Download Chapter 3 PowerPoint

Atomic Mass
Atoms are so small, it is difficult to
discuss how much they weigh in grams.
 Use atomic mass units.
 an atomic mass unit (amu) is one twelth
the mass of a carbon-12 atom.
 This gives us a basis for comparison.
 The decimal numbers on the table are
atomic masses in amu.

They are not whole numbers
Because they are based on averages of
atoms and of isotopes.
 can figure out the average atomic mass
from the mass of the isotopes and their
relative abundance.
 add up the percent as decimals times
the masses of the isotopes.

Examples
There are two isotopes of carbon 12C with
a mass of 12.00000 amu(98.892%), and 13C
with a mass of 13.00335 amu (1.108%).
 There are two isotopes of nitrogen , one
with an atomic mass of 14.0031 amu and
one with a mass of 15.0001 amu. What is
the percent abundance of each?

The Mole
The mole is a number.
 A very large number, but still, just a
number.
 6.022 x 1023 of anything is a mole
 A large dozen.
 The number of atoms in exactly 12
grams of carbon-12.

The Mole

Makes the numbers on the table the
mass of the average atom.
Representative particles
The smallest pieces of a substance.
 For a molecular compound it is a
molecule.
 For an ionic compound it is a formula
unit.
 For an element it is an atom.

More Stoichiometry
Molar mass
Mass of 1 mole of a substance.
 Often called molecular weight.
 To determine the molar mass of an
element, look on the table.
 To determine the molar mass of a
compound, add up the molar masses of
the elements that make it up.

Find the molar mass of

CH4

Mg3P2
Ca(NO3)3
 Al2(Cr2O7)3
 CaSO4 · 2H2O

Examples
How much would 2.34 moles of carbon
weigh?
 How many moles of magnesium in
24.31 g of Mg?
 How many atoms of lithium in 1.00 g of
Li?
 How much would 3.45 x 1022 atoms of
U weigh?

Percent Composition
One can find the percentage of the mass
of a compound that comes from each of
the elements in the compound by using
this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100
© 2009, Prentice-Hall, Inc.
Percent Composition
So the percentage of carbon in ethane
is…
(2)(12.0 amu)
%C =
(30.0 amu)
=
24.0 amu
x 100
30.0 amu
= 80.0%
© 2009, Prentice-Hall, Inc.

Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound

For iron in iron(III) oxide, (Fe2O3):
mass % Fe =
2(55.85 g)
111.70 g
=
2(55.85 g)
+ 3(16.00 g) 159.70 g
Copyright © Cengage Learning. All rights reserved
× 100% = 69.94%
Percent Composition
Percent of each element a compound is
composed of.
 Find the mass of each element, divide by
the total mass, multiply by a 100.
 Easiest if you use a mole of the compound.
 Find the percent composition of CH4
 Al2(Cr2O7)3
 CaSO4 · 2H2O

Finding
Empirical
Formulas
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
© 2009, Prentice-Hall, I
Determining Empirical
Formulas
A compound is comprised of 40.01% carbon, 6.72%
hydrogen, and 53.27% oxygen. Calculate the
empirical formula of the compound.
1) Given the percent composition, assume a mass of
sample
- use 100.00 g for convenience
6.72%
6.72 g
H
H
53.27%
40.01%
O
C
53.27 g
40.01 g
O
C
EOS
Chapter 3: Stoichiometry
18
Calculating Empirical
Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
© 2009, Prentice-Hall, Inc.
Calculating Empirical
Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
61.31 g x
H:
5.14 g x
N:
10.21 g x
O:
23.33 g x
1 mol
12.01 g
1 mol
1.01 g
1 mol
14.01 g
1 mol
16.00 g
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
© 2009, Prentice-Hall, Inc.
Calculating Empirical
Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol = 7.005  7
0.7288 mol
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
© 2009, Prentice-Hall, Inc.
Calculating Empirical
Formulas
These are the subscripts for the empirical formula:
C7H7NO2
© 2009, Prentice-Hall, Inc.
Working backwards
From percent composition, you can
determine the empirical formula.
 Empirical Formula the lowest ratio of
atoms in a molecule.
 Based on mole ratios.
 A sample is 59.53% C, 5.38%H,
10.68%N, and 24.40%O what is its
empirical formula.

Pure O2 in
Sample is burned
completely to
form CO2 and
H2O
CO2 is absorbed
H2O is absorbed

A 0.2000 gram sample of a compound
(vitamin C) composed of only C, H, and
O is burned completely with excess O2 .
0.2998 g of CO2 and 0.0819 g of H2O are
produced. What is the empirical
formula?
Empirical To Molecular
Formulas
Empirical is lowest ratio.
 Molecular is actual molecule.
 Need Molar mass.
 Ratio of empirical to molar mass will
tell you the molecular formula.
 Must be a whole number because...

