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CHAPTER 11Molecular Composition of Gases
Section 1
Volume-Mass Relationships of Gases
Measuring and Comparing the Volumes of Reacting Gases Early 1800s, Gay-Lussac studied gas
volume relationships with chemical reaction between H and O
Observed 2 L H can react with 1 L O to form 2 L of water vapor at constant temperature and pressure
Hydrogen gas + oxygen gas water vapor
2 L 1 L 2 L
2 volumes 1 volume 2 volumes
Reaction shows 2:1:2 relationship between volumes of reactants and product
Ratio applies to any proportions (mL, L, cm3)
Gay-Lussac also noticed ratios by volume between other reactions of gases
Hydrogen gas + chlorine gas hydrogen chloride gas
1 L 1 L 2 L
Law of Combining Volumes of Gases
1808 Gay-Lussac summarized results in 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
Avogadro’s Law
Important point of Dalton’s atomic theory: atoms are indivisible
Dalton also thought particles of gaseous elements exist in form of single atoms
Believed one atom of one element always combines with one atom of another element to form single particle of product
Gay-Lussac’s results presented problem for Dalton’s theory
Ex. Reactions like formation of water
Hydrogen gas + oxygen gas water vapor
2 L 1 L 2 L
Seems that oxygen involved would have to divide into two parts
1811 Avogadro found way to explain Gay-Lussac’s simple ratios of combining volumes without violating Dalton’s idea of indivisible atoms
Rejected Dalton’s idea that reactant elements are always in monatomic form when they combine to form products
Reasoned these molecules could contain more than 1 atom
Avogadro’s law equal volumes of gases at the same temperature and pressure contain equal numbers of molecules
At the same temp and pressure, volume of any given gas varies directly with the number of molecules
1 mol CO2 atSTP = 22.4 L
1 mol O2 atSTP = 22.4 L
1 mol H2 atSTP = 22.4 L
Consider reaction of H and Cl to produce HCl
According to Avogadro’s law, equal volumes of H and Cl contain same number of molecules
b/c he rejected Dalton’s theory that elements are always monatomic, he concluded H and Cl components must each consist of 2 or more atoms joined together
Simplest assumption was that H and Cl molecules had 2 atoms each
Leads to following balanced equation:
H2(g) + Cl2(g) → 2HCl(g)
1 volume 1 volume 2 volumes
1 molecule 1 molecule 2 molecules
If the simplest formula for hydrogen chloride, HCl indicates molecule contains 1 H and 1 Cl
Then the simplest formulas for hydrogen and chlorine must be H2 and Cl2
Avogadro’s law also indicates that gas volume is directly proportional to the amount of gas, at given temp and pressure
V = kn
k = constant n = amount of gas in moles
Molar Volume of Gases
Remember 1 mol of substance contains 6.022 x 1023
According to Avogadro’s law, 1 mol of any gas occupies same volume as 1 mol of any other gas at same temperature and pressure, even though masses are different
Standard molar volume of gas volume occupied by 1 mol of gas at STP
= 22.4 L
Knowing volume of gas, you can use 1mol/22.4 L as conversion factor
Can find number of moles
Can find mass
Can also use molar volume to find volume if you have number of moles or mass
Sample Problem
A chemical reaction produces 0.0680 mol of oxygen gas. What volume in liters is occupied by this gas sample at STP?
1. Analyze
Given: moles of O2 = 0.0680 mol
Unknown: volume of O2 in liters at STP
2. Plan
moles of O2 → liters of O2 at STP
The standard molar volume can be used to find the volume of a known molar amount of a gas at STP
3. Compute
Practice Problems At STP, what is the volume of 7.08 mol of
nitrogen gas? 159 L N2
A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?
0.629 mol H2
At STP, a sample of neon gas occupies 550. cm3. How many moles of neon gas does this represent?
0.0246 mol Ne
Sample Problem
A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO2, at STP. What was the mass (in grams) of the gas produced?
1. Analyze
Given: volume of SO2 at STP = 98.0 mL
Unknown: mass of SO2 in grams
2. Plan liters of SO2 at STP→moles of SO2→grams of SO2
3. Compute
= 0.280 g SO2
Practice Problems What is the mass of 1.33 × 104 mL of
oxygen gas at STP? 19.0 g O2
What is the volume of 77.0 g of nitrogen dioxide gas at STP?
37.5 L NO2
At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?
