Chapter three USING CHEMICAL EQUATIONS IN CALCULATIONS 3.1 EQUATIONS AND MASS RELATIONSHIPS

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1 61 Chapter three USING CHEMICAL EQUATIONS IN CALCULATIONS There are a great may circumstaces i which you may eed to use a balaced equatio. For example, you might wat to kow how much air pollutio would occur whe 100 metric tos of coal were bured i a electric power plat, how much heat could be obtaied from a kilogram of atural gas, or how much vitami C is really preset i a 00- mg tablet. I each istace someoe else would probably have determied what reactio takes place, but you would eed to use the balaced equatio to get the desired iformatio..1 EQUATIONS AND MASS RELATIONSHIPS A balaced chemical equatio such as 4NH (g) + 5O (g) 4NO(g) + 6H O(g) (.1) ot oly tells how may molecules of each kid are ivolved i a reactio, it also idicates the amout of each substace that is ivolved. Equatio (.1) says that 4NH molecules ca react with 5O molecules to give 4 NO molecules ad 6H O molecules. It also says that 4 mol NH would react with 5 mol O yieldig 4 mol NO ad 6 mol H O. The balaced equatio does more tha this, though. It also tells us that 4 = 8 mol NH will react with 5 = 10 mol O, ad that

2 6 ½ 4 = mol NH requires oly ½ 5 =.5 mol O. I other words, the equatio idicates that exactly 5 mol O must react for every 4 mol NH cosumed. For the purpose of calculatig how much O is required to react with a certai amout of NH therefore, the sigificat iformatio cotaied i Eq. (.1) is the ratio 5 mol O 4 mol NH We shall call such a ratio derived from a balaced chemical equatio a stoichiometric ratio ad give it the symbol S. Thus, for Eq. (.1), O 5 mol O S = (.) NH 4 mol NH The word stoichiometric comes from the Greek words stoicheio, elemet, ad metro, measure. Hece the stoichiometric ratio measures oe elemet (or compoud) agaist aother. EXAMPLE.1 Derive all possible stoichiometric ratios from Eq. (.1) Solutio Ay ratio of amouts of substace give by coefficiets i the equatio may be used: NH 4 mol NH O 5 mol O S = S = O 5 mol O NH 4 mol NH NH 4 mol NH S = NO 4 mol NO O 5 mol O S = HO 6 mol HO NH 4 mol NH S = HO 6 mol HO NO 4 mol NO S = HO 6 mol HO There are six more stoichiometric ratios, each of which is the reciprocal of oe of these. [Eq. (.) gives oe of them.] Whe ay chemical reactio occurs, the amouts of substaces cosumed or produced are related by the appropriate stoichiometric ratios. Usig Eq. (.1) as a example, this meas that the ratio of the amout of O cosumed to the amout of NH cosumed must be the stoichiometric ratio S(O /NH ): O cosumed NH 4 mol NH = S = NH cosumed O 5 mol O Similarly, the ratio of the amout of HOproduced to the amout of NH cosumed must be S(H O/NH ): H O produced HO 6 mol HO = S = NH cosumed NH 4 mol NH

3 6 I geeral we ca say that Stoichiometric ratio X amout of X cosumed or produced = Y amout of Y cosumed or produced (.a) or, i symbols, X Y X cosumed or produced S = (..b) Y cosumed or produced Note that i the word Eq. (.a) ad the symbolic Eq. (.b), X ad Y may represet ay reactat or ay product i the balaced chemical equatio from which the stoichiometric ratio was derived. No matter how much of each reactat we have, the amouts of reactats cosumed ad the amouts of products produced will be i appropriate stoichiometric ratios. EXAMPLE. Fid the amout of water produced whe.68 mol NH is cosumed accordig to Eq. (.1). Solutio The amout of water produced must be i the stoichiometric ratio S(H O/NH ) to the amout of ammoia cosumed: HO HO produced S = NH NH cosumed Multiplyig both sides NH cosumed, by we have H O produced HO = S NH cosumed NH 6 mol H O =.68 mol NH 4 mol NH = 5.5 mol H O This is a typical illustratio of the use of a stoichiometric ratio as a coversio factor. Example. is aalogous to Examples 1.10 ad 1.11, where desity was employed as a coversio factor betwee mass ad volume. Example. is also aalogous to Examples.4 ad.6, i which the Avogadro costat ad molar mass were used as coversio factors. As i these previous cases, there is o eed to memorize or do algebraic maipulatios with Eq. (.) whe usig the stoichiometric ratio. Simply remember that the coefficiets i a balaced chemical equatio give stoichiometric ratios, ad that the proper choice results i cacellatio of uits. I road-map form stoichiometric ratio X/Y amout of X cosumed amout of Y cosumed or produced or produced

4 64 or symbolically. Whe usig stoichiometric ratios, be sure you always idicate moles of what. You ca oly cacel moles of the same substace. I other words, 1 mol NH cacels 1 mol NH but does ot cacel 1 mol H O. The ext example shows that stoichiometric ratios are also useful i problems ivolvig the mass of a reactat or product. EXAMPLE. Calculate the mass of sulfur dioxide (SO ) produced whe.84 mol O is reacted with FeS accordig to the equatio 4FeS + 11O Fe O + 8SO Solutio The problem asks that we calculate the mass of SO produced. As we leared i Sec..8, Example.6, the molar mass ca be used to covert from the amout of SO to the mass of SO. Therefore this problem i effect is askig that we calculate the amout of SO produced from the amout of O cosumed. This is the same problem as i Example.. It requires the stoichiometric ratio SO 8 mol SO S = O 11 mol O The amout of SO produced is the SO = SO cosumed coversio factor 8 mol SO =.84 mol O =.79 mol SO 11 mol O The mass of SO is g SO m =.79 mol SO SO 1 mol SO = 179 g SO With practice this kid of problem ca be solved i oe step by cocetratig o the uits. The appropriate stoichiometric ratio will covert moles of O to moles of SO ad the molar mass will covert moles of SO to grams of SO. A schematic road map for the oestep calculatio ca be writte as Thus

