1. Calculate the moles of carbon in 0.0265 g of pencil lead
Given:
Find:
g C
mol C
Conceptual Plan:
Relationships:
1 mol C = g
g C
mol C
Solution:
Check:
because the given amount is much less than 1 mol C, the number makes sense

2. How many copper atoms are in a penny weighing 3.10 g?
Given:
Find:
3.10 g Cu
atoms Cu
Conceptual Plan:
Relationships:
g Cu
mol Cu
atoms Cu
1 mol Cu = g, 1 mol = x 1023
Solution:
Check:
because the given amount is much less than 1 mol Cu, the number makes sense

3. Which has more atoms, 10.0 g Mg or 10.0 g Ca?
Magnesium
Calcium
Both have the same number of atoms.

4. Which of the following has the largest mass?
10.0 g Li
10.0 moles of Li
100 g Na
10.0 moles of K
100 g Rb

5. Which of the following has the largest mass?
10.0 g Li
10.0 moles of Li
100 g Na
10.0 moles of K
100 g Rb
Answer: d

6. If a pure copper penny has 2.94 × 1022 atoms, what will be its mass?
1.30 g
0.323 g
2.79 g
3.10 g
1.12 g

7. If a pure copper penny has 2.94 × 1022 atoms, what will be its mass?
1.30 g
0.323 g
2.79 g
3.10 g
1.12 g
Answer: d

8. Your Turn! Formula is Mg3(PO4)2
What is the formula mass of magnesium phosphate?
262.9 g/mol
150.9 g/mol
333.6 g/mol
119.3 g/mol
166.9 g/mol
Formula is Mg3(PO4)2

9. How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)
Given:
Find:
50.0 g mol PbO2
moles PbO2
Conceptual Plan:
Relationships:
1 mol PbO2 = g
g PbO2
mol PbO2
Solution:
Check:
because the given amount is less than g, the moles being < 1 makes sense

10. Find the number of CO2 molecules in 10.8 g of dry ice
Given:
Find:
10.8 g CO2
molecules CO2
g CO2
mol CO2
molec CO2
Conceptual Plan:
Relationships:
1 mol CO2 = g,
1 mol = x 1023
Solution:
Check:
because the given amount is much less than 1 mol CO2, the number makes sense

11. Which sample represents the greatest number of moles?
44.01 g CO2
1.0 moles C3H8
6.022 x 1023 molecules C4H10
18.02 g H2O
All of the samples have the same number of moles.

12. Which sample represents the greatest number of moles?
44.01 g CO2
1.0 moles C3H8
6.022 x 1023 molecules C4H10
18.02 g H2O
All of the samples have the same number of moles.
Answer: e

13. Your Turn!
Calculate the number of moles of calcium in moles of Ca3(PO4)2
2.53 mol Ca
0.432 mol Ca
3.00 mol Ca
D mol Ca
E mol Ca
2.53 moles of Ca3(PO4)2 = ? mol Ca
3 mol Ca  1 mol Ca3(PO4)2
= 7.59 mol Ca

14. Your Turn!
A sample of sodium carbonate, Na2CO3, is found to contain 10.8 moles of sodium. How many moles of oxygen atoms (O) are present in the sample?
10.8 mol O
7.20 mol O
5.40 mol O
D mol O
E mol O
10.8 moles of Na = ? mol O
2 mol Na  3 mol O
= 16.2 mol O

15. Your Turn!
How many g of iron are required to use up all of g of oxygen atoms (O) to form Fe2O3?
A g
B g
C g
D. 134 g
E g
= 59.6 g Fe
mass O  mol O  mol Fe  mass Fe
25.6 g O  ? g Fe
3 mol O  2 mol Fe

16. Determine the mass % composition of CaCl2 (Ca = 40.08, Cl = 35.45)

17. Benzaldehyde is 79. 2% carbon. What mass of benzaldehyde contains 19
Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?
Given:
Find:
19.8 g C, 79.2% C
g benzaldehyde
Conceptual Plan:
Relationships:
100 g benzaldehyde : 79.2 g C
g C
g benzaldehyde
Solution:
Check:
because the mass of benzaldehyde is more than the mass of C, the number makes sense

