3. CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS

Transcription

1 . CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Solutions to Eercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off..1 a. NO 1 AW of N amu AW of O amu MW of NO amu ( s.f.) b. C 6 H 1 O 6 6 AW of C amu 1 AW of H amu 6 AW of O amu MW of C 6 H 1 O amu 180. amu ( s.f.) c. NaOH 1 AW of Na amu 1 AW of O amu 1 AW of H amu MW of NaOH amu 0.0 amu ( s.f.) d. Mg(OH) 1 AW of Mg.05 amu AW of O amu AW of H amu MW of Mg(OH) amu 58. amu ( s.f.)

2 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 61. a. The molecular model represents a molecule made up of one S and three O. The chemical formula is SO. Using the same approach as Eample.1 in the tet, calculating the formula weight yields amu. b. The molecular model represents one S, four O, and two H. The chemical formula is then H SO. The formula weight is amu.. a. The atomic weight of Ca 0.08 amu; thus, the molar mass 0.08 g/mol, and one mol of Ca Ca atoms. Mass of one Ca 0.08 g 1 mol Ca 1 mol atoms g/atom b. The molecular weight of C H 5 OH, or C H 6 O, ( 1.01) + ( ) Its molar mass 6.07 g/mol, and one mol molecules of C H 6 O. Mass of one C H 6 O 6.07 g 1 mol CH6O 1 mol atoms g/molecule. The molar mass of H O is.0 g/mol. Therefore,.0 g HO mol HO g HO 1mol HO.5 The molar mass of HNO is 6.01 g/mol. Therefore, 1mol HNO 8.5 g HNO mol HNO 6.01g HNO.6 Convert the mass of HCN from milligrams to grams. Then, convert grams of HCN to moles of HCN. Finally, convert moles of HCN to the number of HCN molecules. 56 mg HCN 1 g 1000 mg 1 mol HCN 7.0 g HCN HCN molecules 1 mol HCN HCN molecules

3 6 CHAPTER.7 The molecular weight of NH NO 80.05; thus, its molar mass g/mol. Hence Percent N 8.0 g g Percent H.0 g g Percent O 8.00 g g 100% % 100% % 100% %.8 From the previous eercise, NH NO is 5.0 percent N (fraction N 0.50), so the mass of N in 8.5 g of NH NO is 8.5 g NH NO (0.50 g N/1 g NH NO ) g N.9 First, convert the mass of CO to moles of CO. Net, convert this to moles of C (one mol of CO is equivalent to one mol C). Finally, convert to mass of carbon, changing mg to g first: g CO 1mol CO.01g CO g C 1mol C 1mol CO 1.01g C 1mol C Do the same series of calculations for water, noting that one mol H O contains two mol H g H O 1mol HO mol H 18.0 g H O 1mol H O g H g H 1mol H The mass percentages of C and H can be calculated using the masses from the previous calculations: Percent C Percent H 1.58 mg.87 mg mg.87 mg 100% % C 100% % H The mass percentage of O can be determined by subtracting the sum of the above percentages from 100 percent: Percent O % - ( ) % O

4 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 6.10 Convert the masses to moles that are proportional to the subscripts in the empirical formula:. g S 1mol S.07 g S 1.01 mol S (8.5 -.) g O 1mol O g O.1 mol O Net, obtain the smallest integers from the moles by dividing each by the smallest number of moles: For O: For S: The empirical formula is SO..11 For a g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 6. g are O. Using the molar masses, convert these masses to moles: 68.8 g C 5.0 g H 6. g O 1mol C 1.01g C 1mol H g H 1mol O g O 5.79 mol C.96 mol H 1.68 mol O These numbers are in the same ratio as the subscripts in the empirical formula. They must be changed to integers. First, divide each one by the smallest mol number: For C :.97 For H : For O : Rounding off, we obtain C.5 H.0 O 1.0. Multiplying the numbers by two gives whole numbers for an empirical formula of C 7 H 6 O.

5 6 CHAPTER.1 For a g sample of acetaldehyde, 5.5 g are C, 9. g are H, and 6. g are O. Using the molar masses, convert these masses to moles: 5.5 g C 9. g H 6. g O 1mol C 1.01g C 1mol H g H 1mol O g O.57 mol C 9.1 mol H.68 mol O These numbers are in the same ratio as the subscripts in the empirical formula. They must be changed to integers. First, divide each one by the smallest mol number: For C :.000 For H : For O : Rounding off, we obtain C H O, the empirical formula, which is also the molecular formula..1 H + Cl HCl 1 molec. (mol) H + 1 molec. (mol) Cl molec. (mol) HCl (molec., mole interp.).016 g H g Cl 6.5 g HCl (mass interp.).1 Equation: Na + H O 1/H + NaOH, or Na + H O H + NaOH. From this equation, one mol of Na corresponds to one-half mol of H, or two mol of Na corresponds to one mol of H. Therefore, 7.81 g H 1mol H.016 g H mol Na 1mol H.99 g Na 1mol Na g Na.15 Balanced equation: ZnS + O ZnO + SO Convert grams of ZnS to moles of ZnS. Then, determine the relationship between ZnS and O (ZnS is equivalent to O ). Finally, convert to mass O g ZnS 1mol ZnS 97.6 g ZnS.6.6 kg O mol O mol ZnS.00 g O 1mol O 1kg 1000 g

6 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Balanced equation: HgO Hg + O Convert the mass of O to mol of O. Using the fact that one mol of O is equivalent to two mol of Hg, determine the number of mol of Hg, and convert to mass of Hg. 1 mol O 6.7 g O.00 g O mol Hg 1 mol O 81.1 g Hg g Hg 1 mol Hg First, determine the limiting reactant by calculating the moles of AlCl that would be obtained if Al and HCl were totally consumed: 0.15 mol Al mol AlCl mol Al mol AlCl 0.5 mol HCl mol AlCl 6 mol HCl mol AlCl Because the HCl produces the smaller amount of AlCl, the reaction will stop when HCl is totally consumed but before the Al is consumed. The limiting reactant is therefore HCl. The amount of AlCl produced must be , or 0.1 mol..18 First, determine the limiting reactant by calculating the moles of ZnS produced by totally consuming Zn and S 8 : 7.6 g Zn 1 mol Zn 65.9 g Zn 8 mol ZnS 8 mol Zn mol ZnS 6.5 g S 8 1 mol S g S 8 8 mol ZnS 1 mol S mol ZnS The reaction will stop when Zn is totally consumed; S 8 is in ecess, and not all of it is converted to ZnS. The limiting reactant is therefore Zn. Now convert the moles of ZnS obtained from the Zn to grams of ZnS: mol ZnS 97.6 g ZnS 1 mol ZnS g ZnS