5 65 The chemical reactio i this example is of evirometal iterest. Iro pyrite (FeS ) is ofte a impurity i coal, ad so burig this fuel i a power plat produces sulfur dioxide ( SO ), a major air pollutat. Our ext example also ivolves burig a fuel ad its effect o the atmosphere. EXAMPLE.4 What mass of oxyge would be cosumed whe g,. Pg (petagrams), of octae (C 8 H 18 ) is bured to produce CO ad H O? Solutio First, write a balaced equatio C 8 H O 16CO + 18H O The problem gives the mass of C 8 H 18 bured ad asks for the mass of O required to combie with it. Thikig the problem through before tryig to solve it, we realize that the molar mass of octae could be used to calculate the amout of octae cosumed. The we eed a stoichiometric ratio to get the amout of O cosumed. Fially, the molar mass of permits calculatio of the mass of O. Symbolically O Thus 1 Pg (petagrams) of O would be eeded. The large mass of oxyge obtaied i this example is a estimate of how much O is removed from the earth s atmosphere each year by huma activities. Octae, a compoet of gasolie, was chose to represet coal, gas, ad other fossil fuels. Fortuately, the total mass of oxyge i the air ( g) is much larger tha the yearly cosumptio. If we were to go o burig fuel at the preset rate, it would take about years to use up all the O. Actually we will cosume the fossil fuels log before that! Oe of the least of our evirometal worries is ruig out of atmospheric oxyge. The Limitig Reaget Example.4 also illustrates the idea that oe reactat i a chemical equatio may be completely cosumed without usig up all of aother. I the laboratory as well as the eviromet, iexpesive reagets like atmospheric O are ofte supplied i excess. Some portio of such a reaget will be left uchaged after the reactio. Coversely, at least oe reaget is

6 66 usually completely cosumed. Whe it is goe, the other excess reactats have othig to react with ad they caot be coverted to products. The substace which is used up first is the limitig reaget. EXAMPLE.5 Whe g mercury is reacted with g bromie to form mercuric bromide, which is the limitig reaget? Solutio The balaced equatio Hg + Br HgBr tells us that accordig to the atomic theory, 1 mol Hg is required for each mole of Br. That is, the stoichiometric ratio S(Hg/Br ) + 1 mol Hg/ 1 mol Br. Let us see how may moles of each we actually have 1 mol Hg = g = mol Hg Hg g = g Br 1 mol Br g = mol Br Whe the reactio eds, mol Hg will have reacted with mol Br ad there will be ( ) mol Br = 0.17 mol Br left over. Mercury is therefore the limitig reaget. From this example yo6 ca begi to see what eeds to be doe to determie which of two reagets, X or Y, is limitig. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amouts of X ad Y which were iitially mixed together. I Example.5 this ratio of iitial amouts (iitial) Hg mol Hg mol Hg = = (iitial) mol Br 1 mol Br Br was less tha the stoichiometric ratio Hg 1 mol Hg S = Br 1 mol Br This idicated that there was ot eough Hg to react with all the Br ad mercury was the limitig reaget. The correspodig geeral rule, for ay reagets X ad Y, is X If is less tha S, the X is limitig. Y X If is greater tha S, the Y is limitig. Y (iitial) (iitial) (iitial) (iitial) X Y X Y

7 67 (Of course, whe the amouts of X ad Y are i exactly the stoichiometric ratio, both reagets will be completely cosumed at the same time, ad either is i excess.). This geeral rule for determiig the limitig reaget is applied i the ext example. EXAMPLE.6 Iro ca be obtaied by reactig the ore hematite (Fe O ) with coke (C). The latter is coverted to CO. As maager of a blast furace you are told that you have 0.5 Mg (megagrams) of Fe O ad.84 Mg of coke o had. (a) Which should you order first aother shipmet of iro ore or oe of coke? (b) How may megagrams of iro ca you make with the materials you have? Solutio a) Write a balaced equatio Fe O + C CO + 4Fe The stoichiometric ratio coectig C ad Fe O is C mol C 1.5 mol C S = = Fe O mol Fe O 1 mol Fe O The iitial amouts of C ad Fe O are calculated usig appropriate molar masses Fe O 6 5 (iitial) = g =.6 10 mol C 1 mol C 1.01 g 1 mol Fe O (iitial) g mol Fe O g 6 5 = = C Their ratio is (iitial).6 10 mol C 1.84 mol C 5 C = = 5 (iitial) mol Fe O 1 mol Fe O FeO Sice this ratio is larger tha the stoichiometric ratio, you have more tha eough C to react with all the Fe O. Fe O is the limitig reaget, ad you will wat to order more of it first sice it will be cosumed first. b) The amout of product formed i a reactio may be calculated via a appropriate stoichiometric ratio from the amout of a reactat which was cosumed. Some of the excess reactat C will be left over, but all the iitial amout of Fe O will be cosumed. Therefore we use Fe O (iitial) to calculate how much Fe ca be obtaied This is g, or 14. Mg, Fe.

8 68 As you ca see from the example, i a case where there is a limitig reaget, the iitial amout of the limitig reaget must be used to calculate the amout of product formed. Usig the iitial amout of a reaget preset i excess would be icorrect, because such a reaget is ot etirely cosumed. The cocept of a limitig reaget was used by the ieteeth cetury Germa chemist Justus vo Liebig (1807 to 187) to derive a importat biological ad ecological law. Liebig s law of the miimum states that the essetial substace available i the smallest amout relative to some critical miimum will cotrol growth ad reproductio of ay species of plat or aimal life. Whe a group of orgaisms rus out of that essetial limitig reaget, the chemical reactios eeded for growth ad reproductio must stop. Vitamis, protei, ad other utriets are essetial for growth of the huma body ad of huma populatios. Similarly, the growth of algae i atural bodies of water such as Lake Erie ca be ihibited by reducig the supply of utriets such as phosphorus i the form of phosphates. It is for this reaso that may states have regulated or baed the use of phosphates i detergets ad are costructig treatmet plats which ca remove phosphates from muicipal sewage before they eter lakes or streams. Percet Yield Not all chemical reactios are as simple as the oes we have cosidered, so far. Quite ofte a mixture of two or more products cotaiig the same elemet is formed. For example, whe octae (or gasolie i geeral) burs i a excess of air, the reactio is C 8 H O 16CO + 18H O If oxyge is the limitig reaget, however, the reactio does ot ecessarily stop short of cosumig all the octae available. Istead, some carbo mooxide (CO) forms: C 8 H O 14CO + CO + 18H O Burig gasolie i a automobile egie, where the supply of oxyge is ot always as great as that demaded by the stoichiometric ratio, ofte produces carbo mooxide, a poisoous substace ad a major source of air pollutio. I other cases, eve though oe of the reactats is completely cosumed, o further icrease i the amouts of the products occurs. We say that such a reactio does ot go to completio. Whe a mixture of products is produced or a reactio does ot go to completio, the effectiveess of the reactio is usually evaluated i terms of percet yield of the desired product. A theoretical yield is calculated by assumig that all the limitig reaget is coverted to product. The experimetally determied mass of product is the compared to the theoretical yield ad expressed as a percetage: actual yield Percet yield = 100 percet theoretical yield

