Love, Life and Love, Life and StoichiometryStoichiometry

StoichiometryStoichiometry

• Greek for “measuring Greek for “measuring elements”elements”

• The calculations of quantities in The calculations of quantities in chemical reactions based on a chemical reactions based on a balanced equation.balanced equation.

• We can interpret balanced chemical We can interpret balanced chemical equations several ways.equations several ways.

In terms of ParticlesIn terms of Particles

• Element- atomsElement- atoms

• Molecular compound (non- metals)- Molecular compound (non- metals)- moleculemolecule

• Ionic Compounds (Metal and non-Ionic Compounds (Metal and non-metal) - formula unit metal) - formula unit

2H2H2 2 + O+ O22 2H2H22O O

• Two molecules of hydrogen and one Two molecules of hydrogen and one molecule of oxygen form two molecules of molecule of oxygen form two molecules of water.water.

• 2 Al2 Al22OO33 AlAl ++ 3O3O22

2 formula units Al2O3 form 4 atoms Al

and 3 molecules O2

2Na + 2H2O 2NaOH + H2

Look at it differentlyLook at it differently•2H2H2 2 + O+ O22 2H2H22OO

• 2 dozen molecules of hydrogen and 1 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen dozen molecules of oxygen form 2 dozen molecules of water.molecules of water.

• 2 x (6.02 x 102 x (6.02 x 102323) molecules of hydrogen ) molecules of hydrogen and 1 x (6.02 x 10and 1 x (6.02 x 102323) molecules of ) molecules of oxygen form 2 x (6.02 x 10oxygen form 2 x (6.02 x 102323) molecules ) molecules of water.of water.

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

In terms of In terms of MolesMoles

• 2 Al2 Al22OO33 AlAl ++ 3O3O22

• 2Na + 2H2Na + 2H22O O 2NaOH + H 2NaOH + H22

• The coefficients tell us how many The coefficients tell us how many moles of each kindmoles of each kind

In terms of massIn terms of mass• The law of conservation of mass appliesThe law of conservation of mass applies

• We can check using molesWe can check using moles

•2H2H2 2 + O+ O22 2H2H22OO

2 moles H22.02 g H2

1 moles H2= 4.04 g H2

1 moles O232.00 g O2

1 moles O2

= 32.00 g O2

36.04 g H236.04 g Reactants

Therefore…Therefore…

•2H2H2 2 + O+ O22 2H2H22OO

2 moles H2O18.02 g H2O

1 mole H2O= 36.04 g H2O

2H2 + O2 2H2O

36.04 g H2 + O2= 36.04 g H2O

No mass has been created or destroyed

Your turnYour turn

• Show that the following equation Show that the following equation follows the Law of conservation of follows the Law of conservation of mass.mass.

• 2 Al2 Al22OO33 AlAl ++ 3O3O22

Mole to mole conversionsMole to mole conversions• 2 Al2 Al22OO33 AlAl ++ 3O3O22

• every time we use 2 moles of Alevery time we use 2 moles of Al22OO33 we make 3 we make 3 moles of Omoles of O22

2 moles Al2O3

3 mole O2

or2 moles Al2O3

3 mole O2

Mole to Mole conversionsMole to Mole conversions• How many moles of OHow many moles of O22 are produced when 3.34 moles of Al are produced when 3.34 moles of Al22OO33

decompose?decompose?

• 2 Al2 Al22OO33 AlAl ++ 3O3O22

3.34 moles Al2O3 2 moles Al2O3

3 mole O2 = 5.01 moles O2

Your TurnYour Turn

• 2C2C22HH22 + 5 O + 5 O22 4CO 4CO22 + 2 H + 2 H22OO

• If 3.84 moles of CIf 3.84 moles of C22HH2 2 are burned, how are burned, how

many moles of Omany moles of O22 are needed? are needed?

• How many moles of CHow many moles of C22HH2 2 are needed to are needed to

produce 8.95 mole of Hproduce 8.95 mole of H22O?O?

• If 2.47 moles of CIf 2.47 moles of C22HH2 2 are burned, how are burned, how

many moles of COmany moles of CO22 are formed? are formed?

How do you get good at How do you get good at this?this?

Stoichiometry Practice

• Given the following equation:

2 KClO3 2KCl + 3 O2

• How many moles of O2 can be produced by letting 12.00 moles of KClO3 react?

