Click here to download Solutions Manual Chemistry The Central Science 12thEdition Catherine Murphy Stoichiometry: 3 Calculations with Chemical Formulas and Equations Visualizing Concepts 3.1 Reactant A = blue, reactant B = red Overall, 4 blue A2 molecules+ 4 red B atoms=-s 4 A2B molecules Since 4 is a common factor, this equation reduces to equation (a). 3.2 Write the balanced equation for the reaction. 2H2 +CO�Oi30H The combining ratio of H2: CO is 2:1. If we have 8 H2 molecules, 4 CO molecules are required for complete reaction. Alternatively, you could examine the atom ratios in the formula of Oi30H, but the balanced equation is most direct. 3.3 (a) There are twice as many O atoms as N atoms, so the empirical formula of the original compound is N02. (b) No, because we have no way of knowing whether the empirical and molecular formulas are the same. N02 represents the simplest ratio of atoms in a molecule but not the only possible molecular formula. 3.4 The box contains 4 C atoms and 16 H atoms, so the empirical formula of the hydro- carbon is Oi4. 3.5 (a) Analyze. Given the molecular model, write the molecular formula. Plan. Use the colors of the atoms (spheres) in the model to determine the number of atoms of each element. Solve. Observe 2 gray C atoms, 5 white H atoms, 1 blue N atom, 2 red O atoms. C2HsN02 (b) Plan. Follow the method in Sample Exercise 3.9. Calculate formula weight in amu and molar mass in grams. 38 3 Stoichiometry Solutions to Exercises 2 C atoms = 2(12.0 amu) = 24.0 amu 5 H atoms = 5(1.0 amu) = 5.0 amu 1 N atoms = 1(14.0 amu) = 14.0 amu 2 0 atoms = 2(16.0 amu) = 32.0 amu 75.0amu Formula weight= 75.0 amu, molar mass= 75.0 g/mol (c) Plan. The molar mass of a substance provides the factor for converting moles to grams (or grams to moles). . 75.0 g glycine . Solve. 3 mol glycine x 225 g glycine mol (d) Plan. Use the definition of mass% and the results from parts (a) and (b) above to find mass % N in glycine. Solve. mass% N = -----"g_N__ x 100 gC2H5N02 Assume 1 mol C2H5N02• From the molecular formula of glycine [part (a)], there is 1 mol N/mol glycine. _ 1 x (molar mass N) _ 14.0 g _ , mass 01,o N - . x 100 --- x 100 - 18 . 7010 molar mass glycine 75.0 g 3.6 Analyze. Given: 4.0 mol CH4. Find: mol CO and mol H2 Plan. Examine the boxes to determine the CH4:CO mol ratio and CH4:H20 mole ratio. Solve. There are 2 CH4 molecules in the reactant box and 2 CO molecules in the product box. The mole ratio is 2:2 or 1:1. Therefore, 4.0 mol CH4 can produce 4.0 mol CO. There are 2 CH4 molecules in the reactant box and 6 H2 molecules in the product box. The mole ratio is 2:6 or 1:3. So, 4.0 mol CH4 can produce 12:0 mol H2• Check. Use proportions. 