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Chemistry 11, Stoichiometry, Unit 06 1
Lesson 01 to 05: Stoichiometry
Stoichiometry is at the heart of the production of many things you use in your daily life…
plastics
tires
fertilizers
gasoline All chemically engineered commodities all rely on stoichiometry for their production. Stoichiometry is the calculation of quantities in chemical equations. Given a chemical reaction, stoichiometry tells us what quantity of each reactant we need in order to get enough of our desired product.
Chemistry 11, Stoichiometry, Unit 06 2
01 Unit Conversions When converting between mass, gas volume and number of particles (molecules, atoms or ions) you must follow the procedure below…
Important!
Chemistry 11, Stoichiometry, Unit 06 3
Moles to Mass and Mass to Moles When converting from moles to mass or from mass to moles you need to use one of the following conversion factors…
1 mole of substance
molar mass of substance or
molar mass of substance
1 mole of substance
Example
Determine the mass of 5.0 moles of oxygen gas ( 2O ).
2 22
2
32 g O5.0 mol O
1 mo160 g
O
l O
Example
Determine the number of moles in 6.3 grams of OH2 .
2 22
2
1 mol H O6.3 g H O
1 8 g 0.35 m O
Ho H
Ol
Chemistry 11, Stoichiometry, Unit 06 4
Number of Moles and Gas Volume Remember that Avogadro's Hypothesis states that "equal volumes of gases at the same temperature and pressure contain the same number of molecules regardless of their chemical nature and physical properties".
This number, Avogadro's number, is 236.022x10 . It is the number of molecules of any gas present in a volume of 22.4 L and is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon dioxide or bromine. Note this will occur only at standard temperature and pressure (STP), 1 atmosphere and 25 C .
1 mole of ANY gas at STP has a volume of 22.4 L
Conversion Factor
mol 1
L 22.4 OR
L 22.4
mol 1
We are going to use this information to help us solve the following two problems…
Chemistry 11, Stoichiometry, Unit 06 5
Example 01
What volume is occupied by 0.350 moles of carbon dioxide,
2CO , at STP?
2
2
22.4 L0.350 mol CO
1 7.84 L
mol CO
Example 02
Calculate the number of moles for a volume of 5.00 ml of 2SO
gas.
4
2
2
22
1L5.00 ml SO
1000ml
1 mol SO
222.23x10 mol
.4 L SO
O S
Chemistry 11, Stoichiometry, Unit 06 6
Number of Moles and Number of Molecules or Atoms
Conversion Factor
atoms or molecules mol 1
atoms or molecules 6.022x10
OR
atoms or molecules 6.022x10
atoms or molecules mol 1
23
23
Example 03
How many molecules are there in 0.5 moles of water?
2
2336.022x10
0.5 molmo
3x10 moleculesl
Example 04
How many atoms are the in 0.5 moles of water molecules?
23
22
2
2
23
6.022x10 H O molecules0.5 mol H O molecules
1 mol H O molecules
3 atoms
1 H O molecule9x10 atoms
Chemistry 11, Stoichiometry, Unit 06 7
Chemistry 11, Stoichiometry, Unit 06 8
02 Problems Involving Multiple Conversions
Example 05: Mass - Moles
The combustion of propane ( 83HC ) proceeds according to the
following equation…
O(l)4H(g)3CO(g)5O(g)HC 22283
What mass of carbon dioxide is produced by reacting 2.00 moles of oxygen?
2 22 2
2 2
3 mol CO 44.0 g CO2.00 mole O
5 mol O 1 mol CO52.8 g CO
Example 06: Mass - Mass
The combustion of propane ( 83HC ) proceeds according to the
following equation…
O(l)4H(g)3CO(g)5O(g)HC 22283
What mass of propane is required to produce a mass of 100.0 grams of water?
3 8
3 8 3 822
2 2 3 8
1 mol C H 44.0 g C H1 mol H O100.0 g H O
18.0 g H O 4 mol H O 1 mol C H
61.1 g C H
Chemistry 11, Stoichiometry, Unit 06 9
Example 07: Mass – Gas Volume
The combustion of propane ( 83HC ) proceeds according to the
following equation…
O(l)4H(g)3CO(g)5O(g)HC 22283
If a sample of propane is burned what mass of water is produced if the reaction also produces a volume of 50.0 L of carbon dioxide at STP?
2 2 22
2
2
2 2
1 mol CO 4 mol H O 18.0 g H O50.0 L CO
22.4 L CO 3 mol CO 1 mol H
53.6 g H O
O
Chemistry 11, Stoichiometry, Unit 06 10
Example 08: Mass - Molecules
The combustion of propane ( 83HC ) proceeds according to the
following equation…
O(l)4H(g)3CO(g)5O(g)HC 22283
A mass of g 1.35x10 6 of propane is extracted and burned.