2 KClO3 2KCl + 3 O2

The strategy will be to convert to . Remember, conversion factors are what you over what you .

moles of O2

want

moles of KClO3

have

KClO3 2KCl + O2

12.00 moles of KClO3 moles of O2

moles of KClO3

2 3

= 12 x (3/2)

= 18

------------------

------------------

• If ammonia, NH3, is burned in air, the following reaction takes place:

4NH3 + 3O2 2 N2 + 6H2O

Given that you started with 51.0 g of NH3, how many grams of water will be produced?

4NH3 + 3O2 2 N2 + 6H2O

• The overall strategy will be to convert the of ammonia of ammonia, then convert the of ammonia

of water, and then finally convert the of water of water.

grams to molesmoles

to molesmoles to grams

4NH3 + 3O2 2 N2 + 6H2O

grams NH3 moles NH3 moles H2O grams of H2O

51.0 g NH3 1 mole NH3

17 g NH3

6 mole H2O

4 mole NH3

18 g H2O

1 mole H2O

= 81 g of H2O

• 20.0 g of silver(I)nitrate is reacted with an excess of sodium choride to produce silver(I) chloride

AgNO3 + NaCl => AgCl + NaNO3

What mass of silver(I) chloride is produced?

AgNO3 + NaCl => AgCl + NaNO3

• The overall strategy will be to convert the of silver nitrate of silver nitrate , then convert the of silver nitrate of silver chloride, and then finally convert the of silver chloride of silver chloride.

grams to molesmoles

to molesmoles

to grams

AgNO3 + NaCl => AgCl + NaNO3

20.0 g AgNO3 1 mole AgNO3

169.9 g AgNO3

1 mole AgCl

1 mole AgNO3

143.4 g AgCl

1 mole AgCl

= 16.9

grams AgNO3 moles AgNO3moles AgNO3 moles AgClmoles AgCl grams AgCl

------------------

----------------

------------------

------------------

------------------

------------------

g AgCl

Now go to mass to mass problems

Mass in Chemical Mass in Chemical Reactions Reactions

How much do you make?How much do you make?

How much do you need?How much do you need?

We can’t measure moles!!We can’t measure moles!!

• What can we do?What can we do?

• We can convert grams to moles.We can convert grams to moles.

• Periodic TablePeriodic Table

• Then do the math with the moles.Then do the math with the moles.

• Balanced equationBalanced equation

• Then turn the moles back to grams.Then turn the moles back to grams.

• Periodic tablePeriodic table

For example...For example...

• If 10.1 g of Fe are added to a solution If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid of Copper (II) Sulfate, how much solid copper would form?copper would form?

• Fe + CuSOFe + CuSO44 Fe Fe22(SO(SO44))33 + Cu + Cu

• 2Fe + 3CuSO2Fe + 3CuSO44 Fe Fe22(SO(SO44))33 + 3Cu + 3Cu 10.1 g Fe

55.85 g Fe

1 mol Fe= 0.181 mol Fe

2Fe + 3CuSO2Fe + 3CuSO44 Fe Fe22(SO(SO44))33 + 3Cu + 3Cu

0.181 mol Fe2 mol Fe

3 mol Cu= 0.272 mol Cu

0.272 mol Cu1 mol Cu63.55 g Cu

= 17.3 g Cu

Could have done itCould 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

More ExamplesMore Examples• To make silicon for computer chips they use To make silicon for computer chips they use

this reactionthis reaction

• SiClSiCl44 + 2Mg + 2Mg 2MgCl 2MgCl22 + Si + Si

• How many grams of Mg are needed to make How many grams of Mg are needed to make 9.3 g of Si?9.3 g of Si?

• How many grams of SiClHow many grams of SiCl44 are needed to make are needed to make

9.3 g of Si?9.3 g of Si?

• How many grams of MgClHow many grams of MgCl2 2 are produced along are produced along

with 9.3 g of silicon?with 9.3 g of silicon?

For ExampleFor Example• The U. S. Space Shuttle boosters use this The U. S. Space Shuttle boosters use this

reactionreaction

• 3 Al(s) + 3 NH3 Al(s) + 3 NH44ClOClO44

Al Al22OO33 + AlCl + AlCl33 + 3 NO + 6H + 3 NO + 6H22OO

• How much Al must be used to react with How much Al must be used to react with 652 g of NH652 g of NH44ClOClO44 ? ?

• How much water is produced?How much water is produced?

• How much AlClHow much AlCl33??

We can also changeWe can also change

• Liters of a gas to molesLiters of a gas to moles

• At STPAt STP– 25ºC and 1 atmosphere pressure25ºC and 1 atmosphere pressure

• At STP 22.4 L of a gas = 1 moleAt STP 22.4 L of a gas = 1 mole

• If 6.45 moles of water are If 6.45 moles of water are decomposed, how many liters of decomposed, how many liters of oxygen will be produced at STP?oxygen will be produced at STP?

