The%20Mole

1
The Mole
6.02 X 1023
2
Table of Contents

• The Mole Slides 3 - 8
• Molar Mass Slides 9 - 23
• Percent Composition Slides 24 27
• Formulas Slides 28 - 36

3
C.8.A Define and Use the Concept of a mole

• The Mole

4
The Mole

• A counting unit
• Similar to a dozen, except instead of 12, its
  602 billion trillion 602,000,000,000,000,000,000,0
  00
• 6.02 X 1023 (in scientific notation)
• This number is named in honor of Amedeo Avogadro
  (1776 1856), who studied quantities of gases
  and discovered that no matter what the gas was,
  there were the same number of molecules present

5
The Mole

• 1 dozen cookies 12 cookies
• 1 mole of cookies 6.02 X 1023 cookies
• 1 dozen cars 12 cars
• 1 mole of cars 6.02 X 1023 cars
• 1 dozen Al atoms 12 Al atoms
• 1 mole of Al atoms 6.02 X 1023 atoms
• Note that the NUMBER is always the same, but the
  MASS is very different!
• Mole is abbreviated mol (gee, thats a lot
  quicker to write, huh?)

6
Check your knowledge

• Suppose we invented a new collection unit called
  a rapp. One rapp contains 8 objects.
• 1. How many paper clips in 1 rapp?
• a) 1 b) 4 c) 8
• 2. How many oranges in 2.0 rapp?
• a) 4 b) 8 c) 16
• 3. How many rapps contain 40 gummy bears?
• a) 5 b) 10 c) 20

7
A Mole of Particles Contains 6.02 x 1023
particles

• 6.02 x 1023 C atoms
• 6.02 x 1023 H2O molecules
• 6.02 x 1023 NaCl molecules
• (technically, ionics are compounds not molecules
  so they are called formula units)
• 6.02 x 1023 Na ions and
• 6.02 x 1023 Cl ions

1 mole C 1 mole H2O 1 mole NaCl
8
Avogadros Number as Conversion Factor

• 6.02 x 1023 particles
• 1 mole
• or
• 1 mole
• 6.02 x 1023 particles
• Note that a particle could be an atom OR a
  molecule!

9
Learning Check

• 1. Number of atoms in 0.500 mole of Al
• a) 500 Al atoms
• b) 6.02 x 1023 Al atoms
• c) 3.01 x 1023 Al atoms
• 2.Number of moles of S in 1.8 x 1024 S atoms
• a) 1.0 mole S atoms
• b) 3.0 mole S atoms
• c) 1.1 x 1048 mole S atoms

10
Molar Mass
11
Molar Mass

• The Mass of 1 mole (in grams)
• Equal to the numerical value of the average
  atomic mass (get from periodic table)
• 1 mole of C atoms 12.0 g
• 1 mole of Mg atoms 24.3 g
• 1 mole of Cu atoms 63.5 g

12
Your Turn!

• Find the molar mass of Bromine and Tin
  atoms(usually we round to the tenths place)

79.9 g/mole

1. 1 mole of Br atoms
2. 1 mole of Sn atoms

118.7 g/mole
13
Molar Mass of Molecules and Compounds

• Mass in grams of 1 mole equal numerically to the
  sum of the atomic masses
• 1 mole of CaCl2 110.9 g/mol
• 1 mole Ca x 40.1 g/mol 40.1g/mol
• 2 moles Cl x 35.5 g/mol 70.9 g/mol CaCl2
• 1 mole of N2O4 92.0 g/mol

14
Find!

• Molar Mass of K2O ? Grams/mole
• B. Molar Mass of antacid Al(OH)3 ? Grams/mole

15
Real life Connection

• Prozac, C17H18F3NO, is a widely used
  antidepressant that inhibits the uptake of
  serotonin by the brain. Find its molar mass.

16
Calculations with Molar Mass

• molar mass
• Grams Moles

17
Converting Moles and Grams

• Aluminum is often used for the structure of
  light-weight bicycle frames. How many grams of
  Al are in 3.00 moles of Al?
• 3.00 moles Al ? g Al

18

• Molar mass of Al
• 1 mole Al 27.0 g Al
• 2. Conversion factors for Al
• 27.0g Al or 1 mol Al
• 1 mol Al 27.0 g Al
• 3. Setup 3.00 moles Al x 27.0 g Al
• 1 mole Al
• Answer 81.0 g Al

19
Apply your knowledge!

• The artificial sweetener aspartame
  (Nutra-Sweet) formula C14H18N2O5 is used to
  sweeten diet foods, coffee and soft drinks. How
  many moles of aspartame are present in 225 g of
  aspartame?

20
C.8.B Use the mole concept to calculate the
number of atoms, ions, or molecules in a sample
of material.

• Atoms, Molecules and Grams

21
Atoms/Molecules and Grams

• Since 6.02 X 1023 particles 1 mole AND1
  mole molar mass (grams)
• You can convert atoms/molecules to moles and then
  moles to grams! (Two step process)
• You cant go directly from atoms to grams!!!! You
  MUST go thru MOLES.
• Thats like asking 2 dozen cookies weigh how many
  ounces if 1 cookie weighs 4 oz? You have to
  convert to dozen first!

