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