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Chapter three

USING CHEMICAL EQUATIONS IN CALCULATIONS There are a great many circumstances in which you may need to use a balanced equation. For example, you might want to know how much air pollution would occur when 100 metric tons of coal were burned in an electric power plant, how much heat could be obtained from a kilogram of natural gas, or how much vitamin C is really present in a 300-mg tablet. In each instance someone else would probably have determined what reaction takes place, but you would need to use the balanced equation to get the desired information. 3.1 EQUATIONS AND MASS RELATIONSHIPS A balanced chemical equation such as 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) (3.1) not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (3.1) says that 4NH3 molecules can react with 5O2 molecules to give 4 NO molecules and 6H2O molecules. It also says that 4 mol NH3 would react with 5 mol O2 yielding 4 mol NO and 6 mol H2O. The balanced equation does more than this, though. It also tells us that 2 × 4 = 8 mol NH3 will react with 2 × 5 = 10 mol O2, and that

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½ × 4 = 2 mol NH3 requires only ½ × 5 = 2.5 mol O2. In other words, the equation indicates that exactly 5 mol O2 must react for every 4 mol NH3 consumed. For the purpose of calculating how much O2 is required to react with a certain amount of NH3 therefore, the significant information contained in Eq. (3.1) is the ratio

2

3

5 mol O4 mol NH

We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (3.1),

2 2

3 3

O 5 mol OS (3.2)

NH 4 mol NH=

⎛ ⎞⎜ ⎟⎝ ⎠

The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. EXAMPLE 3.1 Derive all possible stoichiometric ratios from Eq. (3.1) Solution Any ratio of amounts of substance given by coefficients in the equation may be used:

3 3

2 2

NH 4 mol NHS

O 5 mol O=

⎛ ⎞⎜ ⎟⎝ ⎠

2 2O 5 mol OS

NH 4 mol NH=⎛ ⎞

⎜ ⎟⎝ ⎠

3 3NH 4 mol NHS

NO 4 mol NO=

⎛ ⎞⎜ ⎟⎝ ⎠

2 2

2 2

O 5 mol OS

H O 6 mol H O=

⎛ ⎞⎜ ⎟⎝ ⎠

3 3

2 2

NH 4 mol NHS

H O 6 mol H O=

⎛ ⎞⎜ ⎟⎝ ⎠

2 2

NO 4 mol NOS

H O 6 mol H O=

⎛ ⎞⎜ ⎟⎝ ⎠

There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (3.2) gives one of them.] When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (3.1) as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3):

2

3

O consumed 3 3

NH consumed 2 2

NH 4 mol NHS

O 5 mol On

n= =

⎛ ⎞⎜ ⎟⎝ ⎠

Similarly, the ratio of the amount of produced to the amount of consumed must be 2H O 3NHS(H2O/NH3):

2

3

H O produced 2 2

NH consumed 3 3

H O 6 mol H OS

NH 4 mol NH

n

n= =

⎛ ⎞⎜ ⎟⎝ ⎠

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In general we can say that

X amount of X consumed or produced

Stoichiometric ratio (3.3a)Y amount of Y consumed or produced

=⎛ ⎞⎜ ⎟⎝ ⎠

or, in symbols,

X consumed or produced

Y consumed or produced

XS (3.3.b)

Y

n

n=⎛ ⎞

⎜ ⎟⎝ ⎠

Note that in the word Eq. (3.3a) and the symbolic Eq. (3.3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios. EXAMPLE 3.2 Find the amount of water produced when 3.68 mol NH3 is consumed according to Eq. (3.1). Solution The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to the amount of ammonia consumed:

2

3

H O produced2

3 NH consumed

H OS

NH

n

n=

⎛ ⎞⎜ ⎟⎝ ⎠

Multiplying both sides nNH3 consumed, by we have

2 3

2H O produced NH consumed

3

23

3

2

H OS

NH

6 mol H O 3.68 mol NH

4 mol NH

5.52 mol H O

n n= ×

= ×

=

⎛ ⎞⎜ ⎟⎝ ⎠

This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 3.2 is analogous to Examples 1.10 and 1.11, where density was employed as a conversion factor between mass and volume. Example 3.2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3.3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form

stoichiometric ratio X/Yamount of X consumed amount of Y consumed

or produced or produced←⎯⎯⎯⎯→

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or symbolically.

When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH2 but does not cancel 1 mol H2O. The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. EXAMPLE 3.3 Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation 4FeS2 + 11O2 → 2Fe2O3 + 8SO2 Solution The problem asks that we calculate the mass of SO2 produced. As we learned in Sec. 2.8, Example 2.6, the molar mass can be used to convert from the amount of SO2 to the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of SO2 produced from the amount of O2 consumed. This is the same problem as in Example 3.2. It requires the stoichiometric ratio

2 2

2 2

SO 8 mol SOS

O 11 mol O=

⎛ ⎞⎜ ⎟⎝ ⎠

The amount of SO2 produced is then

2 2SO SO consumed

22 2

2

× conversion factor

8 mol SO 3.84 mol O 2.79 mol SO

11 mol O

n n=

= × = The mass of SO2 is 2SO 2

2

m 2.79 mol SO

179 g SO

= ×

=

2

2

64.06 g SO1 mol SO

With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as

Thus

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The chemical reaction in this example is of environmental interest. Iron pyrite is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide ( ), a major air pollutant.