2 mol CH4/2 mol CO= 4 mol CH4/4 mol CO; 2 mol CH4/6 mol H2 = 4 mol CH4/12 mol H2• 3.7 Analyze. Given a box diagram and formulas of reactants, draw a box diagram of products. Plan. Write and balance the chemical equation. Determine combining ratios of elements and decide on limiting reactant. Draw a box diagram of products, containing the correct number of product molecules and only excess reactant. N2 = • Solve. N2 + 3H2 __.. 3NH3. .. , NH3 = Each N atom (1/2 of an N2 molecule) reacts with 3 H atoms (1.5 H2 molecules) to form an NH3 molecule. Eight N atoms (4 N2 molecules) require 24 H atoms (12 H2 molecules) for complete reaction. Only 9 H2 molecules are available, so H2 is the limiting reactant. Nine H2 molecules (18 H atoms) determine that 6 NH3 molecules are produced. One N 2 molecule is in excess. 39 3 Stoichiometry Solutions to Exercises � I Check. Verify that mass is conserved in your solution, that the number and kinds of atoms are the same in reactant and product diagrams. In this example, there are 8 N atoms and 18 H atoms in both diagrams, so mass is conserved. Each NO molecule reacts with 1 0 atom (1/2 of an 02 molecule) to produce 1 N02 molecule. Eight NO molecules react with 8 0 atoms (4 02 molecules) to produce 8 N02 molecules. One 02 molecule doesn't react (is in excess). NO is the limiting reactant (b) % ield = actual yield x lOO; ield actua1 yie = %yield · 1 yie x theoretica ield y theoretical yield 100 The theoretical yield from part (a) is 8 N02 molecules. If the percent yield is 75%, then 0.75(8) = 6 N02 would appear in the products box. Balancing Chemical Equations (section 3.1) 3.9 (a) In balancing chemical equations, the law of conservation of mass, that atoms are neither created nor destroyed during the course of a reaction, is observed. This means that the number and kinds of atoms on both sides of the chemical equation must be the same. (b) Subscripts in chemical formulas should not be changed when balancing equations, because changing the subscript changes the identity of the compound (law of constant composition). (c) liquid water = H�(l); water vapor = H20(g); aqueous sodium chloride = NaO(aq); solid sodium chloride= NaO(s) 3.10 (a) In a CO molecule, there is one O atom bound to C. 2CO indicates that there are two CO molecules, each of which contains one C and one O atom. Adding a subscript 2 to CO to form C02 means that there are two O atoms bound to one C 40 3 Stoichiometry Solutions to Exercises in a C02 molecule. The composition of the different molecules, C02 and CO, is different and the physical and chemical properties of the two compounds they constitute are very different. The subscript 2 changes molecular composition and thus properties of the compound. The prefix 2 indicates how many molecules (or moles) of the original compound are under consideration. (b) Yes. There are the same number and kinds of atoms on the reactants side and the products side of the equation. 