18. How many grams of sodium are in 6. 2 g of NaCl. (Na = 22. 99; Cl = 35
Given:
Find:
6.2 g NaCl
g Na
Conceptual Plan:
Relationships:
1 mol NaCl = g, 1 mol Na = g,
1 mol Na : 1 mol NaCl
g NaCl
mol NaCl
mol Na
g Na
Solution:
Check:
because the amount of Na is less than the amount of NaCl, the answer makes sense

19. Which of the following contains the largest mass percent hydrogen?
10.0 g CH4
10.0 g C2H6
10.0 g H2O
10.0 g H2S

20. Which of the following contains the largest mass percent hydrogen?
10.0 g CH4
10.0 g C2H6
10.0 g H2O
10.0 g H2S
Answer: a

21. Calculate the number of carbon atoms in 25
Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (C3H8O).
1.25 C atoms
15.0 C atoms
2.51 x 1023 C atoms
7.52 x 1023 C atoms
2.07 x 10–24 C atoms

22. Calculate the number of carbon atoms in 25
Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (C3H8O).
1.25 C atoms
15.0 C atoms
2.51 x 1023 C atoms
7.52 x 1023 C atoms
2.07 x 10–24 C atoms
Answer: d

23. Percent Composition
One can determine percentage composition based on the chemical analysis of a substance
Example: A sample of a liquid with a mass of g was decomposed into its elements and gave g of carbon, g of hydrogen, and g of oxygen. What is the percentage composition of this compound?
Analysis:
Calculate percentage by mass of each element in sample
Tools:
Equation for percentage by mass
Total mass = g
Mass of each element

24. % Composition of Compound
For C:
For H:
For O:
Percentage composition tells us mass of each element in g of substance
In g of our liquid
60.26 g C, g H, and g O
= 60.26% C
= 11.11% H
= 28.62% O
Sum of percentages: 99.99%

25. Your Turn!
A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound?
1. Calculate total mass of sample
Total sample mass = g g = g
2. Calculate % Composition of N
3. Calculate % Composition of O
= 25.94% N
= 74.06% O

26. Using Percent Composition
Are the mass percentages 30.54% N and 69.46% O consistent with the formula N2O4?
Procedure:
Assume one mole of compound
Subscripts tell how many moles of each element are present
2 mol N and 4 mol O
Use molar masses of elements to determine mass of each element in 1 mole
Molar Mass of N2O4 = g N2O4 / 1 mol
Calculate % by mass of each element

27. Using Percent Composition (cont)
= g N
= g O
= 30.54% N in N2O4
= 69.46% O in N2O4
The experimental values match the theoretical percentages for the formula N2O4.

28. Your Turn!
If a sample containing only phosphorous and oxygen has percent composition 56.34% P and 43.66% O, is this P4O10?
Yes
B. No
4 mol P  1 mol P4O10
10 mol O  1 mol P4O10
4 mol P = 4  g/mol P = g P
10 mol O = 10 16.00 g/mol O = g O
1 mol P4O10 = g P4O10
This compound is not P4O10. It is P2O3.
= 43.64% P
= 56.36% O

29. 1. Empirical Formula from Mass Data
When a g sample of a compound was analyzed, it was found to contain g of C, g of H, and g of N. Calculate the empirical formula of this compound.
Step 1: Calculate moles of each substance
3.722  10–3 mol C
1.860  10–2 mol H
3.723  10–3 mol N

30. 1. Empirical Formula from Mass Data
Step 2: Select the smallest number of moles
Smallest is × 10–3 mole
C =
H =
Step 3: Divide all number of moles by the smallest one
Mole ratio
Integer ratio
1.000
= 1
4.997
= 5
N =
1.000
= 1
Empirical formula = CH5N

31. 1. Empirical Formula from Mass Data
One of the compounds of iron and oxygen, “black iron oxide,” occurs naturally in the mineral magnetite. When a g sample was analyzed it was found to have g of Fe and g of O. Calculate the empirical formula of this compound. Assembling the tools: 1 mol Fe = g Fe 1 mol O = g O Find the moles of each element, then find the ratio of the moles of the elements. 1. Calculate moles of each substance
mol Fe
mol O

