Unit 2

Moles, Chemical Reactions, and Stoichiometry

The Mole

• A mole (mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12.

• It is a counting unit, similar to a dozen. In a dozen, there are 12 things. In a mole, there are 6.02 x 1023 things.

Visual Concept

Molar Mass• Molar Mass is the mass of one mole of a pure

substance.• Molar mass units

are g/mol.• The molar mass of

an element is the same number as its atomic mass, only the units are different.

• Examples: Molar mass of H = 1.0 g/mol Molar mass of O = 16.0 g/mol

Formula Mass

• The formula mass of any compound is the sum of the masses of all the atoms in its formula.Example: Formula mass of water, H2O:

H2 = 1.0 amu x 2 = 2.0 amu. O = + 16.0 amu. 18.0 amu

• A compound’s molar mass is numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H2O = 18.0 g.)

Molar MassesSample Problem

Determine the molar mass of each of the following compounds:

a.Al2S3

b.Ba(OH)2

Solution:Al2 = 27.0 x 2 = 54.0 g S3 = 32.1 x 3 = + 96.3 g

150.3 g

Ba = 137.3 g O2 = 16.0 x 2 = 32.0 g H2 = 1.0 x 2 = + 2.0 g

171.3 g

Percent Composition

• The percentage by mass of each elementin a compound is known as the percent composition of the compound.

% of element = mass of element in compoundmolar mass of compound

x 100

Percent CompositionSample Problem

Find the percentage composition of copper(I) sulfide, Cu2S.

1.Find the molar mass of Cu2S:

2.Find the percentage by mass of each element:

Solution: Cu2 = 63.5 x 2 = 127.0 g S = + 32.1 g

159.1 g

% Cu = 127.0 g159.1 g x 100 = 79.80% Cu

% S = 32.1 g159.1 g x 100 = 20.2% S

Molar Mass as a Conversion Factor

• The molar mass of a compound can be used as a conversion factor to convert between moles and grams for a given substance.

Example:• What is the mass of 2.5 moles of H2O?

molar mass of H2O = 18.0 g/molgiven conversion factor

2.5 mol H2O g H2Omol H2O

45 g H2O=x118.0

2

Molar Mass as a Conversion FactorSample Problem

Calculate the moles in 1170 g of copper (II) nitrate.Solution:

Given Conversion factor

1170 g Cu(NO3)2mol Cu(NO3)2

g Cu(NO3)2

6.24 mol Cu(NO3)2

=x187.5

1

Cu = 63.5 g

O6 = 16.0 x 6 = + 96.0 g

187.5 g

(NO3)Cu2+

1. Determine the correct formula:2. Calculate the molar mass:

-Cu(NO3)2

N2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol:

Avogadro’s Number

• Avogadro’s number: 6.022 1415 × 1023— the number of particles in exactly one mole of a pure substance.

• Named for nineteenth-century Italian scientist Amedeo Avogadro.

Visual Concept

Conversions with Avogadro’s Number

•Avogadro’s number can be used as a conversion factor between number of particles (atoms or molecules) and moles.

Example:• How many moles of silver, Ag, are in 3.01 1023

atoms of silver?

GivenConversion factor

3.01 x 1023 atoms Ag mol Agatoms Ag

0.500 mol Ag=x6.02 x 1023

1

Combining Conversion FactorsSample Problem

What is the mass in grams of 1.20 108 atoms of copper, Cu?Solution:

GivenConversion factor

1.20 x 108 atoms Cu mol Cuatoms Cu

1.27 x 10-14 g Cu=

x g Cumol Cu

x

2nd

Conversionfactor

1st

6.02 x 1023

1 163.5

Standard Molar Volume

• Standard Temperature and Pressure (STP) is 0oC and 1 atm.

• The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

Molar Volume Conversion Factor

• Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

Molar Volume ConversionSample Problem

a. What quantity of gas, in moles, is contained in 5.00 L at STP?

b. What volume does 0.768 moles of a gas occupy at STP?

5.00 LL

mol22.41x = 0.223 mol

0.768 molmol

L1

22.4x = 17.2 L

The Mole Map

• You can now convert between number of particles, mass (g), and volume (L) by going through moles.

• A chemical reaction is the process by which one or more substances are changed into one or more different substances.

• Signs that a chemical reaction is taking place:1. Release of energy as heat and/or light.2. Production of a gas. 3. Formation of a precipitate - a solid

that separates from a liquid solution.4. Color change.

Chemical Reactions

• In any chemical reaction, the original substances are called reactants and the resulting substances are called products.