How many molecules of carbon dioxide are produced if the gas sample is burned in the presence of excess oxygen?
6 3 8 23 8
3 8 3 8
1623
2
2
1 molC H 3 mol CO1.35x10 g C H
44.0 g C H 1 mol C H
6.022x10 molecules CO
1 mol5.54x10 molecules
CO
Chemistry 11, Stoichiometry, Unit 06 11
Example 09: Mass – Gas Volume
Nitromethane, a fuel used by some drag racers burns according to the following reaction…
(g)2NO(l)6H(g)4CO(g)3O(l)NO4CH 222223
What combined volume of gas at STP is produced if 0.316 grams of nitromethane is burned?
3 23 2
3 2 3 2
1 mol CH NO 6 mol gas0.316 g CH NO
61.0 g CH NO 4 mol CH NO
22.4 L
1 mol g0.174 L gas
as
Example 10: Gas Volume – Gas Volume
Pentane ( 125HC ) burns according to the reaction…
O(l)6H(g)5CO(g)8O(l)HC 222125
What volume of oxygen gas at STP is required to produce a 48.0 L of carbon dioxide at STP?
2 2 2
2
2
2 2 2
1 mol CO 8 mol O 22.4 L O48.0 L CO
22.4 L CO 5 mol CO 1 mol
76.8 L O
O
Chemistry 11, Stoichiometry, Unit 06 12
Chemistry 11, Stoichiometry, Unit 06 13
03 Problems Involving Molar Concentrations The concentration of a solution is the strength of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. Often it is necessary to determine how much water to add to a solution to change it to a specific concentration. The concentration of a solution is typically given in molarity. Molarity is defined as the number of moles of solute (what you are dissolving) divided by the liters of solvent (what is being dissolved into).
number of moles solutemolarity of solution
total volume of solvent
Chemistry 11, Stoichiometry, Unit 06 14
Example 11: Molar Concentration
A student wants 50.0 L of hydrogen gas at STP in a plastic bag by reacting excess aluminum metal with 3.00 M (moles per litre) sodium hydroxide solution according to the reaction…
(g)3H(aq)2NaAlOO(l)2H2NaOH(aq)2Al(s) 222
What volume of sodium hydroxide solution is required?
22
2 3
1 mol H 2 mol NaOH 1 L50.0 L H
22.4 L H 3 mol H 3.00
0.4
mol
96 L
Example 12: Molar Concentration
What volume of 0.250 M HCl is required to completely neutralize 25.0mL of 0.318M NaOH?
NaCl(aq)O(l)HNaOH(aq)HCl(aq) 2
1 L 0.318 mol NaOH 1 mol HCl
25.0 ml1000 ml 1 L NaOH 1 mol NaOH
1 L HCl
0.250 mol HCl0.0318 L HCl
Chemistry 11, Stoichiometry, Unit 06 15
Chemistry 11, Stoichiometry, Unit 06 16
04 Titrations A titration is a process where a measured amount of one particular solution with a known concentration is added and reacted with another solution of unknown concentration until neutralized. By completing a simple calculation, the concentration of the unknown can then be found.
Chemistry 11, Stoichiometry, Unit 06 17
The equivalence point is a point reached when you have combined and neutralized certain volumes of solutions in the acid-base ratio present in the chemical reaction you are considering. The following are examples of what is required to reach the equivalence point (point of neutralization) for different combinations of acid and base species...
2
equivalence point
HCl NaOH NaCl H O
1 mmol olees HCl acid
1moles mole eNaOH bas
2 22
equivalence point
2HCl+Ca OH CaCl +2H O
2
2 moles HCl=
2 moles acid
1 1 moles mole Ca OH base
3 4 3 4 2
equivalence point
H PO +3NaOH Na PO +3H O
3 41 mole H PO=
1 mole acid
33 mole moless NaOH base
The moment neutralization is achieved, the titration is stopped, measurements are taken and calculations are made.
Chemistry 11, Stoichiometry, Unit 06 18
The question now becomes… How do you know when you have reached the equivalence point? The end point is the point at which a colour change occurs during a titration indicating you have reached the equivalence point. More about this in the next lesson (Lesson 17) on indicators...
appearance of colour endpoint
acid base salt water
Chemistry 11, Stoichiometry, Unit 06 19
Example
In the reaction 2 4 2 4 2H SO 2NaOH Na SO 2H O an
equivalence point occurs when 23.10 ml of 0.2055 M NaOH is
added to a 25.00 ml portion of 2 4H SO . What is the 2 4H SO ?