9 69 EXAMPLE.7 Whe g N gas ad 5.0 g H gas are mixed at 50 C ad a high pressure, they react to form 8.96 g NH (ammoia) gas. Calculate the percet yield. Calculate the percet yield. Solutio We must calculate the theoretical yield of NH, ad to do this, we must first discover whether N or H is the limitig reaget. For the balaced equatio N + H NH the stoichiometric ratio of the reactats is H mol H S = N 1 mol N Now, the iitial amouts of the two reagets are ad H N 1 mol H (iitial) = 5.0 g H.016 g H 1 mol N (iitial) = g N 8.0 g N The ratio of iitial amouts is thus = 1.4 mol H =.569 mol N (iitial) 1.4 mol H.47 mol H = = (iitial).569 mol N 1 mol N H N H Sice this ratio is greater tha S, there is a excess of H. N is the limitig reaget. N Accordigly we must use.569 mol N (rather tha 1.4 mol H ) to calculate the theoretical yield of NH. We the have so that mol NH (theoretical) =.569 mol N NH 1 mol N = 7.18 mol NH 17.0 g NH m (theoretical) = 7.18 mol NH NH 1 mol NH = 11.6 g NH The percet yield is the actual yield Percet yield = 100 percet theoretical yield 8.96 g = 100 percet =.81 percet 11.6 g

10 70 Combiatio of itroge ad hydroge to form ammoia is a classic example of a reactio which does ot go to completio. Commercial productio of ammoia is accomplished usig this reactio i what is called the Haber process. Eve at the rather uusual temperatures ad pressures used for this idustrial sythesis, oly about oe-quarter of the reactats ca be coverted to the desired product. This is ufortuate because early all itroge fertilizers are derived from ammoia ad the world has come to rely o them i order to produce eough food for its rapidly icreasig populatio. Ammoia raks third [after sulfuric acid (H SO 4 ) ad oxyge (O )] i the list of most-produced chemicals, worldwide. It might rak eve higher if the reactio by which it is made wet to completio. Certaily ammoia ad the food it helps to grow would be less expesive ad would require much less eergy to produce if this were the case.. ANALYSIS OF COMPOUNDS Up to this poit we have obtaied all stoichiometric ratios from the coefficiets of balaced chemical equatios. Chemical formulas also idicate relative amouts of substace, however, ad stoichiometric ratios may be derived from them, too. For example, the formula HgBr tells us that o matter how large a sample of mercuric bromide we have, there will always be mol of bromie atoms for each mole of mercury atoms. That is, from the formula HgBr we have the stoichiometric ratio Br mol Br S = Hg 1 mol Hg We could also determie that for HgBr (The reciprocals of these stoichiometric ratios are also valid for HgBr.) Stoichiometric ratios derived from formulas istead of equatios are ivolved i the most commo procedure for determiig the empirical formulas of compouds which cotai oly C, H, ad O. A weighed quatity of the substace to be aalyzed is placed i a combustio trai (Fig..1) ad Figure.1 A combustio trai. H O ad C O, produced by combiatio of O with H ad C i the sample, are selectively absorbed by tubes cotaiig Dehydrite [Mg(ClO 4 ) H O] ad ascarite (NaOH o asbestos).

11 71 heated i a stream of dry O. All the H i the compoud is coverted to H O(g) which is trapped selectively i a previously weighed absorptio tube. All the C is coverted to CO (g) ad this is absorbed selectively i a secod tube. The icrease of mass of each tube tells, respectively, how much H O ad CO were produced by combustio of the sample. EXAMPLE.8 A 6.49-mg sample of ascorbic acid (vitami C) was bured i a combustio trai mg CO ad.64 mg H O were formed. Determie the empirical formula of ascorbic acid. Solutio We eed to kow the amout of C, the amout of H, ad the amout of O i the sample. The ratio of these gives the subscripts i the formula. The first two may be obtaied from the masses of CO ad H O usig the molar masses ad the stoichiometric ratios C 1 mol C H mol H S = S = CO 1 mol CO HO 1 mol HO Thus - 1 mol CO 1 mol C = g CO C g CO 1 mol CO mol C = 1 mol H O mol H = g H O 18.0 g H O 1 mol H O - H mol H = The compoud may also have cotaied oxyge. To see if it does, calculate the masses of C ad H ad subtract from the total mass of sample Thus we have 6.49 mg sample.65 mg C 0.95 mg H =.54 mg O ad 1 mol O = = O g O g O.1 10 mol O

12 7 The ratios of the amouts of the elemets i ascorbic acid are therefore H C mol H 1. mol H = = mol C 1 mol C 1 1 mol H 4 mol H = = 1 mol C mol C O C mol O 1 mol O mol O = = = mol C 1 mol C mol C Sice C : H : O is mol C:4 mol H: mol O, the empirical formula is C H 4 O. A drawig of a molecule of ascorbic acid is show i Fig... You ca determie by coutig the atoms that the molecular formula is C 6 H 8 O 6 exactly double the empirical formula. It is also evidet that there is more to kow about a molecule tha just how may atoms of each kid are preset. I ascorbic acid, as i other molecules, the way the atoms are coected together ad their arragemet i three-dimesioal space are quite importat. A picture like Fig.., showig which atoms are coected to which, is called a structural formula. Empirical formulas may be obtaied from percet compositio or combustio-trai experimets, ad, if the molecular weight is kow, molecular formulas may be determied from the same data. More complicated experimets are required to fid structural formulas. I Example.8 we obtaied the mass of O by subtractig the masses of C ad H from the total mass of sample. This assumed that oly C, H, ad O were preset. Sometimes such a assumptio may be icorrect. Whe peicilli was first isolated ad aalyzed, the fact that it cotaied sulfur Figure. The structural formula of ascorbic acid (vitami C), C 6 H 8 O 6. Carbo atoms are dark gray, hydroge atoms are light red, ad oxyge atoms are dark red. (Computer-geerated.) (Copyright@1976 by W.G.Davis ad J.W.Moore.)

