CHAPTER 3 STOICHIOMETRY

Transcription

1 CHAPTER 3 STOICHIOMETRY 3.1 Oe atomic mass uit is defied as a mass exactly equal to oe-twelfth the mass of oe carbo-12 atom. We caot weigh a sigle atom, but it is possible to determie the mass of oe atom relative to aother experimetally. The first step is to assig a value to the mass of oe atom of a give elemet so that it ca be used as a stadard amu. O the periodic table, the mass is listed as amu because this is a average mass of the aturally occurrig mixture of isotopes of carbo. 3.3 The value amu is a average value (a average atomic mass). If we could examie gold atoms idividually, we would ot fid a atom with a mass of amu. However, the average mass of a gold atom i a typical sample of gold is amu. 3.4 You eed the mass of each isotope of the elemet ad each isotope s relative abudace. 3.5 ( amu)(0.7553) ( amu)(0.2447) amu 3.6 Strategy: Each isotope cotributes to the average atomic mass based o its relative abudace. Multiplyig the mass of a isotope by its fractioal abudace (ot percet) will give the cotributio to the average atomic mass of that particular isotope. It would seem that there are two ukows i this problem, the fractioal abudace of 6 Li ad the fractioal abudace of 7 Li. However, these two quatities are ot idepedet of each other; they are related by the fact that they must sum to 1. Start by lettig x be the fractioal abudace of 6 Li. Sice the sum of the two abudace s must be 1, we ca write Solutio: Abudace 7 Li (1 x) Average atomic mass of Li amu x( amu) (1 x)( amu) x x x x correspods to a atural abudace of 6 Li of 7.5 percet. The atural abudace of 7 Li is (1 x) or 92.5 percet amu The coversio factor required is 1g 1g? g 13.2 amu g amu

2 CHAPTER 3: STOICHIOMETRY amu The uit factor required is 1g amu 24? amu 8.4 g = amu 1g 3.9 The mole is the amout of a substace that cotais as may elemetary etities (atoms, molecules, or other particles) as there are atoms i exactly 12 grams of the carbo-12 isotope. The uit for mole used i calculatios is mol. A mole is a uit like a pair, doze, or gross. A mole is the amout of substace that cotais particles. Avogadro s umber ( ) is the umber of atoms i exactly 12 g of the carbo-12 isotope The molar mass of a atom is the mass of oe mole, atoms, of that elemet. Uits are g/mol I oe year: days 24 h 3600 s 2 particles 17 ( people) particles/yr 1 yr 1 day 1 h 1 perso particles 6 Total time yr particles/yr 3.12 The thickess of the book i miles would be: i 1 ft 1 mi 16 ( pages) = mi 1 page 12 i 5280 ft The distace, i miles, traveled by light i oe year is: day 24 h 3600 s m 1 mi yr mi 1 yr 1 day 1 h 1 s 1609 m The thickess of the book i light-years is: 16 1light-yr ( mi) mi light - yr It will take light years to travel from the first page to the last oe! S atoms 5.10 mol S 1molS S atoms molCo ( Co atoms) = Co atoms mol Co mol Ca 77.4 g of Ca g Ca 1.93 mol Ca

3 34 CHAPTER 3: STOICHIOMETRY 3.16 Strategy: We are give moles of gold ad asked to solve for grams of gold. What coversio factor do we eed to covert betwee moles ad grams? Arrage the appropriate coversio factor so moles cacel, ad the uit grams is obtaied for the aswer. Solutio: The coversio factor eeded to covert betwee moles ad grams is the molar mass. I the periodic table (see iside frot cover of the text), we see that the molar mass of Au is g. This ca be expressed as 1 mol Au g Au From this equality, we ca write two coversio factors. 1 mol Au g Au ad g Au 1 mol Au The coversio factor o the right is the correct oe. Moles will cacel, leavig the uit grams for the aswer. We write g Au 3? g Au = 15.3 mol Au = g Au 1molAu Check: Does a mass of 3010 g for 15.3 moles of Au seem reasoable? What is the mass of 1 mole of Au? 3.17 (a) g Hg 1 mol Hg 1molHg Hg atoms g/hg atom (b) g Ne 1 mol Ne 1molNe Ne atoms g/ne atom 3.18 (a) Strategy: We ca look up the molar mass of arseic (As) o the periodic table (74.92 g/mol). We wat to fid the mass of a sigle atom of arseic (uit of g/atom). Therefore, we eed to covert from the uit mole i the deomiator to the uit atom i the deomiator. What coversio factor is eeded to covert betwee moles ad atoms? Arrage the appropriate coversio factor so mole i the deomiator cacels, ad the uit atom is obtaied i the deomiator. Solutio: The coversio factor eeded is Avogadro's umber. We have 1 mol particles (atoms) From this equality, we ca write two coversio factors. 1 mol As As atoms ad As atoms 1molAs The coversio factor o the left is the correct oe. Moles will cacel, leavig the uit atoms i the deomiator of the aswer. We write g As 1 mol As 22? g/as atom g/as atom 1molAs As atoms

4 CHAPTER 3: STOICHIOMETRY 35 (b) Follow same method as part (a) g Ni 1 mol Ni? g/ni atom g/ni atom 1molNi Ni atoms Check: Should the mass of a sigle atom of As or Ni be a very small mass? mol Pb g Pb Pb atoms Pb atoms 1molPb g Pb 3.20 Strategy: The questio asks for atoms of Cu. We caot covert directly from grams to atoms of copper. What uit do we eed to covert grams of Cu to i order to covert to atoms? What does Avogadro's umber represet? Solutio: To calculate the umber of Cu atoms, we first must covert grams of Cu to moles of Cu. We use the molar mass of copper as a coversio factor. Oce moles of Cu are obtaied, we ca use Avogadro's umber to covert from moles of copper to atoms of copper. 1 mol Cu g Cu The coversio factor eeded is 1molCu g Cu Avogadro's umber is the key to the secod coversio. We have 1 mol particles (atoms) From this equality, we ca write two coversio factors. 1 mol Cu Cu atoms ad Cu atoms 1molCu The coversio factor o the right is the oe we eed because it has umber of Cu atoms i the umerator, which is the uit we wat for the aswer. Let's complete the two coversios i oe step. grams of Cu moles of Cu umber of Cu atoms 1 mol Cu Cu atoms 22? atoms of Cu 3.14 g Cu Cu atoms g Cu 1 mol Cu Check: Should 3.14 g of Cu cotai fewer tha Avogadro's umber of atoms? What mass of Cu would cotai Avogadro's umber of atoms? 3.21 For hydroge: 1 mol H H atoms 1.10 g H g H 1 mol H H atoms For chromium: 1 mol Cr Cr atoms 14.7 g Cr g Cr 1 mol Cr Cr atoms There are more hydroge atoms tha chromium atoms.