3.11 (a) 2CO(g) + 02(g)-+ 2C02(g) (b) N20s(g) + H20(l)-+ 2HN03(aq) (c) CH4(g) + 402(g)-+ C04(1) + 4HO(g) (d) Al4C3(s) + 12H20(1)-+ 4Al(0Hh(s) + 3CH4(g) (e) 2C5H1002(1) + 1302(g)-+ 10C02(g) + 10H20(g) (f) 2Fe(OHh(s) + 3H2504(aq)-+ Fe2(S04)3(aq) + 6H20(1) (g) Mg3N2(s) + 4H2S04(aq)-+ 3Mg504(aq) + (NH4hS04(aq) 3.12 (a) 6Li(s) + N 2 (g)-+ 2Li3N(s) (b) TiC14(1) + 2H20(1)-+ Ti02(s) + 4HO(aq) (c) 2NH4N03(s) -+2N2(g) + 02(g) + 4H20(g) (d) Ca3P2(s) + 6H20(l)-+ 3Ca(0Hh(aq) + 2PH3(g) (e) 2Al(OH)3(s) + 3H2504(aq)-+ A12(504)3(aq) + 6H20(1) (f) 2AgN03(aq) + Na2C03(aq)-+ Ag2C03(s) + 2NaN03(aq) (g) 4C2HsNH2(g) + 1502(g)-+ 8C02(g) + 14H20(g) + 2N2(g) .1 (b) 2K003(s)-+ 2KCI(s)+302(g) (c) Zn(s) + H2504(aq)-+ H2(g) + Zn504(aq) (d) P03(1) + 3H20(1)-+ H3P03(aq) + 3HO(aq) (e) 3H2S(g) + 2Fe(OH)3(s)-+ Fe253(s) + 6H20(g) 3.14 (a) S03(g) + H20(1)-+ H2S04(aq) (b) B253(s) + 6H20(l)-+ 2H3B03(aq) + 3H2S(g) (c) 4PH3(g) + 802(g)-+ P 401o(s) + 6H20(g) fl. (d) 2Hg(N03h(s)-+ 2Hg0(s) + 4N02(g) + Oig) (e) Cu(s) + 2H2504(aq)-+ CuS04(aq) + S02(g) + 2H20(l) 41 3 Stoichiometry Solutions to Exercises Patterns of Chemical Reactivity (section 3.2) 3.15 (a) When a metal reacts with a nonmetal, an ionic compound forms. The combining ratio of the atoms is such that the total positive charge on the metal cation(s) is equal to the total negative charge on the nonmetal anion(s). Determine the formula by balancing the positive and negative charges in the ionic product. All ionic compounds are solids. 2 Na(s) + Br2(1) � 2NaBr(s) (b) The second reactant is oxygen gas from the air, 02(g). The products are C02(g) and H20(1). 2C6H6(1) + 1502(g) � 12C02(g) + 6H20(1). 3.16 (a) Neutral Al atom loses 3e- to form AP+. Neutral Br2 molecule gains 2e- to form 2Br-. The formula of the product will be A1Br3, because the cationic and anionic charges balance. 2Al(s) + 3Br2(1) � 2 A1Br3(s) (b) The products are C02(g) and H20(1). C3H60(1) + 402(g) � 3C02(g) + 3H20(1) 3.17 (a) Mg(s) + 02(g) � Mg02(s) A (b) BaC03(s)� BaO(s) +C02(g) (c) C8H8(1) + 1002(g) � 8C02(g) + 4H20(l) (d) CH30CH3 is C2H60. C2H60(g) + 302(g) � 2C02(g) + 3H20(l) 3.18 (a) 2Ca(s) + 02(g) � 2Ca0 A (b) Cu(OH) 2(s) � CuO(s) +H20(g) (c) C7H16(1) + 1102(g) � 7C02(g) + 8H20(l) (d) 2C5H120(1) + 1502(g) � lOC02(g) + 12H20(l) 3.19 (a) 2C3H6(g) + 902(g) � 6C02(g) + 6H20(g) combustion (b) NH4N03(s) � N20(g) + 2H20(g) decomposition (c) C5H60(1) + 602(g) � 5C02(g) + 3H20(g) combustion (d) N2(g) + 3H2(g) � 2NH3(g) combination (e) K20(s) + H20(1) � 2KOH(aq) combination 3.20 (a) PbCD3(s) � PbO(s) + COl(g) decomposition (b) C2H4(g) + 302(g) � 2C02(g) + 2H20(g) combustion (c) 3Mg(s) + N 2 (g) � Mg3N2 (s) combination (d) C7Hs02(1) + 802(g) � 7C02(g) + 4H20(g) combustion (e) 2Al(s) + 302(g) � 2A103(s) combination Formula Weights (section 3.3) 3.21 Analyze. Given molecular formula or name, calculate formula weight. Plan. If a name is given, write the correct molecular formula. Then, follow the method in Sample Exercise 3.5. Solve. 