Chapter 11

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.

Stoichiometry Look at it as if your were cooking…1 egg + 3 c. flour + 1c. water 24 biscuitsThe egg, flour, water and biscuits can be

viewed in comparison to each other.To make 24 biscuits 3 c. flour are needed,

how much flour is needed to make 36 biscuits?

36 biscuits x 3 c. flour = 4.5 c. flour 24 biscuits

Remember your particles Elements- atoms Molecular compounds (non- metals)-

molecules Ionic Compounds (Metal and non-metal)

- formula units

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

molecule of oxygen form two molecules of water.

2 H2 + O2 2 H2O

2 Al2O3 Al + 3O2

2 formula units of Al2O3 form 4 atoms of Al and 3 molecules of O2

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 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form

2 x (6.02 x 1023) molecules of water. 2 moles of hydrogen and 1 mole of oxygen form

2 moles of water.

In terms of Moles 2 Al2O3 Al + 3O2 2 moles of aluminum oxide produce 4

moles of aluminum and 3 moles of oxygen

2Na + 2H2O 2NaOH + H2 2 moles of sodium react with 2 moles of

water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen

The coefficients tell us how many moles of each kind

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

conversion 2H2 + O2 2H2O

2 moles H2

2.02 g H2

1 moles H2

= 4.04 g H2

1 moles O2

32.00 g O2

1 moles O2

= 32.00 g O2

36.04 g H2&O2

In terms of mass 2H2 + O2 2H2O

2 moles H2O18.02 g H2O1 mole H2O

= 36.04 g H2O

2H2 + O2 2H2O

36.04 g (H2 + O2) = 36.04 g H2O

Your turn Show that the following equation follows

the Law of conservation of mass. 2 Al2O3 Al + 3O2

g Al2O3 = g Al + g O2

Your turn Show that the following equation follows

the Law of conservation of mass. 2 Al2O3 Al + 3O2

203.92 g Al2O3 =4x26.9815386 g Al + 3x 32.00 g O2

203.92 g = 203.92 g(rounding makes it be a little off)

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

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

Al2O3 decompose? 2 Al2O3 Al + 3O2

3.34 moles x Al2O3 2 moles Al2O3

3 mole O2 = 5.01 moles O2

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

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

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

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

Your Turn 2C2H2 + 5 O2 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of

O2 are needed?

3.84 mol C2H2 x 5 mol O2 = 9.60 mol O2

2 mol C2H2

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

8.95 mol H2O x 2 mol C2H2 = 8.95 mol C2H2

2 mol H2O If 2.47 moles of C2H2 are burned, how many moles of

CO2 are formed?

2.47 mol C2H2 x 4 mol CO2 = 4.94 mol CO2

2 mol C2H2

How do you get good at this?

Mass in Chemical Reactions

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

We can’t measure moles!! What can we do? We can convert grams to moles. Use periodic table Then do the math with the moles. Use balanced equation Then turn the moles back to grams. Use periodic table again

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

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

Fe + CuSO4 Fe2(SO4)3 + Cu

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

10.1 g Fe55.845 g Fe1 mol Fe = 0.181 mol Fe

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

0.181 mol Fe2 mol Fe3 mol Cu

= 0.272 mol Cu

0.272 mol Cu 1 mol Cu63.546 g Cu

= 17.2 g Cu

Could have done it

10.1 g Fe55.845g Fe1 mol Fe

2 mol Fe3 mol Cu

1 mol Cu63.546 g Cu

= 17.2 g Cu

More Examples To make silicon for computer chips they

use this reaction SiCl4 + 2Mg 2MgCl2 + Si How many grams of Mg are needed to

make 9.3 g of Si? How many grams of SiCl4 are needed to

make 9.3 g of Si? How many grams of MgCl2 are produced

along with 9.3 g of silicon?

More Examples answersSiCl4 + 2Mg 2MgCl2 + Si

How many grams of Mg are needed to make 9.3 g of Si?9.3 g Si x 1 mol Si x 2 mol Mg x 24.3050 g Mg =16 g Mg 28.0855 g Si 1 mol Si 1 mol MgHow many grams of SiCl4 are needed to make 9.3 g of Si?

9.3 g Si x 1 mol Si x 1 mol SiCl4 x 169.90 g SiCl4 = 56 g SiCl4

28.0855 g Si 1 mol Si 1 mol SiCl4

How many grams of MgCl2 are produced along with 9.3 g of silicon?

9.3 g Si x 1 mol Si x 2 mol MgCl2 x 95.21 g MgCl2 = 63 g MgCl2

28.0855 g Si 1 mol Si 1 mol MgCl2

For Example The U. S. Space Shuttle boosters use

this reaction 3 Al(s) + 3 NH4ClO4 Al2O3 + AlCl3

+ 3 NO + 6H2O How much Al must be used to react with

652 g of NH4ClO4 ? How much water is produced? How much AlCl3?

For Example answers The U. S. Space Shuttle boosters use this reaction 3 Al(s) + 3 NH4ClO4 Al2O3 + AlCl3 + 3 NO + 6H2O How much Al must be used to react with 652 g of NH4ClO4 ?

652 g NH4ClO4 x 1 mol NH4ClO4 x 3 mol Al x 26.9815386 g Al

117.49 g 3 molNH4ClO4 1 mol Al

= 150. g Al

How much water is produced?652 g NH4ClO4 x 1 mol NH4ClO4 x 6 mol H2O x 18.02 g H2O

117.49 g 3 molNH4ClO4 1 mol H2O

= 200. g H2O

How much AlCl3?

