CoreChem:Conversion Factors

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{reg}1.4 Conversion Factors

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Introduction (Original)

Atoms, Molecules and Chemical Reactions (Original)















16.1 Spontaneous Processes and Molecular Probability

16.2 Thermodynamic Probability and Entropy

16.3 Getting Acquainted with Entropy

16.4 Including the Surroundings

16.5 The Free Energy

16.6 Equilibrium Constants Revisited

Summary 16


17.1 Electrolysis

17.2 Commercial Applications of Electrolysis

17.3 Quantitative Aspects of Electrolysis

17.4 Galvanic Cells

17.5 Electromotive Force of Galvanic Cells

17.6 Commercial Galvanic Cells

17.7 Galvanic Cells and Free Energy

Summary 17


18.1 Experimental Measurement of Rates

18.2 Microscopic View of Chemical Reactions

18.3 Reaction Mechanisms

18.4 Increasing the Rate of a Reaction

18.5 Some Important Types of Catalysts

Summary 18


19.1 Naturally Occurring Nuclear Reactions

19.2 Artificially Induced Nuclear Reactions

19.3 Nuclear Stability

19.4 The Rate of Radioactive Decay

19.5 Detection and Measurement of Radiation

19.6 Uses of Artificial Isotopes in Chemistry

19.7 Mass-Energy Relationships

19.8 Nuclear Fission

19.9 Nuclear Fusion

Summary 19


20.1 The Elements of Life

20.2 The Building Blocks of Biochemistry

20.3 Fats and Lipids

20.4 Carbohydrates

20.5 Proteins

20.6 Nucleic Acids

Summary 20


21.1 The Nature of Electromagnetic Radiation

21.2 Atomic Spectra and the Bohr Theory

21.3 The Spectra of Molecules: Infrared

21.4 The Visible and Ultraviolet Spectra of Molecules: Molecular Orbitals

Summary 21


22.1 Metallic Bonding

22.2 Metallurgy

22.3 Coordination Compounds

22.4 Transitional Metal Ions in Aqueous Solutions

Summary 22

{/reg}

When we are referring to the same object or sample of material, it is often useful to be able to convert one kind of quantity into another. For example, in our discussion of fossil-fuel reserves we could define the quantities unambiguously by stating that 318 Pg (3.18 × 1017 g) of coal, 28.6 km3 (2.68 × 1010 m3) of petroleum, and 2.83 × 103 km3 (2.83 × 1013 m3) of natural gas (measured at normal atmospheric pressure and 15°C) are available. But none of these quantities tells us what we really want to know ― how much heat energy could be released by burning each of these reserves? Only by converting the mass of coal and the volumes of petroleum and natural gas into their equivalent energies can we make a valid comparison. When this is done, we find that the coal could release 7.2 × 1021 J, , the petroleum 1.1 × 1021 J, and the gas 1.1 × 1021 J of heat energy. Thus the reserves of coal are more than three times those of the other two fuels combined. It is for this reason that more attention is being paid to the development of new ways for using coal resources than to oil or gas. Conversion of one kind of quantity into another is usually done with what we shall call a conversion factor. Since we have not yet discussed energy or the units (joules) in which it is measured, an example involving the more familiar quantities mass and volume will be used to illustrate the way conversion factors are employed. The same principles apply to finding how much energy would be released by burning a fuel, and that problem will be encountered later.

Suppose we have a rectangular solid sample of gold which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its volume is 428 cm3 but how much is it worth? The price of gold is about 5 dollars per gram, and so we need to know the mass rather than the volume. It is unlikely that we would have available a scale or balance which could weigh accurately such a large, heavy sample, and so we would have to determine the mass of gold equivalent to a volume of 428 cm3. This can be done by manipulating Eq. (1.1) which defines density. If we multiply both sides by V, we obtain


\text{V}\times \rho =\frac{\text{m}}{\text{V}}\times \text{V}=\text{m}

m = V × ρ or mass = volume × density       (1.2)


Taking the density of gold from Table 1.4, we can now calculate

\text{Mass}=\text{m}=\text{V}\rho =\text{428 cm}^{3}\times \frac{\text{10}\text{.32 g}}{\text{1 cm}^{3}}=8.27\times \text{10}^{3}\text{g}=\text{8}\text{.27 kg}


This is more than 18 lb of gold. At the price quoted above, it would be worth over 40 000 dollars!

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. (1.2) are multiplied by 1/ρ, we have


\begin{align} & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \\ & \text{ V}=\text{m}\times \frac{\text{1}}{\rho } \\ \end{align}       (1.3)


EXAMPLE 1.10 Find the volume occupied by a 4.73-g sample of benzene.

{slide=Solution} According to Table 1.4, the density of benzene is 0.880 g cm–3. Using Eq. (1.3),

\text{Volume = }V\text{ = }m\text{ }\times \text{ }\frac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\frac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}}


(Note that taking the reciprocal of \tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}} simply inverts the fraction ― 1 cm3 goes on top, and 0.880 g goes on the bottom.)

{/slide}


The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because of the mathematical formula, Eq. (1.1), which relates density, mass, and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. (1.2) and (l.3)], and we used them to solve our problems. In practice, however, it is unnecessary to remember all three formulas or do the algebra to derive one from another. It is much easier just to manipulate the quantities involved until the result has the correct units, as shown in the following example.


EXAMPLE 1.11 A student weights 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?

{slide=Solution}

We know that volume is related to mass through density.

Therefore

V = m × conversion factor


Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:


V=m\times \frac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \frac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3}


If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:


V=\text{98}\text{.0 g}\times \frac{\text{13,6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\; (no cancellation!)


It is clear that square grams per cubic centimeter are not the units we want.

{/slide}

Using a conversion factor is very similar to using a unity factor—we know the factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of that relationship [Eqs. (1.1), (1.2), and (1.3) in the case of density, mass, and volume], not because it is equal to one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.


A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:


\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume or }m\overset{\rho }{\longleftrightarrow}V\text{ }


This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.


EXAMPLE 1.12 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).

{slide=Solution} The road map

V\xrightarrow{\rho }m\text{ }


tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:


Mass = m = 47.3 cm3 × \frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}


Since the volume units are different, we need a unity factor to get them to cancel:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}


We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g} {/slide}


In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.

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