Example

A compound is made of only sulfur and
oxygen. It is 69.6% S by mass. Its molar
mass is 184 g/mol. What is its formula?
Exercise
The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). The molar
mass of the compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
Copyright © Cengage Learning. All rights reserved
28
Chemical Equations
Are sentences.
 Describe what happens in a chemical
reaction.
 Reactants  Products
 Equations should be balanced.
 Have the same number of each kind of
atoms on both sides because ...

Meaning
A balanced equation can be used to
describe a reaction in molecules and
atoms.
 Not grams.
 Chemical reactions happen molecules at
a time
 or dozens of molecules at a time
 or moles of molecules.

Stoichiometry
Given an amount of either starting
material or product, determining the
other quantities.
 use conversion factors from
– molar mass (g - mole)
– balanced equation (mole - mole)
 keep track.

Examples
How many moles is 4.56 g of CO2 ?
 How many grams is 9.87 moles of H2O?
 How many molecules in 6.8 g of CH4?
 49 molecules of C6H12O6 weighs how
much?

Examples
One way of producing O2(g) involves the
decomposition of potassium chlorate into
potassium chloride and oxygen gas. A
25.5 g sample of Potassium chlorate is
decomposed. How many moles of O2(g)
are produced?
 How many grams of potassium chloride?
 How many grams of oxygen?

Examples
A piece of aluminum foil 5.11 in x 3.23 in
x 0.0381 in is dissolved in excess HCl(aq).
How many grams of H2(g) are produced?
 How many grams of each reactant are
needed to produce 15 grams of iron form
the following reaction?
Fe2O3(s) + Al(s)  Fe(s) + Al2O3(s)

Examples
K2PtCl4(aq) + NH3(aq) 
Pt(NH3)2Cl2 (s)+ KCl(aq)
 what mass of Pt(NH3)2Cl2 can be
produced from 65 g of K2PtCl4 ?
 How much KCl will be produced?
 How much from 65 grams of NH3?

Gases and the Mole
 Many
Gases
of the chemicals we deal with are
gases.
 They are difficult to weigh.
 Need to know how many moles of gas
we have.
 Two things effect the volume of a gas
 Temperature and pressure
 Compare at the same temp. and
pressure.
Standard Temperature and
Pressure
 0ºC
and 1 atm pressure
 abbreviated STP
 At STP 1 mole of gas occupies 22.4 L
 Called the molar volume
 Avagadro’s Hypothesis - at the same
temperature and pressure equal volumes
of gas have the same number of particles.
Examples
 What
is the volume of 4.59 mole of CO2
gas at STP?
 How many moles is 5.67 L of O2 at
STP?
 What is the volume of 8.8g of CH4 gas
at STP?
Density of a gas
D = m /V
 for a gas the units will be g / L
 We can determine the density of any gas
at STP if we know its formula.
 To find the density we need the mass
and the volume.
 If you assume you have 1 mole than the
mass is the molar mass (PT)
 At STP the volume is 22.4 L.

Examples
 Find
the density of CO2 at STP.
 Find the density of CH4 at STP.
 Given
The other way
the density, we can find the molar
mass of the gas.
 Again, pretend you have a mole at STP,
so V = 22.4 L.
m = D x V
 m is the mass of 1 mole, since you have
22.4 L of the stuff.
 What is the molar mass of a gas with a
density of 1.964 g/L?
 2.86 g/L?
Stoichiometry
Greek for “measuring elements”
 The calculations of quantities in
chemical reactions based on a balanced
equation.
 We can interpret balanced chemical
equations several ways.

Look at it differently
 2H2 +
O2  2H2O
2 dozen molecules of hydrogen and 1
dozen molecules of oxygen form 2 dozen
molecules of water.
 2 x (6.02 x 1023) molecules of hydrogen and
1 x (6.02 x 1023) molecules of oxygen form
2 x (6.02 x 1023) molecules of water.
 2 moles of hydrogen and 1 mole of oxygen
form 2 moles of water.

Mole to mole conversions
2 Al2O3 Al + 3O2
 every time we use 2 moles of Al2O3 we
make 3 moles of O2

2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
Mole to Mole conversions
How many moles of O2 are produced
when 3.34 moles of Al2O3 decompose?
 2 Al2O3 Al + 3O2

3.34 moles
3 mole O2
= 5.01 moles O2
Al2O3 2 moles Al O
2 3
Your Turn
2C2H2 + 5 O2  4CO2 + 2 H2O
 If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed?
 How many moles of C2H2 are needed
to produce 8.95 mole of H2O?
 If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed?

Periodic
Table
Mass
gA
•Decide
Balanced
Equation
Moles
A
Periodic
Table
Moles
B
Mass
gB
where to start based on the units you
are given and stop based on what unit you are
asked for
For example...
If 10.1 g of Fe are added to a solution of
Copper (II) Sulfate, how much solid
copper would form?
 Fe + CuSO4  Fe2(SO4)3 + Cu
 2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu
55.85 g Fe 2 mol Fe 1 mol Cu
17.3 g Cu
=

More Examples
To make silicon for computer chips they
use this reaction
 SiCl4 + 2Mg  2MgCl2 + Si
 How many moles of Mg are needed to
make 9.3 g of Si?
 3.74 mol of Mg would make how many
moles of Si?
 How many grams of MgCl2 are
produced along with 9.3 g of silicon?

