Chemistry 115

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Chemistry 115 Lecture 3 Outline Chapter
3 Chapter 5 gm - mol - molec.
Conversions Ideal Gas Law, STP Balancing
Chemical Reactions Partial Pressure Stoichiometry
Limiting Reagents Extending the Gas
Law HW2 Recitation Chapters 3 and 5 review
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The Mole - What is a mole?
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The Mole

• The mole (mol) is a counting unit for
  objects-atoms, molecules, ions, etc.

• Avogadros number is chosen so that 1 mole of
  12C atoms has a mass of exactly 12 g.

• Since 6.0221023 12 amu 12g, 6.0221023 amu
  1g
• so 1 mol of amu 1 g

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Average Atomic Mass Why does atomic atomic mass
of carbon 12.011 amu? We average the masses of
isotopes to give relative atomic
masses Naturally occurring C consists of
98.892 12C (12 amu) and 1.108 13C (13.00335
amu). The average mass of C is
therefore (0.98892)(12 amu)
(0.0108)(13.00335) 12.011 amu. Atomic weight
(AW) is also known as average atomic mass AW are
listed on the periodic table.
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Molar mass
The molar mass of a substance mass of 1 mole in
grams. It is numerically equal to the formula
weight in amu.
Avg. mass of 1 C atom 12.011amu Mass of 1 mole
of C 12.011 g
Mass of 1 H2O molecule 18.015 amu Mass of 1 mol
of H2O 18.015 g
molar mass of O plus2x the molar mass of H
Multiply by Avogados number to get atoms O
atoms in 18g H2O? 18g 1 mol 2 x 6.0x1023 O
If the moles are other than one, how do I
calculate the mass?
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Be familiar with the following conversion Ex
CO2
2 mol CO2
1.2 x 1024 molec CO2
88 g CO2
2.4 x 1024 atoms O
4 mol O
64 g O
CQ
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Balancing Chemical Equations
A balanced chemical equation has the same type
and number of atoms in the reactants as in the
products
Fuel Cell Rxn
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Stoichiometry (Chemical Equations) Mass is
conserved in a chemical reaction. The
quantitative nature of chemical formulas and
reactions is called stoichiometry. Chemical
equations give a description of a chemical
reaction.
Two types of numbers numbers in front of the
chemical formulas (stoichiometric coefficents)
and numbers in the formulas (they appear as
subscripts). Stoichiometric coefficients give
the ratio in which the reactants and products
exist. The subscripts give the ratio in which the
atoms are found in the molecule. The law of
conservation of mass demands that matter cannot
be lost in any chemical reactions.
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Example Convert the following into a balanced
equation
Liquid propane burns (combusts) in oxygen to form
carbon dioxide and water vapor.
C3H8 (l) O2 (g) CO2 (g)
H2O (g)
C3H8 (l) O2 (g) 3CO2 (g)
H2O (g)
C3H8 (l) O2 (g) 3CO2 (g)
4H2O (g)
C3H8 (l) 5O2 (g) 3CO2 (g)
4H2O (g)
Once you have the balanced equation, you can
calculate many other useful quantities, such as
grams, moles, molecules consumed or converted,
etc.
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• Lets calculate some quantities related to the
• propane combustion reaction
• C3H8 5O2 3CO2 4H2O
• What is the mass of 1 mol of C3H8?
• What is the number of mol in 2.00 g of C3H8?
• Number of H in 2.00 g C3H8?
• Percent composition of O in CO2?
• How many grams of CO2 will 1 mole of C3H8 make
  when combusted?
• f) How many grams of CO2 will the combustion of
  88.2 g of C3H8 make?

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• ANSWERS
• What is the mass of 1.00 mol of C3H8?
• This question asks about the molar mass or
  molecular weight (MW) sum of masses of atoms in
  the molecule
• MW of C3H8 3 ? 12.011 g/mol 8 ? 1.008 g/mol
• 44.067 g/mol 44.1 g/mol
• 44.1 g ? 1.00 mol 44.1 g
• mol

b) What is the number of mol in 2.00 g of
C3H8? 2.00 g ? 1 mol C3H8 0.045351 mol
0.0454 mol 44.1 g (round to 3 digits)
45.4 mmol
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2.184 ? 1023 H atoms round to 2.18 ?
1023 H atoms
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e) How many grams of CO2 will 1 mole of C3H8
make when combusted?
1 mol C3H8 gives 3 mol CO2 3 mol CO2 x 44.0
g/mol 132 g CO2

• Last one is a three step problem
• How many grams of CO2 will the combustion of 88.2
  g of C3H8 make?
• Need to convert g C3H8 ? moles C3H8 then use
  stoichiometry to figure moles of CO2. Finally,
  convert to g CO2.
• 88.2 g C3H8 x 1 mol C3H8 2.00 mol C3H8

44.1 g C3H8
2.00 mol C3H8 6.00 mol CO2 6.00 mol CO2 x
44.0 g/mol 264 g CO2
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Limiting Reactant

• Stoichiometry describes the way in which
  molecular reactantscombine to yield molecular
  products.