5 36 CHAPTER 3: STOICHIOMETRY mol Pb g Pb 22 2Pbatoms = gpb Pb atoms 1 mol Pb g He 22 ( mol He) = g He 1molHe 2 atoms of lead have a greater mass tha mol of helium. 3. Usig the appropriate atomic masses, (a) CH amu 4(1.008 amu) amu (b) NO amu 2(16.00 amu) amu (c) SO amu 3(16.00 amu) amu (d) C 6 H 6 6(12.01 amu) 6(1.008 amu) amu (e) NaI amu amu amu (f) K 2 SO 4 2(39.10 amu) amu 4(16.00 amu) amu (g) Ca 3 (PO 4 ) 2 3(40.08 amu) 2(30.97 amu) 8(16.00 amu) amu 3.24 Strategy: How do molar masses of differet elemets combie to give the molar mass of a compoud? Solutio: To calculate the molar mass of a compoud, we eed to sum all the molar masses of the elemets i the molecule. For each elemet, we multiply its molar mass by the umber of moles of that elemet i oe mole of the compoud. We fid molar masses for the elemets i the periodic table (iside frot cover of the text). (a) (b) (c) (d) (e) (f) molar mass Li 2 CO 3 2(6.941 g) g 3(16.00 g) g molar mass CS g 2(32.07 g) g molar mass CHCl g g 3(35.45 g) g molar mass C 6 H 8 O 6 6(12.01 g) 8(1.008 g) 6(16.00 g) g molar mass KNO g g 3(16.00 g) g molar mass Mg 3 N 2 3(24.31 g) 2(14.01 g) g 3.25 To fid the molar mass (g/mol), we simply divide the mass (i g) by the umber of moles. 152 g mol 409 g/mol 3.26 The molar mass of acetoe, C 3 H 6 O, is g. We use the molar mass ad Avogadro s umber as coversio factors to covert from grams to moles to molecules of acetoe. 1 mol C3H6O molecules C3H6O g C3H6O g C 3 H 6 O 1 mol C 3 H 6 O molecules C3H6O 3.27 We use the molar mass of squaric acid ( g), Avogradro s umber, ad the subscripts i the formula of squaric acid, C 4 H 2 O 4, to covert from grams of squaric acid to moles of squaric acid to molecules of squaric acid, ad fially to atoms of C, H, or O. We first covert to molecules of squaric acid. 1 mol C4H2O molecules C4H2O g C4H2O molecules C4H2O g C4H2O4 1 mol C4H2O4

6 CHAPTER 3: STOICHIOMETRY 37 Next, we covert to atoms of C, H, ad O usig the subscripts i the formula as coversio factors atoms C molecules C4H2O4 1 molecule C 4 H 2 O atomsH molecules C4H2O4 1 molecule C4H2O4 21 4atomsO molecules C4H2O4 1 molecule C4H2O C atoms H atoms O atoms 3.28 Strategy: We are asked to solve for the umber of N, C, O, ad H atoms i g of urea. We caot covert directly from grams urea to atoms. What uit do we eed to obtai first before we ca covert to atoms? How should Avogadro's umber be used here? How may atoms of N, C, O, or H are i 1 molecule of urea? Solutio: Let's first calculate the umber of N atoms i g of urea. First, we must covert grams of urea to umber of molecules of urea. This calculatio is similar to Problem The molecular formula of urea shows there are two N atoms i oe urea molecule, which will allow us to covert to atoms of N. We eed to perform three coversios: grams of urea moles of urea molecules of urea atoms of N The coversio factors eeded for each step are: 1) the molar mass of urea, 2) Avogadro's umber, ad 3) the umber of N atoms i 1 molecule of urea. We complete the three coversios i oe calculatio.? atoms of N 4 1 mol urea urea molecules 2 N atoms = ( g urea) g urea 1 mol urea 1 molecule urea N atoms The above method utilizes the ratio of molecules (urea) to atoms (itroge). We ca also solve the problem by readig the formula as the ratio of moles of urea to moles of itroge by usig the followig coversios: Try it. grams of urea moles of urea moles of N atoms of N Check: Does the aswer seem reasoable? We have g urea. How may atoms of N would g of urea cotai? We could calculate the umber of atoms of the remaiig elemets i the same maer, or we ca use the atom ratios from the molecular formula. The carbo atom to itroge atom ratio i a urea molecule is 1:2, the oxyge atom to itroge atom ratio is 1:2, ad the hydroge atom to itroge atom ratio is 4: Catom 26? atoms of C ( N atoms) C atoms 2Natoms 26 1 O atom 26? atoms of O ( N atoms) O atoms 2Natoms 26 4Hatoms 26? atoms of H ( N atoms) H atoms 2 N atoms

7 38 CHAPTER 3: STOICHIOMETRY 3.29 The molar mass of C 19 H 38 O is g mol molecules g g 1 mol molecules Notice that eve though g is a extremely small mass, it still is comprised of over a billio pheromoe molecules! g Mass of water = 2.56 ml 1.00 ml = 2.56 g Molar mass of H 2 O (16.00 g) 2(1.008 g) g/mol? H2O molecules 1 mol H2O molecules H2O = 2.56 g H2O g H 2 O 1 mol H 2 O molecules 3.31 Please see Sectio 3.4 of the text The relative abudace of each isotope ca be determied from the area of the peak i the mass spectrum for that isotope Sice there are oly two isotopes of carbo, there are oly two possibilities for CF C 9F4 (molecular mass 88 amu) ad 6C 9F4 There would be two peaks i the mass spectrum. (molecular mass 89 amu) 3.34 Sice there are two hydroge isotopes, they ca be paired i three ways: 1 H- 1 H, 1 H- 2 H, ad 2 H- 2 H. There will the be three choices for each sulfur isotope. We ca make a table showig all the possibilities (masses i amu): 32 S 33 S 34 S 36 S 1 H H H H There will be seve peaks of the followig mass umbers: 34, 35, 36, 37, 38, 39, ad 40. Very accurate (ad expesive!) mass spectrometers ca detect the mass differece betwee two 1 H ad oe 2 H. How may peaks would be detected i such a high resolutio mass spectrum? 3.35 The percet compositio is the percet by mass of each elemet i a compoud. For NH 3, we would speak of the mass % of itroge (N) ad the mass % of hydroge (H) i the compoud. What percetage of the mass of a sample of ammoia is due to itroge ad what percetage of the mass is due to hydroge? 3.36 If you kow the percet compositio by mass of a ukow compoud, you ca determie its empirical formula.

