unit 6 stoichiometry chapter 9. chapter 9 stoichiometry
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Unit 6
StoichiometryChapter 9Chapter 9
CHAPTER 9Stoichiometry
What is Stoichiometry?
• Stoichiometry is the branch of chemistry that deals withthe mass relationships of elements in a chemical reaction.
• Stoichiometry calculationsalways start with a balanced chemical equation.
Chapter 9 – Section 1: Introduction to Stoichiometry
Mole Ratio
• A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reactionExample: 2Al2O3(l) → 4Al(s) + 3O2(g)
Mole Ratios:
2 mol Al2O3 2 mol Al2O3 4 mol Al
4 mol Al 3 mol O2 3 mol O2
Chapter 9 – Section 1: Introduction to Stoichiometry
Molar Mass (Review)• Molar Mass is the mass of one mole of a pure
substance.• Molar mass units
are g/mol.• The molar mass of
an element is the same number as its atomic mass, only the units are different.
• Example: Molar mass of water (H2O).• (2 mol H x 1.01) + (1 mol O x 16.00) = 18.02 g/mol.
Chapter 9 – Section 1: Introduction to Stoichiometry
Mole to Mole Conversions• Using the mole ratio, you can convert from moles
of one substance to moles of any other substance in a chemical reaction.
Example:• For the reaction N2 + 3H2 → 2NH3, how many moles
of H2 are required to produce 12 moles of NH3?
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
This is whatwe’re given:
Use mole ratio as a conversion factor
12 mol NH33 mol H2
2 mol NH3
18 mol H2=x
Mole/Mole ConversionsSample Problem 1
In a spacecraft, the CO2 exhaled by astronauts can be removed by the following reaction with LiOH:
CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l)How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?
Solution:Given Conversion factor
20 mol CO22 mol LiOH1 mol CO2
40 mol LiOH=x
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Mole to Mass Conversions• To convert from moles of one substance to grams
of another, you need 2 conversion factors: 1. mole ratio.2. molar mass of the unknown.
• To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Example:Given the equation: 2Mg(s) + O2(g)→ 2MgO(s),
Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium.
mol MgO
mol MgO
Mole to Mass Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
GivenMole Ratio
2.00 mol Mgmol Mg
80.6 g MgO=x x
2nd C.F.Molar Mass
of the Unknown
1st C.F.
2
2
1
g MgO40.3
Mass to Mole Conversions• To convert from grams of one substance to moles of
another, you need 2 conversion factors: 1. molar mass of the given.2. mole ratio.
• Mass to mole conversion factors are the inverse of mole to mass conversion factors.
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many moles of HgO are needed to produce 125g of O2?
mol O2
mol O2
Mass to Mole Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Given
Molar Massof the Given
125 g O2
g O2
7.81 mol HgO=x x
2nd C.F.Mole Ratio
1st C.F.
32
1
1
mol HgO2
Mole/Mass ConversionsSample Problem 1
The balanced equation for photosynthesis is as follows:
6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)What mass, in grams, of glucose (C6H12O6) is produced when 3.00 mol of water react with carbon dioxide?
Solution:
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
mol C6H12O6
mol C6H12O6
GivenMole Ratio
3.00 mol H2Omol H2O
90.1 g C6H12O6
=x x
2nd C.F.Molar Mass
of the Unknown
1st C.F.
6
1
1
g C6H12O6180.2
Mole/Mass ConversionsSample Problem 2
The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
If the reaction is run using 824 g NH3 and excess oxygen, how many moles of NO are formed?
Solution:
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
mol NH3
mol NH3
Given
Molar Massof the Given
824 g NH3
g NH3
48.5 mol NO=x x
2nd C.F.Mole Ratio
1st C.F.
17.0
1
4
mol NO4
Mass to Mass Conversions• To convert from grams of one substance to
grams of another, you need 3 conversion factors:
1. molar mass of the given.2. mole ratio.3. molar mass of the unknown.
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many grams of HgO are needed to produce 45g of O2?
mol O2
mol O2
Mass to Mass Conversions (continued)
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
Given
Molar Massof the Given
45 g O2
g O2
609.2 g HgO
=x x
2nd C.F.Mole Ratio
1st C.F.
32
1
1
mol HgO2
mol HgOx
3rd C.F.Molar Mass ofthe Unknown
1
g HgO216.6
Mass/Mass ConversionsSample Problem 1
Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the following reaction :
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
How many grams of SnF2 are produced from the reaction of 30.0 g HF with Sn?
Solution:
Chapter 9 – Section 2: Ideal Stoichiometric Calculations
mol HF
mol HF
GivenMolar Massof the Given
30.0 g HFg HF
=x x
2nd C.F.Mole Ratio
1st C.F.
20.0
1
2
mol SnF21
mol SnF2
x
3rd C.F.Molar Mass ofthe Unknown
1
g SnF2156.7 117.5 g SnF2
Limiting Reactants
• When combining 2 or more different things to make a product, you have to stop when one of the things is used up.
• For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants (continued)
• The limiting reactant is the reactant that limits the amount of product formed.
• The excess reactant is the substance that is not used up completely.
• Once the limiting reactant is used up, a chemical reaction will stop.
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Limiting Reactants (continued)
Example:Silicon dioxide reacts with hydrogen fluoride according
to the following equation:SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Set up 2 equations, one for eachgiven:
Mole ratio C.F.
4.5 mol SiO21 mol SiF4
1 mol SiO2
4.5 mol SiF4=x
6 mol HF 1 mol SiF4
4 mol HF1.5 mol SiF4=x 1.5 mol SiF46 mol HF
Limiting Reactant is HF
The limitingreactant
makes the least product.
Limiting ReactantsSample Problem 1
According to the following reaction :CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
a.If 1 mol CS2 reacts with 1 mol O2, identify the limiting reactant.
b.How many moles of excess reactant remain?
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
1 mol O2 2 mol SO2
3 mol O2
0.67 mol SO2=x
1 mol CS22 mol SO2
1 mol CS2
2 mol SO2=
1 mol O2
x
0.67 mol SO2
Limiting Reactant is O2
0.67 mol SO2 1 mol CS2
2 mol SO2
0.33 mol CS2=xWork backwards from actual mol of product formed
Mol of excessreactant thatwas used up
1 mol CS2 - 0.33 mol CS20.67 mol CS2=
Molesremaining
Percentage Yield
• The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry).
• The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem).
• The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Now use the formula
mol C6H5Cl
Percentage Yield (continued)
Example:Given the following equation:
C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)When 36.8 g C6H6 react with excess Cl2, the actual yield of
C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl?
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
mol C6H6
mol C6H636.8 g C6H6g C6H6
=x x78.01
11
mol C6H5Clx
1g C6H5Cl112.5 53.1 g
C6H5Cl
Set up a mass to mass conversionThis is the
Theoretical Yield
Percent Yield
=Actual
Theoreticalx 100 =
38.8 g C6H5Cl53.1 g C6H5Cl
x 100 = 73.1%
Percentage YieldSample Problem 1
According to the following reaction :CO(g) + 2H2(g) → CH3OH(l)
If the typical yield is 80%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess hydrogen gas?
Chapter 9 – Section 3: Limiting Reactants and Percentage Yield
Multiply theoreticalyield by percentageto get actual yield
mol CH3OHmol CO
mol CO75.0 g COg CO
=x x28.01
11
mol CH3OHx
1g CH3OH32.0 85.7 g
CH3OH
Set up a mass to mass conversionThis is the
Theoretical Yield
85.7 g CH3OH x 80% = 68.6 g CH3OH
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