13 7 was missed. This mistake was ot discovered for some time because the atomic weight of sulfur is almost exactly twice that of oxyge. Two atoms of oxyge were substituted i place of oe sulfur atom i the formula.. THERMOCHEMISTRY Whe a chemical reactio occurs, there is usually a chage i temperature of the chemicals themselves ad of the beaker or flask i which the reactio is carried out. If the temperature icreases, the reactio is exothermic eergy is give off as heat whe the cotaier ad its cotets cool back to room temperature. (Heat is eergy trasferred from oe place to aother solely because of a differece i temperature.) A edothermic reactio produces a decrease i temperature. I this case heat is absorbed from the surroudigs to retur the reactio products to room temperature. Thermochemistry, a word derived from the Greek thermé heat, is the measuremet ad study of eergy trasferred as heat whe chemical reactios take place. It is extremely importat i a techological world where a great deal of work is accomplished by trasformig ad haressig heat give off durig combustio of coal, oil, ad atural gas. Eergy Eergy is usually defied as the capability for doig work. For example, a billiard ball ca collide with a secod ball, chagig the directio or speed of motio of the latter. I such a process the motio of the first ball would also be altered. We would say that oe billiard ball did work o (trasferred eergy to) the other. Eergy due to motio is called kietic eergy ad is represeted by E k. For a object movig i a straight lie, the kietic eergy is oe-half the product of the mass ad the square of the speed: Where m = mass of the object u = speed of object E k = ½mu (.4) If the two billiard balls metioed above were studied i outer space, where frictio due to their collisios with air molecules or the surface of a pool table would be egligible, careful measuremets would reveal that their total kietic eergy would be the same before ad after they collided. This is a example of the law of coservatio of eergy, which states that eergy caot be created or destroyed uder the usual coditios of everyday life. Wheever there appears to be a decrease i eergy somewhere, there is a correspodig icrease somewhere else. Eve whe there is a great deal of frictio, the law of coservatio of eergy still applies. If you put a milkshake o a mixer ad leave it there for 10 mi, you will have a warm, rather uappetizig drik. The whirlig

14 74 mixer blades do work o (trasfer eergy to) the milkshake, raisig its temperature. The same effect could be produced by heatig the milkshake, a fact which suggests that heatig also ivolves a trasfer of eergy. The first careful experimets to determie how much work was equivalet to a give quatity of heat were doe by the Eglish physicist James Joule (1818 to 1889) i the 1840s. I a experimet very similar to our milkshake example, Joule coected fallig weights through a pulley system to a paddle wheel immersed i a isulated cotaier of water. This allowed him to compare the temperature rise which resulted from the work doe by the weights with that which resulted from heatig. Uits with which to measure eergy may be derived from the SI base uits of Table 1. by usig Eq. (.4). EXAMPLE.9 Calculate the kietic eergy of a Volkswage Beetle of mass 844 kg (1860 lb) which is movig at 1.4 m s 1 (0 miles per hour). Solutio E k = ½mu = ½ 8.44 kg (1.4 m s 1 ) = kg m s I other words the uits for eergy are derived from the SI base uits kilogram for mass, meter for legth, ad secod for time. A quatity of heat or ay other form of eergy may be expressed i kilogram meter squared per secod squared. I hoor of Joule s pioeerig work this derived uit 1 kg m s called the joule, abbreviated J. The Volkswage i questio could do early J of work o aythig it happeed to ru ito. Aother uit of eergy still widely used by chemists is the calorie. The calorie used to be defied as the eergy eeded to raise the temperature of oe gram of water from 14.5 C to 15.5 C but ow it is defied as exactly J. The Calorie used by dieticias ad others for measurig the eergy values of foods is actually a kilocalorie, i.e., kj. This utritioal Calorie is distiguished by a capital C i its ame. Thermochemical Equatios Eergy chages which accompay chemical reactios are almost always expressed by thermochemical equatios, such as CH 4 (g) + O (g) CO (g) + H O(g) (5 C, 1 atm pressure) ΔH m = 890 kj mol 1 (.5)

15 75 Here the ΔH m (delta H subscript m) tells us whether heat eergy is released or absorbed whe the reactio occurs ad also eables us to fid the actual quatity of eergy ivolved. By covetio, if ΔH m is positive, heat is absorbed by the reactio; i.e., it is edothermic. More commoly, ΔH m is egative as i Eq. (.5), idicatig that heat eergy is released rather tha absorbed by the reactio, ad that the reactio is exothermic. This covetio as to whether ΔH m is positive or egative looks at the heat chage i terms of the matter actually ivolved i the reactio rather tha its surroudigs. I the reactio i Eq. (.5), the C, H, ad O atoms have collectively lost eergy ad it is this loss which is idicated by a egative value of ΔH m. It is importat to otice that ΔH m is ot a eergy but rather a eergy per uit amout. Its uits are ot kilojoules but kilojoules per mole (hece the m subscript i ΔH m ). This is ecessary because the quatity of heat released or absorbed by a reactio is proportioal to the amout of each substace cosumed or produced by the reactio. Thus Eq. (.5) tells us that kj of heat eergy is give off for every mole of CH 4 which is cosumed. Alteratively, it tells us that kj is released for every mole of H O produced, i.e., for every mol H O produced. See i this way, ΔH m is a coversio factor eablig us to calculate the heat absorbed whe a give amout of substace is cosumed or produced. If q is the quatity of heat absorbed ad is the amout of substace ivolved, the q Δ H = (.6) m EXAMPLE.10 How much heat eergy is obtaied whe 1 kg of ethae gas, C H 6, is bured i oxyge accordig to the equatio: C H 6 (g) + 7O (g) 4CO (g) + 6H O(l) ΔH m = 08 kj mol 1 (.7) Solutio The mass of C H 6 is easily coverted to the amout of C H 6 from which the heat eergy q is easily calculated by meas of Eq. (.6). The value of ΔH m is 08 kj per mole of mole of C H 6, i.e., per mol C H 6. The road map is Note: By covetio a egative value of q correspods to a release of heat eergy by the matter ivolved i the reactio.