7 66 CHAPTER.19 First, write the balanced equation: CH OH + CO HC H O Convert grams of each reactant to moles of acetic acid: 15.0 g CH 1mol CH OH 1mol HC OH 0.681mol HCHO.0 g CHOH 1mol CHOH H O 10.0 g CO 1mol CO 8.01g CO 1mol HC 1mol CH H O OH mol HC H O Thus, CO is the limiting reactant, and mol HC H O is obtained. The mass of product is mol HC g HC H HO 1. g HCHO 1mol HCHO O The percentage yield is: 19.1 g actual yield 1. g theoretical yield 100% % Mass percentage g HC H 5.00 g vinegar O 100% % Answers to Concept Checks.1 a. Each tricycle has one seat, so you have a total of 1.5 mol of seats. b. Each tricycle has three tires, so you have 1.5 mol.5 mol of tires. c. Each Mg(OH) has two OH - ions, so there are 1.5 mol.0 mol OH - ions.. a. When conducting this type of eperiment, you are assuming that all of the carbon and hydrogen show up in the CO and H O, respectively. In this eperiment, where all of the carbon and hydrogen do not show up, when you analyze the CO for carbon and H O for hydrogen, you find the weights in the products are less than those in the carbon and hydrogen you started with.

8 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 67 b. Since you collected less carbon and hydrogen than were present in the original sample, the calculated mass percentage will be less than the epected (real) value. For eample, say you have a 10.0 g sample that contains 7.5 g of carbon. You run the eperiment on the 10.0 g sample and collect only 5.0 g of carbon. The calculated percent carbon based on your eperimental results would be 50 percent instead of the correct amount of 75 percent.. a. C H 8 O is not an empirical formula because each of the subscripts can be divided by two to obtain a possible empirical formula of CH O. (The empirical formula is not the smallest integer ratio of subscripts.) b. C 1.5 H is not a correct empirical formula because one of the subscripts is not an integer. Multiply each of the subscripts by two to obtain the possible empirical formula C H 8. (Since the subscript of carbon is the decimal number 1.5, the empirical formula is not the smallest integer ratio of subscripts.) c. Yes, the empirical formula and the molecular formula can be the same, as is the case in this problem where the formula is written with the smallest integer subscripts.. a. Correct. Coefficients in balanced equations can represent amounts in atoms and molecules. b. Incorrect. The coefficients in a balanced chemical equation do not represent amounts in grams. One gram of carbon and one gram of oygen represent different molar amounts. c. Incorrect. The coefficients in a balanced chemical equation do not represent amounts in grams. d. Correct. You might initially think this is an incorrect representation; however, 1g of C, g of O, and g of CO each represent one mole of the substance, so the relationship of the chemical equation is obeyed. e. Correct. The coefficients in balanced equations can represent amounts in moles. f. Incorrect. The amount of O present is not enough to react completely with one mol of carbon. Only one-half of the carbon would react, and one-half mol of CO would form. g. Incorrect. In this representation, oygen is being shown as individual atoms of O, not as molecules of O, so the drawings are not correctly depicting the chemical reaction. h. Correct. The molecular models correctly depict a balanced chemical reaction since the same number of atoms of each element appears on both sides of the equation..5 a. X (g) + Y(g) XY(g) (continued)

9 68 CHAPTER b. Since the product consists of a combination of X and Y in a 1:1 ratio, it must consist of two atoms hooked together. If you count the total number of X atoms (split apart the X molecules) and Y atoms present prior to the reaction, there are four X atoms and three Y atoms. From these starting quantities, you are limited to three XY molecules and left with an unreacted Y. Option #1 represents this situation and is therefore the correct answer. c. Since X (g) was completely used up during the course of the reaction, it is the limiting reactant. Answers to Review Questions.1 The molecular weight is the sum of the atomic weights of all the atoms in a molecule of the substance whereas the formula weight is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether the compound is molecular or not. A given substance could have both a molecular weight and a formula weight if it eisted as discrete molecules.. To obtain the formula weight of a substance, sum up the atomic weights of all atoms in the formula.. A mole of N contains Avogadro's number ( ) of N molecules and of N atoms. One mole of Fe (SO ) contains three moles of SO - ions; and it contains twelve moles of O atoms.. A sample of the compound of known mass is burned, and CO and H O are obtained as products. Net, you relate the masses of CO and H O to the masses of carbon and hydrogen. Then, you calculate the mass percentages of C and H. You find the mass percentage of O by subtracting the mass percentages of C and H from The empirical formula is obtained from the percentage composition by assuming for the purposes of the calculation a sample of 100 g of the substance. Then, the mass of each element in the sample equals the numerical value of the percentage. Convert the masses of the elements to moles of the elements using the atomic mass of each element. Divide the moles of each by the smallest number to obtain the smallest ratio of each atom. If necessary, find a whole-number factor to multiply these results by to obtain integers for the subscripts in the empirical formula..6 The empirical formula is the formula of a substance written with the smallest integer (whole number) subscripts. Each of the subscripts in the formula C 6 H 1 O can be divided by two, so the empirical formula of the compound is C H 6 O.