42 3 Stoichiometry Solutions to Exercises (a) HN0:3: 1(1.0) + 1(14.0) + 3(16.0) = 63.0 amu (b) KMn04: 1(39.1) + 1(54.9) + 4(16.0) = 158.0 arnu (c) CciJ(P04)2: 3(40.1) + 2(31.0) + 8(16.0) = 310.3 amu (d) Si02: 1(28.1) + 2(16.0) = 60.1 amu (e) Ga2�: 2(69.7) + 3(32.1) = 235.7 amu (f) Cr2(S04)3: 2(52.0) + 3(32.1) + 12(16.0) = 392.3 amu (g) PCb: 1(31.0) + 3(35.5) = 137.5 amu 3.22 Formula weight in amu to 1 decimal place. (a) N20: FW = 2(14.0) + 1(16.0) = 44.0 amu (b) HC7H502: 7(12.0) + 6(1.0) + 2(16.0) = 122.0 amu (c) Mg(OH)2: 1(24.3) + 2(16.0) + 2(1.0) = 58.3 amu (d) (NH2)2CO: 2(14.0) + 4(1.0) + 1(12.0) + 1(16.0) = 60.0 amu (e) CH3C02C5H11: 7(12.0) + 14(1.0) + 2(16;0) = 130.0 amu 3.23 Plan. Calculate the formula weight (FW), then the mass % oxygen in the compound. Solve. (a) C17H19N03:. FW = 17(12.0) + 19(1.0) + 1(14.0) + 3(16.0) = 285.0 amu % 0 � 3(l6.0) arnu x 100 = 16.842 = 16.8% 285.0amu · (b) C1sfhtN03: FW = 18(12.0) + 21(1.0) + 1(14.0) + 3(16.0) = 299.0 amu % 0 = 3(l6.0) amu x 1()() = 16.054 = 16.1 % 299.0amu (c) C17futN04: FW = 17(12.0) + 21(1.0) + 1(14.0) + 4(16.0) = 303.0 amu %0 = 4(l6.0)amu xlOO = 21.122 = 21.1% 303.0amu (d) C22H2��s: FW = 22(12.0) + 24(1.0) + 2(14.0) + 8(16.0) = 444.0 amu 8 16 0 % 0 = ( · ) arnu x 100 = 28.829 = 28.8% 444.0amu (e) �11-44013: FW = 4(12.0) + 64(1.0) + 13(16.0) = 764.0 amu % 0 = l3(l6.0) arnu x 100 = 27.225 = 27 2% 764arnu (f) Cutt1sChN�4: FW = 66(12.0)+75(1.0)+2(35.5)+9(14.0)+24(16.0) = 1448.0 amu 24 16 0 % 0 = ( · ) arnu x 100 = 26.519 = 26.5% 1448.0amu 3.24 (a) C2H2: FW = 2(12.0) + 2(1.0) = 26.0 amu % C = 2(12.0) arnu x 100 = 92.3% 26.0amu (b) HC6H706: FW = 6(12.0) + 8(1.0) + 6(16.0) = 176.0 amu %H= 8(1.0)amux100=4.5% 176.0amu 43 3 Stoichiometry Solutions to Exercises (c) (NH4hS04: FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu % H = 8(1.0) amu xlOO = 6.1 % 132.lamu (d) PtC12(NH3)2: FW = 1(195.1) + 2(35.5) + 2(14.0) + 6(1.0) = 300.1 amu %Pt= 1(195.l)amu xlOO = 65.01% 300.lamu (e) C18H2402: FW = 18(12.0) + 24(1.0) + 2(16.0) = 272.0 amu % 0 = 2(16.0) amu x 100 = ll.8% 272.0amu (f) C18H27N03: FW = 18(12.0) + 27(1.0) + 1(14.0) + 3(16.0) = 305.0 amu % C = 18(12.0) amu x 100 = 70.8% 305.0amu 3.25 Plan. Follow the logic for calculating mass % C given in Sample Exercise 3.6. Solve. (a) C7H60: FW = 7(12.0) + 6(1.0) + 1(16.0) = 106.0 amu % C = 7(12.0) amu x 100 = 79.2% 106.0amu (b) C8H803: FW = 8(12.0) + 8(1.0) + 3(16.0) = 152.0 amu % C = 8(12.0) amu x 100 = 63.2% 152.0amu (c) C7H1402: FW = 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu % C = 7(12.0) amu x 100 = 64.6% 130.0amu 3.26 (a) C02: FW = 1(12.0) + 2(16.0) = 44.0 amu 12 0 % C = · amu xlOO = 27.3% 44.0amu (b) CH30H: FW = 1(12.0) + 4(1.0) + 1(16.0) = 32.0 amu % C = 12.0 amu x 100 = 37.5% 32.0amu (c) C2H6: FW = 2(12.0) + 6(1.0) = 30.0 amu %C = 2(12.0) amu x 100 = 80.0% 30.0amu (d) CS(NH2 h: FW = 1(12.0) + 1(32.1) + 2(14.0) + 4(1.0) = 76.1 amu 12 0 %C= · amu x 100 = 15.8% 76.lamu Avogadro's Number and the Mole (section 3.4) 3.27 (a) 6.022 x 1023• This is the number of objects in a mole of anything. (b) The formula weight of a substance in amu has the same numerical value as the molar mass expressed in grams. 