16 76 The quatity ΔH m is usually referred to as a ethalpy chage per mole. I this cotext the symbol Δ (delta) sigifies chage i while H is the symbol for the quatity beig chaged, amely the ethalpy. We will deal with the ethalpy i some detail i Chap. 15. For the momet we ca thik of it as a property of matter which icreases whe matter absorbs eergy ad decreases whe matter releases eergy. It is importat to realize that the value of ΔH m give i thermochemical equatios like (.5) or (.7) depeds o the physical state of both the reactats ad the products. Thus, if water were obtaied as a liquid istead of a gas i the reactio i Eq. (.5), the value of ΔH m would be differet from kj mol 1. It is also ecessary to specify both the temperature ad pressure sice the value of ΔH m depeds very slightly o these variables. If these are ot specified [as i Eq. (.7)] they usually refer to 5 C ad to ormal atmospheric pressure. Two more characteristics of thermochemical equatios arise from the law of coservatio of eergy. The first is that writig a equatio i the reverse directio chages the sig of the ethalpy chage. For example, H O(l) H O(g) ΔH m = 44 kj mol 1 (.8a) tells us that whe a mole of liquid water vaporizes, 44 kj of heat is absorbed. This correspods to the fact that heat is absorbed from your ski whe perspiratio evaporates, ad you cool off. Codesatio of 1 mol of water vapor, o the other had, gives off exactly the same quatity of heat. H O(g) H O(l) ΔH m = 44 kj mol 1 (.8b) To see why this must be true, suppose that ΔH m [Eq. (.8a)] = 44 kj mol 1 while ΔH m [Eq. (.8b)] = 50.0 kj mol 1. If we took 1 mol of liquid water ad allowed it to evaporate, 44 kj would be absorbed. We could the codese the water vapor, ad 50.0 kj would be give off. We could agai have 1 mol of liquid water at 5 C but we would also have 6 kj of heat which had bee created from owhere! This would violate the law of coservatio of eergy. The oly way the problem ca he avoided is for ΔH m of the reverse reactio to be equal i magitude but opposite i sig from ΔH m of the forward reactio. That is, ΔH m forward = ΔH m reverse Hess Law Perhaps the most useful feature of thermochemical equatios is that they ca be combied to determie ΔH m values for other chemical reactios. Cosider, for example, the followig two-step sequece. Step 1 is reactio of 1 mol C(s) ad 0.5 mol O (g) to form 1 mol CO(g): C(s) + ½O (g) CO(g) ΔH m = kj mol 1 = ΔH 1 (Note that sice the equatio refers to moles, ot molecules, fractioal coefficiets are permissible.) I step the mole of CO reads with a additioal 0.5 mol O yieldig 1 mol CO : CO(g) + ½O (g) CO (g) ΔH m = 8.0 kj mol 1 = ΔH

17 77 The et result of this two-step process is productio of 1 mol CO from the origial 1 mol C ad 1 mol O (0.5 mol i each step). All the CO produced i step 1 is used up i step. O paper this et result ca be obtaied by addig the two chemical equatios as though they were algebraic equatios. The CO produced is caceled by the CO cosumed sice it is both a reactat ad a product of the overall reactio Experimetally it is foud that the ethalpy chage for the et reactio is the sum of the ethalpy chages for steps 1 ad : ΔH et = kj mol 1 + ( 8.0 kj mol 1 ) kj mol 1 = ΔH 1 + ΔH That is, the thermochemical equatio C(s) + O (g) CO (g) ΔH m = 9.5 kj mol 1 is the correct oe for the overall reactio. I the geeral case it is always true that wheever two or more chemical equatios ca be added algebraically to give a et reactio, their ethalpy chages may also be added to give the ethalpy chage of the et reactio. This priciple is kow as Hess' law. If it were ot true, it would be possible to thik up a series of reactios i which eergy would be created but which would ed up with exactly the same substaces we started with. This would cotradict the law of coservatio of eergy. Hess law eables us to obtai ΔH m values for reactios which caot be carried out experimetally, as the ext example shows. EXAMPLE.11 Acetylee (C H ) caot be prepared directly from its elemets accordig to the equatio C(s) + H (g) C H (g) (.9) Calculate ΔH m for this reactio from the followig thermochemical equatios, all of which ca be determied experimetally: C(s) + O (g) CO (g) H (g) + ½O (g) H O(l) C H (g) + 5 O (g) CO (g) + H O(l) ΔH m = 9.5 kj mol 1 (.10a) ΔH m = 85.8 kj mol 1 (.10b) ΔH m = kj mol 1 (.10c) Solutio We use the followig strategy to maipulate the three experimetal equatios so that whe added they yield Eq. (.9): a) Sice Eq. (.9) has mol C o the left, we multiply Eq. (.10a) by.

18 78 b) Sice Eq. (.9) has 1 mol H o the left, we leave Eq. (.10b) uchaged. c) Sice Eq. (.9) has 1 mol C H o the right, whereas there is 1 mol C H o the left of Eq. (.10c) we write Eq. (.10c) i reverse. We the have C(s) + O (g) CO (g) Δ H = ( 9.5) kj mol m -1 H(g) + O(g) HO() H = 85.8 kj mol m CO (g) + H O( ) C H (g) + O (g) H = ( kj) mol m C(s) + H (g) + O (g) C H (g) + O (g) 1 l Δ 5 l Δ 1 5 Thus the desired result is Δ H = ( ) kj mol = 7.0 kj mol m C(s) + H (g) C H (g) ΔH m = 7.0 kj mol 1.4 STANDARD ENTHALPIES OF FORMATION By ow chemists have measured the ethalpy chages for so may reactios that it would take several large volumes to list all the thermochemical equatios. Fortuately Hess law makes it possible to list a sigle value, the stadard ethalpy of formatio ΔH f, for each compoud. The stadard ethalpy of formatio is the ethalpy chage whe 1 mol of a pure substace is formed from its elemets. Each elemet must be i the physical ad chemical form which is most stable at ormal atmospheric pressure ad a specified temperature (usually 5 C). For example, if we kow that ΔH f [H O(l)] = 85.8 kj mol 1, we ca immediately write the thermochemical equatio H (g) + ½O (g) H O(l) ΔH m = 85.8 kj mol 1 (.11) The elemets H ad O appear as diatomic molecules ad i gaseous form because these are their most stable chemical ad physical states. Note also that 85.8 kj are give off per mole of H O(l) formed. Equatio (.11) must specify formatio of 1 mol H O(l), ad so the coefficiet of O must be ½. I some cases, such as that of water, the elemets will react directly to form a compoud, ad measuremet of the heat absorbed serves to determie ΔH f. Quite ofte, however, elemets do ot react directly with each other to form the desired compoud, ad ΔH f must be calculated by combiig the ethalpy chages for other reactios. A case i poit is the gas acetylee, C H. I Example.11 we used Hess law to show that the thermochemical equatio C(s) + H (g) C H (g) ΔH m = 7.0 kj mol 1