For ExampleFor Example• If 6.45 grams of water are decomposed, how If 6.45 grams of water are decomposed, how

many liters of oxygen will be produced at STP?many liters of oxygen will be produced at STP?

• HH22O O H H22 + O + O22

6.45 g H2O 18.02 g H2O1 mol H2O

2 mol H2O1 mol O2

1 mol O2

22.4 L O2

Your TurnYour Turn

• How many liters of COHow many liters of CO22 at STP will be at STP will be

produced from the complete produced from the complete combustion of 23.2 g Ccombustion of 23.2 g C44HH1010 ? ?

• What volume of oxygen will be What volume of oxygen will be required?required?

Gases and ReactionsGases and Reactions

A few more detailsA few more details

ExampleExample• How many liters of CHHow many liters of CH4 4 at STP are required to at STP are required to

completely react with 17.5 L of Ocompletely react with 17.5 L of O2 2 ??

• CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO

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

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

Avagadro told usAvagadro told us

• Equal volumes of gas, at the same Equal volumes of gas, at the same temperature and pressure contain temperature and pressure contain the same number of particles.the same number of particles.

• Moles are numbers of particlesMoles are numbers of particles

• You can treat reactions as if they You can treat reactions as if they happen liters at a time, as long as happen liters at a time, as long as you keep the temperature and you keep the temperature and pressure the same. pressure the same.

ExampleExample

• How many liters of COHow many liters of CO2 2 at STP are at STP are

produced by completely burning 17.5 produced by completely burning 17.5 L of CHL of CH4 4 ??

• CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO17.5 L CH4

1 L CH4 2 L CO2

= 35.0 L CH4

Limiting ReagentLimiting Reagent• If you are given one dozen loaves of If you are given one dozen loaves of

bread, a gallon of mustard and three bread, a gallon of mustard and three pieces of salami, how many salami pieces of salami, how many salami sandwiches can you make.sandwiches can you make.

• The limiting reagent is the reactant The limiting reagent is the reactant you run out of first.you run out of first.

• The excess reagent is the one you The excess reagent is the one you have left over.have left over.

• The limiting reagent determines how The limiting reagent determines how much product you can makemuch product you can make

How do you find out?How do you find out?

• Do two stoichiometry problems.Do two stoichiometry problems.

• The one that makes the least product The one that makes the least product is the limiting reagent.is the limiting reagent.

• For exampleFor example

• Copper reacts with sulfur to form Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much reacts with 3.83 g S how much product will be formed? product will be formed?

• If 10.6 g of copper reacts with 3.83 g If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be S. How many grams of product will be formed?formed?

• 2Cu + S 2Cu + S Cu Cu22SS

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

Your turnYour turn

• If 10.1 g of magnesium and 2.87 L of If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters HCl gas are reacted, how many liters of gas will be produced?of gas will be produced?

• How many grams of solid?How many grams of solid?

• How much excess reagent remains?How much excess reagent remains?

Your Turn IIYour Turn II

• If 10.3 g of aluminum are reacted If 10.3 g of aluminum are reacted with 51.7 g of CuSOwith 51.7 g of CuSO44 how much how much copper will be produced?copper will be produced?

• How much excess reagent will How much excess reagent will remain?remain?

Yield Yield • The amount of product made in a chemical The amount of product made in a chemical

reaction.reaction.

• There are three typesThere are three types

• Actual yieldActual yield- what you get in the lab when - what you get in the lab when the chemicals are mixedthe chemicals are mixed

• Theoretical yieldTheoretical yield- what the balanced - what the balanced equation tells you you should make.equation tells you you should make.

• Percent yieldPercent yield = Actual x 100 % = Actual x 100 % Theoretical Theoretical

ExampleExample

• 6.78 g of copper is produced when 6.78 g of copper is produced when 3.92 g of Al are reacted with excess 3.92 g of Al are reacted with excess copper (II) sulfate.copper (II) sulfate.

• 2Al + 3 CuSO2Al + 3 CuSO44 Al Al22(SO(SO44))33 + 3Cu + 3Cu

• What is the actual yield?What is the actual yield?

• What is the theoretical yield?What is the theoretical yield?

• What is the percent yield?What is the percent yield?

DetailsDetails

• Percent yield tells us how “efficient” Percent yield tells us how “efficient” a reaction is.a reaction is.

• Percent yield can not be bigger than Percent yield can not be bigger than 100 %.100 %.

Homework!!Homework!!