8 CHAPTER 3: STOICHIOMETRY A empirical formula tells us which elemets are preset ad the simplest whole-umber ratio of their atoms. A defiitio of empirical is somethig that is derived from experimet ad observatio rather tha from theory. Whe chemists aalyze a ukow compoud, the first step is usually the determiatio of the compoud s empirical formula. With additioal iformatio, it is possible to deduce the molecular formula The approximate molar mass Molar mass of SO 2 (118.7 g) 2(16.00 g) g g/mol %S 100% 78.77% g/mol (2)(16.00 g/mol) %O 100% 21.% g/mol 3.40 Strategy: Recall the procedure for calculatig a percetage. Assume that we have 1 mole of CHCl 3. The percet by mass of each elemet (C, H, ad Cl) is give by the mass of that elemet i 1 mole of CHCl 3 divided by the molar mass of CHCl 3, the multiplied by 100 to covert from a fractioal umber to a percetage. Solutio: The molar mass of CHCl g/mol g/mol 3(35.45 g/mol) g/mol. The percet by mass of each of the elemets i CHCl 3 is calculated as follows: g/mol %C 100% 10.06% g/mol g/mol %H 100% % g/mol 3(35.45) g/mol %Cl 100% 89.07% g/mol Check: Do the percetages add to 100%? The sum of the percetages is (10.06% % 89.07%) 99.97%. The small discrepacy from 100% is due to the way we rouded off The molar mass of ciamic alcohol, C 9 H 10 O, is g/mol. (a) (9)(12.01 g/mol) %C 100% 80.56% g/mol (10)(1.008 g/mol) %H 100% 7.513% g/mol g/mol %O 100% 11.93% g/mol (b) 1 mol C9H10O molecules C9H10O g C9H10O g C 9 H 10 O 1 mol C 9 H 10 O molecules C 9 H 10 O

9 40 CHAPTER 3: STOICHIOMETRY 3.42 Compoud Molar mass (g) N% by mass 2(14.01 g) (a) (NH 2 ) 2 CO % = 46.65% g (b) NH 4 NO (c) HNC(NH 2 ) (d) NH Ammoia, NH 3, is the richest source of itroge o a mass percetage basis. 2(14.01 g) 100% = 35.00% g 3(14.01 g) 100% = 71.14% g g 100% = 82.27% g 3.43 Assume you have exactly 100 g of substace. C 1molC 44.4 g C g C 3.70 mol C 1molH H 6.21 g H g H 6.16 mol H 1molS S 39.5 g S g S 1. mol S 1molO O 9.86 g O g O mol O Thus, we arrive at the formula C 3.70 H 6.16 S 1. O Dividig by the smallest umber of moles (0.616 mole) gives the empirical formula, C 6 H 10 S 2 O. To determie the molecular formula, divide the molar mass by the empirical mass. molar mass empirical molar mass 162 g g Hece, the molecular formula ad the empirical formula are the same, C 6 H 10 S 2 O METHOD 1: Step 1: Assume you have exactly 100 g of substace. 100 g is a coveiet amout, because all the percetages sum to 100%. The percetage of oxyge is foud by differece: 100% (19.8% 2.50% 11.6%) 66.1% I 100 g of PAN there will be 19.8 g C, 2.50 g H, 11.6 g N, ad 66.1 g O. Step 2: Calculate the umber of moles of each elemet i the compoud. Remember, a empirical formula tells us which elemets are preset ad the simplest whole-umber ratio of their atoms. This ratio is also a mole ratio. Use the molar masses of these elemets as coversio factors to covert to moles. C H N 1molC = 19.8 g C g C = 1.65 mol C 1molH = 2.50 g H g H = 2.48 mol H 1molN = 11.6 g N g N = mol N

10 CHAPTER 3: STOICHIOMETRY 41 O 1molO = 66.1 g O = 4.13 mol O g O Step 3: Try to covert to whole umbers by dividig all the subscripts by the smallest subscript. The formula is C 1.65 H 2.48 N O Dividig the subscripts by gives the empirical formula, C 2 H 3 NO 5. To determie the molecular formula, remember that the molar mass/empirical mass will be a iteger greater tha or equal to oe. molar mass 1 (iteger values) empirical molar mass I this case, molar mass 120 g 1 empirical molar mass g Hece, the molecular formula ad the empirical formula are the same, C 2 H 3 NO 5. METHOD 2: Step 1: Multiply the mass % (coverted to a decimal) of each elemet by the molar mass to covert to grams of each elemet. The, use the molar mass to covert to moles of each elemet. 1molC C (0.198) (120 g) 1.98 mol C 2molC g C 1molH H (0.0250) (120 g) 2.98 mol H 3molH g H 1molN N (0.116) (120 g) mol N 1molN g N 1molO O (0.661) (120 g) 4.96 mol O 5molO g O Step 2: Sice we used the molar mass to calculate the moles of each elemet preset i the compoud, this method directly gives the molecular formula. The formula is C 2 H 3 NO 5. Step 3: Try to reduce the molecular formula to a simpler whole umber ratio to determie the empirical formula. The formula is already i its simplest whole umber ratio. The molecular ad empirical formulas are the same. The empirical formula is C 2 H 3 NO molFe2O3 2molFe 24.6 g Fe2O mol Fe g Fe 2 O 3 1 mol Fe 2 O Usig uit factors we covert: g of Hg mol Hg mol S g S 1 mol Hg 1 mol S g S?gS 246 g Hg 39.3gS g Hg 1 mol Hg 1 mol S

11 42 CHAPTER 3: STOICHIOMETRY 3.47 The balaced equatio is: 2Al(s) 3I 2 (s) 2AlI 3 (s) Usig uit factors, we covert: g of Al mol of Al mol of I 2 g of I 2 1molAl 3 mol I g I 20.4 g Al g I g Al 2 mol Al 1 mol I Strategy: Ti(II) fluoride is composed of S ad F. The mass due to F is based o its percetage by mass i the compoud. How do we calculate mass percet of a elemet? Solutio: First, we must fid the mass % of fluorie i SF 2. The, we covert this percetage to a fractio ad multiply by the mass of the compoud (24.6 g), to fid the mass of fluorie i 24.6 g of SF 2. The percet by mass of fluorie i ti(ii) fluoride, is calculated as follows: mass of F i 1 mol SF mass % F 2 100% molar mass of SF2 2(19.00 g) 100% = 24.25% F g Covertig this percetage to a fractio, we obtai 24.25/ Next, multiply the fractio by the total mass of the compoud.? g F i 24.6 g SF 2 (0.2425)(24.6 g) 5.97 g F Check: As a ball-park estimate, ote that the mass percet of F is roughly 25 percet, so that a quarter of the mass should be F. Oe quarter of approximately 24 g is 6 g, which is close to the aswer. Note: This problem could have bee worked i a maer similar to Problem You could complete the followig coversios: g of SF 2 mol of SF 2 mol of F g of F 3.49 I each case, assume 100 g of compoud. (a) 1molH 2.1 g H 2.1 mol H g H 1molO 65.3 g O 4.08 mol O g O 1molS 32.6 g S 1.02 mol S g S This gives the formula H 2.1 S 1.02 O Dividig by 1.02 gives the empirical formula, H 2 SO 4. (b) 1molAl 20.2 g Al mol Al g Al 1molCl 79.8 g Cl 2.25 mol Cl g Cl This gives the formula, Al Cl Dividig by gives the empirical formula, AlCl 3.