9 g Cl2
Section 2The Ideal Gas Law
A gas sample can be characterized by 4 quantities
1. Pressure
2. Volume
3. Temperature
4. Number of moles
Number of moles present always affects at least one of the other 3 quantities
Collision rate per unit area of container wall depends on number of moles
Increase moles, increase collision rate, increase pressure
Pressure, volume, temperature, moles are all interrelated
A mathematical relationship exists to describe behavior of gas for any combination of these conditions
Ideal gas law mathematical relationship among pressure, volume, temperature, and number of moles of a gas
Derivation of Ideal Gas Law Derived by combining the other gas laws Boyle’s law: at constant temp, the
volume of a given mass of gas is inversely proportional to the pressure.
Charles’s law: At constant pressure, volume of given mass of gas is directly proportional to Kelvin temperature
V α T
Avogadro’s law: at constant temp and pressure, volume of given mass of gas is directly proportional to the number of moles
V α n
Volume is proportional to pressure, temp and moles in each equation
Combine the 3:
Can change proportion to equality by adding constant, this time R
This equation says the volume of a gas varies directly with the number of moles of gas and its Kelvin temperature
Volume also varies inversely with pressure
Ideal gas law combines Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws
Ex: PV = nRT n and T are constant, nRT is constant b/c R is also constant
This makes PV = constant which is Boyle’s law
The Ideal Gas Constant Ideal gas constant R Value depends on units for volume, pressure, temp
Unit of R Value of R
Unit of P Unit of V Unit of T Unit of n
62.4 mmHg L K Mol
0.0821 Atm L K Mol
8.314 Pa m3 K Mol
8.314 kPa L K Mol
Sample Problem
What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?
1. Analyze
Given: V of N2 = 10.0 L n of N2 = 0.500 mol T of N2 = 298 K
Unknown: P of N2 in atm
2. Plan
n,V,T → P The gas sample undergoes no change
in conditions Therefore, the ideal gas law can be
rearranged and used to find the pressure as follows
3. Compute
= 1.22 atm
Practice Problems What pressure, in atmospheres, is
exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C?
2.01 atm
A gas sample occupies 8.77 L at 20°C.What is the pressure, in atmospheres, given that there are 1.45 mol of gas in the sample?
3.98 atm
What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0°C and 0.974 atm pressure?
6.17 L O2
A sample that contains 4.38 mol of a gas at 250 K has a pressure of 0.857 atm. What is the volume?
105 L How many liters are occupied by 0.909 mol of
nitrogen at 125°C and 0.901 atm pressure? 33.0 L N2
What mass of chlorine gas, Cl2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?
101 g Cl2 How many grams of carbon dioxide gas
are there in a 45.1 L container at 34°C and 1.04 atm?
81.9 g CO2
What is the mass, in grams, of oxygen gas in a 12.5 L container at 45°C and 7.22 atm?
111 g O2
A sample of carbon dioxide with a mass of 0.30 g was placed in a 250 mL container at 400. K. What is the pressure exerted by the gas?
0.90 atm
Finding Molar Mass or Density from Ideal Gas Law
If P, V, T and mass are known you can calculate number of moles (n) in sample
Can calculate molar mass (g/mol)
Equation shows relationship between density, P, T, molar mass
Mass divided by molar mass gives moles Substitute m/M for n in equation PV=nRT
Density (D) = mass (m) per unit volume (V) D = m/V
Sample Problem
At 28°C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?
1. Analyze
Given: P of gas = 0.974 atm V of gas = 1.00 L T of gas = 28°C + 273 = 301 K m of gas = 5.16 g
Unknown: M of gas in g/mol
2. Plan P, V, T, m → M You can use the rearranged ideal gas law
provided earlier to find the answer
3. Compute
= 131 g/mol
Practice Problems
What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm?
83.8 g/mol What is the density of a sample of
ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0°C?
0.572 g/L NH3
The density of a gas was found to be 2.0 g/L at 1.50 atm and 27°C. What is the molar mass of the gas?
33 g/mol What is the density of argon gas,Ar,
at a pressure of 551 torr and a temperature of 25°C?
