Moles and mixtures

 

Contents of this page

 

Molar volume of a gas: standard temperature and pressure

Molecular weight and density of a gas

Mixtures of gases: mole fractions

Dalton's law of partial pressures

Molar mass of a gas mixture

Collecting a gas over water

Scuba diving

Summary

 

Although all gases closely follow the ideal gas law PV = nRT under appropriate conditions, each gas is also a unique chemical substance consisting of molecular units that have definite masses. In this lesson we will see how these molecular masses affect the properties of gases that conform to the ideal gas law.

Following this, we will look at gases that contain more than one kind of molecule— in other words, mixtures of gases.

Molar volume of a gas: standard temperature and pressure

 

 

 

 

You will recall that the molar mass of a pure substance is the mass of 6.02 x 1023 (Avogadro's number) of particles or molecular units of that substance. Molar masses are commonly expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights.

As was explained in the preceding lesson, equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules (this is the "EVEN" principle, more formally known as Avogadro's law.)

This means that one mole of any gas will occupy the same volume as one mole of any other gas at a given temperature and pressure. The magnitude of this volume will of course depend on the temperature and pressure, so as a means of convenient comparison it is customary to define a set of conditions T = 273K and P = 1 atm as standard temperature and pressure, usually denoted STP. Substituting these values into the ideal gas equation of state and solving for V yields a volume of 22.414 litres for 1 mole

The standard molar volume 22.4 L mol–1 is a value worth memorizing, but remember that it is valid only at STP. The molar volume at other temperatures and pressures can easily be found by simple proportion.

 

Problem Example 1

What would the volume of one mole of air be at 20°C on top of Muana Kea, Haw'aii (altitude 4.2 km; click on picture for enlarged view) where the air pressure is approximately 60 kPa?

Solution: Apply Boyle's and Charles' laws as successive correction factors to the standard sea-level pressure of 101.3 kPa:

The molar volume of a substance can tell us something about how much space each molecule occupies, as the following example shows.

Problem Example 2

Estimate the average distance between the molecules in a gas at 1 atm pressure and 0°C.

Solution. Consider a 1-cm3 volume of the gas, which will contain

(6.02E23 mol–1)/(22400 cm3 mol–1) = 2.69E19 cm–3.

The volume per molecule (not the same as the volume of a molecule, which for an ideal gas is zero!) is just the reciprocal of this, or 3.72E–20 cm3. Assume that the molecules are evenly distributed so that each occupies an imaginary box having this volume. The average distance between the centers of the molecules will be defined by the length of this box, which is the cube root of the volume per molecule:

(3.72 × 10–20)1/3 = 3.38 × 10–7 cm = 3.4 nm

 

 

Molecular weight and density of a gas

 

 

 

 

The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar masses of different gases will vary. This means that different gases will have different densities (different masses per unit volume). If we know the molecular weight of a gas, we can calculate its density.

 

Problem Example 3

Uranium hexafluoride UF6 gas is used in the isotopic enrichment of natural uranium. Calculate its density at STP.

Solution: The molecular weight of UF6 is 352.

(353 g mol–1) ÷ (22.4 mol L–1) = 15.7 g L–1

Note: there is no need to look up a "formula" for this calculation; simply combine the molar mass and molar volume in such a way as to make the units come out correctly.

 

 

More importantly, if we can measure the density of an unknown gas, we have a convenient means of estimating its molecular weight. This is one of many important examples of how a macroscopic measurement (one made on bulk matter) can yield microscopic information (that is, about molecular-scale objects.)

Determination of the molecular weight of a gas from its density is known as the Dumas method, after the French chemist Jean Dumas (1800-1840) who developed it. One simply measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle's and Charles' laws. The weight of the gas divided by its STP volume yields the density of the gas, and the density multiplied by 22.4 mol–1 gives the molecular weight. Pay careful attention to the examples of gas density calculations shown in your textbook. You will be expected to carry out calculations of this kind, converting between molecular weight and gas density.

Gas densities are now measured in industry by electro-mechanical devices such as vibrating reeds which can provide continuous, on-line records at specific locations, as within pipelines.

 

 

Problem Example 4

Calculate the approximate molar mass of a gas whose measured density is 3.33 g/L at 30°C and 780 torr.

Solution. From the ideal gas equation, the number of moles contained in one litre of the gas is

The molecular weight is therefore (33 g L–1)/(.0413 mol L–1) = 80.6 g mol–1

 

Gas density measurements can be a useful means of estimating the composition of a mixture of two different gases:

Mixtures of gases

 

 

 

 

Because most of the volume occupied by a gas consists of empty space, there is nothing to prevent two or more kinds of gases from occupying the same volume. Homogeneous mixtures of this kind are generally known as solutions, but it is customary to refer to them simply as gaseous mixtures.