• According to the law of conservation of mass, the total mass of reactants must equal the total mass of products for any given chemical reaction.

The Law of Conservation of Mass

• A chemical equation represents a chemical reaction using symbols and formulas.

Example:

Chemical Equations

2H2O(l) 2H2(g) + O2(g)Reactant Products

• Oxygen gas (O2) is an example of an element that normally exists as a diatomic molecule. You need to memorize all seven:

Diatomic Molecules

• The final step in writing correct chemical equations is to make sure the law of conservation of mass is satisfied.– The numbers and types of atoms on both sides

of the equation must be the same – this is called balancing an equation.

– Equations are balanced by inserting coefficients - whole numbers that appear in front of formulas in a chemical equation.

Balancing Equations

Balance the following equation: CH4(g) + O2(g) CO2(g) + H2O(g)Solution:•Start with the easiest element…carbon.

– Carbon is already balanced.•Next count the hydrogen atoms.

– Two more hydrogen atoms are needed on the right.•Finally, count oxygen atoms.

– There are 4 oxygens on the right side of the equation, but only two on the left.

– Add a coefficient 2 in front of the O2 on the left.

Balancing EquationsSample Problem A

22

Balance the following equation: Al4C3(s) + H2O(l) CH4(g) + Al(OH)3(s)Solution:•Let’s start with aluminum.

– Add a coefficient 4 to Al(OH)3 on the right.

•Next count the carbon atoms.– Add a coefficient 3 to CH4 on the right.

•Balance the oxygen atoms.– Add a 12 to the H2O on the left.

•Lastly, count the hydrogen atoms.– Hydrogen is already balanced.

Balancing EquationsSample Problem B

3 412

Write an equation for the reaction that occurs when solid copper metal reacts with aqueous silver nitrate to produce

solid silver metal and aqueous copper(II) nitrate.Solution:•First, use correct formulas and symbols to write a chemical equation.•Then, balance your equation.

Writing Chemical EquationsSample Problem

Cu(s) + AgNO3(aq) Ag(s) + Cu(NO3)2(aq)22

• There are 5 basic types of chemical reactions:1. Synthesis2. Decomposition3. Single-Displacement4. Double-Displacement5. Combustion

Types of Chemical Reactions

• In a synthesis reaction (also called a composition reaction) two or more substances combine to form a new compound.

• This type of reaction is represented by the following general equation:

A + X AX

Synthesis Reactions

• In a decomposition reaction, a single compound breaks apart to form 2 or more simpler substances.

• Decomposition is the opposite of synthesis.• This type of reaction is represented by the following

general equation:

AX A + X

Decomposition Reactions

• In a single-displacement reaction (also called single-replacement) one element replaces a similar element in a compound.

• They often take place in aqueous solution.• This type of reaction is represented by the

following general equation:

A + BX AX + B

Single-Displacement Reactions

• In double-displacement reactions, the ionsof 2 compounds exchange places in an aqueous solution to form 2 new compounds.

• One of the compounds formed is usually either a precipitate, a gas, or water.

• Represented by the following general equation:

AX + BY AY + BX

Double-Displacement Reactions

• In a combustion reaction, a hydrocarbon fuel combines with oxygen, releasing a large amount of energy in the form of light and heat.

• Products of combustion reactions are always carbon dioxide and water vapor.

• Example: Combustion of propaneC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

Combustion Reactions

Classify each of the following reactions as a synthesis, decomposition, single-displacement, double-displacement, or combustion reaction.

a.N2(g) + 3H2(g) → 2NH3(g)

b.2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

c.2NaNO3(s) → 2NaNO2(s) + O2(g)

d.2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l)

Types of ReactionsSample Problem

synthesis

single-displacement

decomposition

combustion

What is Stoichiometry?

• Stoichiometry is the branch of chemistry that deals withthe mass relationships of elements in a chemical reaction.

• Stoichiometry calculationsalways start with a balanced chemical equation.

Mole Ratio

• A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reactionExample: 2Al2O3(l) → 4Al(s) + 3O2(g)

Mole Ratios:

2 mol Al2O3 2 mol Al2O3 4 mol Al

4 mol Al 3 mol O2 3 mol O2

Mole to Mole Conversions• Using the mole ratio, you can convert from moles

of one substance to moles of any other substance in a chemical reaction.

