1. Chapter 9 Calculations from Chemical Equations
Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients throughout the world.
Introduction to General, Organic, and Biochemistry 10e
John Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena

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Chapter Outline
9.1 A Short Review
9.2 Introduction to Stoichiometry
9.3 Mole-Mole Calculations
9.4 Mole-Mass Calculations
9.5 Mass-Mass Calculations
9.6 Limiting Reactant and Yield Calculations
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Molar Mass
Molar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound ; the units are g/mol x1023 molecules 6.022x1023 formula units 6.022x1023 atoms 6.022x1023 ions
1 mole =
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Molar Mass
What is the molar mass of Al(ClO3)3? Al 1(26.98 g) 3Cl 3(35.45 g) 9O 9(16.00 g) Al(ClO3) g/mol
atomic mass Al
26.98 Cl
35.45 O
16.00
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Molar Mass
Calculate the mass of 2.5 moles of aluminum chlorate.
Plan
2.5 mol Al(ClO3)3  g Al(ClO3)3
1 mol Al(ClO3)3 = g Al(ClO3)3
Calculate
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Molar Mass
Calculate the moles of 3.52g of aluminum chlorate.
Plan
3.52 g Al(ClO3)3  mol Al(ClO3)3
1 mol Al(ClO3)3 = g Al(ClO3)3
Calculate
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Molar Mass
Calculate the number of formula units contained in g aluminum chlorate.
Plan
12.4 g Al(ClO3)3  formula units Al(ClO3)3
1 mol Al(ClO3)3 = g = 6.022x1023 formula units
Calculate
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Your Turn!
What is the mass of 3.61 moles of CaCl2?
3.61 g
272 g
2.17 × g
401 g
atomic mass Ca
40.08 Cl
35.45
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You Turn!
How many moles of HCl are contained in 18.2 g HCl?
1.00 mol
0.500 mol
0.250 mol
0.125 mol
atomic mass H
1.01 Cl
35.45
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Your Turn!
What is the mass of 1.60×1023 molecules of HCl?
9.69 g
137 g
0.729 g
36.5 g
atomic mass H
1.01 Cl
35.45
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Stoichiometry
Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N2(g) + 3I2(s)  2NI3(s) 1 mol N2 + 3 mol I2  2 mol NI3 Mole ratios come from the coefficients in the balanced equation: The 3 other possibilities are the inverse of these ratios.
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Your Turn!
Which of these statements is not true about the reaction?
1N2(g) + 3I2(s)  2NI3(s)
1 mole of nitrogen is needed for every 3 moles of iodine
1 gram of nitrogen is needed for every 3 grams of iodine
Both statements are true
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Using the mole ratio
Calculate the number of moles of NI3 that can be made from 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s)
Plan
5.50 mol N2  mol NI3
Set-Up
Calculate
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Using the mole ratio
Calculate the number of moles of I2 needed to react with 5.50 mol N2 in the reaction: 1N2(g) + 3I2(s)  2NI3(s)
Plan
5.50 mol N2  mol I2
Set-Up
Calculate
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Your Turn!
How many moles of HF will be produced by the complete reaction of 1.42 moles of H2 in the following equation?
H2 + F2  2HF
0.710
1.42
2.00
2.84
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Stoichiometry
Problem Solving Strategy for stoichiometry problems:
Convert starting substance to moles.
Convert the moles of starting substance to moles of desired substance.
Convert the moles of desired substance to the units specified in the problem.
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Stoichiometry
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18. Mole-Mole Calculations
How many moles of Al are needed to make mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
mol H2  mol Al
Set-Up
Calculate
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19. Mole-Mole Calculations
How many moles of HCl are needed to make mol of H2? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
mol H2  mol HCl
Set-Up
Calculate
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20. Your Turn!
How many moles of H2 are made by the reaction of 1.5 mol HCl with excess aluminum?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
0.75 mol
3.0 mol
6.0 mol
4.5 mol
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Your Turn!
How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation?
C3H8 + 5O2  3CO2 + 9H2O
5.00 mol
3.00 mol
1.80 mol
1.50 mol
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Your Turn!
How many moles of C3H8 are consumed when 1.81x1023 molecules of CO2 are produced in the following equation?
 C3H8 + 5O2  3CO2 + 4H2O
0.100
0.897
6.03 × 1022
5.43 × 1023
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23. Mole-Mass Calculations
What mass of H2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with excess aluminum? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
3.0 mol HCl mol H2  g H2
Calculate
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24. Mole-Mass Calculations
How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
2.00 g Al  mol Al  mol HCl
Calculate
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25. Mole-Mass Calculations
What mass of Al(NO3)3 (213g/mol) is needed to react with .093 mol Na2CO3?
3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)
Plan
0.093 mol Na2CO3  mol Al(NO3)3  g Al(NO3)3
Calculate
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26. Mole-Mass Calculations
How many moles of Al2(CO3)3 are made by the reaction of 3.45g Na2CO3 ( g/mol) with excess Al(NO3)3?
3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)
Plan
3.45g Na2CO3  mol Na2CO3  g Al2(CO3)3
Calculate
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Your Turn!
How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation?