32. 1. Empirical Formula from Mass Data
2. Divide both by smallest number mol to get smallest whole number ratio.
=1.000 Fe
× 3 = Fe
=1.33 O
× 3 = 3.99 O Or
Empirical Formula = Fe3O4

33. 2. Empirical Formula from Percentage Composition
New compounds are characterized by elemental analysis, from which the percentage composition can be obtained
Use percentage composition data to calculate empirical formula
Must convert percentage composition to grams
Assume g sample
Convenient
Sum of percentage composition = 100%
Sum of masses of each element = 100 g

34. 2. Empirical Formula from Percentage Composition
Calculate the empirical formula of a compound whose percentage composition data is 43.64% P and 56.36% O. If the molar mass is determined to be g/mol, what is the molecular formula?
Step 1: Assume 100 g of compound. Calculate moles.
43.64 g P
56.36 g O
1 mol P = g
1 mol O = g
= mol P
= mol P

35. 2. Empirical Formula from Percentage Composition
Step 2: Divide by smallest number of moles
 2 = 2
 2 = 5
Step 3: Multiple to get integers
1.000  2 = 2
2.500  2 = 5
Empirical formula = P2O5

36. 3. Empirical Formulas from Indirect Analysis:
In practice, compounds are not broken down into elements, but are changed into other compounds whose formula is known
Combustion Analysis
Compounds containing carbon, hydrogen, and oxygen, can be burned completely in pure oxygen gas
Only carbon dioxide and water are produced
e.g., Combustion of methanol (CH3OH)
2CH3OH + 3O2  2CO2 + 4H2O

37. Combustion Analysis
Classic
Modern CHN analysis

38. 3. Empirical Formulas from Indirect Analysis:
Carbon dioxide and water are separated and weighed separately
All C ends up as CO2
All H ends up as H2O
Mass of C can be derived from amount of CO2
mass CO2  mol CO2  mol C  mass C
Mass of H can be derived from amount of H2O
mass H2O  mol H2O  mol H  mass H
Mass of oxygen is obtained by difference
mass O = mass sample – (mass C + mass H)

39. Ex. Indirect or Combustion Analysis
The combustion of a g sample of a compound of C, H, and O in pure oxygen gave g CO2 and g of H2O. Calculate the empirical formula of the compound. C H
H2O
CO2
MM (g/mol)
12.011
1.008
18.015
44.01
1. Calculate mass of C from mass of CO2.
mass CO2  mole CO2  mole C  mass C
= g C

40. Ex. Indirect or Combustion Analysis
The combustion of a g sample of a compound of C, H, and O gave g CO2 and g of H2O. Calculate the empirical formula of the compound.
2. Calculate mass of H from mass of H2O.
mass H2O  mol H2O  mol H  mass H
= g H
3. Calculate mass of O from difference.
5.217 g sample – g C – g H
= g O

41. Ex. Indirect or Combustion AnalysisH O
At. mass
12.011
1.008
15.999 g
2.021
0.5049
2.691
4. Calculate mol of each element
= mol C
= mol H
= mol O

42. Ex. Indirect or Combustion Analysis
Preliminary empirical formula
C0.1683H0.5009O0.1682
5. Calculate mol ratio of each element
Because all values are close to integers, round to
= C1.00H2.97O1.00
Empirical Formula = CH3O

43. Your Turn!
The combustion of a g sample of a compound of C, H, and S in pure oxygen gave g CO2 and g of H2O. Calculate the empirical formula of the compound.
C4H12S
CH3S
C2H6S
D. C2H6S3
E. CH3S2

44. Your Turn! Solution 1. Calculate mass of C from mass of CO2.
mass CO2  mole CO2  mole C  mass C
= g C
2. Calculate mass of H from mass of H2O.
mass H2O  mol H2O  mol H  mass H
= g H

45. Your Turn! Solution Continued
3. Calculate mass of S from difference.
4. Convert individual elements to moles to obtain a preliminary empirical formula.
13.66 g sample – g C – g H
= g S
= mol C
= mol H
= mol S

