Solutions to Supplementary Check for Understanding Problems

Chapter 5 s to Supplementary Check for Understanding Problems Moles and Molar Mass 1. Indicate the appropriate quantity for each of the following. a) A mole of N atoms contains atoms. b) A mole of N molecules contains molecules. c) A mole of N molecules contains atoms. d) A mole of N atoms has a mass of grams. e) A mole of N molecules has a mass of grams. Answers: a) 6.0 10 atoms b) 6.0 10 molecules c) 1.044 10 4 atoms d) 14.01 g e) 8.0 g s a) A mole of anything always contains 6.0 10 items of that material. b) A mole of anything always contains 6.0 10 items of that material. c) Since a mole of N molecules contains 6.0 10 molecules of N and there are atoms of N per molecule of N, the total number of N atoms is given by: 6.0 10 N molecules mol N atomsn 1 molecule N = 4 1.044 10 atoms N mol N d) The mass in grams of a mole of atoms of any element (its molar mass) is numerically equally to the weighted average atomic mass in atomic mass units. Since the atomic weight of nitrogen is 14.01, its weighted atomic mass is 14.01 u and a mole of nitrogen atoms weighs 14.01 g. S.5.5

S.5.6 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS e) The mass in grams of a mole of N molecules (its molar mass) is obtained by summing the molar masses of the atoms in the chemical formula. Since the molar mass of N is 14.01 g/mol (see part d), the molar mass of N is: 14.01g mol N moln 1molN 8.01g = moln. a) What is the mass of 75 sodium atoms in atomic mass units? 1.67 10 u The numerical value of the atomic weight of sodium (.99) refers to the mass of a single sodium atom in atomic mass units. Applying this yields: 75 Na atoms.99 u Na atom =1.6710 u b) What is the mass of 75 sodium atoms in grams?.77 10-0 g What we know: number of Na atoms g Na The solution map for this calculation is: atom Na ö mol Na ö g Na The conversion factor needed in the first step is the Avogadro constant epressed in the form 1molNa. 6.0 10 Na atoms The conversion factor needed in the second step is the molar mass of sodium. The numerical value for the molar mass is obtained from the atomic weight of sodium (.99) and is epressed.99g Na in the form. 1molNa

S.5.7 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 75 atoms Na 1molNa 6.0 10 atoms Na.99g Na -0 =.7710 gna 1molNa. How many atoms of an element are present in a sample of that element if the sample has a mass in grams numerically equal to the atomic weight of the element? 6.0 10 atoms A mass in grams numerically equal to the atomic weight of an element is its molar mass and thus contains one mole of atoms of that element. Therefore, this sample will contain 6.0 10 atoms. 4. Blackboard chalk is mostly calcium sulfate. How would you determine how many moles of calcium sulfate it takes to write your name in chalk on a blackboard? The moles of a pure substance can be determined from the mass of the substance and its molar mass. The mass of the chalk used can be obtained by weighing the chalk before and after writing your name and calculating the difference between the masses. The molar mass of calcium sulfate can be obtained from its chemical formula (CaSO 4 ). The calculations will be: g CaSO 4 used = initial mass of chalk - final mass of chalk gcaso 4 1molCaSO 4 used 4 16.15 g CaSO 4 = mol CaSO used 5. What mass of zinc metal contains the same number of atoms as 16.1 grams of silver? 9.76 g What we know: g Ag; number of Ag atoms equals number of Zn atoms g Zn

S.5.8 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The solution map for this calculation is: g Ag ö mol Ag ö mol Zn ö g Zn 1molAg The conversion factor needed in the first step is the molar mass of Ag in the form. 107.9g Ag The problem indicates that the number of atoms of Zn is the same as the number of atoms of Ag. Therefore, mol Zn = mol Ag. This relationship can be applied as the conversion factor in the 1molZn second step in the form. 1molAg The conversion factor needed in the last step is the molar mass of Zn epressed in the form 65.9g Zn. mol Zn 16.1 g Ag 1molAg 107.9 g Ag 1molZn 1molAg 65.9g Zn 1molZn =9.76gZn 6. One atom of an element is found to weigh.107 10 - g. What is the atomic weight of this element? 16.9 What we know: g/atom atomic weight of element We know that the atomic weight of this element is numerically equal to the molar mass of this element. The solution map for calculating the molar mass is: g atom ö g mol The conversion factor needed is the Avogadro constant epressed in the form 6.0 10 atoms. mol