2(FeS )

2SOOur next example also involves burning a fuel and its effect on the atmosphere. EXAMPLE 3.4 What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O? Solution First, write a balanced equation 2C8H18 + 25O2 → 16CO2 + 18H2O The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of permits calculation of the mass of O2. Symbolically 2O

Thus 12 Pg (petagrams) of O2 would be needed. The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen. The Limiting Reagent Example 3.4 also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the laboratory as well as the environment, inexpensive reagents like atmospheric O2 are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is

66

usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent. EXAMPLE 3.5 When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric bromide, which is the limiting reagent? Solution The balanced equation Hg + Br2 → HgBr2 tells us that according to the atomic theory, 1 mol Hg is required for each mole of Br2. That is, the stoichiometric ratio S(Hg/Br2) + 1 mol Hg/ 1 mol Br2. Let us see how many moles of each we actually have

2

Hg

2Br

1 mol Hg100.0 g

200.59 g

1 mol Br100.0 g

159.80 g

n

n

= ×

= × 2

0.4985 mol Hg

0.6258 mol Br

=

=

When the reaction ends, 0.4985 mol Hg will have reacted with 0.4985 mol Br2 and there will be

(0.6258 – 0.4985) mol Br2 = 0.1273 mol Br2 left over. Mercury is therefore the limiting reagent. From this example yo6 can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 3.5 this ratio of initial amounts

2

Hg

Br 2 2

(initial) 0.4985 mol Hg 0.7966 mol Hg(initial) 0.6258 mol Br 1 mol Br

n

n= =

was less than the stoichiometric ratio

2 2

Hg 1 mol HgS

Br 1 mol Br=

⎛ ⎞⎜ ⎟⎝ ⎠

This indicated that there was not enough Hg to react with all the and mercury was the limiting reagent. The corresponding general rule, for any reagents X and Y, is

2Br

X

Y

X

Y

(initial) XIf is less than S , then X is limiting.

(initial) Y

(initial) XIf is greater than S , then Y is limiting.

(initial) Y

nn

nn

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

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(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example. EXAMPLE 3.6 Iron can be obtained by reacting the ore hematite (Fe2O3) with coke (C). The latter is converted to CO2. As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have? Solution a) Write a balanced equation 2Fe2O3 + 3C → 3CO2 + 4Fe The stoichiometric ratio connecting C and Fe2O3 is

2 3 2 3 2 3

C 3 mol C 1.5 mol CS

Fe O 2 mol Fe O 1 mol Fe O= =

⎛ ⎞⎜ ⎟⎝ ⎠

The initial amounts of C and Fe2O3 are calculated using appropriate molar masses

2 3

6 5C

6 52 3Fe O

1 mol C (initial) 2.84 10 g 2.36 10 mol

12.01 g

1 mol Fe O(initial) 20.5 10 g 1.28 10

159.69 g

n

n

= × × = ×

= × × = × 2 3

C

mol Fe O

Their ratio is

2 3

5C

5Fe O 2 3 2 3

(initial) 2.36 10 mol C 1.84 mol C(initial) 1.28 10 mol Fe O 1 mol Fe O

nn

×= =

×

Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed. Therefore we use nFe2O3

(initial) to calculate how much Fe can be obtained

This is 1.43 × 106 g, or 14.3 Mg, Fe.

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As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. Percent Yield Not all chemical reactions are as simple as the ones we have considered, so far. Quite often a mixture of two or more products containing the same element is formed. For example, when octane (or gasoline in general) burns in an excess of air, the reaction is 2C8H18 + 25O2 → 16CO2 + 18H2O If oxygen is the limiting reagent, however, the reaction does not necessarily stop short of consuming all the octane available. Instead, some carbon monoxide (CO) forms: 2C8H18 + 24O2 → 14CO2 + 2CO + 18H2O Burning gasoline in an automobile engine, where the supply of oxygen is not always as great as that demanded by the stoichiometric ratio, often produces carbon monoxide, a poisonous substance and a major source of air pollution. In other cases, even though none of the reactants is completely consumed, no further increase in the amounts of the products occurs. We say that such a reaction does not go to completion. When a mixture of products is produced or a reaction does not go to completion, the effectiveness of the reaction is usually evaluated in terms of percent yield of the desired product. A theoretical yield is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage:

actual yieldPercent yield 100 percent

theoretical yield= ×

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EXAMPLE 3.7 When 100.0 g N2 gas and 25.0 g H2 gas are mixed at 350°C and a high pressure, they react to form 28.96 g NH3 (ammonia) gas. Calculate the percent yield. Calculate the percent yield. Solution We must calculate the theoretical yield of NH3, and to do this, we must first discover whether N2 or H2 is the limiting reagent. For the balanced equation N2 + 3H2 → 2NH3 the stoichiometric ratio of the reactants is

2 2

2 2

H 3 mol HS

N 1 mol N=

⎛ ⎞⎜ ⎟⎝ ⎠

Now, the initial amounts of the two reagents are

2

2

2H 2

2

2N 2

2

1 mol H(initial) 25.0 g H

2.016 g H

1 mol N(initial) 100.0 g N

28.02 g N

n

n

= ×

= ×

2

2

12.4 mol H

3.569 mol N

=

=

and The ratio of initial amounts is thus

2

2

H 2 2

N 2

(initial) 12.4 mol H 3.47 mol H(initial) 3.569 mol N 1 mol N

n

n= =

2

Since this ratio is greater than 2

2

HS , there is an excess of H2. N2 is the limiting reagent.