1. First thing you do is to figure out how many moles of the base you added.
1L 0.2055 molmoles NaOH 23.10 ml
1000ml L
0.004747 moles NaOH
2. The second thing you need to do is to figure out how many moles of the acid you have based on the number of moles of base that were added. Make sense?
2 42 4
2 4
1 mole H SOmoles H SO 0.004747 moles NaOH
2 mole NaOH
0.002374 moles H SO
3. Lastly. You just need to calculate 2 4H SO .
2 42 4
0.002374 moles H SOH SO
0.02500L0.09494 M
Chemistry 11, Stoichiometry, Unit 06 20
Chemistry 11, Stoichiometry, Unit 06 21
05 Problems Involving Limiting or Excess Reactants Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the limiting reactant. Sometimes a reaction will occur between two or more substances where one substance is in excess. This substance is referred to as being an excess reactant. Often, it is necessary to identify the limiting or excess reactant in a problem.
Chemistry 11, Stoichiometry, Unit 06 22
Example 14: Limiting / Excess Reactant
What mass of 2Br is produced when 25.0 grams of 72CrOK ,
55.0 grams of KBr and 66.0 grams of 42SOH are reacted
according to the equation below? How many grams of each excess reactant will remain unreacted?
O7H3BrSOCrSO4K
SO7H6KBrCrOK
2234242
4272
Mass of Product
Based on 72CrOK
2 72 7
2 7
2 2
2 7 2
2
1 mol K CrO25.0 g K CrO
294.2 g K CrO
3 mol Br 159.8 g Br
1 mol K CrO 1 mol
40.7 g Br
Br
Mass of Product Based on KBr
2 2
2
2
1 mol KBr55.0 g KBr
119.0 KBr
3 mol Br 159.8 g Br
6 mol KBr 1 mol Br
36.9 g Br
Mass of Product
Based on 42SOH
2 42 4
2 4
2 2
2 4 2
2
1 mol H SO66.0 g H SO
98.1gH SO
3 mol Br 159.8 g Br
7 mol H SO 1 mol
41.9 g Br
Br
Chemistry 11, Stoichiometry, Unit 06 23
And so…
Limiting Reactant
The limiting reactant is thus KBr because KBr produces the least amount of (in this case) 2Br .
Mass of
72CrOK based
on KBr
2 2 7
2 2 7
2 2
2 2 7
7
1 mol K Cr O1 mol KBr55.0 g KBr
119.0 g KBr 6 mol KBr
294.2 g K Cr O
122.7 g K Cr O
mol K Cr O
Mass of
42SOH based
on KBr
2 4
4
2 4
2 42
7 mol H SO1 mol KBr55.0 g KBr
119.0 g KBr 6 mol KBr
98.1 g H SO
1 mol H52.9 H SO
SOg
And finally…
Mass of
72CrOK
in excess
2 7mass of K CrO in excess 25.0 g 22.7 g
2.3 g
Mass of
42SOH
in excess
2 4mass of H SO in excess 60.0 g 52.9 g
7.1 g
Chemistry 11, Stoichiometry, Unit 06 24
Chemistry 11, Stoichiometry, Unit 06 25
06 Problems Involving Percentage Yield and Percentage Purity
Usually 100% of the expected amount of products cannot be obtained from a reaction. In these cases chemists usually refer to two terms…
percentage yield and
100expected product of mass
actual product of mass yieldpercentage
percentage purity
100reactant impure of mass
reactant pure of masspurity percentage
Chemistry 11, Stoichiometry, Unit 06 26
Example 15: Percentage Yield
What mass of 32COK is produced when 1.50 grams of 2KO is
reacted with an excess of 2CO according to the reaction
below if the reaction has a 76.0% yield?
(g)3O(s)CO2K(g)2CO(s)4KO 23222
2 3 2 322
2 2
2 3
2 3
2 mol K CO 138.2 g K CO1 mol KO1.50 g KO
71.1 g KO 4 mol KO 1 mol K CO
expecte1.4 d5 mass8 g K CO
actual masspercentage yield
100 expected mass
actual mass percentage yield expected mass
2 32 3actual mass 0.760 1.458 1.11 g g K CO K CO
Chemistry 11, Stoichiometry, Unit 06 27
Example 16: Percentage Purity
If 100.0 grams of FeO produce 12.9 grams of pure Fe according to the reaction…
22 2CO2FeO2C2FeO
What is the percentage purity of the FeO used?
1 mol Fe 2 mol FeO 71.8 g FeO
12.9 g Fe55.8 g Fe 2 mol Fe 1 mol FeO
16.6 g FeO
100reactant impure of mass
reactant pure of masspurity percentage
16.6g FeO purepercentage purity 100
100.
16.6 % pure FeO
0g FeO impure