S.5.9 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: -.107 10 g atom 6.0 10 atoms mol = 16.9g mol Since the molar mass of this element is 16.9 g/mol, then the atomic weight of the element is 16.9. 7. Which has the larger mass, 1.0 mmol of calcium or 1.5 mmol of sulfur? Justify your choice. 1.5 mmol sulfur The most direct way to make this determination is to calculate the mass of each and compare. The general solution map for this calculation is: mmol element ö mol element ö g element The conversion factor needed in the first step is that between mmol and mol in the form - 10 mol. 1mmol The conversion factor needed in the second step is the molar mass of the element epressed in g the form. mol Applying these yields: 1.0 mmolca - 10 mol Ca 1mmolCa 40.08g Ca molca =0.040gCa 1.5 mmols - 10 mols 1 mmols.07 gs mols =0.048gS ³ larger mass 8. Which has the larger number of atoms, 0.045 μg of nickel or 0.0 μg of potassium? Justify your choice. 0.0 μg potassium

S.5.10 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The most direct way to make this determination is to calculate the moles of each and compare. The general solution map for this calculation is: μg element ö g element ö mol element -6 10 g The conversion factor needed in the first step is that between μg and g in the form. 1μg The conversion factor needed in the second step is the molar mass of the element epressed in mol the form. g Applying these yields: 0.045 μgni -6 10 g Ni 1 μgni 1molNi -10 =7.710 molni 58.69g Ni 0.0 μgk -6 10 g K 1 μgk 1molK -10 =8.10 molk 9.10g K ³ larger mol 9. Calculate the molar mass for each of the following compounds. a) potassium hydrogen phosphate b) Pb(C H O ) Answers: a) 174.18 g/mol b) 5.9 g/mol s a) What we know: potassium hydrogen phosphate; atomic weights of elements g potassium hydrogen phosphate/mol

S.5.11 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The chemical formula for potassium hydrogen phosphate is K HPO 4. The formula indicates that in 1 mole of potassium hydrogen phosphate there are moles of potassium, 1 mole of hydrogen, 1 mole of phosphorus and 4 moles of oygen. From the periodic table we see that 1 mole of potassium atoms weighs 9.10 g, 1 mole of hydrogen atoms weighs 1.008 g, 1 mole of phosphorus atoms weighs 0.97 g and 1 mole of oygen atoms weighs 16.00 g. Thus, one mole of K HPO 4 will weigh: K molk 9.10g mol K = 78.0g 1 H 1molH 1.008g mol H = 1.008g 1 P 1molP 0.97 g mol P = 0.97 g 4 O 4molO 16.00g 1molO = 64.00g 174.18 g The molar mass of potassium hydrogen phosphate is 174.18 g/mol. b) What we know: Pb(C H O ) ; atomic weights of elements g Pb(C H O ) /mol The chemical formula indicates that in 1 mole of lead(ii) acetate there are 1 mole of lead, 4 moles of carbon, 6 moles of hydrogen and 4 moles of oygen. From the periodic table we see that 1 mole of lead atoms weighs 07. g, 1 mole of carbon atoms weighs 1.01 g, 1 mole of hydrogen atoms weighs 1.008 g and 1 mole of oygen atoms weighs 16.00 g. Thus, one mole of K HPO 4 will weigh:

S.5.1 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 1 Pb 1molPb 07.g 1molPb = 07.g 4 C 4molC 1.01 1molC g = 48.04g 6 H 6molH 1.008g 1molH = 6.048g 4 O 4molO 16.00g 1molO = 64.00g 5.9 g The molar mass of lead(ii) acetate is 5.9 g/mol. 10. Calculate the number of moles of compound in each of the following samples. a).9 g C H 5 OH b) 6.1 ng sulfur trioide c) 1.48 10 kg potassium permanganate Answers: a) 0.04860 mol b) 7.88 10-10 mol c) 9.6 10 mol s a) What we know: g C H 5 OH mol C H 5 OH The solution map for this calculation is g C H 5 OH ö mol C H 5 OH

S.5.1 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed is the molar mass of C H 5 OH. From the periodic table we can get the molar masses of carbon, hydrogen and oygen and add them as follows. (1.01 g) C + 6(1.008 g) H + 16.00 g O = 46.07 g. Thus the molar mass of C H 5 OH is 46.07 g/mol. This is 1molCH5OH used in the form to convert units properly. Applying this yields: 46.07 g C H OH.9 g CH5OH 5 1molC H OH 5 5 46.07 g C H 5 OH b) What we know: ng sulfur trioide = 0.04860 molc H OH mol sulfur trioide The solution map for this calculation is ng sulfur trioide ö g sulfur trioide ö mol sulfur trioide The formula for sulfur trioide is SO. The conversion factor needed in the first step is that -9 10 g between ng and g in the form. The conversion factor needed in the second step is the 1ng molar mass of SO. From the periodic table we can get the molar masses of sulfur and oygen and add them as follows..07 g S + (16.00 g) O = 80.07 g. Thus the molar mass of SO is 1molSO 80.07 g/mol. This is used in the form to convert units properly. 80.07gSO 6.1 ngso -9 10 gso 1ngSO 1molSO -10 80.07 gso =7.8810 molso c) What we know: kg potassium permanganate mol potassium permanganate The solution map for this calculation is kg potassium permanganate ö g potassium permanganate ö mol potassium permanganate