1.18 g/L Ar
Section 3Stoichiometry of Gases
You can apply gas laws to calculate stoichiometry of reactions involving gases
Coefficients in balanced equations represent mole AND volume ratios
Volume-Volume Calculations Propane, C3H8, is a gas that is sometimes
used as a fuel for cooking and heating. The complete combustion of propane occurs according to the following equation.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) (a) What will be the volume, in liters, of
oxygen required for the complete combustion of 0.350 L of propane? (b) What will be the volume of carbon dioxide produced in the reaction? Assume that all volume measurements are made at the same temperature and pressure.
1. Analyze
Given: balanced chemical equation V of propane = 0.350 L
Unknown: a. V of O2 in L;
b. V of CO2 in L
2. Plan
a. V of C3H8 → V of O2;
b. V of C3H8 → V of CO2
All volumes are to be compared at the same temperature and pressure
Therefore, volume ratios can be used like mole ratios to find the unknowns.
3. Compute
= 0.175 L O2
= 1.05 L CO2
Practice Problem
Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?
9.10 L H2
What volume of oxygen gas is needed to react completely with 0.626 L of carbon monoxide gas, CO, to form gaseous carbon dioxide? Assume all volume measurements are made at the same temperature and pressure.
0.313 L O2
Volume-Mass and Mass-VolumeCalculations gas volume A → moles A → moles B → mass B or mass A → moles A → moles B → gas volume B
You must know the conditions under which both the known and unknown gas volumes have been measured
The ideal gas law is useful for calculating values at standard and nonstandard conditions
SAMPLE PROBLEM
Calcium carbonate, CaCO3, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses. The balanced equation for the reaction follows.
Δ
CaCO3(s) → CaO(s) + CO2(g) How many grams of calcium carbonate
must be decomposed to produce 5.00 L of carbon dioxide gas at STP?
1. Analyze
Given: balanced chemical equation desired volume of CO2 produced at STP
= 5.00 L
Unknown: mass of CaCO3 in grams
2. Plan
The known volume is given at STP This tells us the pressure and temperature The ideal gas law can be used to find the
moles of CO2
The mole ratios from the balanced equation can then be used to calculate the moles of CaCO3 needed
(Note that volume ratios do not apply here because calcium carbonate is a solid)
3. Compute
= 0.223 mol CO2
Practice Problem
What mass of sulfur must be used to produce 12.61 L of gaseous sulfur dioxide at STP according to the following equation?
S8(s) + 8O2(g) → 8SO2(g)
18.0 g S8
How many grams of water can be produced from the complete reaction of 3.44 L of oxygen gas, at STP, with hydrogen gas?
5.53 g H2O
Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten oxide with hydrogen.
WO3(s) + 3H2(g) →W(s) + 3H2O(l)
How many liters of hydrogen gas at 35°C and 0.980 atm are needed to react completely with 875 g of tungsten oxide?
292 L H2
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 How many liters of gaseous carbon
monoxide at 27°C and 0.247 atm can be produced from the burning of 65.5 g of carbon according to the following equation?
2C(s) + O2(g) → 2CO(g) 544 L CO
Section 4Effusion and Diffusion
Graham’s Law of Effusion Rates of effusion and diffusion depend
on relative velocities of gas molecules
Velocity varies inversely with mass (lighter molecules move faster)
Average kinetic energy ½ mv2
For two gases, A and B, at same temp:
½ MAvA2 = ½ MBvB
2
MA and MB = molar masses of A and B Multiply by 2
MAvA2 = MBvB
2
MAvA2 = MBvB
2
Suppose you wanted to compare the velocities of the two gases
You would first rearrange the equation above to give the velocities as a ratio
Take square root of each side
This equation shows that velocities of two gases are inversely proportional to the square roots of their molar masses
b/c rates of effusion are directly proportional to molecular velocities, can write
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
Application of Graham’s Law Graham’s experiments dealt with
densities of gases Density varies directly with molar mass So…square roots of molar masses from
equation can be replaced with square roots of densities
Sample Problem
Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
1. Analyze
Given: identities of two gases, H2 and O2
Unknown: relative rates of effusion
2. Plan
molar mass ratio → ratio of rates of effusion
The ratio of the rates of effusion of two gases at the same temperature and pressure can be found from Graham’s law
3. Compute
Hydrogen effuses 3.98 times faster than oxygen.
Practice Problems A sample of hydrogen effuses through a
porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
160 g/mol Compare the rate of effusion of carbon
dioxide with that of hydrogen chloride at the same temperature and pressure.
CO2 will effuse about 0.9 times as fast as HCl
If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C4H10, at the same temperature.
about 235 m/s