Accordingly we must use 3.569 mol N2 (rather than 12.4 mol H2) to calculate the theoretical yield of NH3. We then have

N⎛ ⎞⎜ ⎟⎝ ⎠

3

3NH 2

2

3

2 mol NH(theoretical) 3.569 mol N

1 mol N

7.138 mol NH

n = ×

=

so that

3

3NH 3

3

3

17.03 g NHm (theoretical) 7.138 mol NH

1 mol NH

121.6 g NH

= ×

=

The percent yield is then

actual yieldPercent yield 100 percent

theoretical yield28.96 g

100 percent 23.81 percent121.6 g

= ×

= × =

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Combination of nitrogen and hydrogen to form ammonia is a classic example of a reaction which does not go to completion. Commercial production of ammonia is accomplished using this reaction in what is called the Haber process. Even at the rather unusual temperatures and pressures used for this industrial synthesis, only about one-quarter of the reactants can be converted to the desired product. This is unfortunate because nearly all nitrogen fertilizers are derived from ammonia and the world has come to rely on them in order to produce enough food for its rapidly increasing population. Ammonia ranks third [after sulfuric acid (H2SO4) and oxygen (O2)] in the list of most-produced chemicals, worldwide. It might rank even higher if the reaction by which it is made went to completion. Certainly ammonia and the food it helps to grow would be less expensive and would require much less energy to produce if this were the case. 3.2 ANALYSIS OF COMPOUNDS Up to this point we have obtained all stoichiometric ratios from the coefficients of balanced chemical equations. Chemical formulas also indicate relative amounts of substance, however, and stoichiometric ratios may be derived from them, too. For example, the formula HgBr2 tells us that no matter how large a sample of mercuric bromide we have, there will always be 2 mol of bromine atoms for each mole of mercury atoms. That is, from the formula HgBr2 we have the stoichiometric ratio

Br 2 mol BrS

Hg 1 mol Hg=

⎛ ⎞⎜ ⎟⎝ ⎠

We could also determine that for 2HgBr

(The reciprocals of these stoichiometric ratios are also valid for HgBr2.) Stoichiometric ratios derived from formulas instead of equations are involved in the most common procedure for determining the empirical formulas of compounds which contain only C, H, and O. A weighed quantity of the substance to be analyzed is placed in a combustion train (Fig. 3.1) and

Figure 3.1 A combustion train. H2O and C2O, produced by combination of O2 with H and C in the sample, are selectively absorbed by tubes containing Dehydrite [Mg(ClO4)2•3H2O] and ascarite (NaOH on asbestos).

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heated in a stream of dry O2. All the H in the compound is converted to H2O(g) which is trapped selectively in a previously weighed absorption tube. All the C is converted to CO2(g) and this is absorbed selectively in a second tube. The increase of mass of each tube tells, respectively, how much H2O and CO2 were produced by combustion of the sample. EXAMPLE 3.8 A 6.49-mg sample of ascorbic acid (vitamin C) was burned in a combustion train. 9.74 mg CO2 and 2.64 mg H2O were formed. Determine the empirical formula of ascorbic acid. Solution We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric ratios

2 2

C 1 mol CS

CO 1 mol CO=

⎛ ⎞⎜ ⎟⎝ ⎠

2 2

H 2 mol HS

H O 1 mol H O=

⎛ ⎞⎜ ⎟⎝ ⎠

Thus -3 2

C 22 2

-4

1 mol CO 1 mol C9.74 10 g CO

44.01 g CO 1 mol CO

2.21 10 mol C

n = × ×

= ×

×

-3 2H 2

2 2

-4

1 mol H O 2 mol H2.64 10 g H O

18.02 g H O 1 mol H O

2.93 10 mol H

n = × ×

= ×

× The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample

Thus we have

6.49 mg sample – 2.65 mg C – 0.295 mg H = 3.54 mg O and -3 -4

O

1 mol O3.54 10 g O 2.21 10 mol O

16.00 g On = × × = ×

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The ratios of the amounts of the elements in ascorbic acid are therefore

-4H

-4C

13

2.93 10 mol H 1.33 mol H2.21 10 mol C 1 mol C

1 mol H 4 mol H

1 mol C 3 mol C

nn

×= =

×

= =

-4

O-4

C

2.21 10 mol O 1 mol O 3 mol O2.21 10 mol C 1 mol C 3 mol C

nn

×= = =

×

Since nC:nH:nO is 3 mol C:4 mol H:3 mol O, the empirical formula is C3H4O3. A drawing of a molecule of ascorbic acid is shown in Fig. 3.2. You can determine by counting the atoms that the molecular formula is C6H8O6—exactly double the empirical formula. It is also evident that there is more to know about a molecule than just how many atoms of each kind are present. In ascorbic acid, as in other molecules, the way the atoms are connected together and their arrangement in three-dimensional space are quite important. A picture like Fig. 3.2, showing which atoms are connected to which, is called a structural formula. Empirical formulas may be obtained from percent composition or combustion-train experiments, and, if the molecular weight is known, molecular formulas may be determined from the same data. More complicated experiments are required to find structural formulas. In Example 3.8 we obtained the mass of O by subtracting the masses of C and H from the total mass of sample. This assumed that only C, H, and O were present. Sometimes such an assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that it contained sulfur

Figure 3.2 The structural formula of ascorbic acid (vitamin C), C6H8O6. Carbon atoms are dark gray, hydrogen atoms are light red, and oxygen atoms are dark red. (Computer-generated.) (Copyright@1976 by W.G.Davis and J.W.Moore.)