S.5.14 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for potassium permanganate is KMnO 4. The conversion factor needed in the first 10 g step is that between kg and g in the form. The conversion factor needed in the second step 1kg is the molar mass of KMnO 4. From the periodic table we can get the molar masses of potassium, manganese and oygen and add them as follows. 9.10 g K + 54.94 g Mn + 4(16.00 g) O = 158.04 g. Thus the molar mass of KMnO 4 is 158.04 g/mol. This is used in the 1mol KMnO4 form to convert units properly. 158.04 g KMnO 4 1.48 10 kg KMnO 4 10 g KMnO 4 1kgKMnO 4 1molKMnO 4 158.04 g KMnO 4 =9.610 molkmno 4 11. How many CO molecules are present in 18.4 metric tons of carbon monoide? One metric tons equals 1000 kg..96 10 9 molecules What we know: metric tons CO number of molecules CO The solution map for this calculation is metric tons CO ö kg CO ö g CO ö mol CO ö molecules CO The conversion factor needed in the first step is that between metric tons and kg in the form 1000 kg. The conversion factor needed in the second step is that between kg and g in the 1metricton 10 g form. The conversion factor needed in the third step is the molar mass of CO. From the 1kg periodic table we can get the molar masses of carbon and oygen and add them as follows. 1.01 g C + 16.00 g O = 8.01 g. Thus the molar mass of CO is 8.01 g/mol. This is used in the 1molCO form to convert units properly. 8.01g CO

S.5.15 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the Avogadro constant in the form 6.0 10 molecules CO. molco 18.4 metric tons CO 1000 kg CO 10 g CO 1metrictonCO 1kgCO 1molCO 6.0 10 molecules CO 8.01 g CO 1molCO 9 =.96 10 molecules CO 1. Calculate the mass in grams of each of the following samples. a) 9.44 mol copper(ii) sulfate b) 7.11 mmol Li CO Answers: a) 1.51 10 g b) 0.47 g s a) What we know: mol copper(ii) sulfate g copper(ii) sulfate The solution map for this calculation is mol copper(ii) sulfate ö g copper(ii) sulfate The formula for copper(ii) sulfate is CuSO 4. The conversion factor needed is the molar mass of CuSO 4. From the periodic table we can get the molar masses of copper, sulfur and oygen and add them as follows. 6.55 g Cu +.07 g S + 4(16.00 g) O = 159.6 g. Thus the molar mass 159.6g CuSO4 of CuSO 4 is 159.6 g/mol. This is used in the form to convert units properly. 1molCuSO Applying this yields: 4 9.44 molcuso 4 159.6g CuSO 4 1molCuSO 4 =1.5110 gcuso 4

S.5.16 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS b) What we know: mmol Li CO g Li CO The solution map for this calculation is mmol Li CO ö mol Li CO ö g Li CO The conversion factor needed in the first step is that between mmol and mol in the form - 10 mol. The conversion factor needed in the second step is the molar mass of Li CO. From 1mmol the periodic table we can get the molar masses of lithium, carbon and oygen and add them as follows. (6.941) g Li + 1.01 g C + (16.00 g) O = 60.01 g. Thus the molar mass of Li CO is 60.01g LiCO 60.01 g/mol. This is used in the form to convert units properly. 1molLi CO 7.11 mmol LiCO - 10 mol LiCO 1molLi CO 60.01g Li CO = 0.47g LiCO 1molLi CO 1. Calculate the moles of sulfur atoms in each of the following samples. a) 4.6 g sodium thiosulfate b) 5.81 μg Na S Answers: a) 0.0586 mol b) 7.44 10-8 mol s a) What we know: g sodium thiosulfate mol sulfur atoms The solution map for this calculation is: g sodium thiosulfate ö mol sodium thiosulfate ö mol S