3.28 (a) exactly 12 g (b) 6.0221421 x 1023, Avogadro's number 44 3 Stoichiometry Solutions to Exercises 3.29 Plan. Since the mole is a counting unit, use it as a basis of comparison; determine the total moles of atoms in each given quantity. S.olve. 23 g Na contains 1 mol of atoms 0.5 mol H20 contains (3 atoms x 0.5 mol) = 1.5 mol atoms 6.0 x 1023 N 2 molecules contains (2 atoms x 1 mol) = 2 mol atoms 3.30 16 g Oa contains (2 atoms x 0.5 mol) = 1 mol atoms 9.0 x 1023 H202 molecules contains (4 atoms x 1.5 mol) = 6 mol atoms 2.0 mol CH4 contains (5 atoms x 2 mol) = 10 mol atoms 3.31 Analyu. Given: 160 lb/person; Avogadro's number of people, 6.022 x 1023 people. Find: mass in kg of Avogadro's number of people; compare with mass of Earth. Plan. people � mass in lb � mass in kg; mass of people / mass of Earth Solve. 6.022 x 1023 people x l60lb x lkg =4.370x1025 =4.37x1025 or4.4x1025 kg person 2.2046 lb 4.370 x 1025 kg of people 7_31 or 7_3 5.98x1024 kg Earth One mole of people weighs 7.31 times as much as Earth. Check. This mass of people is reasonable since Avogadro's number is large. Estimate: 160 lb e 70 kg; 6 x 1023 x 70 = 420 x 1023 = 4.2 x 1025 kg 3.32 300 million= 300 x 106 = 3.00 x 108 or 3 x 108 people (The number 300 million has an ambiguous number of sig figs.) 6.022 x 1023 rt x � = $6.022 x 1021 = 2.007 x 1013 = $2.01 x 1013 /person 3.00 x 108 people 100 rt 3.00 x 108 people 2 007 1013 $14.4 trillion= $1.44 x 1013 $ · x = 1.394 = 1.39 or 1 $1.44x1013 Each person would receive an amount that is 1.39 (or 1) times the dollar amount of the national debt. 3.33 (a) Analyu. Given: 0.105 mol sucrose, C12H22011. Find: massing. Plan. Use molar mass (g/mol) of C12H22011 to find g C12H22011. Solve. molar mass= 12(12.0107) + 22(1.00794) + 11(15.9994) = 342.296 = 342.30 342 30 0.105 mol sucrose x · g = 35.942 = 35.9 g C12H22011 lmol Check. 0.1(342) = 34.2 g. The calculated result is reasonable. (b) Analyu. Given: mass. Find: moles. Plan. Use molar mass of Zn(N03)2. Solve. molar mass= 1(65.39) + 2(14.0067) + 6(15.9994) = 189.3998 = 189.40 lmol 143.50 g Zn(N03h x = 0.75766 mol Zn(N03h 189.40 gZn(N03h Check. 140/180 � 7/9 = 0.78 mol 45 3 Stoichiometry Solutions to Exercises (c) Analyze. Given: moles. Find: molecules. Plan. Use Avogadro's number. 1023 Solve. 1.0xlO-{;molCH3CH20H x 6·022x molecules = 6.022xl017 lmol Check. (1.0 x 10-6)(6 x 1023) = 6 x 1017 (d) Analyze. Given: mol NH3. Find: N atoms. Plan. mol NHJ --+ mol N atoms --+ N atoms 1 mol N atoms 6.022 x 1023 atoms Sove. l O . 