Basic Stoichiometry

Pisgah High SchoolM. Jones

Revision history: 5/16/03, 02/04/12, 04/27/12

Part One

The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

Stoichiometry deals with the amounts of reactants and products in a chemical reaction.

Stoichiometry deals with moles.

Recall that …

1 mole = 22.4 L of any gas at STP

1 mole = the molar mass

molecules1 mole = 6.022 x 1023 atoms or

The word mole is one that represents a very large number.

… “mole” means 6.022 x 1023

Much like “dozen” means 12,

The key to doing stoichiometry is the balanced chemical equation.

2 H2 + O2 2 H2O22

The coefficients give the relative number of atoms or molecules of each reactant or product …as well as the number

of moles.

2 H2 + O2 2 H2O

2 H2 + O2 2 H2O2 moleculesof hydrogen

1 molecule of oxygen

2 moleculesof water

Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.

2 H2 + O2 2 H2O

The balanced chemical equation also gives the smallest integer number of moles.

2 moleculesof hydrogen

1 molecule of oxygen

2 moleculesof water

2 H2 + O2 2 H2O

The balanced chemical equation also gives the smallest integer number of moles.

2 molesof hydrogen

1 moleof oxygen

2 molesof water

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

Two moles of hydrogen combine with

one mole of oxygen to make two moles of water.

How can we show that this is really true?

Consider the combustion of hydrogen in oxygen …

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

What do each of the reactants and product weigh?

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

2 x 2.0 g 1 x 32.0 g 2 x 18.0 g

4.0 g 32.0 g 36.0 g+ =

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

4.0 g 32.0 g 36.0 g+ =

The Law of Conservation of Matter

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

4.0 g 32.0 g 36.0 g+ =

Matter can neither be created nor destroyed, only changed in form.

The oxidation of iron

Consider the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

4 moles 3 moles 2 moles

The coefficients give the ratio of moles.

Consider the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2If these react … then we have the following:

8 moles 6 moles 4 moles2 moles 1.5 moles 1 mole0.50 mol 0.375 mol 0.25 mol

4 moles 3 moles 2 moles

Consider the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

Fe mole 4

O mole 3 Fe mole 5.0 2 0.375 mole O2

The 0.375 moles was not as easy to predict.

Fe mole 4

O mole 3 Fe mole 5.0 2

Consider the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

Use a conversion factor to determine the number of moles of an unknown.

0.375 mole O2

The 0.375 moles was not as easy to predict.

Fe mole 4

O mole 3 Fe mole 5.0 2

Consider the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

The conversion factor comes from the coefficients in the balanced equation.

0.375 mole O2

Back to the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

4 moles 3 moles 2 moles

Calculate the masses of these moles.

Back to the oxidation of iron:

Fe(s) + O2(g) Fe2O3(s)4 3 2

4 moles 3 moles 2 moles4 x 55.8 g 3 x 32.0 g 2 x 159.6 g

319.2 g96.0 g223.2 g + =

Mass is conserved.

The decomposition of ammonium carbonate

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

Suppose 96.0 grams of ammonium carbonate decomposes. How many grams of each of the gases will be produced?

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g

The coefficients tell moles, not grams.

Convert 96.0 g (NH4)2CO3 to moles.

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g

Find the molar mass of ammonium carbonate.

2x14 + 8 +12 + 3x16 = 96.0 g/mol

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g

Isn’t that convenient! We have one mole of ammonium

carbonate.

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g1 mol 1 mol1 mol2 mol

2 moles of ammonia are produced, along with 1 mole of carbon dioxide

and 1 mole of water vapor.

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g

How many grams of each product are formed?

1 mol 1 mol1 mol2 mol

2 x 17.0g 44.0 g 18.0 g

Now consider the decomposition of solid ammonium carbonate.

(NH4)2CO3 2 NH3 + CO2 + H2O

96.0 grams ? g ? g ? g

34.0 g + 44.0 g + 18.0 g =

1 mol 1 mol1 mol2 mol

2 x 17.0g 44.0 g 18.0 g

96.0 g

The reaction between dinitrogen pentoxide and

water

Consider the reaction between dinitrogen pentoxide and water.

What kind of reaction is it?Is it a …

Double replacement reaction?Combustion reaction?Decomposition reaction?Single replacement reaction?

So it must be a synthesis reaction

Consider the reaction between dinitrogen pentoxide and water.

Which kind of synthesis reaction is it?

1. Hydrogen + nonmetal = binary acid2. Metal + nonmetal = salt3. Metal + water = base4. Nonmetal + water = ternary acid

N2O5 + HOH 2 HNO3

HNO3 is a ternary acid; HNO3 is nitric acid

N2O5 + HOH 2 HNO3

Suppose we needed to make 100.0 grams of nitric acid.