19 79 TABLE.1 Some Stadard Ethalpies of Formatio at 5 C. ΔH kcal ΔH kcal ΔH kcal ΔH kcal f f f f Compoud mol mol Compoud mol mol AgCl(s) HO(g) AgN (s) HO(l) Ag O(s) HO(l) Al O (s) HS(g) Br (l) HgO(s) Br (g) I(s) C(s), graphite I(g) C(s), diamod CH (g) 4 CO(g) CO (g) CH(g) CH(g) 4 CH(g) 6 CH(l) 6 6 CaO(s) CaCO (s) CuO(s) Fe O (s) HBr(g) HCl(g) HI(g) KCl(s) KBr(s) MgO(s) NH (g) NO(g) NO (g) NO(g) 4 NF (g) NaBr(s) NaCl(s) O(g) SO (g) SO (g) ZO(s) was valid. Sice it ivolves 1 mol C H ad the elemets are i their most stable forms, we ca say that ΔH f.[c H (g)] = 7.0 kj mol 1. Oe further poit arises from the defiitio of ΔH f. The stadard ethalpy of formatio for a elemet i its most stable state must be zero. If we form mercury from its elemets, for example, we are talkig about the reactio Hg(l) Hg(l) Sice the mercury is uchaged, there ca be o ethalpy chage, ad ΔH f = 0 kj mol 1. Stadard ethalpies of formatio for some commo compouds are give i Table.1. These values may be used to calculate ΔH m for ay chemical reactio so log as all the compouds ivolved appear i the tables. To see how ad why this may be doe, cosider the followig example. EXAMPLE.1 Use stadard ethalpies of formatio to calculate ΔH m for the reactio CO(g) + O (g) CO (g)

20 80 Solutio We ca imagie that the reactio takes place i two steps, each of which ivolves oly a stadard ethalpy of formatio. I the first step CO (carbo mooxide) is decomposed to its elemets: CO(g) C(s) + O (g) ΔH m = ΔH f (.1) Sice this is the reverse of formatio of mol CO from its elemets, the ethalpy chage is ΔH 1 = { ΔH f [CO(g)]} = [ ( kj mol 1 )] = +1.0 kj mol 1 I the secod step the elemets are combied to give mol CO (carbo dioxide): C(s) O (g) + CO (g) ΔH m = ΔH (.1) I this case ΔH = ΔH f [CO(g)] = (9.5 kj mol 1 ) = kj mol 1 You ca easily verify that the sum of Eqs. (.1) ad (.1) is CO(g) + O (g) CO (g) ΔH m = ΔH et Therefore ΔH et = ΔH 1 + ΔH = 1.0 kj mol kj mol 1 = 17.5 kj mol 1 Note carefully how Example.1 was solved. I step 1 the reactat compoud CO(g) was hypothetically decomposed to its elemets. This equatio was the reverse of formatio of the compoud, ad so ΔH 1 was opposite i sig from ΔH f. Step 1 also ivolved mol CO(g) ad so the ethalpy chage had to be doubled. I step we had the hypothetical formatio of the product CO (g) from its elemets. Sice mol were obtaied, the ethalpy chage was doubled but its sig remaied the same. Ay chemical reactio ca be approached similarly. To calculate ΔH m we add all the ΔH f values for the products, multiplyig each by the appropriate coefficiet, as i step above. Sice the sigs of ΔH f for the reactats had to be reversed i step 1, we subtract them, agai multiplyig by appropriate coefficiets. This ca he summarized by the equatio ΔH m = ΔH f (products) ΔH f (reactats) (.14) The symbol Σ meas the sum of. Sice ΔH f values are give per mole of compoud, you must be sure to multiply each ΔH f by a appropriate coefficiet derived from the equatio for which ΔH m is beig calculated.

21 81 EXAMPLE.1 Use Table.1 to calculate ΔH m for the reactio 4NH (g) 5O (g) 6H O(g) + 4NO(g) Solutio Usig Eq. (.14), we have ΔH m = ΔH f (products) ΔH f (reactats) = [6 ΔH f (H O) + 4 ΔH f (NO)] [4 ΔH f (NH ) + 5 ΔH f (O )] = 6( 41.8) kj mol 1 + 4(90.) kj mol 1 4( 46.1 kj mol 1 ) 5 0 = kj mol kj mol kj mol 1 = 905. kj mol 1 Note that we were careful to use ΔH f [H O(g)] ot ΔH f [H O(l)]. Eve though water vapor is ot the most stable form of water at 5 C, we ca still use its ΔH f value. Also the stadard ethalpy of formatio of the elemet O (g) is zero by defiitio. Obviously it would be a waste of space to iclude it i Table.1..5 SOLUTIONS I the laboratory, i your body, ad i the outside eviromet, the majority of chemical reactios take place i solutios. Macroscopically a solutio is defied as a homogeeous mixture of two or more substaces, that is, a mixture which appears to be uiform throughout. O the microscopic scale a solutio ivolves the radom arragemet of oe kid of atom or molecule with respect to aother. There are a umber of reasos why solutios are so ofte ecoutered both i ature ad i the laboratory. The most commo type of solutio ivolves a liquid solvet which dissolves a solid solute. (The term solvet usually refers to the substace preset i greatest amout. There may be more tha oe solute dissolved i it.) Because a liquid adopts the shape of its cotaier but does ot expad to fill all space available to it, liquid solutios are coveiet to hadle. You ca easily pour them from oe cotaier to aother, ad their volumes are readily measured usig graduated cyliders, pipets, burets, volumetric flasks, or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved i a liquid are close together but still able to move past oe aother. They cotact each other more frequetly tha if two solids were placed ext to each other. This itimacy i liquid solutios ofte facilitates chemical reactios. Cocetratio Sice solutios offer a coveiet medium for carryig out chemical reactios, it is ofte ecessary to kow how much of oe solutio will react