12 CHAPTER 3: STOICHIOMETRY (a) Strategy: I a chemical formula, the subscripts represet the ratio of the umber of moles of each elemet that combie to form the compoud. Therefore, we eed to covert from mass percet to moles i order to determie the empirical formula. If we assume a exactly 100 g sample of the compoud, do we kow the mass of each elemet i the compoud? How do we the covert from grams to moles? Solutio: If we have 100 g of the compoud, the each percetage ca be coverted directly to grams. I this sample, there will be 40.1 g of C, 6.6 g of H, ad 53.3 g of O. Because the subscripts i the formula represet a mole ratio, we eed to covert the grams of each elemet to moles. The coversio factor eeded is the molar mass of each elemet. Let represet the umber of moles of each elemet so that C H O 1molC 40.1 g C 3.34 mol C g C 1molH 6.6 g H 6.5 mol H g H 1molO 53.3 g O 3.33 mol O g O Thus, we arrive at the formula C 3.34 H 6.5 O 3.33, which gives the idetity ad the mole ratios of atoms preset. However, chemical formulas are writte with whole umbers. Try to covert to whole umbers by dividig all the subscripts by the smallest subscript (3.33). C: H: O: This gives the empirical formula, CH 2 O. Check: Are the subscripts i CH 2 O reduced to the smallest whole umbers? (b) Followig the same procedure as part (a), we fid: 1molC C 18.4 g C 1.53 mol C g C N K 1molN 21.5 g N 1.53 mol N g N 1molK 60.1 g K 1.54 mol K g K Dividig by the smallest umber of moles (1.53 mol) gives the empirical formula, KCN The molar mass of CaSiO 3 is g/mol g %Ca 100% 34.50% g g %Si 100% 24.18% g (3)(16.00 g) %O 100% 41.32% g Check to see that the percetages sum to 100%. (34.50% 24.18% 41.32%) %

13 44 CHAPTER 3: STOICHIOMETRY 3.52 The empirical molar mass of CH is approximately g. Let's compare this to the molar mass to determie the molecular formula. Recall that the molar mass divided by the empirical mass will be a iteger greater tha or equal to oe. I this case, molar mass 1 (iteger values) empirical molar mass molar mass empirical molar mass 78 g g Thus, there are six CH uits i each molecule of the compoud, so the molecular formula is (CH) 6, or C 6 H Fid the molar mass correspodig to each formula. For C 4 H 5 N 2 O: For C 8 H 10 N 4 O 2 : 4(12.01 g) 5(1.008 g) 2(14.01 g) (16.00 g) g 8(12.01 g) 10(1.008 g) 4(14.01 g) 2(16.00 g) g The molecular formula is C 8 H 10 N 4 O METHOD 1: Step 1: Assume you have exactly 100 g of substace. 100 g is a coveiet amout, because all the percetages sum to 100%. I 100 g of MSG there will be g C, 4.77 g H, g O, 8.29 g N, ad g Na. Step 2: Calculate the umber of moles of each elemet i the compoud. Remember, a empirical formula tells us which elemets are preset ad the simplest whole-umber ratio of their atoms. This ratio is also a mole ratio. Let C, H, O, N, ad Na be the umber of moles of elemets preset. Use the molar masses of these elemets as coversio factors to covert to moles. 1molC C g C g C mol C H 1molH 4.77 g H g H 4.73 mol H 1molO O g O g O mol O 1molN N 8.29 g N g N mol N 1 mol Na Na g Na g Na mol Na Thus, we arrive at the formula C H 4.73 O N Na , which gives the idetity ad the ratios of atoms preset. However, chemical formulas are writte with whole umbers. Step 3: Try to covert to whole umbers by dividig all the subscripts by the smallest subscript. C: = H: =8.00 O: = N: = Na : = This gives us the empirical formula for MSG, C 5 H 8 O 4 NNa.

14 CHAPTER 3: STOICHIOMETRY 45 To determie the molecular formula, remember that the molar mass/empirical mass will be a iteger greater tha or equal to oe. molar mass 1 (iteger values) empirical molar mass I this case, molar mass 169 g 1 empirical molar mass g Hece, the molecular formula ad the empirical formula are the same, C 5 H 8 O 4 NNa. It should come as o surprise that the empirical ad molecular formulas are the same sice MSG stads for moosodiumglutamate. METHOD 2: Step 1: Multiply the mass % (coverted to a decimal) of each elemet by the molar mass to covert to grams of each elemet. The, use the molar mass to covert to moles of each elemet. 1molC C (0.3551) (169 g) 5.00 mol C g C 1molH H (0.0477) (169 g) 8.00 mol H g H 1molO O (0.3785) (169 g) 4.00 mol O g O 1molN N (0.0829) (169 g) 1.00 mol N g N 1 mol Na Na (0.1360) (169 g) 1.00 mol Na g Na Step 2: Sice we used the molar mass to calculate the moles of each elemet preset i the compoud, this method directly gives the molecular formula. The formula is C 5 H 8 O 4 NNa A chemical reactio is a process i which a substace (or substaces) is chaged ito oe or more ew substaces. I this reactio, hydroge ad oxyge, the reactats, are chaged ito the product, water A chemical equatio uses chemical symbols to show what happes durig a chemical reactio. A chemical reactio is a process i which a substace (or substaces) is chaged ito oe or more ew substaces To accurately show what happes durig a chemical reactio, a chemical equatio must be balaced. The Law of Coservatio of Mass is obeyed by a balaced chemical equatio (g), (l), (s), (aq) The balaced equatios are as follows: (a) 2C O 2 2CO (f) 2O 3 3O 2 (b) 2CO O 2 2CO 2 (g) 2H 2 O 2 2H 2 O O 2 (c) H 2 Br 2 2HBr (h) N 2 3H 2 2NH 3 (d) 2K 2H 2 O 2KOH H 2 (i) Z 2AgCl ZCl 2 2Ag (e) 2Mg O 2 2MgO (j) S 8 8O 2 8SO 2

15 46 CHAPTER 3: STOICHIOMETRY (k) 2NaOH H 2 SO 4 Na 2 SO 4 2H 2 O (m) 3KOH H 3 PO 4 K 3 PO 4 3H 2 O (l) Cl 2 2NaI 2NaCl I 2 () CH 4 4Br 2 CBr 4 4HBr 3.60 The balaced equatios are as follows: (a) 2N 2 O 5 2N 2 O 4 O 2 (b) 2KNO 3 2KNO 2 O 2 (c) NH 4 NO 3 N 2 O 2H 2 O (d) NH 4 NO 2 N 2 2H 2 O (e) 2NaHCO 3 Na 2 CO 3 H 2 O CO 2 (f) P 4 O 10 6H 2 O 4H 3 PO 4 (g) 2HCl CaCO 3 CaCl 2 H 2 O CO 2 (h) 2Al 3H 2 SO 4 Al 2 (SO 4 ) 3 3H 2 (i) CO 2 2KOH K 2 CO 3 H 2 O (j) CH 4 2O 2 CO 2 2H 2 O (k) Be 2 C 4H 2 O 2Be(OH) 2 CH 4 (l) 3Cu 8HNO 3 3Cu(NO 3 ) 2 2NO 4H 2 O (m) S 6HNO 3 H 2 SO 4 6NO 2 2H 2 O () 2NH 3 3CuO 3Cu N 2 3H 2 O 3.61 Stoichiometry is the quatitative study of reactats ad products i a chemical reactio; therefore, it is based o the Law of Coservatio of Mass. A balaced chemical equatio is essetial to solvig stoichiometric problems so that the mole method ca be applied correctly The steps of the mole method are show i Figure 3.8 of the text O the reactats side there are 8 A atoms ad 4 B atoms. O the products side, there are 4 C atoms ad 4 D atoms. Writig a equatio, 8A 4B 4C 4D Chemical equatios are typically writte with the smallest set of whole umber coefficiets. Dividig the equatio by four gives, 2A B C D The correct aswer is choice (c) O the reactats side there are 6 A atoms ad 4 B atoms. O the products side, there are 4 C atoms ad 2 D atoms. Writig a equatio, 6A 4B 4C 2D Chemical equatios are typically writte with the smallest set of whole umber coefficiets. Dividig the equatio by two gives, 3A 2B 2C D The correct aswer is choice (d) The mole ratio from the balaced equatio is 2 moles CO 2 : 2 moles CO. 2molCO 3.60 mol CO mol CO 2 2molCO