S.5.17 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for sodium thiosulfate is Na S O. The conversion factor needed in the first step is the molar mass of Na S O. From the periodic table we can get the molar masses of sodium, sulfur and oygen and add them as follows. (.99 g) Na + (.07 g) S + (16.00 g) O = 158.1 g. Thus the molar mass of Na S O is 158.1 g/mol. This is needed in the form 1molNaSO. 158.1g Na S O The formula of Na S O indicates that in 1 mole of Na S O there are moles of sulfur atoms. This relationship can be applied as the conversion factor in the second step in the form mols. 1molNa S O 4.6 g Na SO 1molNaSO mols 158.1 g Na SO 1molNa SO = 0.0586molS b) What we know: μg Na S mol S atoms The solution map for this calculation is: μg Na S ö g Na S ö mol Na S ö mol S The conversion factor needed in the first step is that between μg and g in the form -6 10 g 1μg. The conversion factor needed in the second step is the molar mass of Na S. From the periodic table we can get the molar masses of sodium and sulfur and add them as follows. (.99 g) Na +.07 g S = 78.05 g. Thus the molar mass of Na S is 78.05 g/mol. This is needed in the form 1molNaS. 78.05g Na S The formula of Na S indicates that in 1 mole of Na S there is 1 mole of sulfur atoms. This relationship can be applied as the conversion factor in the last step in the form 1molS. 1molNa S 5.81 μgnas -6 10 g Na S 1 μgna S 1molNaS 1molS -8 =7.4410 mols 78.05 g Na S 1molNa S

S.5.18 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 14. Calculate the number of carbon atoms in a.9-g sample of C 6 H 4 Cl. 9.64 10 atoms What we know: g C 6 H 4 Cl number of C atoms The solution map for this calculation is: g C 6 H 4 Cl ö mol C 6 H 4 Cl ö mol C ö atoms C The conversion factor needed in the first step is the molar mass of C 6 H 4 Cl. From the periodic table we can get the molar masses of carbon, hydrogen and chlorine and add them as follows. 6(1.01 g) C + 4(1.008 g) H + (5.45 g) Cl = 146.99 g. Thus the molar mass of C 6 H 4 Cl is 1molC6H4Cl 146.99 g/mol. This is needed in the form. 146.99g C H Cl 6 4 The formula of C 6 H 4 Cl indicates that in 1 mole of C 6 H 4 Cl there are 6 moles of carbon atoms. This relationship can be applied as the conversion factor in the second step in the form 6molC. 1molC H Cl 6 4 The conversion factor needed in the last step is the Avogadro constant in the form 6.0 10 atoms C. 1molC.9 g C6H4Cl 1mol C6H4Cl 146.99 g C H Cl 6 4 6molC 1molC6H4Cl 6.0 10 atoms C 1molC =9.6410 atomsc 15. How many moles of oygen atoms are present in 4.40 mmol calcium phosphate? 0.05 mol

S.5.19 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS What we know: mmol calcium phosphate mol O The solution map for this calculation is: mmol calcium phosphate ö mol calcium phosphate ö mol O The formula for calcium phosphate is Ca (PO 4 ). The conversion factor needed in the first step - 10 mol is that between mmol and mol in the form. 1mmol The formula of Ca (PO 4 ) indicates that in 1 mole of Ca (PO 4 ) there are 8 moles of oygen atoms. This relationship can be applied as the conversion factor in the second step in the form 8molO. 1molCa (PO ) 4 4.40 mmolca (PO 4) - 10 molca (PO 4) 8molO 1 mmolca (PO 4) 1molCa (PO 4 ) = 0.05molO Mass Percent 1. Calculate the mass percent of each element in the following compounds. a) barium chloride b) sodium sulfate Answers: a) 65.95 % Ba and 4.05 % Cl b).7 % Na,.58 % S and 45.05 % O s a) What we know: barium chloride; atomic weights of elements mass % of each element present