10 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 69.7 The number of empirical formula units in a compound, n, equals the molecular weight divided by the empirical formula weight. n.0 amu 17.0 amu.00 The molecular formula of hydrogen peroide is therefore (HO), or H O..8 The coefficients in a chemical equation can be interpreted directly in terms of molecules or moles. For the mass interpretation, you will need the molar masses of CH, O, CO, and H O, which are 16.0,.0,.0, and 18.0 g/mol, respectively. A summary of the three interpretations is given below the balanced equation: CH + O CO + H O 1 molecule + molecules 1 molecule + molecules 1 mole + moles 1 mole + moles 16.0 g +.0 g.0 g g.9 A chemical equation yields the mole ratio of a reactant to a second reactant or product. Once the mass of a reactant is converted to moles, this can be multiplied by the appropriate mole ratio to give the moles of a second reactant or product. Multiplying this number of moles by the appropriate molar mass gives mass. Thus, the masses of two different substances are related by a chemical equation..10 The limiting reactant is the reactant that is entirely consumed when the reaction is complete. Because the reaction stops when the limiting reactant is used up, the moles of product are always determined by the starting number of moles of the limiting reactant..11 Two eamples are given in the book. The first involves making cheese sandwiches. Each sandwich requires two slices of bread and one slice of cheese. The limiting reactant is the cheese because some bread is left unused. The second eample is assembling automobiles. Each auto requires one steering wheel, four tires, and other components. The limiting reactant is the tires, since they will run out first..1 Since the theoretical yield represents the maimum amount of product that can be obtained by a reaction from given amounts of reactants under any conditions, in an actual eperiment you can never obtain more than this amount.

11 70 CHAPTER Answers to Conceptual Problems.1 a. H (g) + N (g) NH (g) b. Since there is no H present in the container, it was entirely consumed during the reaction, which makes it the limiting reactant. c. According to the chemical reaction, three molecules of H are required for every molecule of N. Since there are two moles of unreacted N, you would need si additional moles of H to complete the reaction..1 a. The limiting reactant when cooking with a gas grill would be the propane. This makes sense since propane is the material you must purchase in order to cook your food. b. Since the chemical reaction only requires propane and oygen, if the grill will not light with ample propane present, the limiting reactant must be the oygen. c. Once again, here is a case where you have adequate propane, so you can conclude that a yellow flame indicates that not enough oygen is present to combust all of the propane. If there is not enough O available for complete combustion, a reasonable assumption is that some of the products will have fewer oygen atoms than CO. Therefore, a miture of products would be obtained in this case, including carbon monoide (CO) and soot (carbon particles)..15 a. This answer is unreasonable because g is too small a weight for 0. mole of an element. For eample, 0. mol of hydrogen, the lightest element, would have a mass of 0. g. b. This answer is unreasonable because g is too large for one water molecule. (The mass of one water molecule is g) c. This answer is reasonable because is one-half of Avogadro s number. d. This answer is unreasonable because the units for molar mass should be g/mol, so this quantity is 1000 times too large..16 a. In order to have a complete reaction, a ratio of two moles of hydrogen to every mole of oygen is required. In this case, there is not enough oygen in the air outside of the bubble for the complete reaction of hydrogen. b. In this case, you have a ratio of one mole of H to one mole of O. According to the balanced chemical reaction, every one mole of O can react with two moles of H. In this case, when 0.5 mole of O has reacted, all of the H (one mole) will be consumed, leaving behind 0.5 mole of unreacted O. (continued)

12 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 71 c. In this case, you have a ratio of two moles of H to one mole of O, which is the correct stoichiometric amount, so all of the hydrogen and all of the oygen react completely. d. In order for reaction to occur, both oygen and hydrogen must be present. Oygen does not combust, and there is no hydrogen present to burn, so no reaction occurs..17 a. The limiting reactant would be the charcoal because the air would supply as much oygen as needed. b. The limiting reactant would be the magnesium because the beaker would contain much more water than is needed for the reaction (approimately 18 ml of water is one mole). c. The limiting reactant would be the H because the air could supply as much nitrogen as is needed..18 a. Since the balanced chemical equation for the reaction is H + O H O in order to form the water, you need two molecules of hydrogen for every one molecule of oygen. Given the quantities of reactants present in the container and applying the :1 ratio, you can produce a maimum of twelve molecules of water. b. The drawing of the container after the reaction should contain twelve H O molecules and two O molecules..19 a. The problem is that Avogadro s number was inadvertently used for the molar mass of calcium, which should be 0.08 g/mol. The correct calculation is 7.0 g Ca 1 mol Ca 0.08 g Ca mol Ca b. The problem here is an incorrect mole ratio. There are two mol K + ions per one mol K SO. The correct calculation is.5 mol K SO + mol K ions 1 mol K SO K ions + 1 mol K ions K + ions c. The problem here is an incorrect mole ratio. The result should be 0.50 mol Na mol HO mol Na 0.50 mol H O

13 7 CHAPTER.0 a. The missing concept is the mole ratio. His reasoning would only be correct if the reactants reacted in a one-to-one mole ratio. Here, the ratio is 5 mol O / mol C H. This means that, for every two moles of C H, five moles of O are required. Since there are only four moles here, there is insufficient O, and it is the limiting reactant. b. The missing concept again is the mole ratio. Since in this problem the reactants react in a one-to-one mole ratio, the reactant with the fewest moles is the limiting reactant, and the lucky guess works. Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off..1 a. Formula weight of CH OH AW of C + (AW of H) + AW of O. Using the values of atomic weights in the periodic table (inside front cover) rounded to four significant figures and rounding the answer to three significant figures, we have FW 1.01 amu + ( amu) amu.0.0 amu b. FW of NO AW of N + (AW of O) 1.01 amu + ( amu) amu c. FW of K CO (AW of K) + AW of C + (AW of O) ( 9.10 amu) amu + ( amu) amu d. FW of Ni (PO ) (AW of Ni) + (AW of P) + 8(AW of O) ( amu) + ( 0.97 amu) + ( amu) amu. a. FW of HC H O (AW of C) + (AW of H) + (AW of O) ( 1.01) + ( 1.008) + ( 16.00) amu (continued)

14 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 7 b. FW of PCl 5 AW of P + 5(AW of Cl) 0.97 amu + (5 5.5 amu) amu c. FW of K SO (AW of K) + AW of S + (AW of O) ( 9.10 amu) +.07 amu + ( amu) amu ( s.f.) d. FW of Ca(OH) AW of Ca + (AW of H) + (AW of O) 0.08 amu + ( amu) + ( amu) amu ( s.f.). a SO 1 AW of S.07 amu AW of O amu MW of NO 6.07 amu 6.1 amu ( s.f.) b. PCl 1 AW of P 0.97 amu AW of Cl amu MW of PCl 17. amu 17 amu ( s.f.). a HNO 1 AW of H amu 1 AW of N 1.01 amu AW of O amu MW of HNO amu 7.0 amu ( s.f.) b. CO 1 AW of C 1.01 amu 1 AW of O amu MW of CO 8.01 amu 8.0 amu ( s.f.).5 First, find the formula weight of NH NO by adding the respective atomic weights. Then convert it to the molar mass: FW of NH NO (AW of N) + (AW of H) + (AW of O) ( 1.01 amu) + ( amu) + ( amu) amu The molar mass of NH NO g/mol.