How many grams of dinitrogen pentoxide would we need to

react with excess water?

N2O5 + HOH 2 HNO3100.0 g??? g

3

33 HNO g 63.0

HNO mole HNO g 0.100

= 1.59 mole HNO3

1.59 mole0.794 mole

52

5252 ON mole 1

ON g .0108 ON mole 794.0 = 85.7 g N2O5

1.59 mol/2

N2O5 + HOH 2 HNO3100.0 g85.7 g

1.59 mole0.794 mole1.59 mol/2

But. Suppose that we find out that there are only 60.0 grams of dinitrogen pentoxide.

How many grams of nitric acid could we actually make?

N2O5 + HOH 2 HNO3??? g60.0 g

3

33 HNO mol 1.00

HNO g 0.63 HNO mol 11.1 = 70.0 g HNO3

1.11 mole0.556 mole

52

5252 ON g 108

ON mol 1 ON g 0.06 = 0.556 mol N2O5

0.556 mol x 2

Description of stoichiometry problems

Stoichiometry problems will usually take one of the following forms:

1. Mole-mole problem where you might be given moles and asked to find moles of another substance.

2. Mole-mass problem where you might be given moles and asked find the mass of another substance.

3. Mass-mass problem where you might be given a mass and asked to find the mass of another substance.

4. Mass-volume problem where you might be given a mass and asked to find the volume of a gas.

5. Volume-volume problem where you might be given a volume and asked to find another volume.

Volume-volume stoichiometry problems are easiest when you use Gay-Lussac’s Law.

“The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.”

Simply put, Gay-Lussac’s Law says this:

The volumes of the gases are in the same ratio as the coefficients in the balanced equation.

2 H2 + O2 2 H2O2 moles 1 mole 2 moles2 L 1 L 2 L

Applications of Gay-Lussac’s Law

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

1 mol 5 mols 3 mols 4 mols

Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required?

17.5 L O2

1 L 5 L 3 L 4 L

HC L 1

O L 5 HC L .53

83

283

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

1 mol 5 mols 3 mols 4 mols

Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O2

How many liters of carbon dioxide gas and water vapor at STP would be produced?

10.5 L CO2 and 14.0 L H2O

1 L 5 L 3 L 4 L

CH4(g) + 4 Cl2(g) CCl4(g) + 4 HCl(g)

When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced.

How many L of methane will react with 0.800 L of chlorine gas at STP?

Cl L 4

CH L 1 Cl L 800.0

2

42 0.200 L Cl2

Stoichiometry problems involving gases

Find the mass of HOCl that can be produced when 2.80 L of chlorine gas at STP reacts with excess hydrogen peroxide.

Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g

Convert 2.80 L of Cl2 gas at STP to moles

Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g

Cl L 22.4

Cl mol 1 Cl L .802

2

22 0.125 mol Cl2

0.125 mol.125 x 2

0.250 mol

HOCl mol 1

HOCl g 52.5 HOCl mol 250.0 13.1g

HOCl

The reaction between copper and nitric acid

Will copper dissolve in acids?

Cu + 2HCl CuCl2 + 2H2 (g) No Reaction

Most metals react with HCl to produce a metal chloride solution and H2 gas.

Not copper

Consider hydrochloric acid

Will copper dissolve in acids?

Cu + 2HCl CuCl2 + 2H2 (g) No Reaction

Consider hydrochloric acid

Copper is below hydrogen in the activity series. Copper metal will only replace elements that are below it in the activity series.

What about other acids?

The same is true for all acids except nitric acid

Cu + HBr NRCu + HI NRCu + HF NRCu + H2SO4 NRCu + HC2H3O2 NR

Nitric acid is the only acid that will dissolve copper.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

A beaker contains a penny and some nitric acid is added.

Nitric acid is the only acid that will dissolve copper.

The penny begins to disappear and the solution turns blue-green and a brown gas is given off.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid is the only acid that will dissolve copper.

The gas produced in the reaction is NO, which is colorless. Reddish brown NO2 forms when NO reacts with the oxygen in the air.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid is the only acid that will dissolve copper.

The penny is gone and the solution turns a dark blue. The brown NO2 gas escapes from the open beaker.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid is the only acid that will dissolve copper.

Calculate the volume of NO gas at STP when 20.0 grams of copper dissolves.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid is the only acid that will dissolve copper.

20.0 g ??? L

Cu g 63.5

Cu mol 1Cu g 0.20 0.315 mol Cu

0.315 mol 0.210 mol x 0.667

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid is the only acid that will dissolve copper.

20.0 g

NO mol 1

NO L 22.4 NO mol 0.210 4.70 L NO

0.315 mol

??? L

0.210 mol

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

x 0.667

Part Two of Basic Stoichiometry

will include the gas laws.