S.5.0 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The formula for barium chloride is BaCl. Since the mass percent values apply to any sample of BaCl, it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of barium and chlorine and add them as follows. 17. g Ba + (5.45 g) Cl = 08. g. Thus the molar mass of BaCl is 08. g/mol, so 08. g represents the total mass of material. The formula indicates that 1 mole of BaCl contains 1 mole of barium. Therefore the mass of barium (=component mass) present is 17. g. These values are used to obtain the mass percent for barium. 17.g Ba mass % Ba = 100 = 65.95% 08.g BaCl Since chlorine is the only other element present in the compound, the mass % Cl = 100.00 % - 65.95 % = 4.05 %. b) What we know: sodium sulfate; atomic weights of elements mass % of each element present The formula for sodium sulfate is Na SO 4. Since the mass percent values apply to any sample of Na SO 4, it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of sodium, sulfur and oygen and add them as follows. (.99 g) Na +.07 g S + 4(16.00 g) O = 14.05 g. Thus the molar mass of Na SO 4 is 14.05 g/mol, so 14.05 g represents the total mass of material. The formula indicates that 1 mole of Na SO 4 contains moles of sodium. Therefore the mass of sodium (=component mass) present is (.99 g) = 45.98 g. These values are used to obtain the mass percent for sodium. 45.98g Na mass % Na = 100 =.7 % 14.05g Na SO 4 The formula indicates that 1 mole of Na SO 4 contains 1 mole of sulfur. Therefore the mass of sulfur (=component mass) present is.07 g. These values are used to obtain the mass percent for sulfur..07 gs mass%s = 100 =.58% 14.05g Na SO 4 Since oygen is the only other element present in the compound, the mass % O = 100.00 % -.7 % -.58% = 45.05 %.

S.5.1 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS. Which of the following compounds contains the largest mass percent of nitrogen? Justify your choice. a) NH 4 NO b) HNO c) N O 4 d) Al(NO ) NH 4 NO The most direct way to answer this is to determine the mass percent N in each compound. This is calculated from the mass of one mole of a compound and the mass of nitrogen present in one mole of that compound. The required data are shown below. Compound Mass of one mole of compound Mass of N in one mole of compound Mass % N NH 4 NO 80.05 g 8.0 g 5.00 % HNO 6.0 g 14.01 g. % N O 4 9.0 8.0 g 0.45 % Al(NO ) 1.01 g 4.0 g 19.7 % Thus, NH 4 NO has the largest mass % N.. In a particular molecular compound the mass percent sulfur is 50.0% and the mass percent oygen is 50.0%. What is the ratio of oygen atoms to sulfur atoms in a molecule of this compound? O atoms/s atom What we know: mass % of each element present; atomic weights of elements number of O atoms/s atom in compound The ratio of O atoms to S atoms is the same as the ratio of mol O to mol S. In order to obtain the moles of each element, choose an arbitrary mass of the compound (100 g simplifies the math) and use the following solution map: g compound ö g element ö mol element

S.5. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor for the first step is the mass percent of the element used in the form gelement. The conversion factor for the second step is the molar mass of the element in 100 g compound mol element the form. gelement Applying these to each element yields: 100 g compound 50.0 g O 100 g compound 1molO 16.00 g O =.1molO 100 g compound 50.0 gs 100 g compound 1molS.07 gs =1.56molS The mole ratio of O to S is.1 mol O/1.56 mol S = which is the same as the atom ratio. 4. If a type of stainless steel contains 18% chromium by mass, how many moles of chromium are present in a bar of this material weighing 1.5 kg? 5. mol What we know: mass of sample; mass % Cr mol Cr The solution map for this calculation is: kg sample ö g sample ö g Cr ö mol Cr 10 g The conversion factor needed in the first step is that between kg and g in the form. 1kg The conversion factor for the second step is the mass percent of Cr used in the form 18g Cr. 100g sample 1molCr The conversion factor for the last step is the molar mass of Cr in the form. 5.00g Cr

S.5. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 1.5 kg sample 10 gsample 1kgsample 18 g Cr 100 g sample 1molCr 5.00 g Cr =5.molCr Stoichiometric Calculations (mole-to-mole) 1. Balance the following equation and state the meaning of the equation in terms of individual units of reactants and products and in terms of moles of reactants and products. Al(s) + MnO (s) Mn(s) + Al O (s) Answers: 4Al(s) + MnO (s) Mn(s) + Al O (s) 4 atoms formula units atoms formula units 4 mol mol mol mol To balance this equation, place a coefficient of in front of MnO and a coefficient of in front of Al O to balance oygen atoms. This requires a coefficient of in front of Mn to balance manganese atoms. Finally, a coefficient of 4 in front of Al will balance aluminum atoms. The balanced equation is: 4Al(s) + MnO (s) Mn(s) + Al O (s) At the most fundamental level this equation suggests that 4 atoms of aluminum react with formula units of MnO to produce atoms of Mn and formula units of Al O. Multiplying each reactant amount and each product amount by the Avogadro constant yields 4 mol Al reacting with mol MnO to form mol Mn and mol Al O.. How many moles of CO are needed to react completely with 0.675 mol LiOH? LiOH(aq) + CO (g) Li CO (aq) + H O(l) (unbalanced) 0.8 mol