But

• Balanced chemical equations tell us only the
  relative amounts of substances consumed or
  produced in a reaction (ideal proportions).

• In the real world, the reaction ends when one
  reactant is totally consumed. This is the
  limiting reactant. The other reactants are excess
  reactants.

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Example 1 Many metals react with oxygen gas to
form the metal oxide. Calcium reacts as follows
Calculate the mass of calcium oxide that can be
prepared from 4.20 g of calcium and 2.80 g of
oxygen.
Notice When the amounts of two or more reactants
is given, it is a limiting reactant problem.
To solve, first identify the limiting reactant.
Calculate the number of moles of each reactant
available.
2.80 g O2
2.80 g of O2
0.088 mol O2
32.00 g/mol O2
4.20 g Ca
0.105 mol Ca
4.20 g of Ca
40.08 g/mol Ca
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From the stoichiometry of the balanced equation
0.088 mol O2 0.105 mol Ca
We see that 2 mol Ca requires 1 mol O2
0.105 mol Ca
x 1 mol O2
So, 0.105 mol Ca will require
O2 in the amount
0.052 mol O2
2 mol Ca
(We have 0.088 mol O2 available)
Similarly,
x 2 mol Ca
0.088 mol O2
0.088 mol of O2 will require Ca
in the amount
0.176 mol Ca
1 mol O2
(We have only 0.105 mol Ca)
Therefore, Ca is the limiting reactant .
So, what mass of calcium oxide can we produce?
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From the balanced equation,
0.105 mol Ca
0.105 mol Ca will yield
0.105 mol CaO
Convert mol CaO to mass
0.105 mol CaO x 56.08 g/mol CaO
5.89 g CaO
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Chapter 5
Importance of knowing gas-phase chemistry
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Gas laws in action
The operation of a four-stroke engine can be
divided into four cycles intake, compression,
power, and exhaust stages.
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Chapter 5. Gases Characteristics of
Gases There are three phases for all
substances solid, liquid, and gases. Gases
are highly compressible and occupy the full
volume of their containers. When a gas is
subjected to pressure, its volume decreases
liquids, solids dont compress much. Gases
always form homogeneous mixtures with other gases
liquids separate (oil/vinegar). Gases only
occupy about 0.1 of the volume of their
containers. Density 1/1000 of liquids
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Ideal Gas Law
Note that
V const2 ? T
V const3 ? n
R depends on units. Another popular set is
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STP Conditions
Lets take 1 mol of N2 at 0C (or 273.15 K).
Calculate volume
22.41 L
note always use K in calculations. 273.15 K, 1
atm are standard conditions STP.
So, the volume of 1 mol at STP is 22.41 L.
You need to be able to solve for V, or P, or n,
or T
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Standard Molar Volume
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Notice that the gas laws like Boyles Law are just
special cases of PV nRT,
if n, T are constant P1V1 constant
P2V2 P1V1 P2V2 We can derive other laws
similarly. It may be easiest to set up problem
using PV nRT.
Lets try an example A compressed gas cylinder
contains 50 L volume at 15 atm pressure. What is
the volume at atmospheric pressure?
We could just use P1V1 P2V2
15 atm 50 L 1 atm V2 , so V2 750 L.
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Alternately, use full Ideal Gas Law You can make
table of changing variables constant parameters
Or just write
V2 750 L
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Example 2 H2 2O2 ? 2H2O What happens to P if
V, T are constant?
If n1 3, n2 2
pressure goes down by 1/3
so P2 2/3 P1
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Example 3 Now consider the following
compression of a gas by a piston
P 2.2 atm T2 ?
P 1 atm T1 0C
final
initial

T2 300 K 27C. gas heats up!
Well actually come back to this type of problem
when we discuss thermodynamics.
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Concept Questions 1 mole of a gas at STP has a
volume of 22.4 L. If we raise the temperature to
273 C, what is the pressure? A) 0 torr B) 380
torr C) 760 torr D) 1520 torr If, at the same
temperature, I now halve the pressure, what is
the new volume? A) 11.2 L B) 22.4 L C) 44.8
L D) Cant be determined
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Partial Pressures gas mixtures Our atmosphere
consists of 78 N2 mol (dry air) 21 O2 1
Ar What are the partial pressures of those gases?
Note that the sum of partial pressures is equal
to total pressure PT PA PB
We can therefore write that the partial pressures
are related to the number of moles of each gas by
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We can also determine mole fraction
Note this is not the same as mass fraction (
mass)
ie N2 mole fraction XN2 .78
An important application of partial pressure is
the effect of H2O in the atmosphere
Example If the pressure of water vapor is 22
torr at 24 C, what is the mole fraction of H2O,
N2 and O2? (assume other gas pressures are
negligible).
PN2 PO2 PH2O 760 torr PN2 PO2 738
torr PN2 78 x 738 576 torr PO2 162 torr.