15 7 CHAPTER.6 First, find the formula weight of H PO by adding the respective atomic weights. Then convert it to the molar mass: FW of H PO (AW of H) + AW of P + (AW of O) ( 1.008) ( 16.00) amu The molar mass of H PO g/mol..7 a. The atomic weight of Na equals.99 amu; thus, the molar mass equals.99 g/mol. Because one mol of Na atoms equals Na atoms, we calculate Mass of one Na atom.99 g/mol atom/mol g/atom b. The atomic weight of N equals 1.01 amu; thus, the molar mass equals 1.01 g/mol. Because one mol of N atoms equals N atoms, we calculate Mass of one N atom 1.01 g/mol atom/mol g/atom c. The formula weight of CH Cl [ ( 1.008) + 5.5] 50.8 amu; thus, the molar mass equals 50.8 g/mol. Because one mol of CH Cl molecules equals CH Cl molecules, we calculate Mass of one CH Cl molecule 50.8 g/mol molecules/mol g/molecule d. The formula weight of Hg(NO ) ( 1.01) + ( )].61 amu; thus, the molar mass equals.61 g/mol. Because one formula weight of Hg(NO ) equals Hg(NO ) formula units, we calculate Mass of one Hg(NO ).61g/mol units/mol g/unit.8 As in the previous problem, the atomic weights are used with units of g/mol. In addition, the formula weights have been found by addition of the atomic weights and are epressed in g/mol. a. Mass of one Fe atom g/mol g/atom atom/mol (continued)

16 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 75 b. Mass of one F atom g/mol atom/mol g/atom c. Mass of one N O molecule.0 g/mol molecules/mol g/molecule d. Mass of one Al(OH) unit g/mol unit/mol g/unit.9 First, find the formula weight (in amu) using the periodic table (inside front cover): FW of (CH CH ) O ( 1.01 amu) + ( amu) amu 7.1 amu Mass of one (CH CH ) O molecule 7.1 g/mol molecules/mol g/molecule.0 First, find the formula weight (in amu) using the periodic table (inside front cover): FW of glycerol ( 1.01 amu) + ( amu) + ( amu) 9.09 amu Mass of one glycerol molecule 9.09 g/mol molecules/mol g/molecule.1 From the table of atomic weights, we obtain the following molar masses for parts a through d: Na.99 g/mol; S.07 g/mol; C 1.01 g/mol; H g/mol; Cl 5.5 g/mol; and N 1.01 g/mol. a mol Na.99 g 1mol Na.8. g Na b mol S.07 g S 1mol S g S (continued)

17 76 CHAPTER c. Using molar mass 8.9 g/mol for CH Cl, we obtain.78 mol CH Cl 8.9 g CHCl 1mol CHCl g CH Cl d. Using molar mass 68.1 g/mol for (NH ) S, we obtain 68.1 g (NH 8 mol (NH ) S ) S 1 mol (NH ) S g (NH ) S. From the table of atomic weights, we obtain the following molar masses for parts a through d: C 1.01 g/mol; O g/mol; K 9.10 g/mol; Cr 5.00 g/mol; Fe g/mol; and F g/mol. a mol Fe g Fe 1mol Fe g Fe b mol F g F 1mol F g F c. Using molar mass.01 g/mol for CO, we obtain 5.8 mol CO.01g CO 1mol CO g CO d. Using molar mass 19.0 g/mol for K CrO, we obtain 19.0 g K CrO 8.1mol K CrO g K CrO 1mol K CrO. First, find the molar mass of H BO : ( amu) amu + ( amu) Therefore, the molar mass of H BO 61.8 g/mol. The mass of H BO is calculated as follows: 61.8 g HBO 0.5 mol HBO.57.6 g HBO 1mol HBO. First, find the molar mass of CS : 1.01 amu + (.07 amu) amu. Therefore, the molar mass of CS g/mol. The mass of CS is calculated as follows: g CS mol CS g CS 1mol CS

18 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 77.5 From the table of atomic weights, we obtain the following rounded molar masses for parts a through d: C 1.01 g/mol; Cl 5.5 g/mol; H g/mol; Al 6.98 g/mol; and O g/mol. a..86 g C 1mol C 1.01g C mol C b g Cl 1mol Cl g Cl mol Cl c. The molar mass of C H 10 ( 1.01) + ( ) 58.1 g C H 10 /mol C H 10. The mass of C H 10 is calculated as follows: 1mol CH10 76 g CH mol CH g CH10 d. The molar mass of Al (CO ) ( 6.98) + ( 1.01) + ( g).99 g/mol Al (CO ). The mass of Al (CO ) is calculated as follows: 6. g Al (CO ) 1 mol Al (CO ).99 g Al (CO ) mol Al (CO ).6 From the table of atomic weights, we obtain the following rounded molar masses for parts a through d: As 7.9 g/mol; S.07 g/mol; N 1.01 g/mol; H g/mol; Al 6.98 g/mol; and O g/mol. a..57 g As b. 7.8 g S 8 1mol As 7.9 g As 1mol S g S mol As mol S 8 c. The molar mass of N H ( 1.01) + ( 1.008).05 g N H /mol N H. The mass of N H is calculated as follows: 1 mol N 6.5 g H.05 g N H g d. The molar mass of Al (SO ) ( 6.98) + (.07) + ( ).17 g Al (SO ) /mol Al (SO ). The mass of Al (SO ) is calculated as follows: 7 g Al (SO ) 1 mol Al (SO ).17 g Al (SO ) mol Al (SO )