S.5.4 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS What we know: mol LiOH; equation relating CO and LiOH mol CO The solution map for this calculation is: mol LiOH ö mol CO The conversion factor needed is the mole ratio for these two substances from the balanced equation. To balance this equation place a coefficient of in front of LiOH. The balanced equation is: Applying the mole ratio yields: LiOH(aq) + CO (g) Li CO (aq) + H O(l) 0.675 mol LiOH 1molCO mollioh =0.8molCO. Given the reaction 4FeS(s) + 7O (g) Fe O (s) + 4SO (g) how many moles of O are needed to: a) produce 0.69 mol Fe O? b) react completely with 9.14 mol FeS? c) form 1.51 mol SO? Answers: a).4 mol b) 16.0 mol c).64 mol s a) What we know: mol Fe O ; balanced equation relating O and Fe O mol O

S.5.5 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The solution map for this problem is: mol Fe O ö mol O The conversion factor needed is the mole ratio for these two substances from the balanced 7molO equation in the form. molfe O Applying this yields: 0.69 mol FeO 7molO molfeo =.4molO b) What we know: mol FeS; balanced equation relating O and FeS mol O The solution map for this problem is: mol FeS ö mol O The conversion factor needed is the mole ratio for these two substances from the balanced 7molO equation in the form. 4molFeS Applying this yields: 9.14 mol FeS 7molO = 16.0molO 4molFeS c) What we know: mol SO ; balanced equation relating O and SO mol O The solution map for this problem is: mol SO ö mol O The conversion factor needed is the mole ratio for these two substances from the balanced 7molO equation in the form. 4molSO

S.5.6 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: 1.51 molso 7molO 4molSO =.64molO Stoichiometric Calculations (mole-to-mass & mass-to-mole) 1. How many moles of each product can be formed from the decomposition of 1.00 g of the rocket fuel hydrazine (N H 4 )? N H 4 (l) 4NH (g) + N (g) Answers: 0.0416 mol NH and 0.0104 mol N What we know: g N H 4 ; balanced equation relating NH, N and N H 4 mol NH and mol N The solution maps for these calculations are: g N H 4 ö mol N H 4 ö mol NH g N H 4 ö mol N H 4 ö mol N In each case, the conversion factor needed in the first step is the molar mass of N H 4. From the periodic table we can get the molar masses of nitrogen and hydrogen and add them as follows. (14.01 g) N + 4(1.008 g) H =.05 g. Thus, the molar mass of N H 4 is.05 g/mol. It is used 1molNH4 in the form in order to cancel units properly..05g N H 4 The conversion factor needed in the second step is the mole ratio for the particular product and 4molNH 1molN N H 4 : and moln H moln H 4 4

S.5.7 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 1.00 g NH4 1.00 g NH4 1molNH4.05 g N H 4 1molNH4.05 g N H 4 4molNH moln H 1molN 4 =0.0416molNH moln H 4 = 0.0104mol N Note that the number of moles of NH is four(4) times that of N as suggested by the balanced equation.. How many moles of oygen gas are needed for the complete combustion of 19.6 g acetylene, C H (g)? 1.88 mol What we know: g C H mol O First, a balanced equation is needed for the reaction between C H and O. Since this is a combustion reaction, the products are CO (g) and H O(l). C H (g) + O (g) CO (g) + H O(l) Notice that the hydrogen atoms are balanced so first, balance the carbon atoms. Since there are C atoms on the reactant side and only 1 C atom on the product side, a coefficient of in front of CO will balance carbon atoms. C H (g) + O (g) CO (g) + H O(l) (H and C atoms balanced) Finally, balance the oygen atoms. Since there are O atoms on the reactant side and 5 O atoms on the product side, multiplying O by ½, or 5/, will balance oygen atoms. C H (g) + 5/O (g) CO (g) + H O(l) (all atoms balanced)

S.5.8 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The fractional coefficient in front of O can be converted to the smallest whole number by multiplying by. This requires that all other coefficients be multiplied by in order to retain the atom balance. The resulting balanced equation is: C H (g) + 5O (g) 4CO (g) + H O(l) (all atoms balanced using whole-number coefficients) The solution map for this problem is: g C H ö mol C H ö mol O The conversion factor needed in the first step is the molar mass of C H. From the periodic table we can get the molar masses of carbon and hydrogen and add them as follows. (1.01 g) C + (1.008 g) H = 6.04 g. Thus, the molar mass of C H is 6.04 g/mol. It is used in the form 1molCH in order to cancel units properly. 6.04g C H The conversion factor needed in the second step is the mole ratio for these two substances from 5molO the balanced equation in the form. molc H 19.6 g CH 1molCH 6.04 g C H 5molO molch =1.88molO. How many kilograms of Li O are needed to react completely 4.17 10 mol H O? Li O(s) + H O(g) LiOH(s) 15 kg What we know: mol H O; balanced equation relating Li O and H O kg Li O The solution map for this problem is: mol H O ö mol Li O ö g Li O ö kg Li O