46. Your Turn! Solution Continued
Preliminary empirical formula
C0.4497H1.319S0.2198
5. Calculate mol ratio of each element
Because all values are close to integers, round to
= C2.03H6.00S1.00
Empirical Formula = C2H6S

47. Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%

48. Example: Find the empirical formula of aspirin with the given mass percent composition
Information Given:C = 60.00%, H = 4.48 %, O =35.53% \ g C, 4.48 g H, g O Find: empirical formula, CxHyOz
g C
mol C
conceptual plan
whole
number
ratio
mole
ratio
empirical
formula
pseudo-
formula
g H
mol H
g O
mol O
C4.996H4.44O2.220
{C2.25H2O1} x 4
C9H8O4

49. Learning Check
The empirical formula of a compound containing phosphorous and oxygen was found to be P2O5. If the molar mass is determined to be g/mol, what is the molecular formula?
Step 1: Calculate empirical mass
Step 2: Calculate ratio of molecular to empirical mass
= 2
Molecular formula = P4O10

50. Molecular formula = {C5H3} x 4 = C20H12
Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)
C5 = 5(12.01 g) = g
H3 = 3(1.01 g) = g
C5H3 = g
molar mass
empirical formula
Molecular formula = {C5H3} x 4 = C20H12

51. What are the coefficients for the decomposition of nitroglycerin?
__ C3H5N3O9  __ N2 + __ CO2 + __ H2O + __ O2
2,3,6,2,1
2,3,6,5,1
4,6,12,10,12
4,3,12,10,1
4,6,12,10,1

52. What are the coefficients for the decomposition of nitroglycerin?
__ C3H5N3O9  __ N2 + __ CO2 + __ H2O + __ O2
2,3,6,2,1
2,3,6,5,1
4,6,12,10,12
4,3,12,10,1
4,6,12,10,1
Answer: e

53. Which of the following statements is true about a balanced chemical reaction?
The mass of the reactants is equal to the mass of the products.
The number of each type of atom is the same in the reactants and in the products.
The moles of each type of atom is the same in the reactants and in the products.
All of the above.

54. Which of the following statements is true about a balanced chemical reaction?
The mass of the reactants is equal to the mass of the products.
The number of each type of atom is the same in the reactants and in the products.
The moles of each type of atom is the same in the reactants and in the products.
All of the above.
Answer: d

55. Write a balanced equation for the combustion of octane, C8H18

56. Write a balanced equation for the combustion of octane, C8H18
Write a skeletal equation
16  C  16; 36  H  36; 50  O  50
{C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(g)}x 2
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
25/2 x 2  O  25
C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(g)
8  C  1 x 8
C8H18(l) + O2(g)  8 CO2(g) + H2O(g)
18  H  2 x 9
C8H18(l) + O2(g)  8 CO2(g) + 9 H2O(g)
C8H18(l) + O2(g)  CO2(g) + H2O(g)
Balance atoms in complex substances first
Balance free elements by adjusting coefficient in front of free element
If fractional coefficients, multiply thru by denominator
Check

57. Your Turn! Balance each of the following equations.
What are the coefficients in front of each compound?
__ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 1 2
___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3
__H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(l ) 2 3 1 6

58. Your Turn!
How many moles of hydrochloric acid are required to completely react with 1.43 moles of iron(II) chloride according to the reaction below?
14HCl + Na2Cr2O7 + 6FeCl2 →
2CrCl3 + 7H2O + 6FeCl3 + 2NaCl
20.0 mole
1.43 mole
0.613 mole
3.34 mole
8.58 mole
Assembling the tools
1.43 mole FeCl2 = ? moles HCl
6 moles FeCl2 = 14 mole HCl
= 3.34 mol HCl

59. The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) How many grams of N2 are produced from 6.00 g of NaN3?
3.88 g
1.72 g
0.138 g
2.59 g

60. The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) How many grams of N2 are produced from 6.00 g of NaN3?
3.88 g
1.72 g
0.138 g
2.59 g
Answer: a

61. The overall equation involved in photosynthesis is
The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of glucose (C6H12O6, g/mol) form when 4.40 g of CO2 react?
18.0 g
3.00 g
108 g g