Example:• For the reaction N2 + 3H2 → 2NH3, how many moles

of H2 are required to produce 12 moles of NH3?This is whatwe’re given:

Use mole ratio as a conversion factor

12 mol NH33 mol H2

2 mol NH3

18 mol H2=x

Mole to Mass Conversions• To convert from moles of one substance to grams

of another, you need 2 conversion factors: 1. mole ratio.2. molar mass of the unknown.

• To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.

Example:Given the equation: 2Mg(s) + O2(g)→ 2MgO(s),

Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium.

mol MgO

mol MgO

Mole to Mass Conversions (continued)

GivenMole Ratio

2.00 mol Mgmol Mg

80.6 g MgO=x x

2nd C.F.Molar Mass

of the Unknown

1st C.F.

2

2

1

g MgO40.3

Mass to Mole Conversions• To convert from grams of one substance to moles of

another, you need 2 conversion factors: 1. molar mass of the given.2. mole ratio.

• Mass to mole conversion factors are the inverse of mole to mass conversion factors.

Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),

How many moles of HgO are needed to produce 125g of O2?

mol O2

mol O2

Mass to Mole Conversions (continued)

Given

Molar Massof the Given

125 g O2

g O2

7.81 mol HgO=x x

2nd C.F.Mole Ratio

1st C.F.

32.0

1

1

mol HgO2

Mass to Mass Conversions• To convert from grams of one substance to

grams of another, you need 3 conversion factors:

1. molar mass of the given.2. mole ratio.3. molar mass of the unknown.

Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),

How many grams of HgO are needed to produce 45.0 g of O2?

mol O2

mol O2

Mass to Mass Conversions (continued)

Given

Molar Massof the Given

45.0 g O2

g O2

609 g HgO

=x x

2nd C.F.Mole Ratio

1st C.F.

32.0

1

1

mol HgO2

mol HgOx

3rd C.F.Molar Mass ofthe Unknown

1

g HgO216.6

Volume Ratios

• You can use the volume ratios as conversion factors just like you would use mole ratios.

2CO(g) + O2(g) → 2CO2(g)2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes

• Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2?

0.626 L COL COL O2

21x = 0.313 L O2

Gas StoichiometrySample Problem

Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation:

CuO(s) + H2(g) → Cu(s) + H2O(g)

a.How many moles of H2 react?

b.How many grams of Cu are produced?

5.61 L H2L H2

mol H2

22.41x = 0.250 mol H2

5.61 L H2L H2

mol H2

22.41x

mol H2

mol Cu1

1xmol Cu

g Cu1

63.5x = 15.9 g Cu

Limiting Reactants

• When combining 2 or more different things to make a product, you have to stop when one of the things is used up.

• For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.

Limiting Reactants (continued)

• The limiting reactant is the reactant that limits the amount of product formed.

• The excess reactant is the substance that is not used up completely.

• Once the limiting reactant is used up, a chemical reaction will stop.

Limiting Reactants (continued)

Example:Silicon dioxide reacts with hydrogen fluoride according

to the following equation:SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?

Set up 2 equations, one for eachgiven:

Mole ratio C.F.

4.5 mol SiO21 mol SiF4

1 mol SiO2

4.5 mol SiF4=x

6.0 mol HF 1 mol SiF4

4 mol HF1.5 mol SiF4=x 1.5 mol SiF4

Limiting Reactant is HF

The limitingreactant

makes the least product.

6.0 mol HF

Percentage Yield

• The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry).

• The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem).

• The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

Now use the formula

mol C6H5Cl

Percentage YieldSample Problem A

Given the following equation:C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)

When 36.8 g C6H6 react with excess Cl2, the actual yield of C6H5Cl is 38.8 g. What is the % yield of C6H5Cl?

mol C6H6

mol C6H636.8 g C6H6g C6H6

=x x78.01

11

mol C6H5Clx

1g C6H5Cl112.5 53.1 g

C6H5Cl

Set up a mass to mass conversionThis is the

Theoretical Yield

Percent Yield

=Actual

Theoreticalx 100 =

38.8 g C6H5Cl53.1 g C6H5Cl

x 100 = 73.1%

Percentage YieldSample Problem B

According to the following reaction :CO(g) + 2H2(g) → CH3OH(l)

If the typical yield is 80.0%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess H2 gas?

Multiply theoreticalyield by percentageto get actual yield

mol CH3OHmol CO

mol CO75.0 g COg CO

=x x28.01

11

mol CH3OHx

1g CH3OH32.0 85.7 g

CH3OH

Set up a mass to mass conversionThis is the

Theoretical Yield

85.7 g CH3OH x 80.0% = 68.6 g CH3OH