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was missed. This mistake was not discovered for some time because the atomic weight of sulfur is almost exactly twice that of oxygen. Two atoms of oxygen were substituted in place of one sulfur atom in the formula. 3.3 THERMOCHEMISTRY When a chemical reaction occurs, there is usually a change in temperature of the chemicals themselves and of the beaker or flask in which the reaction is carried out. If the temperature increases, the reaction is exothermic—energy is given off as heat when the container and its contents cool back to room temperature. (Heat is energy transferred from one place to another solely because of a difference in temperature.) An endothermic reaction produces a decrease in temperature. In this case heat is absorbed from the surroundings to return the reaction products to room temperature. Thermochemistry, a word derived from the Greek thermé “heat,” is the measurement and study of energy transferred as heat when chemical reactions take place. It is extremely important in a technological world where a great deal of work is accomplished by transforming and harnessing heat given off during combustion of coal, oil, and natural gas. Energy Energy is usually defined as the capability for doing work. For example, a billiard ball can collide with a second ball, changing the direction or speed of motion of the latter. In such a process the motion of the first ball would also be altered. We would say that one billiard ball did work on (transferred energy to) the other. Energy due to motion is called kinetic energy and is represented by Ek. For an object moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: Ek = ½mu2 (3.4) Where m = mass of the object

u = speed of object If the two billiard balls mentioned above were studied in outer space, where friction due to their collisions with air molecules or the surface of a pool table would be negligible, careful measurements would reveal that their total kinetic energy would be the same before and after they collided. This is an example of the law of conservation of energy, which states that energy cannot be created or destroyed under the usual conditions of everyday life. Whenever there appears to be a decrease in energy somewhere, there is a corresponding increase somewhere else. Even when there is a great deal of friction, the law of conservation of energy still applies. If you put a milkshake on a mixer and leave it there for 10 min, you will have a warm, rather unappetizing drink. The whirling

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mixer blades do work on (transfer energy to) the milkshake, raising its temperature. The same effect could be produced by heating the milkshake, a fact which suggests that heating also involves a transfer of energy. The first careful experiments to determine how much work was equivalent to a given quantity of heat were done by the English physicist James Joule (1818 to 1889) in the 1840s. In an experiment very similar to our milkshake example, Joule connected falling weights through a pulley system to a paddle wheel immersed in an insulated container of water. This allowed him to compare the temperature rise which resulted from the work done by the weights with that which resulted from heating. Units with which to measure energy may be derived from the SI base units of Table 1.2 by using Eq. (3.4). EXAMPLE 3.9 Calculate the kinetic energy of a Volkswagen Beetle of mass 844 kg (1860 lb) which is moving at 13.4 m s–1 (30 miles per hour). Solution Ek = ½mu2 = ½ × 8.44 kg × (13.4 m s–1) 2 = 7.58 × 104 kg m2 s–2 In other words the units for energy are derived from the SI base units kilogram for mass, meter for length, and second for time. A quantity of heat or any other form of energy may be expressed in kilogram meter squared per second squared. In honor of Joule’s pioneering work this derived unit 1 kg m2 s–2 called the joule, abbreviated J. The Volkswagen in question could do nearly 76 000 J of work on anything it happened to run into. Another unit of energy still widely used by chemists is the calorie. The calorie used to be defined as the energy needed to raise the temperature of one gram of water from 14.5°C to 15.5°C but now it is defined as exactly 4.184 J. The Calorie used by dieticians and others for measuring the energy values of foods is actually a kilocalorie, i.e., 4.184 kJ. This nutritional Calorie is distinguished by a capital C in its name. Thermochemical Equations Energy changes which accompany chemical reactions are almost always expressed by thermochemical equations, such as CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) (25°C, 1 atm pressure) ΔHm = –890 kJ mol–1 (3.5)

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Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. (3.5), indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (3.5), the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of ΔHm. It is important to notice that ΔHm is not an energy but rather an energy per unit amount. Its units are not kilojoules but kilojoules per mole (hence the m subscript in ΔHm). This is necessary because the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (3.5) tells us that 890.4 kJ of heat energy is given off for every mole of CH4 which is consumed. Alternatively, it tells us that 890.4 kJ is released for every mole of 2H2O produced, i.e., for every 2 mol H2O produced. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed and n is the amount of substance involved, then

m

qH

nΔ = (3.6)

EXAMPLE 3.10 How much heat energy is obtained when 1 kg of ethane gas, C2H6, is burned in oxygen according to the equation: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ΔHm = –3083 kJ mol–1 (3.7) Solution The mass of C2H6 is easily converted to the amount of C2H6 from which the heat energy q is easily calculated by means of Eq. (3.6). The value of ΔHm is –3083 kJ per mole of 2 mole of 2 C2H6, i.e., per 2 mol C2H6. The road map is

Note: By convention a negative value of q corresponds to a release of heat energy by the matter involved in the reaction.

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The quantity ΔHm is usually referred to as an enthalpy change per mole. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy. It is important to realize that the value of ΔHm given in thermochemical equations like (3.5) or (3.7) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a liquid instead of a gas in the reaction in Eq. (3.5), the value of ΔHm would be different from -890.4 kJ mol–1. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified [as in Eq. (3.7)] they usually refer to 25°C and to normal atmospheric pressure. Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change. For example, H2O(l) → H2O(g) ΔHm = 44 kJ mol–1 (3.8a) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H2O(g) → H2O(l) ΔHm = –44 kJ mol–1 (3.8b) To see why this must be true, suppose that ΔHm [Eq. (3.8a)] = 44 kJ mol–1 while ΔHm [Eq. (3.8b)] = –50.0 kJ mol–1. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is, ΔHm forward = –ΔHm reverse Hess’ Law Perhaps the most useful feature of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is reaction of 1 mol C(s) and 0.5 mol O2(g) to form 1 mol CO(g): C(s) + ½O2(g) → CO(g) ΔHm = –110.5 kJ mol–1 = ΔH1 (Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) In step 2 the mole of CO reads with an additional 0.5 mol O2 yielding 1 mol CO2: CO(g) + ½O2(g) → CO2(g) ΔHm = –283.0 kJ mol–1 = ΔH2

77

The net result of this two-step process is production of 1 mol CO2 from the original 1 mol C and 1 mol O2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2. On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction

Experimentally it is found that the enthalpy change for the net reaction is the sum of the enthalpy changes for steps 1 and 2: ΔHnet = –110.5 kJ mol–1 + (–283.0 kJ mol–1) + –393.5 kJ mol–1 = ΔH1 + ΔH2 That is, the thermochemical equation C(s) + O2(g) → CO2(g) ΔHm = –393.5 kJ mol–1 is the correct one for the overall reaction. In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction. This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows. EXAMPLE 3.11 Acetylene (C2H2) cannot be prepared directly from its elements according to the equation 2C(s) + H2(g) → C2H2(g) (3.9) Calculate ΔHm for this reaction from the following thermochemical equations, all of which can be determined experimentally:

C(s) + O2(g) → CO2(g) ΔHm = –393.5 kJ mol–1 (3.10a) H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ mol–1 (3.10b) C2H2(g) + 5

2 O2(g) → 2CO2(g) + H2O(l) ΔHm = –1299.8 kJ mol–1 (3.10c) Solution We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (3.9): a) Since Eq. (3.9) has 2 mol C on the left, we multiply Eq. (3.10a) by 2.

78

b) Since Eq. (3.9) has 1 mol H2 on the left, we leave Eq. (3.10b) unchanged. c) Since Eq. (3.9) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of Eq. (3.10c) we write Eq. (3.10c) in reverse. We then have

2 2

2 2 2

2 2 2 2 2

2 2 2 2 2

-1 -1m

12

52

512 2

2 C(s) + O (g) CO (g)

H (g) + O (g) H O( )

2 CO (g) + H O( ) C H (g) + O (g)

2 C(s) + H (g) + 2 O (g) C H (g) + O (g)

( 787.0 285.8 + 1299.8) kJ mol 227.0 kJ mol

l

l

H

→

→

→

→

Δ = − − =

-1m

-1m

-1m

2 ( 393.5) kJ mol

285.8 kJ mol

( 1299.8 kJ) mol

H

H

H

Δ = −

Δ = −

Δ = − −

Thus the desired result is 2C(s) + H2(g) → C2H2(g) ΔHm = 227.0 kJ mol–1 3.4 STANDARD ENTHALPIES OF FORMATION By now chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess’ law makes it possible to list a single value, the standard enthalpy of formation ΔHf, for each compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C). For example, if we know that ΔHf[H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation

H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ mol–1 (3.11) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed. Equation (3.11) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½. In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine ΔHf. Quite often, however, elements do not react directly with each other to form the desired compound, and ΔHf must be calculated by combining the enthalpy changes for other reactions. A case in point is the gas acetylene, C2H2. In Example 3.11 we used Hess’ law to show that the thermochemical equation 2C(s) + H2(g) → C2H2(g) ΔHm = 227.0 kJ mol–1

79

TABLE 3.1 Some Standard Enthalpies of Formation at 25°C.

3

2

2 3

2

2

4

2

2 2

2 4

2 6

6 6

3

2 3

AgCl(s)AgN (s)Ag O(s)Al O (s)Br (l)Br (g)C(s), graphiteC(s), diamondCH (g)CO(g)CO (g)C H (g)C H (g)C H (g)C H (l)CaO(s)CaCO (s)CuO(s)Fe O (s)HBr(g)HCl(g)HI(g)

- 127.0 + 310.3 - 31.0- 1675.7 0.0 + 31.0 0.0 + 1.9 - 74.8 - 110.5 - 393.5 + 226.9 + 52.6 - 84.5 + 49.1 - 635.5- 1207.8 - 157.3 - 822.2 - 36.4 - 92.3 + 26.4

- 30.35 + 66.79 - 7.41 - 400.50 0.00 + 7.41 0.00 + 0.45 - 17.88 - 26.41 - 94.05 + 54.23 + 12.57 - 20.20 + 11.74 - 151.89 - 288.67 - 37.60 - 196.51 - 8.70 - 22.06 + 6.31

2

2

2 2

2

2

2

3

2

2 4

3

3

2

3

H O(g)H O(l)H O (l)H S(g)HgO(s)I (s)I (g)KCl(s)KBr(s)MgO(s)NH (g)NO(g)NO (g)N O (g)NF (g)NaBr(s)NaCl(s)O (g)SO (g)SO (g)ZnO(s)

- 241.8- 285.8- 187.8 - 20.2 - 90.8 0.0 + 62.3- 435.9- 392.0- 601.8 - 46.1 + 90.3 + 33.8 + 9.2- 124.7- 359.9- 411.0

+ 142.7- 296.8- 395.8- 348.1

- 57.79 - 68.31 - 44.84 - 4.83 - 21.70 0.0+ 14.89

- 104.18 - 93.69

- 143.83 - 11.02+ 21.58 + 8.08 + 2.20 - 29.80 - 86.02 - 98.23+ 34.11 - 70.94 - 94.60 - 83.20

Δ Δ Δ Δ-1 -1 -1

Compound Compound

kcal kcal kcal kcal mol mol mol

f f f fH H H H

-1 mol

was valid. Since it involves 1 mol C2H2 and the elements are in their most stable forms, we can say that ΔHf.[C2H2(g)] = 227.0 kJ mol–1. One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. If we form mercury from its elements, for example, we are talking about the reaction Hg(l) → Hg(l) Since the mercury is unchanged, there can be no enthalpy change, and ΔHf = 0 kJ mol–1. Standard enthalpies of formation for some common compounds are given in Table 3.1. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example. EXAMPLE 3.12 Use standard enthalpies of formation to calculate ΔHm for the reaction 2CO(g) + O2(g) → 2CO2(g)

