Download The Mole - C405 Chemistry

The Mole
23
6.02 X 10
Table of Contents
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•
•
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The Mole
Molar Mass
Percent Composition
Formulas
Slides 3 - 8
Slides 9 - 23
Slides 24 – 27
Slides 28 - 36
The Mole
C.8.A DEFINE AND USE THE
CONCEPT OF A MOLE
The Mole
• A counting unit
• Similar to a dozen, except instead
of 12, it’s 602 billion trillion
602,000,000,000,000,000,000,000
• 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
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, that’s a lot
quicker to write, huh?)
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
A Mole of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole NaCl = 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
Avogadro’s 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!
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
Molar Mass
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
Your Turn!
Find the molar mass of Bromine and Tin
atoms(usually we round to the tenths place)
A. 1 mole of Br atoms =
B. 1 mole of Sn atoms =
79.9 g/mole
118.7 g/mole
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
1 mole of N2O4 = 92.0 g/mol
= 70.9 g/mol CaCl2
Find!
A. Molar Mass of K2O = ? Grams/mole
B. Molar Mass of antacid Al(OH)3 = ?
Grams/mole
Real life Connection
Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain. Find its molar
mass.
Calculations with Molar Mass
molar mass
Grams
Moles
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
1. Molar mass of Al
1 mole Al = 27.0 g Al
2. Conversion factors for Al
27.0g Al
1 mol Al
or
1 mol Al
27.0 g Al
3. Setup 3.00 moles Al
x
Answer = 81.0 g Al
27.0 g Al
1 mole Al
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?
Atoms, Molecules and Grams
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
• Since 6.02 X 1023 particles = 1 mole
AND
1 mole = molar mass (grams)
• You can convert atoms/molecules to
moles and then moles to grams! (Two step
process)
• You can’t go directly from atoms to
grams!!!! You MUST go thru MOLES.
• That’s like asking 2 dozen cookies weigh
how many ounces if 1 cookie weighs 4 oz?
You have to convert to dozen first!
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
Atoms/Molecules and Grams
How many atoms of Cu are
present in 35.4 g of Cu?
35.4 g Cu
1 mol Cu
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
Let’s try!
How many atoms of K are
present in 78.4 g of K?
= 1.2
x 1024 atoms
Let’s 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
Percent Composition
C.8.C Calculate percent composition
and empirical and molecular
formulas.
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
Mass of compound
100
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 12.01
O is 16.00
• Next, determine how many grams of each element are
present in one mole of CO2:
• 12.01 g (1 mol) of C
32.00 g (2 mole x 16.00 gram per mole) of O
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 %
Formulas
Empirical
Ionic
Molecular
Formulas
C.8.C CALCULATE PERCENT COMPOSITION AND
EMPIRICAL AND MOLECULAR FORMULAS
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.
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.
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.
4. 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
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
(Amount given in the problem)
moles of N = 2.34g of N = 0.167 moles of N
14.01 g/mole
(Weight of N from periodic table x 2,
diatomic element)
moles of O = 5.34 g = 0.334 moles of O
16.00 g/mole
Writing an Empirical Formula
To obtain the simplest ratio, divide both
numbers of moles by the smaller number of
moles (0.167 mol).
Formula:
N0.167 O0.334
N 0.167 O 0.334  NO 2
0.167
0.167
Molecular Formula
• Definition: a chemical formula based on
analysis and molecular weight
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.
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
Stoichiometry
C.8.E Perform stoichiometric calculations,
including determination of mass
relationships between reactants and
products, calculation of limiting reagents,
and percent yield
Table of Contents
• Slides 39 - 50 Stoichiometry
• Slides 51 - 57 Limiting Reagent
• Slides 58 - 62 Percent Yield
Definition of Stoichiometry
Stoichiometry is: Calculations of
quantities of substances involved
in chemical reactions
What you need to know
1. How to balance an equation.
2. How to find molecular masses.
3. How to set up conversions.
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.
Conversion T-bar
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
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
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
STEP 4: Set up Conversion T-bar
(27gNH3) (1mol NH3) (1 molN2) (28 gN2) =
(1) (17gNH3) (2 molNH3) (1molN2)
ANSWER = 22.2 g N2
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?
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
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
Example 2
Step 4: Conversion and answer
(80.5 gO2)
(1)
(1 molO2)
(2 mol KCl)
(74 g KCl)=
(32 gO2 ) (3 molO2 ) (1 mol KCl)
Answer = 124.1 g of KCl
Limiting Reagents
C.8.E Perform stoichiometric calculations,
including determination of mass
relationships between reactants and
products, calculation of limiting reagents,
and percent yield
What is limiting reagent?
• Limiting reagent is the reactant that is
consumed completely in a chemical
reaction.
Calculate
• How many grams of CO2 are formed if
10.0g of carbon are burned in 20.0 dm3 of
oxygen? (Assume STP)
• Step 1 Write a balanced equation
• C + O2 → CO2
• Step 2 Change both quantities to moles
10g C
1mole C
12g C
0.833 mol C
20dm3 O2
1 mol
0.893 mol O2
22.4 dm3 O2
• 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
• 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.
• Step 4 Complete the problem on the basis of
the limiting reactant.
0.833 mol C
1 mol CO2
1 mol C
44 g CO2
1 mol CO2
• Answer: 36.7 g CO2
• We concluded that 10g of C will react with
excess O2 to form 36.7 g of CO2
Percent Yield
C.8.E Perform stoichiometric calculations,
including determination of mass
relationships between reactants and
products, calculation of limiting reagents,
and percent yield
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
Theoretical amount of product
X 100
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)
Calculate!
• Step 2 Set up the problem using
dimensional analysis and solve.
5.50g H2
1 mol H2
2 mol NH3
17 g NH3
2.02g H2
3 mol H2
1 mol NH3
• Answer: 30.9 g NH3 theoretical yield
Calculate!
• Step 3 Solve for % Yield
• Percent Yield =
20.4g NH3
30.9 g NH3
• Answer: 66.0%
X 100