62. The overall equation involved in photosynthesis is
The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of glucose (C6H12O6, g/mol) form when 4.40 g of CO2 react?
18.0 g
3.00 g
108 g g
Answer: b

63. Your Turn!
How many grams of sodium dichromate are required to produce 24.7 g iron(III) chloride from the following reaction?
14HCl + Na2Cr2O7 + 6FeCl2 →
2CrCl3 + 7H2O + 6FeCl3 + 2NaCl
6.64 g Na2Cr2O7
0.152 g Na2Cr2O7
8.51 g Na2Cr2O7
39.9 g Na2Cr2O7
8.04 g Na2Cr2O7
= 6.64 g Na2Cr2O7

64. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
How much CO2 is pumped into the atmosphere by combustion of a tankful of gasoline?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Given:
Find:
15.0 gallon tank of gas, 1 gallon = L, d = g/mL, 1 mol octane = g, 1 mol CO2 = g
g CO2
gallons octane → L → mL → g → moles octane → moles CO2 → g CO2
Plan:
Relation-ships:
3.785 L/1 gal
1000 mL/L
0.703 g/mL
1 mol C8H18 = g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
Soltn:
15.0 gal x 3.785L/gal x 1000 mL/L x g/mL x 1 mol oct/ g oct x 16 mol CO2/1 mol oct x g CO2/ mol CO2
= 123,030 g kg
Check:
15.0 gal of octane is 39.9 kg; because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

65. Stoichiometry Problem
When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.
conceptual plan
identify limiting reactant
calculate theoretical yield of Ti from TiO2 and C (g → mols)
calculate theoretical yield from limiting reagent (mols → g)
calculate percent yield (actual yield/theoretical yield)

66. Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld.
Write a conceptual plan }
smallest
amount is
from
limiting
reactant kg
TiO2 C
smallest
mol Ti

67. Collect needed relationships
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti
Collect needed relationships
1000 g = 1 kg
Molar Mass TiO2 = g/mol
Molar Mass Ti = g/mol
Molar Mass C = g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C : 1 mol Ti (from the chem. equation)

68. Apply the conceptual plan
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan
limiting reactant
smallest moles of Ti

69. Apply the conceptual plan
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan
theoretical yield

70. Apply the conceptual plan
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan

71. limiting reactant = TiO2 theoretical yield = 52.9 kg
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Check the solutions
limiting reactant = TiO2
theoretical yield = 52.9 kg
percent yield = 80.9%
Because Ti has lower molar mass than TiO2, the T.Y. makes
sense and the percent yield makes sense as it is less than 100%

72. Ammonia is produced using the Haber process:
Ammonia is produced using the Haber process: H2 + N2  2 NH3 Calculate the mass of ammonia produced when 35.0 g of nitrogen react with 12.5 g of hydrogen.
47.5 g
42.6 g
35.0 g
63.8 g
70.5 g

73. Ammonia is produced using the Haber process:
Ammonia is produced using the Haber process: H2 + N2  2 NH3 Calculate the mass of ammonia produced when 35.0 g of nitrogen react with 12.5 g of hydrogen.
47.5 g
42.6 g
35.0 g
63.8 g
70.5 g
Answer: b

74. Ammonia is produced using the Haber process:
Ammonia is produced using the Haber process: 3 H2 + N2  2 NH3 What percent yield of ammonia is produced from 15.0 kg each of H2 and N2, if 13.7 kg of product are recovered? Assume the reaction goes to completion.
7.53 × 10–2 %
1.50 × 10–1 %
75.3%
15.0%
16.2%

75. Ammonia is produced using the Haber process:
Ammonia is produced using the Haber process: 3 H2 + N2  2 NH3 What percent yield of ammonia is produced from 15.0 kg each of H2 and N2, if 13.7 kg of product are recovered? Assume the reaction goes to completion.
7.53 × 10–2 %
1.50 × 10–1 %
75.3%
15.0%
16.2%
Answer: c

76. What mass of TiCl4 is needed to produce 25
What mass of TiCl4 is needed to produce 25.0 g of Ti if the reaction proceeds with an 82% yield? TiCl4 + 2Mg  Ti + 2 MgCl2
30.5 g
121 g
99.1 g
81.2 g