Mole fractions and volumes

We can specify the composition of gaseous mixtures in many different ways, but the most common ones are by volumes and by mole fractions. From Avogadro's Law we know that "equal volumes contains equal numbers of molecules", so when we say that air, for example, is 21 percent oxygen and 78 percent nitrogen by volume, this is the same as saying that these same percentages of the molecules in air consist of O2 and N2. Similarly, in 1.0 mole of air, there is 0.21 mol of O2 and 0.78 mol of N2 (the other 0.1 mole consists of various trace gases, but is mostly neon.)

Note that you could never assume a similar equivalance with mixtures of liquids or solids,
to which the E.V.E.N. principle does not apply.

These last two numbers (.21 and .78) also express the mole fractions of oxygen and nitrogen in air. Mole fraction means exactly what it says: the fraction of the molecules that consist of a specific substance. This is expressed algebraically by

so in the case of oxygen in the air, its mole fraction is

 

Don't let this type of notation put you off! The summation sign Σ (Greek Sigma) simply means to add up the n's (number of moles) of every substance present. Thus if O2 is the "i-th" substance as in the expression immediately above, the summation runs from i=1 through i=3.

 

Problem Example 5

A mixture of O2 and nitrous oxide, N2O, is sometimes used as a mild anesthetic in dental surgery. A certain mixture of these gases has a density of 1.482 g L–1 at 25 and 0.980 atm. What was the mole-percent of N2O in this mixture?

Solution: First, find the density the gas would have at STP:

The molar mass of the mixture is (5 g L–1)(22.4 L mol–1) = 37.0 g mol–1. The molecular weights of O2 and N2 are 32 and 44, respectively. Thus the mole fraction of the heavier gas in the mixture is

 

Problem Example 6

What is the mole fraction of carbon dioxide in a mixture consisting of equal masses of CO2 (MW=44) and neon (MW=20.2)?

Solution: Assume any arbitrary mass, such as 100 g, find the equivalent numbers of moles of each gas, and then substitute into the definition of mole fraction:

nCO2 = (100 g) ÷ (44 g mol–1) = 2.3 mol

nNe= (100 g) ÷ (20.2 g mol–1) = 4.9 mol

XNe = (2.3 mol) ÷ (2.3 mol + 4.9 mol) = 0.32

 

 

Dalton's law of partial pressures

 

The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently.

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"Every gas is a vacuum to every other gas."

John Dalton

The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas. Dalton himself stated this law in the simple and vivid way shown at the left. The usual way of stating Dalton's Law of Partial Pressures is

The total pressure of a gas is the sum of the partial pressures of its components

Expressed algebraically,

 

or, equivalently,

(If you feel a need to memorize these formulas, you probably don't really understand Dalton's Law!)

 

Problem Example 7

Calculate the mass of each component present in a mixture of fluorine (MW 19.0) and xenon (MW 131.3) contained in a 2.0-L flask. The partial pressure of Xe is 350 torr and the total pressure is 724 torr at 25°C.

Solution: From Dalton's law, the partial pressure of F2 is (724 – 350) = 374 torr:

The mole fractions are XXe = 350/724 = .48 and XF2 = 374/724 = .52 . The total number of moles of gas is

The mass of Xe is (131.3 g mol–1) × (.48 × .078 mol) = 4.9 g

 

 

Molar mass of a gas mixture

A gas made up of more than one kind of molecule will have a molar mass that is a weighted average of the molar masses of its components. (By the way, the older term "molecular weight" implies a certain mass possessed by individual molecules, so it is commonly modified by the prefix “average” in the context of a gas mixture.) The molar mass of a mixture of gases is just the sum of the mole fractions of each gas, multiplied by the molar mass of that substance.

 

Problem Example 8

Find the average molar mass of dry air whose volume-composition is O2 (21%), N2 (78%) and Ar (1%).

Solution: The average molecular weight is the mole-fraction-weighted sum of the molecular weights of its components. The mole fractions, of course, are the same as the volume-fractions (E.V.E.N. principle.)

m = (.21 x 32) + (.78 x 28) + (.01 x 20) = 28

 

 

 

Some applications of Dalton's Law

Collecting gases over water
A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part.

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The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H2O. The partial pressure of H2O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas.

 

Problem Example 9

Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of O2 produced.

Solution. From Dalton's law, PO2= Ptotal – PH2O = 754 – 19.8 = 734 torr.

Scuba Diving

Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure.

But at a total pressure of 2 atm, the partial pressure of O2 in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the O2 pressure of .8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of O2. This can be achieved most simply by raising the nitrogen content, but high partial pressures of N2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures.

The Chemistry of Scuba Diving

Diving physics and "fizzyology"

Scuba diving and hyperbaric medicine

Scuba Diving Medicine

 

Summary

 

 

 

 

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.

 

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