4Al(s) + 3O2(g)  2Al2O3(s)
 0.248
0.559
1.50
3.00
atomic mass Al
26.98 O
16.00
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Your Turn!
What mass of HCl is produced when 1.81x1024 molecules of H2 react completely in the following equation?
H2(g) + Cl2(g)  2HCl(g)
54.7g
72.9g
109g
219g
atomic mass H
1.01 Cl
35.45
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29. Mass-Mass Calculations
Now we will put it all together. What mass of Br2 ( g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s)
7.00 g Al mol Al  mol Br2  g Br2
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30. Mass-Mass Calculations
What mass of Br2 ( g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al(s) + 3Br2(l)  2AlBr3(s)
Plan
7.00 g Al mol Al  mol Br2  g Br2
Calculate
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31. Mass-Mass Calculations
What mass of Fe2S3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl3 ( g/mol) with excess Na2S? 2FeCl3(aq) + 3Na2S(aq)  Fe2S3(s) + 6NaCl(aq)
Plan
9.34 g FeCl3  mol FeCl3  mol Fe2S3  g Fe2S3
Calculate
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Your Turn!
What mass of oxygen is consumed when 54.0g of water is produced in the following equation?
 2H2 + O2  2H2O
0.167 g
0.667 g
1.50 g
47.9 g
atomic mass H
1.01 O
16.00
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Your Turn!
What mass of H2O is produced when 12.0g of HCl react completely in the following equation?
 6HCl + Fe2O3  2FeCl3 + 3H2O
2.97 g
39.4 g
27.4 g
110. g
atomic mass H
1.01 O
16.00 Cl
35.45
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Limiting Reactant
Determine the number of that can be made given these quantities of reactants and the reaction equation:  + + 
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Limiting Reactant
The limiting reactant is the reactant that limits the amount of product that can be made. The reaction stops when the limiting reactant is used up. What was the limiting reactant in the reaction: The small blue balls. + 
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Excess Reactant
The excess reactant is the reactant that remains when the reaction stops. There is always left over excess reactant. What was the excess reactant in the reaction: The excess reactant was the larger blue ball. + 
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Limiting reactant
Figure 9.2
The number of bicycles that can be built from these parts is determined by the “limiting reactant” (the pedal assemblies).
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38. Limiting Reactant Calculations
Technique for solving limiting reactant problems:
Convert reactant 1 to moles or mass of product
Convert reactant 2 to moles or mass of product
Compare answers. The smaller answer is the maximum theoretical yield.
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39. Limiting Reactant Calculation
Calculate the number of moles of water that can be made by the reaction of 1.51 mol H2 with mol O2.
2H2(g) + O2(g)  2H2O(g)
Calculate the theoretical yield of H2O assuming H2 is the limiting reactant and that O2 is the excess reactant.
Calculate the theoretical yield of H2O assuming that O2 is the limiting reactant and that H2 is the excess reactant.
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40. Limiting Reactant Calculation continued
Assuming that H2 is limiting and O2 is excess:
Assuming that O2 is limiting and H2 is excess:
So what is the maximum yield of H2O?
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41. Limiting Reactant Calculation continued
How much H2 and O2 remain when the reaction stops? H2: Limiting Reactant – None remains. It was used up in the reaction. O2: Excess Reactant – Calculate the amount of O2 used in the reaction with H2. Then subtract that from the original amount.
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42. Limiting Reactant Calculation
Calculate the mass of copper that can be made from the combination of 15.0 g aluminum with 25.0 g copper(II) sulfate.
2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s)
Plan
15 g Al  mol Al  mol Cu  g Cu
25 g CuSO4  mol CuSO4  mol Cu  g Cu
Compare answers. The smaller number is the right answer.
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43. Limiting Reactant Calculation continued
2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s)
1. Assume Al is limiting and CuSO4 is in excess.
53.0 g Cu
2. Assume CuSO4 is limiting and Al is in excess.
9.96 g Cu
3. Compare answers.
CuSO4 is the limiting reagent. The theoretical yield of Cu is 9.96 g.
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Your Turn!
True/False:
You can compare the quantities of reactant when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant.
True
False
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Your Turn!
Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation?
  Cu + 2AgNO3  Cu(NO3)2 + 2Ag Cu
AgNO3
Cu(NO3)2 Ag
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Your Turn!
What is the mass of silver ( g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?
  Cu + 2AgNO3  Cu(NO3)2 + 2Ag
162g
216g
324g
647g
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Percent Yield
To determine the efficiency of a process for making a compound, chemists compute the percent yield of the reaction.
The theoretical yield is the result calculated using stoichiometry.
The actual yield of a chemical reaction is the experimental result, which is often less than the theoretical yield due to experimental losses and errors along the way.
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Percent Yield
Calculate the % yield of PCl3 that results from reacting 5.00 g P with excess Cl2 if only 17.2 g of PCl3 were recovered.
2P + 3Cl2 2PCl3
Compute the expected yield of PCl3 from 5.00 g P with excess Cl2.
22.2g PCl3
Compute the % Yield.
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Your Turn!
In a reaction to produce ammonia, the theoretical yield is 420. g. What is the percent yield if the actual yield is 350. g? A. 83.3% B. 20.0% C. 16.7% D. 120.%
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