chem ch 11 - wbm

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Molecular Composition of Gases
Chapter 11
Chemistry Chapter 11
1
Gay-Lussac’s law of combining
volumes of gases
• At constant temperature and pressure,
the volumes of gaseous reactants and
products can be expressed as ratios of
small whole numbers
Chemistry Chapter 11
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Example
• When 2 L of hydrogen react with 1 L of
oxygen 2 L of water vapor are
produced.
• Write the balanced chemical equation:
Chemistry Chapter 11
3
You try
• When 1 L of hydrogen gas reacts with
1 L of chlorine gas, 2 L of hydrogen
chloride gas are produced.
• Write the balanced chemical equation:
Chemistry Chapter 11
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Avogadro's Law
• Equal volumes of gases at the same
pressure and temperature contain the
same number of molecules
• Atoms can’t split  diatomic molecules
• Gas volume is proportional to the
number of molecules
V  kn
Chemistry Chapter 11
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Molar Volume
• 1 mole of any gas contains
6.022 x 1023 molecules.
• According to Avogadro’s law, 1 mole of
any gas must have the same volume.
• Standard molar volume: volume of 1
mole of any gas at STP
• 22.4 L
Chemistry Chapter 11
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Example
• You are planning an experiment that
requires 0.0580 mol of nitrogen
monoxide gas. What volume in liters is
occupied by this gas at STP?
• 1.30 L NO
Chemistry Chapter 11
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You try
• A chemical reaction produces 2.56 L of
oxygen gas at STP. How many moles
of oxygen are in this sample?
• 0.114 mol O2
Chemistry Chapter 11
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Example
• Suppose you need 4.22 g of chlorine
gas. What volume at STP would you
need to use?
• 1.33 L Cl2
Chemistry Chapter 11
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You try
• What is the mass of 1.33 x 104 mL of
oxygen gas at STP?
• 19.0 g O2
Chemistry Chapter 11
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Discuss
• Explain Gay-Lussac’s law of combining
volumes
• State Avogadro’s law and explain its
significance.
Chemistry Chapter 11
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Review
• Boyles Law:
1
V
P
• Charles Law:
V T
• Avogadro’s Law:
V n
Chemistry Chapter 11
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Math
• A quantity that is proportional to each
of several quantities is also proportional
to their product. Therefore:
1
V  T  n
P
Chemistry Chapter 11
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More math
• Convert a proportionality
yx
• to an equality by multiplying by a
constant
y  kx
Chemistry Chapter 11
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Therefore
• We can covert
• to
1
V  T  n
P
1
V  R  T  n
P
Chemistry Chapter 11
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More neatly
nRT
V
P
or
PV  nRT
Chemistry Chapter 11
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This means….
• The volume of a gas varies directly with
the number of moles and the
temperature in Kelvin.
• The volume varies indirectly with
pressure.
Chemistry Chapter 11
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What if…
• n and T are constant?
• nRT is a constant, k
PV  k
• Boyle’s Law
• n and P are constant?
• nR/P is a constant, k
V  kT
• Charles’s Law
Chemistry Chapter 11
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What if…
• P and T are constant?
• RT/P is a constant, k
V  kn
• Avogadro’s law
Chemistry Chapter 11
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The ideal gas constant
•R
• Value depends on units
• SI units:
J
R  8.314
mol  K
Chemistry Chapter 11
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Other units
Chemistry Chapter 11
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Solving ideal gas problems
• Make sure the R you use matches the
units you have.
• Make sure all your units cancel out
correctly.
Chemistry Chapter 11
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Example
• A 2.07 L cylinder contains 2.88 mol of
helium gas at 22 °C. What is the
pressure in atmospheres of the gas in
the cylinder?
• 33.7 atm
Chemistry Chapter 11
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You try
• A tank of hydrogen gas has a volume of
22.9 L and holds 14.0 mol of the gas at
12 °C. What is the reading on the
pressure gauge in atmospheres?
• 14.3 atm
Chemistry Chapter 11
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Example
• A reaction yields 0.00856 mol of oxygen
gas. What volume in mL will the gas
occupy if it is collected at 43 °C and
0.926 atm pressure?
• 240. mL
Chemistry Chapter 11
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You try
• A researcher collects 9.09 x 10-3 mol of
an unknown gas by water displacement
at a temperature of 16 °C and
0.873 atm pressure (after the partial
pressure of the water vapor has been
subtracted). What volume of gas in mL
does the researcher have?
• 247 mL
Chemistry Chapter 11
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Finding mass
• Number of moles (n) equals mass (m)
divided by molar mass (M).
PV  nRT
mRT
PV 
M
mRT
M 
PV
Chemistry Chapter 11
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Example
• What mass of ethene gas, C2H4, is
contained in a 15.0 L tank that has a
pressure of 4.40 atm at a temperature
of 305 K?
• 74.0 g
Chemistry Chapter 11
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You try
• NH3 gas is pumped into the reservoir of
a refrigeration unit at a pressure of 4.45
atm. The capacity of the reservoir is
19.4 L. The temperature is 24 °C.
