Review

Stoichiometry Greek for “measuring elements” The calculations of quantities in

chemical reactions based on a balanced equation.

We can interpret balanced chemical equations several ways.

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In terms of Particles Element- atoms Molecular compound (non- metals)-

molecule Ionic Compounds (Metal and non-metal)

- formula unit (ion ratio)

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2H2 + O2 2H2O Two molecules of hydrogen and one

molecule of oxygen form two molecules of water.

2 Al2O3 Al + 3O2

2 formula units Al2O3 form 4 atoms Al

And 3molecules O2

2Na + 2H2O 2NaOH + H2

2 sodium reacts with 2 water to form 2 sodium hydroxide and 1 hydrogen

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Counting in Chemistry We use certain words to indicate number.

Examples:Dozen means a collection of 12Gross means a collection of 144Ream a package of 500 sheets of paperIn Chemistry the particles that we deal with

are very small. To collect a usable sample we need a very large number of particles – 6.02 x 1023. This number is given the name mole (mol).

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Look at it differently 2H2 + O2 2H2O 2 dozen molecules of hydrogen and 1

dozen molecules of oxygen form 2 dozen molecules of water.

2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

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In terms of Moles 2 Al2O3 Al + 3O2

2Na + 2H2O 2NaOH + H2

The coefficients tell us how many moles of each kind

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We can’t count to measure moles!! The particles are too small. What can we do? If you job was to roll pennies, there are three

ways to complete this task:

1. We can count the pennies directly

2. Count by using mass (use average mass of a penny to count 50).

3. Use volume (Plastic rolls are sized to hold 50).

Number 2 & 3 count indirectly by using mass and volume. This is how a chemist counts – uses mass and volume.

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Counting Atoms by mass On the periodic table the average atomic mass for

carbon is 12.01 u. 1 mole of carbon atoms has a mass of 12.01g The mass of I mole is called the molar mass. The molar mass of 1 mole of atoms for any

element is the same numeric value as the atomic mass expressed in grams.

Atomic mass H = 1.01u Molar mass H = 1.01g

Atomic mass O = 16.00u Molar mass O = 16.00g

Atomic mass N = 14.01u Molar mass N = 14.01g9

Conversion factors

“A ratio of equivalent measurements” Start with two things that are the same one meter is one hundred centimeters As an equation 1 m = 100 cm can divide by each side by 100cm to equal the number 1 Called conversion factors because they allow us to

convert (change) units. really just multiplying by one, in a creative way.

100 cm1 m =

100 cm 100 cm = 1

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Conversion Factors 75 cm x 1 m = 0.75 m

100cm

36.03 g C x 1 mole C = 3.000 mole C

12.01g C

2.50 mole C x 12.01 g C = 30.0g C

1 mole C

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In terms of mass The law of conservation of mass applies We can check using moles

2H2 + O2 2H2O

2 moles H22.02 g H2

1 mole H2

= 4.04 g H2

1 moles O232.00 g O2

1 mole O2

= 32.00 g O2

36.04 g H2O2 moles H2O 18.02 g H2O1 mole H2O

=

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Mole to mole conversions 2 Al2O3 Al + 3O2

every time we use 2 moles of Al2O3 we make 3 moles of O2

2 moles Al2O3

3 mole O2

or2 moles Al2O3

3 mole O2

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Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of

Al2O3 decompose?

2 Al2O3 Al + 3O2

3.34 mol Al2O3 2 mol Al2O3

3 mole O2

5.01 moles O2

x mole O2 =

x mole O2 =x 3 mole O2

2 mol Al2O3

3.34 mol Al2O3

=x mole O2

O2

Al2O3

=

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Your Turn 2C2H2 + 5 O2 4CO2 + 2 H2O

If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?

O2 = 5 = x x = 9.60 mol O2

C2H2 2 3.84 mol

How many moles of C2H2 are needed to produce 8.95 mole of H2O?

C2H2 = 2 = x x = 8.95 mol C2H2

H2 O 2 8.95 mol

If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

CO2 = 4 = x x = 4.94 mol CO2

C2H2 2 2.47 mol15

How do you get good at this?

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Mass in Chemical Reactions

How much do you make?

How much do you need?

For example... If 10.1 g of Fe are added to a solution of

Copper (II) Sulfate, how much solid copper would form?

Fe + CuSO4 Fe2(SO4)3 + Cu

2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu

10.1 g Fe55.85 g Fe

1 mol Fe= 0.181 mol Fe

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2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu

Fe 0.181 mol Fe 2 mol Fe3 mol Cu

=

0.272 mol Cu

0.272 mol Cu1 mol Cu

63.55 g Cu= 17.3 g Cu

Cu = x mol Cu =

x mol Cu 0.181 mol Fe x 3 mol Cu

2 mol Fex mol Cu =

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Could have done it

10.1 g Fe55.85 g Fe1 mol Fe

2 mol Fe3 mol Cu

1 mol Cu63.55 g Cu

= 17.3 g Cu

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We can also change Liters of a gas to moles At STP (0ºC and 101.3 kPa) At STP 22.4 L of a gas = 1 mole If 6.45 grams of water are decomposed,

how many liters of oxygen will be produced at STP?

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For Example If 6.45 grams of water are decomposed, how many

liters of oxygen will be produced at STP? 2H2O 2H2 + 1O2

6.45 g H2O 18.02 g H2O1 mol H2O

2 mol H2O1 mol O2

1 mol O2

22.4 L O2

= 0.3579 mol H2O

x mol O2

0.3579 mol H2O= x mol O2 = 0.1790 mol O2

0.1790 mol O2 = 4.01 L O2 at STP

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Your Turn How many liters of CO2 at STP will be

produced from the complete combustion of 23.2 g C4H10 ?

(58.14g/mol) What volume of oxygen will be required?

2C4H10+ 13O2 → 8CO2 + 10H2O

0.399mole C4H10 = 1.60 mol CO2 = 35.8L @STP

= 2.59 mol O2 = 58.1L @STP

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Gases and Reactions

A few more details

Example How many liters of CH4 at STP are required to

completely react with 17.5 L of O2 ?

1CH4 + 2O2 1CO2 + 2H2O

17.5 L O2 22.4 L O2

1 mol O2

2 mol O2

1 mol CH4

1 mol CH4

22.4 L CH4 = 8.75 L CH4 at STP

= 0.7812 mol O2

x mol CH4

0.7812 mol O2 = = 0.3906 mol CH4

0.3906 mol CH4

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Avogadro told us Equal volumes of gas, at the same

temperature and pressure contain the same number of particles.

Moles are numbers of particles You can treat reactions as if they

happen liters at a time, as long as you keep the temperature and pressure the same.

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Example How many liters of CO2 at STP are

produced by completely burning 17.5 L of CH4 ?

1CH4 + 2O2 1CO2 + 2H2O

17.5 L CH4 1 L CH4

1 L CO2 = 17.5 L CO2 at STP

x L CO2 =

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