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