410 mol NH 3x x------ 1 mol NH3 1 mol = 2.47 x 1023 N atoms Check. (0.4)(6 x 1023) = 2.4 x 1023. 3.34 (a) molar mass= 1(112.41) + 1(32.07) = 144.48 g 144·48g 1.50x10-2 molCdS x = 2.17gCdS lmol (b) molar mass= 1(14.01) + 4(1.008) + 1(35.45) = 53.49 g/mol lmol 86.6g NH4Clx -1.6190 = 1.62 mol NH4Cl 53.49g 1023 (c) 8.447 x 10-2 mol C6H6 x 6·02214 x molecules = 5.087 x 1022 C6H6 molecules · 1 mol 23 (d) 6.25 x 10-3 mol Al(NO 3 ) 3 x 9 mol O x 6.022 x 10 0 atoms · 1 mol Al(N03h 1 mol = 3.39 x 1022 0 atoms 3.35 Analyze/Plan. See Solution 3.33 for stepwise problem-solving approach. Solve. (a) (NH4)3P04 molar mass= 3(14.007) + 12(1.008) + 1(30.974) + 4(16.00) = 149.091 = 149.1 g/mol 49 2.50xl0-3 mol(NH4hP04 x l ·lg(NH4hP04 = 0.373g(NH4hP04 lmol (b) AlCh molar mass= 26.982 + 3(35.453) = 133.341 = 133.34 g/mol 0.2550gAl03 x lmol x 3molCl- = 5.737x10-3 mol Cl- 133.34 g Al03 1 mol AlCl3 (c) C8Hu1N402 molar mass= 8(12.01) + 10(1.008) + 4(14.01) + 2(16.00) = 194.20 = 194.2 g/mol lmol 194.2gC8H10N402 . x 1020 mo1ecules x 770 x ----'"--=---::.;;.._--=--� 6.022 x 1023 molecules 1 mol caffeine = 0.248 g CsH10N4Qi (d) _ ___;;,___ 0.406 g cholesterol = 387 cho estero l/mo g I 1 0.00105mol 46 3 Stoichiometry Solutions to Exercises 3.36 (a) Fe2 (S04h molar mass= 2(55.845) + 3(32.07) + 12(16.00) = 399.900 = 399.9 g/mol 399.9gFe2(S04)J 1.223mo1Fe2(S04h x 489.077 = 489.lg Fe2(S04h lmol (b) (NH4)2C03 molar mass= 2(14.007) + 8(1.008) + 12.011 + 3(15.9994) = %.0872 = 96.087 g/ mol 2molNH/ 6.955g(NH4hC03 x lmol x = 0.1448molNH/ · %.087 g (NH4 h C03 1 mol (NH4 h C03 (c) C9H804 molar mass= 9(12.01) + 8(1.008) + 4(16.00) = 180.154 = 180.2 g/mol 1.50 x 1021 molecules x 123mol x 180·2 g C9HsO 4 = 0.449 g C 9 H 8 O 4 6.022 x 10 molecules 1 mol aspirin 15·86 g Valium = 284.7 (d) Valium/mol 0.05570 mol g 3.37 (a) C6H100S2 molar mass= 6(12.01) + 10(1.008) + 1(16.00) + 2(32.07) = 162.28 = 162.3 g/mol (b) Plan. mg � g � mol Solve. . . lxl0-3 g lmol 5.00 mg allicin x x -- = 3.081 x 10-5 = 3.08 x 10-5 mol allicin lmg 162.3g Check. 5.00 mg is a small mass, so the small answer is reasonable. (5 x 10-3)/200 = 2.5 x 10-5 (c) Plan. Use mol from part (b) and Avoga�o's number to calculate molecules. . 6.022 x 1023 molecules Solve. 3.081 x 10-5 mol allicin x 1.855x1019 mol = 1.86 x 1019 allicin molecules Check. (3 x 10-5)(6 x 1023) = 18 x 1018 = 1.8 x 1019 (d) Plan. Use molecules from part (c) and molecular formula to calculate S atoms. 2 Solve. 1.855x1019 allicinmolecules x Satoms =3.71x1019 Satoms 1 allicin molecule Check. Obvious. 3.38 (a) C14H1gN205 molar mass= 14(12.01) + 18(1.008) + 2(14.01) + 5(16.00) = 294.30 g/ mol lxl0-3g 1 mol (b) 1.00mgaspartamex x 3.398x10-6 =3.40x10-6 molaspartame lmg 294.3g (c) 3.398 x 10-6 molaspartarne x 6.022x1023molecules = 2.046x1018 lmol = 2.05 x 1018 aspartame molecules 47 3 Stoichiometry Solutions to Exercises 3.41 Analyze. Given: g C2H3Cl/L. Find: mol/L, molecules/L. Plan. The /L is constant throughout the problem, so we can ignore it. Use molar mass for g � mol, Avogadro's number for mol � molecules. Solve. 