19 78 CHAPTER.7 Calculate the formula weight of calcium sulfate: 0.08 amu +.07 amu + ( amu) amu. Therefore, the molar mass of CaSO is g/mol. Use this to convert the mass of CaSO to moles: 1mol CaSO g CaSO mol CaSO g CaSO Calculate the molecular weight of water: ( amu) amu 18.0 amu. Therefore, the molar mass of H O equals 18.0 g/mol. Use this to convert the rest of the sample to moles of water: 1mol HO 0.09 g H O mol H O 18.0 g HO Because mol is about twice mol, both numbers of moles are consistent with the formula, CaSO H O..8 Calculate the formula weight of copper(ii) sulfate: 6.55 amu +.07 amu + ( amu) amu. Thus, the molar mass of CuSO is g/mol. From the previous problem, the molar mass of H O is 18.0 g/mol. Use this to convert the rest of the sample to moles of water: 1mol HO gh O mol HO 18.0 g HO Calculate the moles of CuSO to be able to compare relative molar amounts of CuSO and H O. Then, divide the moles of H O by the moles of CuSO : 1mol CuSO g CuSO mol CuSO g CuSO mol H O mol CuSO.99/1, or about 5:1 (consistent with CuSO 5H O).9 The following rounded atomic weights are used: Li 6.9 g/mol; Br g/mol; N 1.01 g/mol; H g/mol; Pb 07. g/mol; Cr 5.00 g/mol; O g/mol; and S.07 g/mol. Also, Avogadro s number is atoms, so a atoms No. Li atoms 8.1g Li atoms 6.91g Li b. No. Br atoms.0 g Br atoms.1 10 atoms ( 79.90) g Br (continued)

20 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 79 c. No. NH molecules 5 g NH molecules 17.0 g NH molecules d. e units No. PbC ro units 01g PbC ro units. g PbC ro ions No. SO ions 1. g Cr (SO ) ions 9.1g Cr (SO ).0 These rounded atomic weights are used: Al 6.98 g/mol; I g/mol; N 1.01 g/mol; O g/mol; Na.99 g/mol; Cl 5.5 g/mol; Ca 0.08 g/mol; and P 0.97 g/mol. Also, Avogadro s number is atoms, so atoms a. No. A l atoms 5.7 g A l atoms 6.98 g A l b. No. I atoms 8.71 g I atoms g I.1 10 atoms c. No. N O 5 molecules 1.9 g N O molecules g NO molecules d. No. NaClO units.1g NaClO units 1. g NaC lo units e. No. Ca + ions.71 g Ca (PO ) ions g Ca (PO ).7 10 ions.1 Calculate the molecular weight of CCl : 1.01 amu + ( 5.5 amu) amu. Use this and Avogadro's number to epress it as g/n A to calculate the number of molecules: 7.58 mg CCl 1 g 1000 mg molecules g CCl molecules

21 80 CHAPTER. Calculate the molecular weight of ClF : 5.5 amu + ( amu) 9.5 amu. Use this and Avogadro's number to epress it as 9.5 g/n A to calculate the number of molecules: 5.88 mg ClF 1 g 1000 mg molecules 9.5 g ClF molecules Mass percentage carbon mass of C in sample mass of sample 100% Percent carbon 1.58 g 1.86 g 100% %. Mass percentage alcohol mass of alcohol in solution mass of sample 100% Percent alcohol.98 g 6.01 g 100% %.5 Mass percentage phosphorus oychloride mass of POCl in sample mass of sample 100% Percent POCl 1.7 mg 100% % 8.5 mg.6 Mass percentage sulfur mass of S in sample mass of sample 100% Percent sulfur 1.6 mg.17 mg 100% %

22 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 81.7 Start with the definition for percentage nitrogen, and rearrange this equation to find the mass of N in the fertilizer. Mass percentage nitrogen mass of N in fertilizer mass of fertilizer 100% Mass N mass % N 100% mass of fertilizer 1.0% 100%.15 kg kg N.8 Start by finding the mass of L of seawater using the density of 1.05 g/cm. Mass seawater 1.00 L 10 cm 1L 1.05 g 1cm g Continue with the definition for percentage of bromine in the seawater, and rearrange this equation to find the mass of Br in the seawater. Mass percentage Br mass of Br in seawater mass of seawater 100% Mass Br mass % Br 100% mass seawater g Br % 105 g %.9 Convert moles to mass using the molar masses from the respective atomic weights. Then, calculate the mass percentages from the respective masses mol Al 6.98 g Al 1mol Al. g Al mol Mg.1g Mg 1mol Mg 0.96 g Mg Percent Al mass of Al mass of alloy. g Al.9 g alloy 100% 7. 7.% Al Percent Mg mass of Mg mass of alloy 0.96 g Mg.9 g alloy 100% % Mg

23 8 CHAPTER.50 Convert moles to mass using the molar masses from the respective atomic weights of 0.18 g/mol for Ne and 8.80 g/mol for Kr. Then, calculate the mass percentages from the respective masses mol Ne 0.05 mol Kr 0.18 g Ne 1mol Ne 8.80 g Kr 1mol Kr 1.77 g Ne.19 g Kr Percent Ne mass of Ne mass of mi 1.77 g Ne.856 g mi 100%.79.8% Ne Percent Kr mass of Kr mass of mi.19 g Kr.856 g mi 100% % Kr.51 In each part, the numerator consists of the mass of the element in one mole of the compound; the denominator is the mass of one mole of the compound. Use the atomic weights of C 1.01 g/mol; O g/mol; Na.99 g/mol; H g/mol; P 0.97 g/mol; Co 58.9 g/mol; and N 1.01 g/mol. a. Percent C mass of C mass of CO 1.01 g C 8.01 g CO 100%.878.9% Percent O % -.878%C % b. Percent C mass of C mass of CO 1.01 g C.01 g CO 100% % Percent O % % C % c. Percent Na Percent H Percent P mass of Na.99 g Na mass of NaHPO g NaHPO % mass of H.016 g H mass of NaHPO g NaHPO mass of P 0.97 g P mass of NaHPO g NaHPO 100% 100% % 100% % Percent O % - ( ) % (continued)

24 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 8 d. Percent Co mass of Co mass of Co(NO ) 58.9 g Co g Co(NO ) 100%.11.% Percent N mass of N mass of Co(NO ) 1.01 g N g Co(NO ) 100% % Percent O % - ( ) %.5 In each part, the numerator consists of the mass of the element in one mole of the compound; the denominator is the mass of one mole of the compound a. Percent N mass of N mass of NO 1.01 g N 0.01 g NO 100% % Percent O % % N % b. Percent H mass of H mass of HO.016 g H.0 g HO 100% % Percent O % % H % c. Percent K Percent Cl mass of K mass of KClO 9.10 g K g KClO mass of Cl mass of KClO 5.5 g Cl g KClO 100% % 100% % Percent O % - ( ) % d. Percent Mn mass of Mn mass of Mn(NO ) 5.9 g Mn g Mn(NO ) % Percent N mass of N 1.01 g N mass of Mn(NO ) g Mn(NO ) % Percent O % - ( ).550.5%