77. What mass of TiCl4 is needed to produce 25
What mass of TiCl4 is needed to produce 25.0 g of Ti if the reaction proceeds with an 82% yield? TiCl4 + 2Mg  Ti + 2 MgCl2
30.5 g
121 g
99.1 g
81.2 g
Answer: b

78. Learning Check: Percentage Yield
A chemist set up a synthesis of solid phosphorus trichloride by mixing 12.0 g of solid phosphorus with g chlorine gas and obtained 42.4 g of solid phosphorus trichloride. Calculate the percentage yield of this compound.
Analysis:
Write balanced equation
P(s) + Cl2(g)  PCl3(s) Is this balanced?
Determine Limiting Reagent
Determine Theoretical Yield
Calculate Percentage Yield

79. Learning Check: Percentage Yield
Assembling the Tools:
1 mol P = g P
1 mol Cl2 = g Cl2
3 mol Cl2  2 mol P
Solution
Determine Limiting Reactant
But you only have 35.0 g Cl2, so Cl2 is limiting reactant
= 41.2 g Cl2

80. Learning Check: Percentage Yield
Solution
Determine Theoretical Yield
Determine Percentage Yield
Actual yield = 42.4 g
= 45.2 g PCl3
= 93.8 %

81. Calculate the mass in grams of one mole of footballs if one football has a mass of 0.43 kg.

82. Calculate the mass in grams of one mole of footballs if one football has a mass of 0.43 kg.
Answer: c

83. Your Turn! How many moles of CO2 are there in 10.0 g? 1.00 mol
E mol
Molar mass of CO2
1 × g = g C
2 × g = g O
1 mol CO2 = g CO2
= mol CO2

84. Macroscopic to Microscopic
How many silver atoms are in a 85.0 g silver bracelet?
What do we want to determine?
85.0 g silver = ? atoms silver
What do we know?
g Ag = 1 mol Ag
1 mol Ag = 6.022×1023 Ag atoms
g Ag  mol Ag  atoms Ag
= 4.75 × 1023 Ag atoms

85. Your Turn! How many atoms are in 1.00 × 10–9 g of U?
Molar mass U = g/mole.
6.02 × 1014 atoms
4.20 × 1011 atoms
C × 1012 atoms
D × 10–31 atoms
E × 1021 atoms
= 2.53 × 1012 atoms U

86. Learning Check: Using Molar Mass
Example: How many moles of iron (Fe) are in g Fe?
What do we want to determine?
15.34 g Fe = ? mol Fe
What do we know?
1 mol Fe = g Fe
Set up ratio so that what you want is on top and what you start with is on the bottom
Start
End
= mol Fe

87. Your Turn! How many grams of platinum (Pt) are in 0.475 mole Pt? 195 g
Molar mass of Pt = g/mol
= 92.7 g Pt

88. Your Turn!
Calculate the mass in grams of FeCl3 in 1.53 × 1023 formula units. (molar mass = g/mol)
162.2 g
0.254 g
1.661 ×10–22 g
41.2 g
2.37 × 10–22
= 41.2 g FeCl3

89. Your Turn! How much, in grams, do 8.85 × 1024 atoms of zinc weigh?
= 961 g Zn

90. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)
Given: 75.7% Sn, (100 – 75.3) = 24.3% F 
in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F
Find: SnxFy
Rel: 1 mol Sn = g; 1 mol F = g
Conceptual plan:
whole
number
ratio
g Sn
mol Sn
mole
ratio
pseudo-
formula
empirical
formula
g F
mol F

91. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)
Apply the conceptual plan
Sn0.638F1.28
SnF2

92. Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Given: 72.4% Fe, (100 – 72.4) = 27.6% O 
in 100 g magnetite there are 72.4 g Fe and 27.6 g O
Find: FexOy
Rel: 1 mol Fe = g; 1 mol O = g
Conceptual plan:
whole
number
ratio
g Fe
mol Fe
mole
ratio
pseudo-
formula
empirical
formula
g O
mol O

93. Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Apply the conceptual plan
Fe1.30O1.73