What is the mass of the gas in kg?
• 6.03 x 10-2 kg
Chemistry Chapter 11
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Example
• A chemist determines the mass of a
sample of gas to be 3.17 g. Its volume
is 942 mL at a temperature of 14 °C
and a pressure of 1.09 atm. What is
the molar mass of the gas?
• 72.7 g/mol
Chemistry Chapter 11
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Density
mRT
M 
PV
m
D
V
DRT
M 
P
Chemistry Chapter 11
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You try
• The density of dry air at sea level (1
atm) is 1.225 g/L at 15 °C. What is the
average molar mass of the air?
• 29.0 g/mol
Chemistry Chapter 11
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Stoichiometry
• Involves mass relationships between
reactants and products in a chemical
reaction
• For gases, the coefficients in the
balanced chemical equation show
volume ratios as well as mole ratios
• All volumes must be measured at the same
temperature and pressure
Chemistry Chapter 11
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Volume-Volume calculations
• From volume of one gas to volume of
another gas
• Use volume ratios just like mole ratios
in chapter 9
Chemistry Chapter 11
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Example
• Xenon gas reacts with fluorine gas to
produce the compound xenon
hexafluoride, XeF6. Write the balanced
equation for this reaction.
• Xe(g) + 3F2(g)  XeF6(g)
• If a researcher needs 3.14 L of XeF6 for
an experiment, what volumes of xenon
and fluorine should be reacted?
• 3.14 L of Xe and 9.42 L of F2
Chemistry Chapter 11
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Example
• Nitric acid can be produced by the
reaction of gaseous nitrogen dioxide
with water.
3NO2(g) + H2O(l)  2HNO3(l) + NO(g)
• If 708 L of NO2 gas react with water,
what volume of NO gas will be
produced?
• 236 L
Chemistry Chapter 11
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You try
• What volume of hydrogen gas is
needed to react completely with 4.55 L
of oxygen gas to produce water vapor?
• 9.10 L
Chemistry Chapter 11
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You try
• At STP, what volume of oxygen gas is
needed to react completely with 2.79 x
10-2 mol of carbon monoxide gas, CO,
to form gaseous carbon dioxide?
• 0.312 L
Chemistry Chapter 11
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You try
• Fluorine gas reacts violently with water
to produce hydrogen fluoride and ozone
according to the following equation:
3F2(g) + 3H2O(l)  6HF(g) + O3(g)
• What volumes of O3 and HF gas would
be produced by the complete reaction
of 3.60 x 104 mL of fluorine gas?
• 1.20 x 104 mL O3 and 7.20 x 104 mL HF
Chemistry Chapter 11
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You try
• Ammonia is oxidized to make nitrogen
monoxide and water
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
• At STP, what volume of oxygen will be
used in a reaction of 125 mol of NH3?
What volume of NO will be produced?
• 3.50 x 103 L O2 and 2.80 x 103 L NO
Chemistry Chapter 11
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Volume-mass and mass-volume
• Converting from volume to mass or
from mass to volume
• Must convert to moles in the middle
• Ideal gas law may be useful for finding
standard conditions
Chemistry Chapter 11
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Example
• Aluminum granules are a component of some
drain cleaners because they react with
sodium hydroxide to release both heat and
gas bubbles, which help clear the drain clog.
The reaction is:
2NaOH(aq) + 2Al(s) + 6H2O (l) 
2NaAl(OH)4(aq) + 3 H2(g)
• What mass of aluminum would be needed to
produce 4.00 L of hydrogen gas at STP?
• 3.21 g
Chemistry Chapter 11
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Example
• Air bags in cars are inflated by the
sudden decomposition of sodium azide,
NaN3 by the following reaction:
2NaN3(s)  3N2(g) + 2Na(s)
• What volume of N2 gas, measured at
1.30 atm and 87 °C, would be produced
by the reaction of 70.0 g of NaN3?
• 36.6 L
Chemistry Chapter 11
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You try
• What volume of chlorine gas at 38°C
and 1.63 atm is needed to react
completely with 10.4 g of sodium to
form NaCl?
• 3.54 L Cl2
Chemistry Chapter 11
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Example
• A sample of ethanol burns in O2 to form CO2
and H2O according to the following reaction.
C2H5OH + 3O2  2CO2 + 3H2O
• If the combustion uses 55.8 mL of oxygen
measured at 2.26 atm and 40.°C, what
volume of CO2 is produced when measured at
STP?
• 73.3 mL CO2
Chemistry Chapter 11
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You try
• Dinitrogen pentoxide decomposes into
nitrogen dioxide and oxygen. If 5.00 L
of N2O5 reacts at STP, what volume of
NO2 is produced when measured at
64.5 °C and 1.76 atm?
• 7.02 L NO2
Chemistry Chapter 11
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Review
• Diffusion: the gradual mixing of gases
due to their random motion
• Effusion: gases in a container randomly
pass through a tiny opening in the
container
Chemistry Chapter 11
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Rate of effusion
• Depends on relative velocities of gas
molecules.
• Velocity varies inversely with mass
• Lighter particles move faster
Chemistry Chapter 11
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Kinetic energy
• Depends only on temperature
• Equals 1
2
2
mv
• For two gases, A and B, at the same
temperature
1
1
2
2
M Av A  M B vB
2
2
• Each M stands for molar mass
Chemistry Chapter 11
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Algebra time
1
1
2
2
M Av A  M B vB
2
2
M AvA  M BvB
2
2
2
vA
MB