25 8 CHAPTER.5 The molecular model of toluene contains seven carbon atoms and eight hydrogen atoms, so the molecular formula of toluene is C 7 H 8. The molar mass of toluene is 9.1 g/mol. The mass percentages are Percent C mass of C mass of C7H g 9.1 g Percent H 100% % 100% %.5 The molecular model of -propanol contains three carbon atoms, eight hydrogen atoms, and one oygen atom, so the molecular formula of -propanol is C H 8 O. The molar mass of -propanol is g/mol. The mass percentages are Percent C Percent H mass of C mass of CH8O mass of H mass of CH8O 1.01 g g g g 100% % 100% % Percent O 100% - ( ) %.55 Find the moles of C in each amount in one-step operations. Calculate the moles of each compound using the molar mass; then multiply by the number of moles of C per mole of compound: Mol C (glucose) 6.01 g 1mol 180. g 6 mol C 1mol glucose 0.00 mol Mol C (ethanol) 5.85 g 1 mol 6.07 g mol C 1 mol ethanol 0.5 mol (more C).56 Find the moles of S in each amount in one-step operations. Calculate the moles of each compound using the molar mass; then multiply by the number of moles of S per mole of compound. Mol S (CaSO ) 0.8 g 1mo l CaSO g 1mol S 1mol CaSO mol (more S) Mol S (NaSO ) 5. g 1mo l NaSO g 1mol S 1mol NaSO 0.79 mol

26 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS First, calculate the mass of C in the glycol by multiplying the mass of CO by the molar mass of C and the reciprocal of the molar mass of CO. Then, calculate the mass of H in the glycol by multiplying the mass of H O by the molar mass of H and the reciprocal of the molar mass of H O. Then, use the masses to calculate the mass percentages. Calculate O by difference mg CO 1mol CO.01g CO 1.01g C 1mol C. 7 mg C 5.58 mg H O 1mol HO 18.0 g HO H 1HO g H 1mol H 0.6 mg H Mass O 6.8 mg - ( ).8 mg O Percent C (.7 mg C/6.8 mg glycol) 100% % Percent H (0.6 mg H/6.8 mg glycol) 100% % Percent O (.8 mg O/6.8 mg glycol) 100% %.58 First, calculate the mass of C in the phenol by multiplying the mass of CO by the molar mass of C and the reciprocal of the molar mass of CO. Then, calculate the mass of H in the phenol by multiplying the mass of H O by the molar mass of H and the reciprocal of the molar mass of H O. Then, use the masses to calculate the mass percentages. Calculate O by difference mg CO 1mol CO.01g CO 1.01g C 1mol C.00 mg C.01mg H O 1mol HO 18.0 g CO H 1HO g H 1mol H mg H Mass O 5. mg - ( ) mg O Percent C (.00 mg/5. mg) 100% % Percent H (0.68 mg/5. mg) 100% % Percent O ( mg/5. mg) 100% %

27 86 CHAPTER.59 Start by calculating the moles of Os and O; then divide each by the smaller number of moles to obtain integers for the empirical formula. Mol Os.16 g Os 1 mol Os 190. g Os mol (smaller number) Mol O ( ) g O 1 mol O g O mol Integer for Os Integer for O Within eperimental error, the empirical formula is OsO..60 Start by calculating the moles of W and O; then divide each by the smaller number of moles to obtain integers for the empirical formula. Mol W. g W 1 mol W g W mol (smaller number) Mol O (5. -.) g O 1 mol O g O mol Integer for W Integer for O Because within eperimental error, the empirical formula is WO.61 Assume a sample of g of potassium manganate. By multiplying this by the percentage composition, we obtain 9.6 g of K, 7.9 g of Mn, and.5 g of O. Convert each of these masses to moles by dividing by molar mass. Mol K 9.6 g K 1 mol K 9.10 g K 1.01 mol Mol Mn 9.7 g Mn 1 mol Mn 5.9 g Mn mol (smallest number) Mol O.5 g O 1 mol O g O.01 mol (continued)

28 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 87 Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for K , or Integer for Mn , or 1 Integer for O , or The empirical formula is thus K MnO..6 Assume a sample of g of hydroquinone. By multiplying this by the percentage composition, we obtain 65. g of C, 5.5 g of H, and 9.1 g of O. Convert each of these masses to moles by dividing by the molar mass. Mol C 65. g C Mol H 5.5 g H Mol O 9.1 g O 1mol C 1.01g C 1mol H g H 1mol O g O 5.5 mol 5.6 mol mol (smallest number) Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C , or Integer for H , or Integer for O , or 1 The empirical formula is thus C H O..6 Assume a sample of g of acrylic acid. By multiplying this by the percentage composition, we obtain 50.0 g C, 5.6 g H, and. g O. Convert each of these masses to moles by dividing by the molar mass. Mol C 50.0 g C 1mol C 1.01g C.16 mol Mol H 5.6 g H 1mol H g H 5.56 mol (continued)

29 88 CHAPTER Mol O.0 g O 1mol O g O.775 mol (smallest number) Now, divide each number of moles by the smallest number to obtain the smallest number of moles and the tentative integers for the empirical formula. Tentative integer for C , or 1.5 Tentative integer for H , or Tentative integer for O , or 1 Because 1.5 is not a whole number, multiply each tentative integer by two to obtain the final integer for the empirical formula: C: 1.5 H: O: 1 The empirical formula is thus C H O..6 Assume a sample of g of malonic acid. By multiplying this by the percentage composition, we obtain.6 g C,.9 g H, and 61.5 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C.6 g C Mol H.9 g H Mol O 61.5 g O 1mol C 1.01g C 1mol H g H 1mol O g O.881 mol (smallest number).87 mol.8 mol Now, divide each number of moles by the smallest number to obtain the smallest number of moles and the tentative integers for the empirical formula. Tentative integer for C , or 1 Tentative integer for H , or 1-1/ Tentative integer for O , or 1-1/ (continued)

