Download 6.022 x 10 23 - Fort Bend ISD

Stoichiometry
“In solving a problem of this sort,
the grand thing is to be able to
reason backward. This is a very
useful accomplishment, and a very
easy one, but people do not practice
it much.”
Sherlock Holmes, in Sir Arthur Conan
Doyle’s A Study in Scarlet
The Mole
23
6.02 X 10
The Mole
1 dozen = 12
1 gross = 144
1 ream = 500
1 mole = 6.022 x 1023
There are exactly 12 grams of
carbon-12 in one mole of carbon-12.
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
6.022 x 1023 is called “Avogadro’s Number” in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
I didn’t discover it. Its
just named after me!
Amadeo Avogadro
23
6.02 X 10
Just How Big is a Mole?



Enough soft drink cans to cover the
surface of the earth to a depth of
over 200 miles.
If you had Avogadro's number of
unpopped popcorn kernels, and
spread them across the United
States of America, the country would
be covered in popcorn to a depth of
over 9 miles.
If we were able to count atoms at the
rate of 10 million per second, it
would take about 2 billion years to
count the atoms in one mole.
Everybody Has Avogadro’s Number!
But Where Did it Come From?



It was NOT just picked!
It was MEASURED.
One of the better
methods of measuring
this number was the
Millikan Oil Drop
Experiment
Since then we have
found even better ways
of measuring using xray technology
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?)
Learning Check
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
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

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
Other Names Related to Molar Mass

Molecular Mass/Molecular Weight: If you have a single
molecule, mass is measured in amu’s instead of grams. But,
the molecular mass/weight is the same numerical value as 1
mole of molecules. Only the units are different. (This is the
beauty of Avogadro’s Number!)

Formula Mass/Formula Weight: Same goes for
compounds. But again, the numerical value is the same.
Only the units are different.

THE POINT: You may hear all of these terms
which mean the SAME NUMBER… just different units
Learning Check!
Find the molar mass
(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
= 111.1 g/mol
1 mole Ca x 40.1 g/mol
+ 2 moles Cl x 35.5 g/mol
1 mole of N2O4
= 111.1 g/mol CaCl2
= 92.0 g/mol
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
When the equation is balanced it has quantitative
significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles
of water
Solving a Stoichiometry Problem
1.
2.
3.
4.
5.
Balance the equation.
Convert masses to moles.
Determine which reactant is
limiting.
Use moles of limiting reactant and
mole ratios to find moles of
desired product.
Convert from moles to grams.
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed.
1. Identify reactants and products and write
the balanced equation.
4 Al
+ 3 O2
2 Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al
6.50 g Al
X
1 mol Al
26.98 g Al
+
X
3 O2  2Al2O3
2 mol Al2O3 101.96 g Al2O3
4 mol Al
X
1 mol Al2O3
= ? g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts
of products formed.
Calculations with Molar Mass
molar mass
Grams
Moles
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.94 g Li
X
1 mol Li
=
45.1
g Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
X
6.94 g Li
=
2.62
mol Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol Li 6.022 x 1023 atoms Li
X
1 mol Li
= 2.11 x 1024 atoms Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li 1 mol Li
X
X
6.94 g Li
6.022 x 1023 atoms Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
= 1.58 x 1024 atoms Li
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
Answer
x
27.0 g Al
1 mole Al
= 81.0 g Al
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
Learning Check!
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
Percent Composition
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG
monosodium glutamate), a compound used
to flavor foods and tenderize meats?
a) 8.22 %C
b) 24.3 %C
c) 41.1 %C
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
100.00
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
molecular formula =
(empirical formula)n [n =
integer]
 molecular formula = C6H6
= (CH)6
 empirical formula = CH

Formulas (continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1.
2.
3.
4.
Base calculation on 100 grams of
compound.
Determine moles of each element
in 100 grams of compound.
Divide each value of moles by the
smallest of the values.
Multiply each number by an
integer to obtain all whole
numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2
=
C6H10O4
Chemical Formulas of Compounds
(HONORS only)

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
(HONORS only)

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.
To obtain an Empirical Formula
(HONORS only)
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
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.
require 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
N 0.167 O 0.334  NO 2
Formula: N O
0.167
0.334
0.167
0.167
(HONORS only)
Calculation of the Molecular Formula
(HONORS only)
A compound has an empirical formula
of NO2. The colourless liquid, used in
rocket engines has a molar mass of
92.0 g/mole. What is the molecular
formula of this substance?
Empirical Formula from % Composition
(HONORS only)
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 28.60 grams of B and
10.60 grams H
Determine the number of moles of each
Determine the simplest whole number ratio