S.5.9 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the mole ratio for these two substances from the 1molLiO balanced equation in the form. 1molHO The conversion factor needed in the second step is the molar mass of Li O. From the periodic table we can get the molar masses of lithium and oygen and add them as follows. (6.941 g) Li + 16.00 g O = 9.88 g. Thus, the molar mass of Li O is 9.88 g/mol. It is used in the form 9.88g LiO in order to cancel units properly. 1molLiO 1kg The conversion factor needed in the last step is that between g and kg in the form. 10 g 4.17 10 mol HO 1molLiO 1molH O 9.88 g LiO 1molLi O 1kgLi O = 15kg Li O 10 g Li O 4. Carbon dioide is produced in the reaction H PO 4 (aq) + MgCO (s) Mg (PO 4 ) (s) + CO (g) + H O(l) (unbalanced) How many grams of MgCO are needed to produce 14.8 moles of CO? 1.5 10 g What we know: mol CO ; equation relating MgCO and CO g MgCO First, a balanced equation is needed. To balance this equation, place a coefficient of in front of MgCO to balance magnesium atoms. Then place a coefficient of in front of H PO 4 to balance PO 4 groups. Net, a coefficient of in front of CO will balance carbon atoms. Finally, a coefficient of in front of H O will balance hydrogen and oygen atoms. The balanced equation is: H PO 4 (aq) + MgCO (s) Mg (PO 4 ) (s) + CO (g) + H O(l) The solution map for this problem is: mol CO ö mol MgCO ö g MgCO

S.5.0 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the mole ratio for these two substances from the molmgco balanced equation in the form. molco The conversion factor needed in the second step is the molar mass of MgCO. From the periodic table we can get the molar masses of magnesium, carbon and oygen and add them as follows. 4.1 g Mg + 1.01 g C + (16.00 g) O = 84. g. Thus, the molar mass of MgCO is 84. 84.g MgCO g/mol. It is used in the form in order to cancel units properly. 1molMgCO 14.8 molco molmgco molco 84.g MgCO 1molMgCO = 1.510 g MgCO Stoichiometric Calculations (mass-to-mass) 1. How many grams of sulfur can react with 1.79 g of copper according to the following equation? Cu(s) + S(s) CuS(s) 0.90 g What we know: g Cu; balanced equation relating S and Cu g S The solution map for this problem is: g Cu ö mol Cu ö mol S ö g S The conversion factor needed in the first step is the molar mass of Cu. From the periodic table 1molCu we see that the molar mass of copper is 6.55 g/mol. It is used in the form in order 6.55g Cu to cancel units properly. The conversion factor needed in the second step is the mole ratio for these two substances from 1molS the balanced equation in the form. 1molCu

S.5.1 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the molar mass of S. From the periodic table we.07 gs see that the molar mass of sulfur is.07 g/mol. It is used in the form in order to 1molS cancel units properly. 1.79 g Cu 1molCu 6.55 g Cu 1molS 1molCu.07 gs 1molS =0.90gS. How many grams of chlorine gas are required to react completely with 0.455 g iron to form iron(iii) chloride? 0.867 g What we know: g Fe g Cl First, a balanced equation is needed for the reaction between Fe and Cl. The unbalanced equation for this combination reaction is: Fe(s) + Cl (g) FeCl (s) To balance this equation, place a coefficient of in front of FeCl and a coefficient of in front of Cl to balance chlorine atoms. This requires a coefficient of in front of Fe to balance iron atoms. The balanced equation is: Fe(s) + Cl (g) FeCl (s) The solution map for this problem is: g Fe ö mol Fe ö mol Cl ö g Cl The conversion factor needed in the first step is the molar mass of Fe. From the periodic table 1molFe we see that the molar mass of iron is 55.84 g/mol. It is used in the form in order to 55.84g Fe cancel units properly.