22
Calculations

• molar mass
  Avogadros number Grams
  Moles particles
• Everything must go through Moles!!!

23
Atoms/Molecules and Grams

• How many atoms of Cu are present in 35.4 g of Cu?

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms
Cu 63.5 g Cu 1 mol Cu
3.4 X 1023 atoms Cu
24
Lets try!

• How many atoms of K are
  present in 78.4
  g of K?

1.2 x 1024 atoms
25
Lets try Continued!

• How many atoms of O are present in 78.1 g of
  oxygen?

78.1 g O2 1 mol O2 6.02 X 1023 molecules O2 2
atoms O 32.0 g O2 1 mol O2
1 molecule O2
2.9 x 1024 atoms
26
Percent Composition

• C.8.C Calculate percent composition and empirical
  and molecular formulas.

27
Definition of Percent Composition C.8.C
Calculate percent composition and empirical and
molecular formulas.

• The percentage composition of a compound is a
  statement of the relative mass each element
  contributes to the mass of the compound as a
  whole.
• composition

Mass of element
100
Mass of compound
28
Percent Composition

• What are the mass of carbon and oxygen in
  carbon dioxide, CO2?
• First, look up the atomic masses for carbon and
  oxygen from the Periodic Table. The atomic masses
  are found to be
• C is 12.01O is 16.00
• Next, determine how many grams of each element
  are present in one mole of CO2
• 12.01 g (1 mol) of C32.00 g (2 mole x 16.00 gram
  per mole) of O

29
Percent Composition

• The mass of one mole of CO2 is
• 12.01 g 32.00 g 44.01 g
• And the mass percentages of the elements are
• mass C 12.01 g / 44.01 g x 100 27.29 mass
  O 32.00 g / 44.01 g x 100 72.71
• Answer
• mass C 27.29 mass O 72.71

30
Formulas

• Empirical
• Ionic
• Molecular

31
C.8.C Calculate percent composition and empirical
and molecular formulas

• Formulas

32
Chemical Formulas of Compounds

• Formulas give the relative numbers of atoms or
  moles of each element in a formula unit - always
  a whole number ratio (the law of definite
  proportions).
• NO2 2 atoms of O for every 1 atom of N
• 1 mole of NO2 2 moles of O atoms to every 1
  mole of N atoms
• If we know or can determine the relative number
  of moles of each element in a compound, we can
  determine a formula for the compound.

33
Types of Formulas

• Empirical Formula
• The formula of a compound that expresses the
  smallest whole number ratio of the atoms present.
• Ionic formula are always empirical formula
• Molecular Formula
• The formula that states the actual number of
  each kind of atom found in one molecule of the
  compound.

34
Writing an Empirical Formula

• 1. Determine the mass in grams of each element
  present, if necessary.
• 2. Calculate the number of moles of each
  element.
• 3. Divide each by the smallest number of moles to
  obtain the simplest whole number ratio.
• If whole numbers are not obtained in step 3,
  multiply through by the smallest number that will
  give all whole numbers
• Be careful! Do not round off numbers
  prematurely

35
Writing an Empirical Formula

• A sample of a brown gas, a major air pollutant,
  is found to contain 2.34 g N and 5.34g O.
  Determine a formula for this substance.
• Requires mole ratios so convert grams to moles
• moles of N 2.34g of N 0.167 moles of N
• 14.01 g/mole
• moles of O 5.34 g 0.334 moles of O
• 16.00 g/mole

(Amount given in the problem)
(Weight of N from periodic table x 2, diatomic
element)
36
Writing an Empirical Formula
To obtain the simplest ratio, divide both numbers
of moles by the smaller number of moles (0.167
mol).
Formula
37
Molecular Formula

• Definition a chemical formula based on analysis
  and molecular weight

38
Molecular Formula

• What is the molecular formula of a substance that
  has an empirical formula of AgCO2and a molecular
  mass of 304?
• The formula mass of the empirical unit, AgCO2, is
  152. If we divide the formula mass 304 by 152,
  we get 2. Therefore, the molecular formula must
  be 2 times the empirical formula, or Ag2C2O4.

39
Empirical Formula from Composition

• A substance has the following composition by
  mass 60.80 Na 28.60 B 10.60 H
• What is the empirical formula of the substance?
• Consider a sample size of 100 grams
• This will contain 60.80 grams of Na, 28.60
• grams of B and 10.60 grams H
• Determine the number of moles of each
• Determine the simplest whole number ratio

40
Stoichiometry

• C.8.E Perform stoichiometric calculations,
  including determination of mass relationships
  between reactants and products, calculation of
  limiting reagents, and percent yield

41
Table of Contents

• Slides 39 - 50 Stoichiometry
• Slides 51 - 57 Limiting Reagent
• Slides 58 - 62 Percent Yield

42
Definition of Stoichiometry

• Stoichiometry is Calculations of quantities of
  substances involved in chemical reactions

43
What you need to know

1. How to balance an equation.
2. How to find molecular masses.
3. How to set up conversions.

44
Steps for solving mass-mass problems

1. Write and balance the chemical equation.
2. Set up the mole equivalency using the balanced
   equation.
3. Find the molecular mass for the Given and for the
   substance Asked For.
4. Set up the conversion T-bar.