30 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 89 Because 1-1/ is not a whole number, multiply each tentative integer by three to give the final integer for the empirical formula: C: 1 H: 1-1/ O: 1-1/ The empirical formula is thus C H O..65 a. Assume for the calculation that you have g; of this quantity, 9.5 g is C and 7.75 g is H. Now, convert these masses to moles: 9.5 g C 1mol C 1.01g C mol C 7.75 g H 1mol H g H mol H Usually, you divide all the mole numbers by the smaller one, but in this case both are equal, so the ratio of the number of C atoms to the number of H atoms is 1:1. Thus, the empirical formula for both compounds is CH. b. Obtain n, the number of empirical formula units in the molecule, by dividing the molecular weight of 5.0 amu and amu by the empirical formula weight of amu: For 5.0 : n 5.0 amu amu.9968, or For : n amu amu , or 6 The molecular formulas are: for 5.0, (CH) or C H ; and for 78.05, (CH) 6 or C 6 H a. Assume for the calculation that you have g; of this quantity, 85.6 g is C and 1.8 g is H. Now convert these masses to moles: 85.6 g C 1mol C 1.01g C mol C 1.8 g H 1mol H g H 1.6 mol H (continued)

31 90 CHAPTER Divide both mole numbers by the smaller one: For C : 7.19 mol 7.19 mol 1.00 For H : 1.6 mol 7.19 mol.000 The empirical formula is obviously CH. b. Obtain n, the number of empirical formula units in the molecule, by dividing the molecular weights of 8.0 amu and amu by the empirical formula weight of 1.06 amu: For 8.0 : n 8.0 amu 1.06 amu 1.998, or For : n amu 1.06 amu.9968, or The molecular formulas are: for 8.0, (CH ) or C H ; and for 56.06, (CH ) or C H The formula weight corresponding to the empirical formula C H 6 N may be found by adding the respective atomic weights. Formula weight ( 1.01 amu) + ( amu) amu.08 amu Dividing the molecular weight by the formula weight gives the number of times the C H 6 N unit occurs in the molecule. Because the molecular weight is an average of 88.5 ([ ] ), this quotient is 88.5 amu.1 amu.006, or Therefore, the molecular formula is (C H 6 N), or C H 1 N..68 The formula weight corresponding to the empirical formula BH may be found by adding the respective atomic weights. Formula weight amu + ( amu) 1.8 amu Dividing the molecular weight by the formula weight gives the number of times the BH unit occurs in the molecule. Because the molecular weight is 8 amu, this quotient is 8 amu 1.8 amu.0 Therefore, the molecular formula is (BH ), or B H 6.

32 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Assume a sample of g of oalic acid. By multiplying this by the percentage composition, we obtain 6.7 g C,. g H, and 71.1 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C 6.7 g C Mol H. g H Mol O 71.1 g O 1mol C 1.01g C 1mol H g H 1mol O g O. mol.18 mol (smallest number). mol Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Integer for C , or 1 Integer for H , or 1 Integer for O , or The empirical formula is thus CHO. The formula weight corresponding to this formula may be found by adding the respective atomic weights: Formula weight 1.01 amu amu + ( amu) 5.0 amu Dividing the molecular weight by the formula weight gives the number of times the CHO unit occurs in the molecule. Because the molecular weight is 90 amu, this quotient is 90 amu 5.0 amu.00, or The molecular formula is thus (CHO ), or C H O..70 Assume a sample of g of adipic acid. By multiplying this by the percentage composition, we obtain 9. g C, 6.9 g H, and.8 g O. Convert each of these masses to moles by dividing by the molar mass. Mol C 9. g C 1mol C 1.01g C.105 mol Mol H 6.9 g H 1mol H g H 6.85 mol (continued)

33 9 CHAPTER Mol O.8 g O 1mol O g O.78 mol (smallest number) Now, divide each number of moles by the smallest number to obtain the smallest set of integers for the empirical formula. Tentative integer for C , or 1.5 Tentative integer for H , or.5 Tentative integer for O , or 1 Because 1.5 and.5 are not whole numbers, multiply each tentative integer by two to give the final integers for the empirical formula: C: 1.5 H:.5 5 O: 1 The empirical formula is thus C H 5 O. The formula weight corresponding to this formula may be found by adding the respective atomic weights: Formula weight ( 1.01 amu) + ( amu) + ( amu) 7.1 amu Dividing the molecular weight by the formula weight gives the number of times the CHO unit occurs in the molecule. Because the molecular weight is 16 amu, this quotient is 16 amu 7.1 amu.00, or The molecular formula is thus (C H 5 O ), or C 6 H 10 O..71 C H + O CO + H O 1 molecule C H + molecules O molecules CO + molecules H O 1 mole C H + moles O moles CO + moles H O 8.05 g C H +.00 g O.01 g CO g H O

34 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 9.7 H S + O SO + H O molecules H S + molecules O molecules SO + molecules H O moles H S + moles O moles SO + moles H O.06 g H S +.00 g O 6.06 g CO g H O.7 By inspecting the balanced equation, obtain a conversion factor of eight mol CO to two mol C H 10. Multiply the given amount of 0.0 moles of C H 10 by the conversion factor to obtain the moles of H O. 8 mol CO 0.0 mol C H 10 mol C H mol CO.7 By inspecting the balanced equation, obtain a conversion factor of three mol H O to one mol C H 5 OH. Multiply the given amount of 0.69 mol of C H 5 OH by the conversion factor to obtain the moles of H O. mol H 0.69 mol C H 5 OH O 1 mol CHOH.07.1 mol H O 5.75 By inspecting the balanced equation, obtain a conversion factor of three mol O to two mol Fe O. Multiply the given amount of.91 mol Fe O by the conversion factor to obtain moles of O. mol O.91 mol Fe O mol Fe O mol O.76 By inspecting the balanced equation, obtain a conversion factor of three mol NiCl to one mol Ni (PO ). Multiply the given amount of 0.79 mol Ni (PO ) by the conversion factor to obtain moles of NiCl. mol NiCl 0.79 mol Ni (PO ) mol NiCl 1mol Ni (PO )