16 CHAPTER 3: STOICHIOMETRY Si(s) 2Cl 2 (g) SiCl 4 (l) Strategy: Lookig at the balaced equatio, how do we compare the amouts of Cl 2 ad SiCl 4? We ca compare them based o the mole ratio from the balaced equatio. Solutio: Because the balaced equatio is give i the problem, the mole ratio betwee Cl 2 ad SiCl 4 is kow: 2 moles Cl 2 1 mole SiCl 4. From this relatioship, we have two coversio factors. 2 mol Cl2 1 mol SiCl ad 4 1 mol SiCl4 2 mol Cl2 Which coversio factor is eeded to covert from moles of SiCl 4 to moles of Cl 2? The coversio factor o the left is the correct oe. Moles of SiCl 4 will cacel, leavig uits of "mol Cl 2 " for the aswer. We calculate moles of Cl 2 reacted as follows: 2molCl? mol Cl mol SiCl 2 2reacted mol Cl 2 1molSiCl4 Check: Does the aswer seem reasoable? Should the moles of Cl 2 reacted be double the moles of SiCl 4 produced? 3.67 Startig with the amout of ammoia produced (6.0 moles), we ca use the mole ratio from the balaced equatio to calculate the moles of H 2 ad N 2 that reacted to produce 6.0 moles of NH 3. 3H 2 (g) N 2 (g) 2NH 3 (g) 3molH?molH molNH3 9.0 mol H 2 2 mol NH3 1molN?molN molNH3 3.0 mol N 2 2molNH Startig with the 5.0 moles of C 4 H 10, we ca use the mole ratio from the balaced equatio to calculate the moles of CO 2 formed. 2C 4 H 10 (g) 13O 2 (g) 8CO 2 (g) 10H 2 O(l) 8molCO? mol CO mol C4H10 20 mol CO2 2molC4H mol CO It is coveiet to use the uit to-mol i this problem. We ormally use a g-mol. 1 g-mol SO 2 has a mass of g. I a similar maer, 1 to-mol of SO 2 has a mass of tos. We eed to complete the followig coversios: tos SO 2 to-mol SO 2 to-mol S to S. 7 1 to-mol SO2 1 to-mol S to S ( tos SO 2) to SO2 1 to-mol SO2 1 to-mol S tos S 3.70 (a) 2NaHCO 3 Na 2 CO 3 H 2 O CO 2

17 48 CHAPTER 3: STOICHIOMETRY (b) Molar mass NaHCO g g g 3(16.00 g) g Molar mass CO g 2(16.00 g) g The balaced equatio shows oe mole of CO 2 formed from two moles of NaHCO 3. mass NaHCO 3 1molCO2 2 mol NaHCO g NaHCO3 = 20.5 g CO g CO2 1 mol CO2 1 mol NaHCO g NaHCO The balaced equatio shows a mole ratio of 1 mole HCN : 1 mole KCN. 1 mol KCN 1 mol HCN g HCN g KCN g HCN g KCN 1 mol KCN 1 mol HCN 3.72 C 6 H 12 O 6 2C 2 H 5 OH 2CO 2 glucose ethaol Strategy: We compare glucose ad ethaol based o the mole ratio i the balaced equatio. Before we ca determie moles of ethaol produced, we eed to covert to moles of glucose. What coversio factor is eeded to covert from grams of glucose to moles of glucose? Oce moles of ethaol are obtaied, aother coversio factor is eeded to covert from moles of ethaol to grams of ethaol. Solutio: The molar mass of glucose will allow us to covert from grams of glucose to moles of glucose. The molar mass of glucose 6(12.01 g) 12(1.008 g) 6(16.00 g) g. The balaced equatio is give, so the mole ratio betwee glucose ad ethaol is kow; that is 1 mole glucose 2 moles ethaol. Fially, the molar mass of ethaol will covert moles of ethaol to grams of ethaol. This sequece of three coversios is summarized as follows: grams of glucose moles of glucose moles of ethaol grams of ethaol?gc2h5oh 1 mol C6H12O6 2 mol C2H5OH g C2H5OH g C6H12O g C 6 H 12 O 6 1 mol C 6 H 12 O 6 1 mol C 2 H 5 OH g C 2 H 5 OH Check: Does the aswer seem reasoable? Should the mass of ethaol produced be approximately half the mass of glucose reacted? Twice as may moles of ethaol are produced compared to the moles of glucose reacted, but the molar mass of ethaol is about oe-fourth that of glucose. The liters of ethaol ca be calculated from the desity ad the mass of ethaol. volume mass desity Volume of ethaol obtaied g = = 324 ml = g/ml L 3.73 The mass of water lost is just the differece betwee the iitial ad fial masses. Mass H 2 O lost g 9.60 g 5.41 g 1molH2O moles of H2O 5.41 g H2O mol H2O g H 2 O

