lesson 01 to 05: stoichiometry - wordpress.com · 2018-03-19 · lesson 01 to 05: stoichiometry...

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.


01 Unit Conversions When converting between mass, gas volume and number of particles (molecules, atoms or ions) you must follow the procedure below…

Important!


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


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…


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


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


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



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


Example 07: Mass – Gas Volume



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


Example 08: Mass - Molecules



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


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


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

http://www.shodor.org/UNChem/glossary.html#molarity

http://www.shodor.org/UNChem/glossary.html#solute

http://www.shodor.org/UNChem/glossary.html#solution


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


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.


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.


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


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


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.


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


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


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


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


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

lesson 01 to 05: stoichiometry - wordpress.com · 2018-03-19 · lesson 01 to 05: stoichiometry...

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