22 8 with a give quatity of aother. Examples earlier i this chapter have show that the amout of substace is the quatity which determies how much of oe material will react with aother. The ease with which solutio volumes may be measured suggests that it would be very coveiet to kow the amout of substace dissolved per uit volume of solutio. The by measurig a certai volume of solutio, we would also be measurig a certai amout of substace. The cocetratio c of a substace i a solutio (ofte called molarity) is the amout of the substace per uit volume of solutio: amout of solute Cocetratio of solute = c = solute volume of solute V solute solute Usually the uits moles per cubic decimeter (mol dm ) or moles per liter (mol liter 1 ) are used to express cocetratio. If a pure substace is soluble i water, it is easy to prepare a solutio of kow cocetratio. A cotaier with a sample of the substace is weighed accurately, ad a appropriate mass of sample is poured through a fuel ito a volumetric flask, as show i Fig... The cotaier is the reweighed. Ay solid adherig to the fuel is rised ito the flask, ad water is added util the flask is about three-quarters full. After swirlig the flask to dissolve the solid, water is added carefully util the bottom of the meiscus coicides with the calibratio mark o the eck of the flash. EXAMPLE.14 A solutio of KI was prepared as described above. The iitial mass of the cotaier plus KI was g, ad the fial mass after pourig was g. The volume of the flask was ml. What is the cocetratio of the solutio? Solutio The cocetratio ca be calculated by dividig the amout of solute by the volume of solutio [Eq. (.15)]: KI c = KI V We obtai from the mass of KI added to the flask: KI m = g 0,1544 g = 1.10 g KI 1 mol - = 1.10 g = mol KI g The volume of solutio is ml, or 1 dm -1 V = cm = dm solutio 10 cm Thus c mol KI KI -1 V dm solutio -1 = = = mol dm -

23 8 Figure. The preparatio of a sodium chloride solutio of cocetratio mol dm. (a) A sample of pure solid NaCl is firs weighed with its cotaier. (b) Most of the sample is ow trasferred ito the volumetric flask. (c) The cotaier ad remaiig solid NaCl are agai weighed. Subtractio yields m NaCl = g. Thus NaCl = mol. (d) Ay solid remaiig i the fuel is washed dow ito the body of the flask. (e) The flask is ow filled to about 80 percet capacity ad the solid allowed to dissolve. The flask is shake to achieve a uiform solutio. (f) Solvet is ow added carefully util the bottom of the me- iscus coicides with the mark o the flask. The flask is agai shake to achieve a uiform solutio. Sice the volume of the solutio is dm, the cocetratio is mol/1.000 dm = mol dm. Note that the defiitio of cocetratio is etirely aalogous to the defiitios of desity, molar mass, ad stoichiometric ratio that we have previously ecoutered. Cocetratio will serve as a coversio factor relatig the volume of solutio to the amout of dissolved solute. cocetratio c Volume of solutio amout of solute V Because the volume of a liquid ca be measured quickly ad easily, cocetratio is a much-used quatity. The ext two examples show how this coversio factor may be applied to commoly ecoutered solutios i which water is the solvet (aqueous solutios). EXAMPLE.15 A aqueous solutio of HCl [represeted or writte HCl(aq)] has a cocetratio of mol dm. If 4.71 cm³ (4.71 ml) of this solutio is delivered from a buret, what amout of HCl has bee delivered?

24 84 Solutio Usig cocetratio as a coversio factor, we have V c HCl 4.71 cm = mol 1 dm The volume uits will cacel if we supply a uity factor to covert cubic cetimeters to cubic decimeters: mol 1 dm = 4.71 cm HCl 1 dm 10 cm mol 1 dm = 4.71 cm 1 dm 10 cm = mol The cocetratio uits of moles per cubic decimeter are ofte abbreviated M, proouced molar. That is, a 0.1-M (oe-teth molar) solutio cotais 0.1 mol solute per cubic decimeter of solutio. This abbreviatio is very coveiet for labelig laboratory bottles ad for writig textbook problems; however, whe doig calculatios, it is difficult to see that 1 dm 1 M = 1 mol Therefore we recommed that you always write the uits i full whe doig ay calculatios ivolvig solutio cocetratios. That is, mol 1 dm 1 = 1mol dm Problems such as Example.15 are easier for some persos to solve if the solutio cocetratio is expressed i millimoles per cubic cetimeter (mmol cm ) istead of moles per cubic decimeter. Sice the SI prefix m meas 10, 1 mmol = 10 mol, ad 1 mol 1 mol 1 dm 1 mmol 1 mmol - = 1 dm 1 dm 10 cm 10 mol 1 cm Thus a cocetratio of mol dm (0.196 M) ca also be expressed as mmol cm. Expressig the cocetratio this way is very coveiet whe dealig with laboratory glassware calibrated i milliliters or cubic cetimeters. EXAMPLE.16 Exactly 5.0 ml NaOH solutio whose cocetratio is M was delivered from a pipet. (a) What amout of NaOH was preset? (b) What mass of NaOH would remai if all the water evaporated?

25 85 Solutio a) Sice M meas mol dm, or mmol cm, we choose the latter, more coveiet quatity as a coversio factor: mol = 5.0 cm =.44 mmol NaOH 1 cm - = mol b) Usig molar mass, we obtai g - m = mol = g NaOH 1 mol Note: The symbols NaOH ad m NaOH refer to the amout ad mass of the solute NaOH, respectively. They do ot refer to the solutio. If we wated to specify the mass of aqueous NaOH solutio, the symbol m NaOH(aq) could be used. Dilutig ad Mixig Solutios Ofte it is coveiet to prepare a series of solutios of kow cocetratios by first preparig a sigle stock solutio as described i Example.14. Aliquots (carefully measured volumes) of the stock solutio ca the be diluted to ay desired volume. I other cases it may be icoveiet to weigh accurately a small eough mass of sample to prepare a small volume of a dilute solutio. Each of these situatios requires that a solutio be diluted to obtai the desired cocetratio. EXAMPLE.17 A pipet is used to measure 50.0 ml of M HCl ito a ml volumetric flask. Distilled water is carefully added up to the mark o the flask. What is the cocetratio of the diluted solutio? Solutio To calculate cocetratio, we first obtai the amout of HCI i the 50.0 ml (50.0 cm ) of solutio added to the volumetric flask: Dividig by the ew volume gives the cocetratio c HCl 5.14 mol = = = mmol cm V cm HCl Thus the ew solutio is M. - EXAMPLE.18 What volume of the solutio prepared i Example.14 would be required to make ml of M KI?

26 86 Solutio Usig the volume ad cocetratio of the desired solutio, we ca calculate the amout of KI required. The the cocetratio of the origial solutio ( M) ca be used to covert that amout of KI to the ecessary volume. Schematically Thus we should dilute a 7.90-ml aliquot of the stock solutio to ml. This could be doe by measurig 7.90 ml from a buret ito a ml volumetric flask ad addig water up to the mark. Titratios Whe solutios are used quatitatively i the laboratory, titratio is usually ivolved. Titratio is a techique used to determie the volume of oe solutio ecessary to cosume exactly some reactat i aother solu- Figure.4 The techique of titratio.