18 CHAPTER 3: STOICHIOMETRY The balaced equatio shows that eight moles of KCN are eeded to combie with four moles of Au.?molKCN 1molAu 8molKCN 29.0 g Au = g Au 4 mol Au 0.294molKCN 3.75 The balaced equatio is: CaCO 3 (s) CaO(s) CO 2 (g) 1000 g 1 mol CaCO3 1 mol CaO g CaO 1.0 kg CaCO3 1 kg g CaCO3 1 mol CaCO3 1 mol CaO g CaO 3.76 (a) NH 4 NO 3 (s) N 2 O(g) 2H 2 O(g) (b) Startig with moles of NH 4 NO 3, we ca use the mole ratio from the balaced equatio to fid moles of N 2 O. Oce we have moles of N 2 O, we ca use the molar mass of N 2 O to covert to grams of N 2 O. Combiig the two coversios ito oe calculatio, we have: mol NH 4 NO 3 mol N 2 O g N 2 O 1 mol N2O g N2O 1?gN2O 0.46 mol NH4NO gn2o 1 mol NH 4 NO 3 1 mol N 2 O 3.77 The quatity of ammoia eeded is: 8 1mol(NH 4) 2SO4 2 mol NH g NH3 1kg g (NH 4) 2SO g (NH 4) 2SO4 1 mol (NH 4) 2SO4 1 mol NH g kg NH The balaced equatio for the decompositio is : 2KClO 3 (s) 2KCl(s) 3O 2 (g) 1 mol KClO3 3 mol O g O?gO g KClO3 18.0gO g KClO 3 2 mol KClO 3 1 mol O The reactat used up first i a reactio is called the limitig reaget. Excess reagets are the reactats preset i quatities greater tha ecessary to react with the quatity of the limitig reaget. The maximum amout of product formed depeds o how much of the limitig reaget is preset. Whe this reactat is used up, o more product ca be formed. I a reactio with oe reactat, the oe reactat is by defiitio the limitig reaget If you are makig sadwiches ad have four pieces of bread, you ca oly make two sadwiches, o matter how much mayoaise, mustard, sadwich meat, etc. that you have. The bread limits the umber of sadwiches that you ca make A B C (a) The umber of B atoms show i the diagram is 5. The balaced equatio shows 2 moles A 1 mole B. Therefore, we eed 10 atoms of A to react completely with 5 atoms of B. There are oly 8 atoms of A preset i the diagram. There are ot eough atoms of A to react completely with B. A is the limitig reaget.

19 50 CHAPTER 3: STOICHIOMETRY (b) There are 8 atoms of A. Sice the mole ratio betwee A ad B is 2:1, 4 atoms of B will react with 8 atoms of A, leavig 1 atom of B i excess. The mole ratio betwee A ad C is also 2:1. Whe 8 atoms of A react, 4 molecules of C will be produced. B C 3.82 N 2 3H 2 2NH 3 9 moles of H 2 will react with 3 moles of N 2, leavig 1 mole of H 2 i excess. The mole ratio betwee N 2 ad NH 3 is 1:2. Whe 3 moles of N 2 react, 6 moles of NH 3 will be produced. H 2 NH This is a limitig reaget problem. Let's calculate the moles of NO 2 produced assumig complete reactio for each reactat. 2NO(g) O 2 (g) 2NO 2 (g) 2molNO mol NO mol NO2 2molNO 2 mol NO mol O mol NO2 1molO2 NO is the limitig reaget; it limits the amout of product produced. The amout of product produced is mole NO Strategy: Note that this reactio gives the amouts of both reactats, so it is likely to be a limitig reaget problem. The reactat that produces fewer moles of product is the limitig reaget because it limits the amout of product that ca be produced. How do we covert from the amout of reactat to amout of product? Perform this calculatio for each reactat, the compare the moles of product, NO 2, formed by the give amouts of O 3 ad NO to determie which reactat is the limitig reaget. Solutio: We carry out two separate calculatios. First, startig with g O 3, we calculate the umber of moles of NO 2 that could be produced if all the O 3 reacted. We complete the followig coversios. grams of O 3 moles of O 3 moles of NO 2

20 CHAPTER 3: STOICHIOMETRY 51 Combiig these two coversios ito oe calculatio, we write 1molO3 1molNO? mol NO g O mol NO g O3 1 mol O3 Secod, startig with g of NO, we complete similar coversios. grams of NO moles of NO moles of NO 2 Combiig these two coversios ito oe calculatio, we write 1molNO 1 mol NO? mol NO g NO mol NO g NO 1 mol NO. The iitial amout of O 3 limits the amout of product that ca be formed; therefore, it is the limitig reaget. The problem asks for grams of NO 2 produced. We already kow the moles of NO 2 produced, mole. Use the molar mass of NO 2 as a coversio factor to covert to grams (Molar mass NO g) g NO? g NO mol NO g NO 2 1 mol NO2 Check: Does your aswer seem reasoable? mole of product is formed. What is the mass of 1 mole of NO 2? Strategy: Workig backwards, we ca determie the amout of NO that reacted to produce mole of NO 2. The amout of NO left over is the differece betwee the iitial amout ad the amout reacted. Solutio: Startig with mole of NO 2, we ca determie the moles of NO that reacted usig the mole ratio from the balaced equatio. We ca calculate the iitial moles of NO startig with g ad usig molar mass of NO as a coversio factor. 1molNO mol NO reacted mol NO mol NO 1 mol NO2 1molNO mol NO iitial g NO 0.02 mol NO g NO mol NO remaiig mol NO iitial mol NO reacted.. mol NO remaiig 0.02 mol NO mol NO mol NO 3.85 (a) The balaced equatio is: C 3 H 8 (g) 5O 2 (g) 3CO 2 (g) 4H 2 O(l) (b) The balaced equatio shows a mole ratio of 3 moles CO 2 : 1 mole C 3 H 8. The mass of CO 2 produced is: 3 mol CO g CO 3.65 mol C 2 3H8 482 g CO 2 1molC H 1molCO This is a limitig reaget problem. Let's calculate the moles of Cl 2 produced assumig complete reactio for each reactat. 1molCl 0.86 mol MO 2 2 = 0.86 mol Cl2 1molMO2

21 52 CHAPTER 3: STOICHIOMETRY 1molHCl 1molCl 48.2 g HCl 2 = mol Cl g HCl 4 mol HCl HCl is the limitig reaget; it limits the amout of product produced. It will be used up first. The amout of product produced is mole Cl 2. Let's covert this to grams g Cl? g Cl mol Cl =.4 g Cl 2 1molCl The theoretical yield of a reactio is the amout of product that would result if all the limitig reaget reacted. Whe the limitig reactat is used up, o more product ca be formed There are may reasos why the actual yield is less tha the theoretical yield. Some reactios are reversible, so they do ot proceed 100% from reactats to product.. There could be impurities i the startig materials. Sometimes it is difficult to recover all the product. There may be side reactios that lead to additioal products The balaced equatio is give: CaF 2 H 2 SO 4 CaSO 4 2HF The balaced equatio shows a mole ratio of 2 moles HF : 1 mole CaF 2. The theoretical yield of HF is: 3 1molCaF2 2 mol HF g HF 1 kg ( g CaF 2) 3.08 kg HF g CaF2 1 mol CaF2 1 mol HF 1000 g The actual yield is give i the problem (2.86 kg HF). actual yield % yield 100% theoretical yield 2.86 kg % yield 100% 92.9% 3.08 kg 3.90 (a) Start with a balaced chemical equatio. It s give i the problem. We use NG as a abbreviatio for itroglyceri. The molar mass of NG g/mol. 4C 3 H 5 N 3 O 9 6N 2 12CO 2 10H 2 O O 2 Map out the followig strategy to solve this problem. g NG mol NG mol O 2 g O 2 Calculate the grams of O 2 usig the strategy above. 2 1molNG 1 mol O g O?gO g NG gO g NG 4 mol NG 1 mol O 2 (b) The theoretical yield was calculated i part (a), ad the actual yield is give i the problem (6.55 g). The percet yield is: actual yield % yield 100% theoretical yield 6.55 g O %yield 2 100% = 92.9% 7.05 g O2