2
MA
vB
vA

vB
MB
MA
Chemistry Chapter 11
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Rate of effusion
• Depends on relative velocities of gas
molecules.
rateof effusion of A

rateof effusion of B
Chemistry Chapter 11
MB
MA
51
Graham’s law of effusion
• The rates of effusion of gases at the
same temperature and pressure are
inversely proportional to the square
roots of their molar masses.
Chemistry Chapter 11
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Graham’s law
• Graham experimented with densities of
gases, not molar masses.
• Density and molar mass are directly
proportional
• So we can replace molar mass with
density in the equation
densityB
rateof effusion of A

rateof effusion of B
densityA
Chemistry Chapter 11
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Use of Graham’s law
• Finding the molar mass
• Compare rates of effusion of a gas with
known molar mass and a gas with
unknown molar mass
• Use Graham’s law equation to solve for the
unknown M
• Used to separate isotopes of uranium
Chemistry Chapter 11
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Example
• Compare the rates of effusion of
hydrogen and helium at the same
temperature and pressure.
• Hydrogen diffuses about 1.41 times
faster
Chemistry Chapter 11
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Example
• Nitrogen effuses through a pinhole 1.7
times as fast as another gaseous
element at the same conditions.
Estimate the other element’s molar
mass and determine its probable
identity.
• 81 g/mol, krypton
Chemistry Chapter 11
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You try
• Estimate the molar mass of a gas that
effuses at 1.6 times the effusion rate of
carbon dioxide.
• 17 g/mol
Chemistry Chapter 11
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