80

Solution We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step CO (carbon monoxide) is decomposed to its elements: 2CO(g) → 2C(s) + O2(g) ΔHm = ΔHf (3.12) Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is ΔH1 = 2 × {–ΔHf [CO(g)]} = 2 × [– (–110.5 kJ mol–1)] = +221.0 kJ mol–1 In the second step the elements are combined to give 2 mol (carbon dioxide): 2CO 2C(s) → 2O2(g) + CO2(g) ΔHm = ΔH2 (3.13) In this case ΔH2 = 2 × ΔHf [CO(g)] = 2 × (393.5 kJ mol–1) = 787.0 kJ mol–1 You can easily verify that the sum of Eqs. (3.12) and (3.13) is 2CO(g) + 2 O2(g) → CO2(g) ΔHm = ΔHnet Therefore ΔHnet = ΔH1 + ΔH2 = 221.0 kJ mol–1 – 393.5 kJ mol–1 = – 172.5 kJ mol–1 Note carefully how Example 3.12 was solved. In step 1 the reactant compound CO(g) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH1 was opposite in sign from ΔHf. Step 1 also involved 2 mol CO(g) and so the enthalpy change had to be doubled. In step 2 we had the hypothetical formation of the product CO2(g) from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same. Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants) (3.14) The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation for which ΔHm is being calculated.

81

EXAMPLE 3.13 Use Table 3.1 to calculate ΔHm for the reaction 4NH3(g) 5O2(g) → 6H2O(g) + 4NO(g) Solution Using Eq. (3.14), we have ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants) = [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)] = 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0 = –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1 = –905.2 kJ mol–1 Note that we were careful to use ΔHf [H2O(g)] not ΔHf [H2O(l)]. Even though water vapor is not the most stable form of water at 25°C, we can still use its ΔHf value. Also the standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it would be a waste of space to include it in Table 3.1. 3.5 SOLUTIONS In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another. There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquid solvent which dissolves a solid solute. (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions. Concentration Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react

82

with a given quantity of another. Examples earlier in this chapter have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance. The concentration c of a substance in a solution (often called molarity) is the amount of the substance per unit volume of solution:

solutesolute

solute

amount of soluteConcentration of solute

volume of soluten

cV

= =

Usually the units moles per cubic decimeter (mol dm–3) or moles per liter (mol liter–1) are used to express concentration. If a pure substance is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in Fig. 3.3. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. EXAMPLE 3.14 A solution of KI was prepared as described above. The initial mass of the container plus KI was 43.2874 g, and the final mass after pouring was 30.1544 g. The volume of the flask was 250.00 ml. What is the concentration of the solution? Solution The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (3.15)]:

KIKI

nc

V=

We obtain from the mass of KI added to the flask: KIn -2

3.1330 g

.9114 10 mol

=

= ×

KI

KI

m 43.2874 g 30,1544 g 1

1 mol13.1330 g 7

166.00 gn

= −

= × The volume of solution is 250.00 ml, or

-13

3 3solution 3 3

1 dm250.00 cm 2.5000 10 dm

10 cm V = × = ×

Thus

-1-2

-3KIKI -1 3

solution

17.9114 10 mol 3.1645 10 mol dm

250.00 10 m d

nc

V×

= = = ××

83

Figure 3.3 The preparation of a sodium chloride solution of concentration 1.000 mol dm–3. (a) A sample of pure solid NaCl is firs weighed with its container. (b) Most of the sample is now transferred into the volumetric flask. (c) The container and remaining solid NaCl are again weighed. Subtraction yields mNaCl = 58.44 g. Thus nNaCl = 1.000 mol. (d) Any solid remaining in the funnel is washed down into the body of the flask. (e) The flask is now filled to about 80 percent capacity and the solid allowed to dissolve. The flask is shaken to achieve a uniform solution. (f) Solvent is now added carefully until the bottom of the me- niscus coincides with the mark on the flask. The flask is again shaken to achieve a uniform solution. Since the volume of the solution is 1.000 dm3, the concentration is 1.000 mol/1.000 dm3 = 1.000 mol dm–

3. Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute.

Volume of solution amount of solute concentration cV n←⎯⎯⎯⎯→ ← →⎯ Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (aqueous solutions). EXAMPLE 3.15 An aqueous solution of HCl [represented or written HCl(aq)] has a concentration of 0.1396 mol dm–3. If 24.71 cm³ (24.71 ml) of this solution is delivered from a buret, what amount of HCl has been delivered?

84

Solution Using concentration as a conversion factor, we have cV n⎯⎯→ The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters: The concentration units of moles per cubic decimeter are often abbreviated M, pronounced molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that 1 dm3 × 1 M = 1 mol Therefore we recommend that you always write the units in full when doing any calculations involving solution concentrations. That is, Problems such as Example 3.15 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm–3) instead of moles per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and Thus a concentration of 0.1396 mol dm–3 (0.1396 M) can also be expressed as 0.1396 mmol cm–3. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters. EXAMPLE 3.16 Exactly 25.0 ml NaOH solution whose concentration is 0.0974 M was delivered from a pipet. (a) What amount of NaOH was present? (b) What mass of NaOH would remain if all the water evaporated?

33

0.1396 mol1 dm

= ×HCl 24.71 cmn

3

3

3

3 3 3

mol 1 dm dm 10 cm

6 mol 1 dmm 10 cm

×

×

⎛ ⎞⎜ ⎟⎝ ⎠

3HCl

3

0.139624.71 cm

1

0.139 24.71 cm

1 d 0.003 450 mol

n = ×

= ×

=

33

mol1 dm 1 1mol

dm× =

3

3 3 3 3 -3

1 mol 1 mol 1 dm 1 mmol 1 mmol1 dm 1 dm 10 cm 10 mol 1 cm

= × × ×3

85

Solution a) Since 0.0974 M means 0.0974 mol dm–3, or 0.0974 mmol cm–3, we choose the latter, more convenient quantity as a conversion factor:

3NaOH 3

-3

0.0974 mol25.0 cm 2.44 mmol

1 cm 2.44 10 mol

n = × =

= ×

b) Using molar mass, we obtain

-3NaOH

40.01 gm 2.44 10 mol 9.