22 CHAPTER 3: STOICHIOMETRY The balaced equatio shows a mole ratio of 1 mole TiO 2 : 1 mole FeTiO 3. The molar mass of FeTiO 3 is g/mol, ad the molar mass of TiO 2 is g/mol. The theoretical yield of TiO 2 is: 6 1molFeTiO3 1 mol TiO g TiO2 1kg g FeTiO g FeTiO3 1 mol FeTiO3 1 mol TiO g kg TiO 2 The actual yield is give i the problem ( kg TiO 2 ). 3 actual yield kg % yield 100% 100% theoretical yield kg 87.2% 3.92 This is a limitig reaget problem. Let's calculate the moles of Li 3 N produced assumig complete reactio for each reactat. 6Li(s) N 2 (g) 2Li 3 N(s) 1molLi 2molLi3N 12.3 g Li mol Li3N g Li 6 mol Li 1molN2 2molLi3N 33.6 g N mol Li3N g N2 1 mol N2 Li is the limitig reaget; it limits the amout of product produced. The amout of product produced is mole Li 3 N. Let's covert this to grams g Li3N?gLi3N 0.591molLi3N 20.6 g Li3N 1molLi3N This is the theoretical yield of Li 3 N. The actual yield is give i the problem (5.89 g). The percet yield is: actual yield 5.89 g % yield 100% 100% theoretical yield 20.6 g 28.6% 3.93 All the carbo from the hydrocarbo reactat eds up i CO 2, ad all the hydroge from the hydrocarbo reactat eds up i water. I the diagram, we fid 4 CO 2 molecules ad 6 H 2 O molecules. This gives a ratio betwee carbo ad hydroge of 4:12. We write the formula C 4 H 12, which reduces to the empirical formula CH 3. The empirical molar mass equals approximately 15 g, which is half the molar mass of the hydrocarbo. Thus, the molecular formula is double the empirical formula or C 2 H 6. Sice this is a combustio reactio, the other reactat is O 2. We write: C 2 H 6 O 2 CO 2 H 2 O Balacig the equatio, 2C 2 H 6 7O 2 4CO 2 6H 2 O H 2 (g) O 2 (g) 2H 2 O(g) We start with 8 molecules of H 2 ad 3 molecules of O 2. The balaced equatio shows 2 moles H 2 1 mole O 2. If 3 molecules of O 2 react, 6 molecules of H 2 will react, leavig 2 molecules of H 2 i excess. The balaced equatio also shows 1 mole O 2 2 moles H 2 O. If 3 molecules of O 2 react, 6 molecules of H 2 O will be produced.

23 54 CHAPTER 3: STOICHIOMETRY After complete reactio, there will be 2 molecules of H 2 ad 6 molecules of H 2 O. The correct diagram is choice (b) First, let's covert to moles of HNO 3 produced lb g 1molHNO to HNO mol HNO3 1 to 11b g HNO3 Now, we will work i the reverse directio to calculate the amout of reactat eeded to produce mol of HNO 3. Realize that sice the problem says to assume a 80% yield for each step, the amout of reactat eeded i each step will be larger by a factor of 100%, compared to a stadard stoichiometry 80% calculatio where a 100% yield is assumed. Referrig to the balaced equatio i the last step, we calculate the moles of NO molNO2 100% 4 ( mol HNO 3) mol NO2 1 mol HNO3 80% Now, let's calculate the amout of NO eeded to produce mol NO 2. Followig the same procedure as above, ad referrig to the balaced equatio i the middle step, we calculate the moles of NO. 4 1 mol NO 100% 4 ( mol NO 2) mol NO 1molNO2 80% Now, let's calculate the amout of NH 3 eeded to produce mol NO. Referrig to the balaced equatio i the first step, the moles of NH 3 is: 4 4 mol NH3 100% 4 ( mol NO) mol NH3 4molNO 80% Fially, covertig to grams of NH 3 : g NH mol NH3 1molNH g NH We assume that all the Cl i the compoud eds up as HCl ad all the O eds up as H 2 O. Therefore, we eed to fid the umber of moles of Cl i HCl ad the umber of moles of O i H 2 O. 1 mol HCl 1 mol Cl mol Cl 0.3 g HCl mol Cl g HCl 1 mol HCl 1molH2O 1molO mol O g H2O mol O 18.02gHO 2 1molHO 2 Dividig by the smallest umber of moles ( mole) gives the formula, ClO 3.5. Multiplyig both subscripts by two gives the empirical formula, Cl 2 O The umber of moles of Y i g of Y is: 1molX 1molY g X mol Y g X 1 mol X

24 CHAPTER 3: STOICHIOMETRY 55 The molar mass of Y is: molar mass Y g Y g/mol mol Y The atomic mass of Y is amu This is a calculatio ivolvig percet compositio. Remember, mass of elemet i 1 mol of compoud percet by mass of each elemet 100% molar mass of compoud The molar masses are: Al, g/mol; Al 2 (SO 4 ) 3, g/mol; H 2 O, g/mol. Thus, usig x as the umber of H 2 O molecules, 2(molar mass of Al) mass % Al 100% molar mass of Al 2(SO 4) 3 x(molar mass of H2O) 2(26.98 g) 8.20% 100% g x(18.02 g) x Roudig off to a whole umber of water molecules, x 18. Therefore, the formula is Al 2 (SO 4 ) 3 18 H 2 O The amout of Fe that reacted is: The amout of Fe remaiig is: g 83.0 g reacted g 83.0 g 581 g remaiig Thus, 83.0 g of Fe reacts to form the compoud Fe 2 O 3, which has two moles of Fe atoms per 1 mole of compoud. The mass of Fe 2 O 3 produced is: 1molFe 1 mol Fe2O g Fe2O g Fe 119 g Fe2O g Fe 2 mol Fe 1 mol Fe2O3 The fial mass of the iro bar ad rust is: 581 g Fe 119 g Fe 2 O g The mass of oxyge i MO is g g 7.76 g O. Therefore, for every g of M, there is 7.76 g of O i the compoud MO. The molecular formula shows a mole ratio of 1 mole M : 1 mole O. First, calculate moles of M that react with 7.76 g O. 1molO 1molM mol M 7.76 g O mol M g O 1 mol O molar mass M g M mol M 65.4 g/mol Thus, the atomic mass of M is 65.4 amu. The metal is most likely Z.