45
Conversion T-bar
46
EXAMPLE 1

• In the synthesis reaction of nitrogen and
  hydrogen to form ammonia, 27 g of ammonia is
  produced. How much nitrogen is required to
  produce this much ammonia?
• STEP 1 Write and balance the equation
• N2 H2 ? NH3
• N2 3H2 ? 2NH3

47
Example 1 cont

• STEP 2 Find the mole equivalency
• use the balanced equation to set up mole
  equivalency equation
• Balanced Equation N2 3H2 ? 2NH3
• Mole Equivalency
• 1 mol N2 3 mol H2 ? 2 mol NH3

48
STEP 3 Find the molecular masses for GIVEN and
ASKED FOR substances

• GIVEN SUBSTANCE
• Ammonia
• N 1 x 14 g 14
• H 3 x 1 g 3
• 17g/mol

• SUBSTANCE ASKED FOR
• Nitrogen
• N 2 x 14 28 g/mol

49
STEP 4 Set up Conversion T-bar

• (27gNH3) (1mol NH3) (1 molN2) (28 gN2)
• (1) (17gNH3) (2 molNH3) (1molN2)
• ANSWER 22.2 g N2

50
Example 2

• Potassium Chlorate (KClO3) reacts as a
  decomposition reaction to form Oxygen and
  Potassium Chloride. If 80.5 g of O2 is produced,
  how many grams of potassium chloride are formed?

51
Example 2

• Equation
• KClO3? KCl O2
• Step 1 Balance the Equation
• 2KClO3? 2KCl 3O2
• Step 2 Mole Equivalency
• 2 mol KClO3? 2 molKCl 3 mol O2

52
Example 2

• Step 3 Molecular Masses
• Given Oxygen
• O 2 x 16 32g/mol
• Asked For Potassium Chloride
• K 1 x 39 39 g
• Cl 1 x 35 35g
• 74 g/mol

53
Example 2

• Step 4 Conversion and answer
• (80.5 gO2) (1 molO2) (2 mol KCl) (74 g
  KCl)
• (1) (32 gO2 ) (3 molO2 ) (1 mol
  KCl)
• Answer 124.1 g of KCl

54
Limiting Reagents

• C.8.E Perform stoichiometric calculations,
  including determination of mass relationships
  between reactants and products, calculation of
  limiting reagents, and percent yield

55
What is limiting reagent?

• Limiting reagent is the reactant that is consumed
  completely in a chemical reaction.

56
Calculate

• How many grams of CO2 are formed if 10.0g of
  carbon are burned in 20.0 dm3 of oxygen? (Assume
  STP)

57

• Step 1 Write a balanced equation
• C O2 ? CO2
• Step 2 Change both quantities to moles

1mole C
10g C
0.833 mol C
12g C
58

• Step 3 Compare the amount of CO2 produced by
  complete reaction of each of the reactants. The
  equation indicates that
• 1 mole C 1 mole O2 ? 1 mole CO2

20dm3 O2
1 mol
0.893 mol O2
22.4 dm3 O2
59

• Therefore, 0.833 mol C would produce 0.833 mol
  CO2 and 0.893 mol O2 would produce 0.893 mol CO2.
  Because carbon would produce fewer moles of
  carbon dioxide, the carbon limits the reaction.
  Some oxygen (0.060 mole) is left unreacted. We
  call carbon the limiting reactant.

60

• Step 4 Complete the problem on the basis of the
  limiting reactant.
• Answer 36.7 g CO2
• We concluded that 10g of C will react with excess
  O2 to form 36.7 g of CO2

1 mol CO2
44 g CO2
0.833 mol C
1 mol C
1 mol CO2
61
Percent Yield

• C.8.E Perform stoichiometric calculations,
  including determination of mass relationships
  between reactants and products, calculation of
  limiting reagents, and percent yield

62
What is Percent Yield?

• The percent yield of a product is the actual
  amount of product expressed as a percentage of
  the calculated theoretical yield of that product.
• Percent Yield

Actual amount of product
X 100
Theoretical amount of product
63
Calculate!

• What is the percentage yield in the following
  reaction if 5.50g of hydrogen react with nitrogen
  to form 20.4g of ammonia?
• Step 1 Write and balance the equation
• N2(g) 3H2(g) ? 2NH3(g)

64
Calculate!

• Step 2 Set up the problem using dimensional
  analysis and solve.
• Answer 30.9 g NH3 theoretical yield

5.50g H2
1 mol H2
17 g NH3
2 mol NH3
2.02g H2
1 mol NH3
3 mol H2
65
Calculate!

• Step 3 Solve for Yield
• Percent Yield
• Answer 66.0

20.4g NH3
X 100
30.9 g NH3