2.0xl0-6 g C2H3Clx lmolC2H3Cl =3.20x10-s =3_2x10-s molC H Cl/L 2 3 lL 62.50gC2H3Cl 3.20xl0-8 molC2H3Cl 6.022x1023 molecules -------'-x = 19. x 1016 mo1ecu 1es/L lL lmol Check. (200 x 10-8)/60 = 2.5 x 10-8 mol (2.5 x 10-8) x (6 x 1023) = 15 x 101s = 1.5 x 1016 3.42 Empirical Formulas (section 3.5) 3.43 (a) Analyze. Given: moles. Find: empirical formula. Plan. Find the simplest ratio of moles by dividing by the smallest number of moles present. Solve. 0.0130 mol C / 0.0065 = 2 0.0390 mol H / 0.0065 = 6 0.0065 mol O / 0.0065 = 1 The empirical formula is C2H60. Check. The subscripts are simple integers. (b) Analyze. Given: grams. Find: empirical formula. Plan. Calculate the moles of each element present, then the simplest ratio of moles. lmolFe Solve. 11.66 g Fe x = 0.2088 mol Fe; 0.2088 I 0.2088 = 1 55.85gFe lmolO 5.01 g Ox = 0.3131 mol O; 0.3131 I 0.2088 fl:$ 1.5 16.00gO Multiplying by two, the integer ratio is 2 Fe: 3 0; the empirical formula is Fe203. Check. The subscripts are simple integers. (c) Analyze. Given: mass%. Firid: empirical formulas. Plan. Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles. 49 3 Stoichiometry Solutions to Exercises lmolC Solve. 40.0 g C x = 3.33 mol C; 3.33 I 3.33 = 1 12.0lgC lmolH 6.7 g Hx 6.65 mol H; 6.65 I 3.33 � 2 1.008molH lmolO 53.3 g Ox = 3.33 mol O; 3.33 I 3.33 = 1 16.00molO The empirical formula is CH20. Check. The subscripts are simple integers. 3.44 (a) Calculate the simplest ratio of moles. 0.104 mol K / 0.052 = 2 0.052 mol C / 0.052 = 1 0.156 mol O / 0.052 = 3 The empirical formula is K2C03• (b) Calculate moles of each element present, then the simplest ratio of moles. 1 5.28 Sn x mol Sn 0.04448 mol Sn; 0.04448 I 0.04448 = 1 g 118.7gSn 1 3.37 g F x mol F 0.1774 mol F; 0.1774 I 0.04448 � 4 19.00gFSn The integer ratio is 1 Sn: 4 F; the empirical formula is SnF4• (c) Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles. lmolN 87.5% N = 87.5 g N x = 6.25 mol N; 6.25 I 6.25 = 1 14.0lg lmol 12.5% H = 12.5 g H x -- = 12.4 mol H; 12.4 I 6.25 � 2 1.008g The empirical formula is NH2• 3.45 Analyze/Plan. The procedure in all these cases is to assume 100 g of sample, calculate the number of moles of each element present in that 100 g, then obtain the ratio of moles as smallest whole numbers. Solve. lmolC (a) 10.4 g C x = 0.866 mol C; 0.866 I 0.866 = 1 12.0lgC lmolS 27.8 g S x = 0.867 mol S; 0.867 I 0.866 � 1 32.07gS lmolCl 61.7gClx =1.74mol0; 1.74/0.866�2 35.45gCl The empirical formula is CS02• lmolC (b) 21.7 g Cx = 1.81 mol C; 1.81 I 0.600 � 3 12.0lgC lmolO 9.6 g Ox = 0.600 mol O; 0.600 I 0.600 = 1 16.00gO 50
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