27 87 tio. As show i Fig..4, a measured volume of the solutio to be titrated is placed i a flask or other cotaier. The titrat (the solutio to be added) is placed i a buret. The volume of titrat added ca be determied by readig the level of liquid i the buret before ad after titratio. This readig ca usually be estimated to the earest hudredth of a milliliter, for example, 5.6 ml. I Fig..4 the solutio to be titrated is colorless aqueous hydroge peroxide, H O (aq), which cotais excess sulfuric acid H SO 4 (aq). The titrat is purple-colored potassium permagaate solutio KMO 4 (aq). The reactio which occurs is KMO 4 (aq) + 5H O + H SO 4 MSO 4 (aq) + 5O (g) + K SO 4 (aq) + 8H O(l) (.16) As the first few cubic cetimeters of titrat flow ito the flask, there is a large excess of.. H O. The limitig reaget KMO 4 is etirely cosumed, ad its purple color disappears almost as soo as it is added. Evetually, though, all the H O is cosumed. Additio of just oe more drop of titrat produces a lastig pik color due to ureacted KMO 4 i the flask. This idicates that all the H O has bee cosumed ad is called the edpoit of the titratio. If more KMO 4 solutio were added, there would be a excess of KMO 4 ad the color of the solutio i the flask would get much darker. The darker color would show that we had overtitrated, or overshot the edpoit, by addig more tha eough KMO 4 to react with all the H O. I the titratio we have just described, the itese purple color of permagaate idicates the edpoit. Usually, however, it is ecessary to add a idicator a substace which combies with excess titrat to produce a visible color or to form a isoluble substace which would precipitate from solutio. No matter what type of reactio occurs or how the edpoit is detected, however, the object of a titratio is always to add just the amout of titrat eeded to cosume exactly the amout of substace beig titrated. I the KMO 4 H O reactio [Eq. (.16)], the edpoit occurs whe exactly mol KMO 4 have bee added from the buret for every 5 mol H O origially i the titratio flask. That is, at the edpoit the ratio of the amout of KMO 4, added to the amout of H O cosumed must equal the stoichiometric ratio KMO4 4 4 HO (added from buret) KMO mol KMO = S = (.17) (iitially i flask) H O 5 mol H O Whe the edpoit has bee reached i ay titratio, the ratio of the amouts of substace of the two reactats is equal to the stoichiometric ratio obtaied from the appropriate balaced chemical equatio. Therefore we ca use the stoichiometric ratio to covert from the amout of oe substace to the amout of aother. EXAMPLE.19 What volume of M KMO 4 would be eeded to reach the edpoit whe titratig 5.00 ml of 0.17 M H O.

28 88 Solutio At the edpoit, Eq. (.17) will apply, ad we ca use it to calculate the amout of KMO 4 which must be added: KMO4 (added) = (i flask) S KMO4 HO HO The amout of H O is obtaied from the volume ad cocetratio: mmol (i flask) = 5.00 cm 0.17 HO cm =.180 mmol H O The - mol KMO 10 4 (added) =.180 mmol H O KMO4-5 mol H O 10 mmol KMO4 =.180 mmol H O 5 mmol H O = 1.7 mmol KMO To obtai V KMO4 (aq) we use the cocetratio as a coversio factor: 1 cm V = 1.7 mmol KMO =.6 cm KMO 4 ( aq) mmol KMO 4 4 Note that overtitratig [addig more tha.6 cm of KMO 4 (aq) would ivolve a excess (more tha 1.7 mmol) of KMO 4. Titratio is ofte used to determie the cocetratio of a solutio. I may cases it is ot a simple matter to obtai a pure substace, weigh it accurately, ad dissolve it i a volumetric flask as was doe i Example.14. NaOH, for example, combies rapidly with H O ad CO from the air, ad so eve a freshly prepared sample of solid NaOH will ot be pure. Its weight would chage cotiuously as CO (g) ad H O(g) were absorbed. Hydroge chloride (HCl) is a gas at ordiary temperatures ad pressures, makig it very difficult to hadle or weigh. Aqueous solutios of both of these substaces must be stadardized; that is, their cocetratios must be determied by titratio. EXAMPLE.0 A sample of pure potassium hydroge phthalate (KHC 8 H 4 O 4 ) weighig 0.41 g is dissolved i distilled water. Titratio of the sample requires 7.0 ml NaOH(aq). The titratio reactio is NaOH(aq) + KHC 8 H 4 O 4 (aq) NaKC 8 H 4 O 4 (aq) + H O What is the cocetratio of NaOH(aq)?

29 89 Solutio To calculate cocetratio, we eed to kow the amout of NaOH ad the volume of solutio i which it is dissolved. The former quatity could be obtaied via a stoichiometric ratio from the amout of KHC 8 H 4 O 4, ad that amout ca be obtaied from the mass m KHC8H4O M S (NaOH/KHC H O ) KHC8H4O4 KHC8H4O4 NaOH 1 mol KHC H O 1 mol NaOH =.180 g NaOH The cocetratio is c 04. g 1 mol KHC H O = = mol NaOH mmol NaOH mmol NaOH = = = mmol c m V 7.0 cm NaOH or M. - 4 NaOH By far the most commo use of titratios is i determiig ukows, that is, i determiig the cocetratio or amout of substace i a sample about which we iitially kew othig. The ext example ivolves a ukow that may persos ecouter every day. EXAMPLE.1 Vitami C tablets cotai ascorbic acid (C 6 H 8 O 6 ) ad a starch filler which holds them together. To determie how much vitami C is preset, a tablet ca be dissolved i water ad with sodium hydroxide solutio, NaOH(aq). The equatio is C 6 H 8 O 6 (aq) + NaOH(aq) Na C 6 H 7 O 6 (aq) + H O(l) If titratio of a dissolved vitami C tablet requires cm³ of M NaOH, how accurate is the claim o the label of the bottle that each tablet cotais 00 mg of vitami C? Solutio The kow volume ad cocetratio allow us to calculate the amout of NaOH(aq) which reacted with all the vitami C. Usig the stoichiometric ratio CHO 1mmol CHO = mmol NaOH S NaOH we ca obtai the amout of C 6 H 8 O 6. The molar mass coverts that amout to a mass which ca be compared with the label. Schematically