652 g NH4ClO4 x 1 mol NH4ClO4 x 1 mol AlCl3 x 133.34 g AlCl3

117.49 g 3 molNH4ClO4 1 mol AlCl3

= 247 g AlCl3

We can also change Liters of a gas to moles At STP At STP 22.4 L of a gas = 1 mole If 6.45 moles of water are decomposed,

how many liters of oxygen will be produced at STP?

For Example If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? H2O H2 + O2

6.45 g H2O 18.02 g H2O1 mol H2O

2 mol H2O1 mol O2

1 mol O2

22.4 L O2

= 4.01 L O2

Your Turn How many liters of CO2 at STP will be

produced from the complete combustion of 23.2 g C4H10 ?

What volume of oxygen will be required?

Your Turn answers How many liters of CO2 at STP will be produced

from the complete combustion of 23.2 g C4H10 ?

2C4H10 + 13O2 8CO2 + 10H2O

23.2 g C4H10 x 1 mol C4H10 x 8 mol CO2 x 22.4L CO2

58.12 g 2 mol C4H10 1 mol O2

= 4.93 L CO2

What volume of oxygen will be required?23.2 g C4H10 x 1 mol C4H10 x 13 mol O2 x 22.4L O2

58.12 g 2 mol C4H10 1 mol O2

= 58.1 L O2

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 ? CH4 + 2O2 CO2 + 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

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

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.

Example How many liters of CO2 at STP are

produced by completely burning 17.5 L of CH4 ?

CH4 + 2O2 CO2 + 2H2O

17.5 L CH4 1 L CH4 2 L CO2 = 35.0 L CH4

Limiting reactant If you are given one dozen loaves of bread, a

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

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

The excess reactant is the one you have left over.

The limiting reactant determines how much product you can make.

How do you find out? Do two stoichiometry problems. Calculate how many moles of each

reactant is needed for the reaction to occur.

For example Copper reacts with sulfur to form copper

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

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

2Cu + S Cu2S

10.6 g Cu 63.546g Cu 1 mol Cu

3.83 g S 32.065g S 1 mol S

= 0.167 mol Cu

= 0.119 mol S

Now, figure out if you have enough S to react with the Cu available.

HAVE NEED 0.167 mol Cu 2 mol Cu 0.119 mol S 1 mol S Simplify the ratio… 1.40 mol Cu 2 mol Cu

1 mol S 1 mol S You have 1.40 mol Cu and you need 2

mol of Cu, so….

Cu is Limiting

Reactant

Limiting reactant Once you figure out which one is the limiting

reactant, use that one to calculate your answer of how much product is formed.

10.6 g Cu x 1 mol Cu x 1 mol Cu2S x 159.16 g Cu2S

63.546 g Cu 2 mol Cu 1 mol Cu2S

= 13.3 g Cu2S

Mg (s) + 2HCl (g) MgCl2 (s) + H2 (g) If 10.1 g of magnesium and 4.66 g of

HCl gas are reacted, what is the limiting reactant?

How many grams of solid will be produced?

Mg (s) + 2HCl (g) MgCl2 (s) + H2 (g) If 10.1 g of magnesium and 4.66 g of HCl gas are reacted, what is

the limiting reactant?10.1 g Mg x 1 mol Mg = .416 mol Mg

24.3050gMg4.66 g HCl x 1 mol HCl = 0.128 mol HCl

36.46 g HClHAVE NEED0.416 mol Mg 1 mol Mg0.128 mol HCl 2 mol HCl= 3.25 mol Mg 0.5 mol Mg 1 mol HCl 1 mol HCl How many grams of solid will be produced?4.66 g HCl x 1 mole HCl x 1 mol MgCl2 x 95.21 g MgCl2 = 6.08 g

36.36 g HCl 2 mol HCl 1 mole MgCl2

MgCl2

>HCl is limiting

Your Turn If 10.3 g of aluminum are reacted with 51.7 g

of CuSO4 how much copper will be produced?

2 Al + 3 CuSO4 3 Cu + Al2(SO4)310.3 g Al x 1 mole Al = 0.381 mol Al 26.9815386 g Al

51.7 g CuSO4 x 1 mole CuSO4 = 0.324 mol CuSO4

159.61 g CuSO4

HAVE NEED0.381 mol Al 2 mol Al0.324 mol CuSO4 3 mol CuSO4

=1.18 mol Al 0.67 mol Al 1 mol CuSO4 1 mol CuSO4

CuSO4 is limiting>

Yield The amount of product made in a

chemical reaction. There are three types Actual yield- what you get in the lab

when the chemicals are mixed Theoretical yield- what the balanced

equation tells you you should make. Percent yieldPercent yield = Actual x 100 %

Theoretical

Example 6.78 g of copper is produced when 3.92

g of Al are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield?

Example answers 6.78 g of copper is produced when 3.92 g of Al

are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? (6.78 g Cu) What is the theoretical yield?3.92 g Al x 1 mol Al x 3 mol Cu x 63.546 g Cu =

26.981538 g Al 2 mol Al 1 mol Cu= 13.8 g

Cu What is the percent yield? 6.78 g Cu x 100 = 49.13% 13.8 g Cu

Details Percent yield tells us how “efficient” a

reaction is. Percent yield can be bigger than 100 %

if your product is contaminated.