1 mol = × × = -276 10 g×

Note: The symbols nNaOH and mNaOH refer to the amount and mass of the solute NaOH, respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous NaOH solution, the symbol mNaOH(aq) could be used. Diluting and Mixing Solutions Often it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution as described in Example 3.14. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be di- luted to obtain the desired concentration. EXAMPLE 3.17 A pipet is used to measure 50.0 ml of 0.1027 M HCl into a 250.00-ml volumetric flask. Distilled water is carefully added up to the mark on the flask. What is the concentration of the diluted solution? Solution To calculate concentration, we first obtain the amount of HCI in the 50.0 ml (50.0 cm3) of solution added to the volumetric flask: Dividing by the new volume gives the concentration -3mol cmHCl

HCl 3

5.14 mol0.0205 m

250.00 cmn

cV

= = = Thus the new solution is 0.0205 M. EXAMPLE 3.18 What volume of the solution prepared in Example 3.14 would be required to make 50.00 ml of 0.0500 M KI?

86

Solution Using the volume and concentration of the desired solution, we can calculate the amount of KI required. Then the concentration of the original solution (0.316 46 M) can be used to convert that amount of KI to the necessary volume. Schematically Thus we should dilute a 7.90-ml aliquot of the stock solution to 50.00 ml. This could be done by measuring 7.90 ml from a buret into a 50.00-ml volumetric flask and adding water up to the mark. Titrations When solutions are used quantitatively in the laboratory, titration is usually involved. Titration is a technique used to determine the volume of one solution necessary to consume exactly some reactant in another solu-

Figure 3.4 The technique of titration.

87

tion. As shown in Fig. 3.4, a measured volume of the solution to be titrated is placed in a flask or other container. The titrant (the solution to be added) is placed in a buret. The volume of titrant added can be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, for example, 25.62 ml. In Fig. 3.4 the solution to be titrated is colorless aqueous hydrogen peroxide, H2O2(aq), which contains excess sulfuric acid H2SO4(aq). The titrant is purple-colored potassium permanganate solution KMnO4(aq). The reaction which occurs is 2KMnO4(aq) + 5H2O2 + 3H2SO4 → 2MnSO4(aq) + 5O2(g) + K2SO4(aq) + 8H2O(l) (3.16)

.. As the first few cubic centimeters of titrant flow into the flask, there is a large excess of H2O2. The limiting reagent KMnO4 is entirely consumed, and its purple color disappears almost as soon as it is added. Eventually, though, all the H2O2 is consumed. Addition of just one more drop of titrant produces a lasting pink color due to unreacted KMnO4 in the flask. This indicates that all the H2O2 has been consumed and is called the endpoint of the titration. If more KMnO4 solution were added, there would be an excess of KMnO4 and the color of the solution in the flask would get much darker. The darker color would show that we had overtitrated, or overshot the endpoint, by adding more than enough KMnO4 to react with all the H2O2. In the titration we have just described, the intense purple color of permanganate indicates the endpoint. Usually, however, it is necessary to add an indicator—a substance which combines with excess titrant to produce a visible color or to form an insoluble substance which would precipitate from solution. No matter what type of reaction occurs or how the endpoint is detected, however, the object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substance being titrated. In the KMnO4—H2O2 reaction [Eq. (3.16)], the endpoint occurs when exactly 2 mol KMnO4 have been added from the buret for every 5 mol H2O2 originally in the titration flask. That is, at the endpoint the ratio of the amount of KMnO4, added to the amount of H2O2 consumed must equal the stoichiometric ratio

4

2 2

KMnO 4 4

H O 2 2 2 2

(added from buret) KMnO 2 mol KMnOS (3

(initially in flask) H O 5 mol H O

n

n= =

⎛ ⎞⎜ ⎟⎝ ⎠

.17) When the endpoint has been reached in any titration, the ratio of the amounts of substance of the two reactants is equal to the stoichiometric ratio obtained from the appropriate balanced chemical equation. Therefore we can use the stoichiometric ratio to convert from the amount of one substance to the amount of another. EXAMPLE 3.19 What volume of 0.05386 M KMnO4 would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 M H2O2.

88

Solution At the endpoint, Eq. (3.17) will apply, and we can use it to calculate the amount of KMnO4 which must be added:

4 2 2

4KMnO H O

2 2

KMnO(added) (in flask) S

H On n= ×

⎛ ⎞⎜ ⎟⎝ ⎠

The amount of H2O2 is obtained from the volume and concentration:

2 2

3H O 3

2 2

mmol(in flask) 25.00 cm 0.1272

cm 3.180 mmol H O

n = ×

=

Then

4

-34

KMnO 2 2 -32 2

42 2

2 2

4

2 mol KMnO 10(added) 3.180 mmol H O

5 mol H O 10

2 mmol KMnO 3.180 mmol H O

5 mmol H O

1.272 mmol KMnO

n = ×

= ×

=

× To obtain VKMnO4(aq) we use the concentration as a conversion factor:

4

3

KMnO ( ) 4 -24

1 cm1.272 mmol KMnO 23.62 c

5.386 10 mmol KMnO aqV = ××

3m= Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4. Titration is often used to determine the concentration of a solution. In many cases it is not a simple matter to obtain a pure substance, weigh it accurately, and dissolve it in a volumetric flask as was done in Example 3.14. NaOH, for example, combines rapidly with H2O and CO2 from the air, and so even a freshly prepared sample of solid NaOH will not be pure. Its weight would change continuously as CO2(g) and H2O(g) were absorbed. Hy- drogen chloride (HCl) is a gas at ordinary temperatures and pressures, making it very difficult to handle or weigh. Aqueous solutions of both of these substances must be standardized; that is, their concentrations must be determined by titration. EXAMPLE 3.20 A sample of pure potassium hydrogen phthalate (KHC8H4O4) weighing 0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml NaOH(aq). The titration reaction is NaOH(aq) + KHC8H4O4(aq) → NaKC8H4O4(aq) + H2O What is the concentration of NaOH(aq) ?

