78 Chapter Summary

79 1.What is the mole, and why is it useful in chemistry? A mole refers to Avogadro’s number × formula units of a substance. One mole of any substance has a mass (a molar mass) equal to the molecular or formula weight of the substance in grams. Because equal numbers of moles contain equal numbers of formula units, molar masses act as conversion factors between numbers of molecules and masses in grams.

80 2.How are molar quantities and mass quantities related? The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction. The ratios of coefficients act as mole ratios that relate amounts of reactants and/or products. By using molar masses and mole ratios in factor-label calculations, unknown masses or molar amounts can be found from known masses or molar amounts.

81 3.What are the limiting reagent, theoretical yield, and percent yield of a reaction? The limiting reagent is the reactant that runs out first. The theoretical yield is the amount of product that would be formed based on the amount of the limiting reagent. The actual yield of a reaction is the amount of product obtained. The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%.

82 The official name for what we are doing is Stoichiometry

83 Copyright © Cengage Learning. All rights reserved 83 amu to grams Recall: in Ch 2 we calculated the masses in amu needed to balance this reaction H 2(g) + I 2(g) → 2HI (g) amu amu – So the mass ratio of I 2 to H 2 needed is actually:


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