35 9 CHAPTER.77 NO + H O HNO + NO three mol of NO are equivalent to two mol of HNO (from equation). one mol of NO is equivalent to 6.01 g NO (from molecular weight of NO ). one mol of HNO is equivalent to 6.0 g HNO (from molecular weight of HNO ). 1mol HNO mol NO 6.01g NO 7.50 g HNO g HNO mol HNO 1mol NO 8.1 g NO.78 Ca (PO ) + 6SiO + 10C P + 6CaSiO + 10CO two mol of Ca (PO ) are equivalent to one mol of P (from equation). one mol of P is equivalent to 1.9 g P (from molecular weight of P ). one mol of Ca (PO ) is equivalent to 10. g Ca (PO ) [from molecular weight of Ca (PO ) ]. 1 mol P 15.0 g P 1.88 g P mol Ca (PO ) 1 mol P g Ca (PO ) 1 mol Ca (PO ) g Ca (PO ).79 WO + H W + H O one mol of W is equivalent to three moles of H (from equation). one mol of H is equivalent to.016 g H (from molecular weight of H ). one mol of W is equivalent to 18.8 g W (from atomic weight of W)..81 kg of H is equivalent to g of H g H 1mol H.016 g H 1mol W mol H g W 1mol W g W

36 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS C H 6 + 6NO C H N + 6H O + N four mol of C H 6 are equivalent to four mol of C H N (from equation). one mol of C H 6 is equivalent to.08 g C H 6 (from molecular weight of C H 6 ). one mol of C H N is equivalent to 5.06 g C H N (from molecular weight of C H N). 651 kg of C H 6 are equivalent to g C H g CH 6 1mol CH 6.08 g CH 6 mol CH N mol CH g CH N 1mol CH N g C H N.81 Write the equation, and set up the calculation below the equation (after calculating the two molecular weights): CS + Cl CCl + S Cl 6.7 g Cl 1mol Cl g Cl 1mol CS mol Cl g CS 1mol CS. 8. g CS.8 From the molecular models, the balanced chemical equation is NH + 5O Pt NO + 6HO The molar mass of NH is 17.0 g/mol, and for O, it is.00 g/mol. This gives 6.1 g NH 1 mol NH 17.0 g NH 5 mol O mol NH.00 g O 1 mol O g O.8 Write the equation, and set up the calculation below the equation (after calculating the two molecular weights): N O 5 NO + O 1.15 g O 1mol O.00 g O mol NO 1mol O 6.01g NO 1mol NO g NO

37 96 CHAPTER.8 Write the equation, and set up the calculation below the equation (after calculating the two formula weights): Cu + 8HNO Cu(NO ) + NO + H O 5.9 g Cu(NO ) 1mol Cu(NO g Cu(NO ) ) mol NO mol Cu(NO ) 0.01g NO 1mol NO g NO.85 First determine whether KO or H O is the limiting reactant by calculating the moles of O that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of O formed. mol O 0.15 mol H O 0.5 mol O mol H O mol O 0.5 mol KO mol O (KO is the limiting reactant) mol KO The moles of O produced 0.19 mol..86 First, determine whether NaOH or Cl is the limiting reactant by calculating the moles of NaClO that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of NaClO formed. 1. mol NaOH 1 mol NaClO mol NaOH mol NaClO (smaller number) 1.7 mol Cl 1 mol NaClO 1 mol Cl 1.7 mol NaClO NaOH is the limiting reactant, and moles of NaClO will form..87 First determine whether CO or H is the limiting reactant by calculating the moles of CH OH that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of CH OH formed. Use the molar mass of CH OH to calculate the mass of CH OH formed. Then, calculate the mass of the unconsumed reactant. CO + H CH OH (continued)

38 CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS 97 1 mol H 10. g H.016 g H 1 mol CH OH mol H.59 mol CH OH 5. g CO 1 mol CO 8.01 g CO 1 mol CHOH 1 mol CO 1.6 mol CH OH CO is the limiting reactant. Mass CH OH formed 1.6 mol CH OH.0 g CH OH 1 mol CH OH g CH OH Hydrogen is left unconsumed at the end of the reaction. The mass of H that reacts can be calculated from the moles of product obtained: 1. 6 mol CH mol H.016 g H OH 5.09 g H 1mol CHOH 1mol H The unreacted H 10. g total H g reacted H g H..88 First, determine whether CS or O is the limiting reactant by calculating the moles of SO that each would form if it were the limiting reactant. Identify the limiting reactant by the smaller number of moles of SO formed. Use the molar mass of SO to calculate the mass of SO formed. Then, calculate the mass of the unconsumed reactant. CS + O CO + SO 1 mol O 0.0 g O.00 g O mol SO mol O mol SO 1 mol CS 5.0 g CS g CS O is the limiting reactant. mol SO 1 mol CS mol SO Mass SO formed mol SO 6.07 g SO 1mol SO g (continued)

39 98 CHAPTER CS is left unconsumed at the end of the reaction. The mass of CS that reacts can be calculated from the moles of product obtained: 1 mol CS mol SO mol SO g CS 1 mol CS.79 g CS The unreacted CS 0.0 g total CS -.79 g reacted CS g CS.89 First, determine which of the three reactants is the limiting reactant by calculating the moles of TiCl that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of TiCl formed. Use the molar mass of TiCl to calculate the mass of TiCl formed. TiO + C + 6Cl TiCl + CO + CO 1 mol TiO.15 g TiO g TiO mol TiCl mol TiO mol TiCl 5.67 g C 1 mol C 1.01 g C mol TiCl mol C mol TiCl 1 mol Cl 6.78 g Cl g Cl Cl is the limiting reactant. mol TiCl 6 mol Cl mol TiCl g TiCl Mass TiCl formed mol TiCl 1 mol TiCl g TiCl.90 First, determine which of the three reactants is the limiting reactant by calculating the moles of HCN that each would form if it were the limiting reactant. Identify the limiting reactant by the smallest number of moles of HCN formed. Use the molar mass of HCN to calculate the mass of HCN formed. NH + O + CH HCN + 6H O 11.5 g NH 1 mol NH 17.0 g NH mol HCN mol NH mol HCN 1 mol CH 10.5 g CH 16.0 g CH mol HCN mol CH 0.65 mol HCN (continued)