89

Solution To calculate concentration, we need to know the amount of NaOH and the volume of solution in which it is dissolved. The former quantity could be obtained via a stoichiometric ratio from the amount of KHC8H4O4, and that amount can be obtained from the mass

KHC H O8 4 4 8 4 4

8 4 4 8 4 4KHC H O KHC H O NaOH(NaOH/KHC H O ) M Sm n⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯⎯→ n

8 4 4

NaOH8 4

-3

1 mol KHC H O 1 mol NaOH3.180 g

204.22 g 1 mol KHC H O

1.674 10 mol NaOH 1.675 mmol

n = × ×

= × =

4

NaOH

The concentration is -3mNaOH

NaOH 3

1.675 mmol NaOH0.06197 mmol c

27.03 cmn

cV

= = = or 0.06197 M. By far the most common use of titrations is in determining unknowns, that is, in determining the concentration or amount of substance in a sample about which we initially knew nothing. The next example involves an unknown that many persons encounter every day. EXAMPLE 3.21 Vitamin C tablets contain ascorbic acid (C6H8O6) and a starch “filler” which holds them together. To determine how much vitamin C is present, a tablet can be dissolved in water and with sodium hydroxide solution, NaOH(aq). The equation is C6H8O6(aq) + NaOH(aq) → Na C6H7O6(aq) + H2O(l) If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 M NaOH, how accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin C? Solution The known volume and concentration allow us to calculate the amount of NaOH(aq) which reacted with all the vitamin C. Using the stoichiometric ratio

6 8 6C H O 1S

NaOH 1=

⎛ ⎞⎜ ⎟⎝ ⎠

6 8 6 mmol C H O mmol NaOH

we can obtain the amount of C6H8O6. The molar mass converts that amount to a mass which can be compared with the label. Schematically

90

C H ONaOH 6 8 6 6 8 6

6 8 6

S(C H O /NaOH)NaOH NaOH C H O

McV n n⎯⎯⎯→ ⎯⎯⎯⎯⎯→ ⎯⎯⎯→6 8 6C H Om⎯

6 8 6

3C H O 3

6 8 6

6 8 6

0.1038 mmol NaOHm 16.85 cm

1 cm1 mmol C H O 176.1 mg

308.0 1 mmol NaOH mmol C H O

= ×

× ×

mg= Note that the molar mass of C6H8O6 -3 -3

-3 -36 8 6 6 8 6 6 8 6 6 8

176.1 g 176.1 g 10 176.1 g 10 176.1 mg 1 mol C H O 1 mol C H O 10 10 mol C H O 1 mmol C H

×= × = =

6O can be expressed in milligrams per millimole as well as in grams per mole. The 308.0 mg obtained in this example is in reasonably close agreement with the manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed individually, and so some variation is expected. SUMMARY This chapter has been concerned with the amounts of substances which participate in chemical reactions, with the quantities of heat given off or absorbed when reactions occur, and with the volumes of solutions which react exactly with one another. These seemingly unrelated subjects have been discussed together because many of the calculations invo1ving them are almost identical in form. The same is true of the density calculations described in Chap. 1 (Secs. 1.3 and 1.4) and of the cal- TABLE 3.2 Summary of Related Quantities and Conversion Factors. Related Quantities Conversion Factor Definition Road Map Volume ↔ mass Density, ρ m

Vρ = V mρ←⎯→

Amount of substance ↔ mass Molar Mass, M mMn

= Mn m←⎯→

Amount of substance ↔ number of particles

Avogadro constant, NA A

NNn

= A Nn N←⎯⎯→

Amount of X consumed or produced ↔ amount of Y consumed or produced

Stoichiometric ratio, S(Y/X)

Y

X

(Y/X) nSn

= (Y/X)

X Y Sn n←⎯⎯→

Amount of X consumed or produced ↔ quantity of heat absorbed during reaction

ΔHm for thermo-chemical equation m

X

qHn

Δ = m

X Hn qΔ←⎯⎯→

Volume of solution ↔ amount of solute

Concentration of solute, cX

XX

ncV

= X

X cV n←⎯→

91

culations involving molar mass and the Avogadro constant in Chap. 2 (Secs. 2.4 and 2.5). In each case one quantity is defined as the ratio of two others. The first quantity serves as a conversion factor relating the other two. A summary of the relationships and conversion factors we have encountered so far is given in Table 3.2. An incredible variety of problems can he solved using the conversion factors in Table 3.2. Sometimes only one factor is needed, but quite often several are applied in sequence, as in Example 3.21. In solving such problems, it is necessary first to think your way through, perhaps by writing down a road map showing the relationships among the quantities given in the problem. Then you can apply conversion factors, making sure that the units cancel, and calculate the result. The examples and end-of-chapter problems in this and the preceding two chapters should give you some indication of the broad applications of the problem-solving techniques we have developed here. Once you have mastered these techniques, you will be able to do a great many useful computations which are related to problems in the chemical laboratory, in everyday life, and in the general environment. You will find that the same type of calculations, or more complicated problems based on them, will he encountered again and again throughout your study of chemistry and other sciences.