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

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 (H2) reacts with one molecule of iodine (I2), 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) =
0.1443
12.01 amu

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

12. Molecular Weight (Mass): Mass in grams of one mole of the substance:
Molar Mass of N = g/mol Molar Mass of N2 = g/mol Molar Mass of H2O = g/mol (2 × g) g Molar Mass of Ba(NO3)2 = g/mol g + (2 × g) + (6 × g)
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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 (H2) is amu
If the atomic weight of one iodine atom (I) is amu
then the molecular weight of one iodine molecule (I2) is amu
Finally, the molecular weight of each HI molecule must be:
1.008 amu amu = amu

14. Counting molecules: by mass

15. Here is where we left our reaction making two HI molecules from H2 and I2

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 (I2) is approximately 126 times heavier than the hydrogen molecule (H2)
H2 = amu
I2 = amu

17. Meaning that this reaction is balanced – in amu units
H2(g) + I2(g) → 2HI(g)
2.016 amu amu :
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18. So by the law of identical ratios, this reaction must also be balanced – in gram units
H2(g) + I2(g) → 2HI(g)
2.016 grams grams :
Now notice
we have gone
from amu to grams
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19. And again by The Law of Identical Ratios, this must also be balanced
H2(g) + I2(g) → 2HI(g)
1.00 g g
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20. Mass / Number relationship for I2 & H2
So in order control the number ratio of I2 to H2 molecules at 1:1 we can measure out any of the following
so long we keep the mass ratio at 1:126
H2(g) + I2(g) → 2HI(g)
2.016 amu amu
2.016 g g
1.00 g g
g g

21. But how many molecules is that?
We know that:
2.016 amu of H2 contains one molecule, and
amu of I2 contains one molecule
Our final question is, how many molecules are contained in:
2.016 g of H2
g of I2

22. That amount we call The Mole

23. The mole (NA) has now been measured
A mole is the amount of substance whose mass in grams is numerically equal to its molecular weight
2.016 g of H2 contains one mole of H2 molecules
g of I2 contains one mole of I2 molecules
The mole (NA) has now been measured
One mole of any molecule contains × 1023 molecules
One mole of any salt contains × 1023 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. How many silicon atoms are in 0.532 moles of silicon?

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 18.02 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
Worked Example 6.3
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?

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

34. What is the mass in grams of 3.2 × 1022 molecules of water?

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. Mole to mole
This reaction represents the reaction of A2 (red) with B2 (blue)
Chapter 6, Unnumbered Figure 2, Page 174
a) Write a balanced equation for the reaction
b) How many moles of product can be made from 1.0 mole of A2? From 1.0 mole of B2?

39. How many moles of NH3 can be produced from 5
How many moles of NH3 can be produced from moles of H2 according to the following equation? N2 + 3 H2  2 NH3
0.67 mol
2.00 mol
3.33 mol
7.50 mol

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
Mole to mole conversions (section 6.3) are carried out using mole ratios as conversion factors
If you have 9.0 moles of H2, how many moles of NH3 can you make?

44. 2) Mole-to-mass and mass-to-mole conversions (from section 6
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. Let’s go back to our hydrogen / iodine reaction giving hydrogen iodide

48. 10.0 grams of HI requires how many grams of H2
I2 = g/mol
H2 = g/mol
HI = g/mol

49. How many grams of oxygen are needed to react with 25
How many grams of oxygen are needed to react with 25.0 g of K according to the following equation? K(s) + O2(g)  2 K2O(s)
10.2 g O2
2.56 g O2
8.66 g O2
5.12 g O2

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
N2(g) + 3H2(g)  2NH3(g)
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56. A stoichiometric mixture of reactants
Contains the relative amounts (in moles) of reactants that matches the coefficients in the balanced equation
N2(g) + 3H2(g)  2NH3(g)
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57. A limiting reactant mixture
on the other hand, contains an excess of one of the reactants
The other reactant is thus limiting
N2(g) + 3H2(g)  2NH3(g)
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58. A limiting reactant mixture
on the other hand, contains an excess of one of the reactants
The other reactant is thus limiting
N2(g) + 3H2(g)  2NH3(g)
Which reactant is in
excess in this
reaction mixture?
Which reactant is
limiting?

59. Definition: Limiting Reactant
N2(g) + 3H2(g)  2NH3(g)
The limiting reactant (reagent) is the reactant that runs out first and thus constrains (limits) the amounts of product(s) that can form
H2 in the previous example is limiting
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60. But the limiting reactant must be identified before the correct amount of product can be calculated
2H2 + O2  H2O
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61. 10.0 g of Chemical A are reacted with 10.0 g of B
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
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62. 10.0 g of Chemical A are reacted with 10.0 g of B
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?

63. 10.0 g of Chemical A are reacted with 10.0 g of B
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

64. 10.0 g of Chemical A are reacted with 10.0 g of B
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
So B is limiting reactant

65. 10.0 g of Chemical A are reacted with 10.0 g of B
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

66. 10.0 g of Chemical A are reacted with 10.0 g of B
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

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

68. 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 K(s) + O2(g)  2 K2O(s)
K2O = 94.2g/mol
67g K2O; K is the limiting reactant.
270 g K2O; K is the limiting reactant.
270 g K2O; O2 is the limiting reactant.
67g K2O; O2 is the limiting reactant.

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. FYI
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

71. Theoretical Yield Actual 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. WORKED EXAMPLE 6.8 Percent Yield
The combustion of acetylene gas (C2H2) produces carbon dioxide and water as indicated in the following reaction 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2 is 88.0 g Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO2 Analysis—The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100

74. WORKED EXAMPLE 6.8 Percent Yield (finished)

75. WORKED EXAMPLE 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield
The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature. B2O3 (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

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

77. If 28. 56 g of K2O is produced when 25
If g of K2O is produced when g K is reacted according to the following equation, what is the percent yield of the reaction? K(s) + O2(g)  2 K2O(s)
87.54%
95%
94.82%
88%

78. Chapter Summary

79. What is the mole, and why is it useful in chemistry?
A mole refers to Avogadro’s number × 1023 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. 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. 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. amu to grams H2(g) + I2(g) → 2HI(g)
Recall: in Ch 2 we calculated the masses in amu needed to balance this reaction
H2(g) + I2(g) → 2HI(g)
2.016 amu amu
So the mass ratio of I2 to H2 needed is actually:
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