94. Example: Find the empirical formula of compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H
Find: empirical formula, CxHyOz
conceptual plan g
CO2, H2O
mol
CO2, H2O
mol
C, H g
C, H g O
mol O
mol
C, H, O
pseudo
formula
mol
ratio
empirical
formula

95. Example: Find the empirical formula of compound with the given amounts of combustion products
Information
Given: g compound,
2.445 g CO2, g H2O
Find: empirical formula, CxHyOz
CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula
needed relationships 1 mole CO2 = g CO2 1 mole H2O = g H2O 1 mole C = g C 1 mole H = g H 1 mole O = g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H

96. An unknown compound contains the following percents by mass: C: 60
An unknown compound contains the following percents by mass: C: 60.86%, H: 5.83%, O: 23.16%, and N: 10.14%. Find the empirical formula.
C6H8O2N2
C7H8O2N
C6H8O2N
C8H8O2N
C8H8ON

97. An unknown compound contains the following percents by mass: C: 60
An unknown compound contains the following percents by mass: C: 60.86%, H: 5.83%, O: 23.16%, and N: 10.14%. Find the empirical formula.
C6H8O2N2
C7H8O2N
C6H8O2N
C8H8O2N
C8H8ON
Answer: b

98. What is the empirical formula of a compound containing C, H, and O if combustion of 3.69 g of the compound yields 5.40 g of CO2 and g of H2O?
CH3O
C2H3O
CHO
C2H3O2
CH2O

99. What is the empirical formula of a compound containing C, H, and O if combustion of 3.69 g of the compound yields 5.40 g of CO2 and g of H2O?
CH3O
C2H3O
CHO
C2H3O2
CH2O
Answer: e

100. Your Turn! 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l )
How many grams of Al2O3 are produced when 41.5 g Al react?
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l )
78.4 g
157 g
314 g
22.0 g
11.0 g
= 78.4 g Al2O3

101. Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to:
4 NH3 + 5 O2  4 NO + 6 H2O
Solution: Step 1
mass NH3  mole NH3  mole O2  mass O2
Assembling the tools
1 mol NH3 = g
1 mol O2 = g
4 mol NH3  5 mol O2
Only have 40.0 g O2, O2 limiting reactant
= 70.5 g O2 needed

102. Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to:
4 NH3 + 5 O2  4 NO + 6 H2O
Solution: Step 2
mass O2  mole O2  mole NO  mass NO
Assembling the tools
1 mol O2 = g
1 mol NO = g
5 mol O2  4 mol NO
Can only form 30.0 g NO
= 30.0 g NO formed

103. Your Turn!
If 18.1 g NH3 is reacted with 90.4 g CuO, what is the maximum amount of Cu metal that can be formed?
2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)
(MM) (17.03) (79.55) (28.01) ( (18.02) (g/mol)
127 g
103 g
72.2 g
108 g
56.5 g
127 g CuO needed
Only have 90.4 g so CuO limiting
= 72.2 g Cu can be formed

104. Your Turn!
If 8.51 g C2H5SH reacts with 22.4 g O2, what is the maximum amount of sulfur dioxide that can form?
2C2H5SH(l) + 9O2(g)  4CO2(g) + 6H2O(g) + 2SO2(g)
(62.13) (32.00) (44.01) (18.02) (64.06)
19.7 g
4.38 g
39.5 g
9.97 g
8.77 g
19.7 g O2 needed
Have 22.4 g so C2H5SH is limiting
= 8.77 g SO2 can form

105. Ex. Percentage Yield Calculation
When 18.1 g NH3 and 90.4 g CuO are reacted, the theoretical yield is 72.2 g Cu. The actual yield is 58.3 g Cu. What is the percent yield?
2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)
= 80.7%

106. Your Turn!
When 6.40 g of CH3OH was mixed with 10.2 g of O2 and ignited, 6.12 g of CO2 was obtained. What was the percentage yield of CO2?
2CH3OH + 3O2  2CO2 + 4H2O
MM(g/mol) (32.04) (32.00) (44.01) (18.02)
6.12%
8.79%
100%
142%
69.6%
= 9.59 g O2 needed; CH3OH limiting
= 8.79 g CO2 in theory