25 56 CHAPTER 3: STOICHIOMETRY (a) Z(s) H 2 SO 4 (aq) ZSO 4 (aq) H 2 (g) (b) We assume that a pure sample would produce the theoretical yield of H 2. The balaced equatio shows a mole ratio of 1 mole H 2 : 1 mole Z. The theoretical yield of H 2 is: 1molZ 1 mol H g H 3.86 g Z g H g Z 1 mol Z 1 mol H g H percet purity 2 100% 64.2% g H2 (c) We assume that the impurities are iert ad do ot react with the sulfuric acid to produce hydroge The wordig of the problem suggests that the actual yield is less tha the theoretical yield. The percet yield will be equal to the percet purity of the iro(iii) oxide. We fid the theoretical yield : g Fe2O3 1 mol Fe2O3 2 mol Fe g Fe 1 kg Fe ( kg Fe2O 3) 1 kg Fe2O g Fe2O3 1 mol Fe2O3 1 mol Fe 1000 g Fe kg Fe actual yield percet yield 100% theoretical yield kg Fe percet yield = 100% = 89.6% purity of Fe 3 2O kg Fe The balaced equatio is: C 6 H 12 O 6 6O 2 6CO 2 6H 2 O g glucose 1 mol glucose 6 mol CO g CO2 365 days 9 ( people) 1 day g glucose 1 mol glucose 1 mol CO2 1 yr g CO 2 /yr The carbohydrate cotais 40 percet carbo; therefore, the remaiig 60 percet is hydroge ad oxyge. The problem states that the hydroge to oxyge ratio is 2:1. We ca write this 2:1 ratio as H 2 O. Assume 100 g of compoud. 1molC 40.0 g C 3.33 mol C g C 1molH2O 60.0 g H2O 3.33 mol H2O g H2O Dividig by 3.33 gives CH 2 O for the empirical formula. To fid the molecular formula, divide the molar mass by the empirical mass. molar mass 178 g = 6 empirical mass g Thus, there are six CH 2 O uits i each molecule of the compoud, so the molecular formula is (CH 2 O) 6, or C 6 H 12 O 6.

26 CHAPTER 3: STOICHIOMETRY The mass of the metal (X) i the metal oxide is 1.68 g. The mass of oxyge i the metal oxide is 2.40 g 1.68 g 0.72 g oxyge. Next, fid the umber of moles of the metal ad of the oxyge. moles X 1molX 1.68 g 55.9 g X mol X moles O 1molO 0.72 g g O mol O This gives the formula X O Dividig by the smallest umber of moles ( moles) gives the formula X 1.00 O 1.5. Multiplyig by two gives the empirical formula, X 2 O 3. The balaced equatio is: X 2 O 3 (s) 3CO(g) 2X(s) 3CO 2 (g) Both compouds cotai oly M ad O. Whe the first compoud is heated, oxyge gas is evolved. Let s calculate the empirical formulas for the two compouds, the we ca write a balaced equatio. (a) Compoud X: Assume 100 g of compoud. 1molM 63.3 g M 1.15 mol M g M 1molO 36.7 g O 2.29 mol O g O Dividig by the smallest umber of moles (1.15 moles) gives the empirical formula, MO 2. Compoud Y: Assume 100 g of compoud. 1molM 72.0 g M 1.31 mol M g M 1molO 28.0 g O 1.75 mol O g O Dividig by the smallest umber of moles gives MO Recall that a empirical formula must have whole umber coefficiets. Multiplyig by a factor of 3 gives the empirical formula M 3 O 4. (b) The ubalaced equatio is: Balacig by ispectio gives: MO 2 M 3 O 4 O 2 3MO 2 M 3 O 4 O We carry a additioal sigificat figure throughout this calculatio to avoid roudig errors. Assume 100 g of sample. The, 1molNa mol Na g Na mol Na g Na 1molO mol O g O mol O g O 1molCl mol Cl g Cl mol Cl g Cl Sice Cl is oly cotaied i NaCl, the moles of Cl equals the moles of Na cotaied i NaCl. mol Na (i NaCl) mol

27 58 CHAPTER 3: STOICHIOMETRY The umber of moles of Na i the remaiig two compouds is: mol mol mol Na. To solve for moles of the remaiig two compouds, let x moles of Na 2 SO 4 y moles of NaNO 3 The, from the mole ratio of Na ad O i each compoud, we ca write 2x y mol Na mol 4x 3y mol O mol Solvig two equatios with two ukows gives x mol Na 2 SO 4 ad y mol NaNO 3 Fially, we covert to mass of each compoud to calculate the mass percet of each compoud i the sample. Remember, the sample size is 100 g. mass % NaCl g NaCl mol NaCl 100% 1 mol NaCl 100 g sample 32.16% NaCl g Na2SO4 1 mass % Na2SO mol Na2SO4 100% 20.21% Na2SO 4 1 mol Na2SO4 100 g sample g NaNO3 1 mass % NaNO mol NaNO3 100% 47.64% NaNO 3 1 mol NaNO3 100 g sample We assume that the icrease i mass results from the elemet itroge. The mass of itroge is: g g g N The empirical formula ca ow be calculated. Covert to moles of each elemet. 1molMg g Mg mol Mg g Mg 1molN g N mol N g N Dividig by the smallest umber of moles gives Mg 1.5 N. Recall that a empirical formula must have whole umber coefficiets. Multiplyig by a factor of 2 gives the empirical formula Mg 3 N 2. The ame of this compoud is magesium itride The balaced equatios are: CH 4 2O 2 CO 2 2H 2 O 2C 2 H 6 7O 2 4CO 2 6H 2 O If we let x mass of CH 4, the the mass of C 2 H 6 is (13.43 x) g. Next, we eed to calculate the mass of CO 2 ad the mass of H 2 O produced by both CH 4 ad C 2 H 6. The sum of the masses of CO 2 ad H 2 O will add up to g. 1 mol CH4 1 mol CO g CO? g CO 2 2 (from CH 4) x g CH x g CO g CH4 1 mol CH4 1 mol CO2

28 CHAPTER 3: STOICHIOMETRY 59 1 mol CH4 2 mol H2O g H2O? g H2O (from CH 4) x g CH x g H2O g CH4 1 mol CH4 1 mol H2O 1molC2H6 4 mol CO g CO? g CO 2 2 (from C2H 6) (13.43 x ) g C2H g C2H6 2 mol C2H6 1 mol CO (13.43 x) g CO 2 1molC2H6 6 mol H2O g H2O? g H2O (from C2H 6) (13.43 x ) g C2H gC2H6 2molC2H6 1molH2O 1.798(13.43 x) g H 2 O Summig the masses of CO 2 ad H 2 O: 2.744x g 2.247x g 2.927(13.43 x) g 1.798(13.43 x) g g 0.266x x 5.20 g The fractio of CH 4 i the mixture is 5.20 g g The molecular formula of cysteie is C 3 H 7 NO 2 S. The mass percetage of each elemet is: %C (3)(12.01 g) 100% g 29.74% %H (7)(1.008 g) 100% g 5.8% %N g 100% g 11.56% %O (2)(16.00 g) 100% g 26.41% g %S 100% g 26.47% Check: 29.74% 5.8% 11.56% 26.41% 26.47% % The molecular formula of isoflurae is C 3 H 2 ClF 5 O. The mass percetage of each elemet is: %C (3)(12.01 g) 100% 19.53% g %H (2)(1.008 g) 100% 1.093% g %Cl g 100% g 19.21% (5)(19.00) g %F 100% 51.49% g