For Example
The U. S. Space Shuttle boosters use this
reaction
 3 Al(s) + 3 NH4ClO4 
Al2O3 + AlCl3 + 3 NO + 6H2O
 How much Al must be used to react
with 652 g of NH4ClO4 ?
 How much water is produced?
 How much AlCl3?

Gases and Reactions
We can also change
Liters of a gas to moles
 At STP
 0ºC and 1 atmosphere pressure
 At STP 22.4 L of a gas = 1 mole
 If 6.45 moles of water are decomposed,
how many liters of oxygen will be
produced at STP?

For Example
If 6.45 grams of water are decomposed,
how many liters of oxygen will be
produced at STP?
 H2O  H2 + O2
 2H2O  2H2 + O2

6.45 g H2O 1 mol H2O
1 mol O2 22.4 L O2
18.02 g H2O 2 mol H2O 1 mol O2
Your Turn
How many liters of CO2 at STP will be
produced from the complete
combustion of 23.2 g C4H10 ?
 What volume of oxygen will be
required?

Yield
How much you get from an
chemical reaction
Limiting Reagent
If you are given one dozen loaves of
bread, a gallon of mustard and three
pieces of salami, how many salami
sandwiches can you make?
 The limiting reagent is the reactant you
run out of first.
 The excess reagent is the one you have left
over.
 The limiting reagent determines how
much product you can make

Limiting Reagent
Reactant that determines the amount of
product formed.
 The one you run out of first.
 Makes the least product.
 Book shows you a ratio method.
 It works.
 So does mine

Limiting reagent
To determine the limiting reagent
requires that you do two stoichiometry
problems.
 Figure out how much product each
reactant makes.
 The one that makes the least is the
limiting reagent.

How do you find out?
Do two stoichiometry problems.
 The one that makes the least product is
the limiting reagent.
 For example
 Copper reacts with sulfur to form
copper ( I ) sulfide. If 10.6 g of copper
reacts with 3.83 g S how much product
will be formed?

If 10.6 g of copper reacts with 3.83 g S.
How many grams of product will be
formed?
Cu is
 2Cu + S  Cu2S
Limiting

1 mol
Cu2S 159.16 g Cu2S
1
mol
Cu
Reagent
10.6 g Cu
63.55g Cu 2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
Example
Ammonia is produced by the following
reaction
N2 + H2  NH3
What mass of ammonia can be
produced from a mixture of 100. g N2
and 500. g H2 ?
 How much unreacted material
remains?

How much excess reagent?
Use the limiting reagent to find out how
much excess reagent you used
 Subtract that from the amount of excess
you started with

Excess Reagent
The reactant you don’t run out of.
 The amount of stuff you make is the
yield.
 The theoretical yield is the amount you
would make if everything went perfect.
 The actual yield is what you make in
the lab.

Your turn
Mg(s) +2 HCl(g)  MgCl2(s) +H2(g)
 If 10.1 mol of magnesium and 4.87 mol
of HCl gas are reacted, how many
moles of gas will be produced?
 How much excess reagent remains?

Your Turn II
If 10.3 g of aluminum are reacted with
51.7 g of CuSO4 how much copper will
be produced?
 How much excess reagent will remain?

Percent Yield

% yield = Actual
x 100%
Theoretical

% yield =
what you got
x 100%
what you could have got
Yield
The amount of product made in a
chemical reaction.
 There are three types
 Actual yield- what you get in the lab
when the chemicals are mixed
 Theoretical yield- what the balanced
equation tells you you should make.
 Percent yield = Actual
x 100 %
Theoretical

Example
6.78 g of copper is produced when 3.92 g
of Al are reacted with excess copper (II)
sulfate.
 2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
 What is the actual yield?
 What is the theoretical yield?
 What is the percent yield?
 If you had started with 9.73 g of Al, how
much copper would you expect?

Examples

Aluminum burns in bromine producing
aluminum bromide. In a laboratory 6.0
g of aluminum reacts with excess
bromine. 50.3 g of aluminum bromide
are produced. What are the three types
of yield.
Examples
Years of experience have proven that the
percent yield for the following reaction is
74.3%
Hg + Br2  HgBr2
If 10.0 g of Hg and 9.00 g of Br2 are
reacted, how much HgBr2 will be
produced?
 If the reaction did go to completion, how
much excess reagent would be left?

Examples

Commercial brass is an alloy of Cu and
Zn. It reacts with HCl by the following
reaction Zn(s) + 2HCl(aq)  ZnCl2 (aq)
+ H2(g)
Cu does not react. When 0.5065 g of
brass is reacted with excess HCl, 0.0985
g of ZnCl2 are eventually isolated.
What is the composition of the brass?