S.5. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS 70.90g Cl 1molCl The conversion factor needed in the second step is the mole ratio for these two substances from molcl the balanced equation in the form. molfe The conversion factor needed in the last step is the molar mass of Cl. From the periodic table we see that the molar mass of chlorine is 5.45 g/mol so the molar mass of Cl is (5.45 g/mol) = 70.90 g/mol. It is used in the form in order to cancel units properly. 0.455 g Fe 1molFe molcl 70.90g Cl =0.867gCl 55.84 g Fe molfe 1molCl. How many grams of each product can be formed from the decomposition of 14.0 g of sodium chlorate? NaClO (s) NaCl(s) + O (g) Answers: 7.69 g NaCl and 6.1 g O What we know: g NaClO ; balanced equation relating NaCl, O and NaClO g NaCl and g O The solution maps for these calculations are: g NaClO ö mol NaClO ö mol NaCl ö g NaCl g NaClO ö mol NaClO ö mol O ö g O In each case, the conversion factor needed in the first step is the molar mass of NaClO. From the periodic table we can get the molar masses of sodium, chlorine and oygen and add them as follows..99 g Na + 5.45 g Cl + (16.00 g) O = 106.44 g. Thus, the molar mass of NaClO 1molNaClO is 106.44 g/mol. It is used in the form in order to cancel units properly. 106.44 g NaClO The conversion factor needed in the second step is the mole ratio for the particular product and molnacl molo N H 4 : and molnaclo molnaclo

S.5. CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the last step is the molar mass of the product: and.00g O 1molO 58.44g NaCl 1molNaCl 14.0 g NaClO 1molNaClO molnacl 58.44g NaCl 1molNaCl =7.69gNaCl 106.44 g NaClO molnaclo 14.0 g NaClO 1molNaClO 106.44 g NaClO molo molnaclo.00g O =6.1gO 1molO Note that the total mass of the products equals the mass of the reactant. 4. How many grams of potassium are needed to produce 16.5 kg K O? KNO (s) + K(s) K O(s) + N (g) (unbalanced) 0.867 g What we know: kg K O; equation relating K and K O g K First, a balanced equation is needed. To balance this equation, place a coefficient of in front of KNO to balance nitrogen atoms. Then place a coefficient of 6 in front of K O to balance oygen atoms. Finally, a coefficient of 10 in front of K will balance potassium atoms. The balanced equation is: KNO (s) + 10K(s) 6K O(s) + N (g) The solution map for this problem is: kg K O ö g K O ö mol K O ö mol K ö g K The conversion factor needed in the first step is that between kg and g in the form. 10 g 1kg

S.5.4 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the second step is the molar mass of K O in the form 1molKO. 94.0g K O The conversion factor needed in the third step is the mole ratio for these two substances from the 10 mol K balanced equation in the form. 6molK O The conversion factor needed in the last step is the molar mass of K. From the periodic table we 9.10g K see that the molar mass of potassium is 9.10 g/mol. It is used in the form in order to 1molK cancel units properly. 16.5 kg KO 10 g KO 1kgK O 1molKO 94.0 g K O 10 mol K 6molKO 9.10g K 1molK 4 =1.1410 gk Theoretical Yield and Limiting Reactant 1. Which is the limiting reactant when 0.68 g magnesium reacts with 17 mmol nitrogen gas to form Mg N? Mg What we know: g Mg; mmol N which reactant, Mg or N, is the limiting reactant The balanced equation for this combination reaction is: Mg(s) + N (g) Mg N (s) First, determine the limiting reactant by calculating how many moles of Mg N can form from each starting amount of reactant. The solution maps for these calculations are: g Mg ö mol Mg ö mol Mg N mmol N ö mol N ö mol Mg N

S.5.5 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS For the first calculation the conversion factor needed in the first step is the molar mass of Mg in 1molMg the form. 4.1g Mg The conversion factor needed for the second step is the Mg N /Mg mole ratio from the balanced equation. Applying these yields: 0.68 g Mg 1molMg 1molMgN = 0.009mol MgN 4.1 g Mg molmg For the second calculation the conversion factor needed in the first step is that between mmol and - 10 mol mol in the form. 1mmol The conversion factor needed in the second step is the Mg N /N mole ratio from the balanced equation. Applying this yields: 17 mmol N - 10 mol N 1mmolN 1molMg N = 0.017 mol MgN 1molN Since the starting amount of Mg produces the smaller amount of Mg N, magnesium is the limiting reactant.. How many moles of AsF 5 can be produced when 14 moles of arsenic react with 9 mol fluorine gas? 1 mol What we know: mol As; mol F mol AsF 5 The balanced equation for this combination reaction is: As(s) + 5F (g) AsF 5 (g)

S.5.6 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS First, determine the limiting reactant by calculating how many moles of AsF 5 can form from each starting amount of reactant. The solution maps for these calculations are: mol As ö mol AsF 5 mol F ö mol AsF 5 For the first calculation the conversion factor needed is the AsF 5 /As mole ratio from the balanced equation. Applying this yields: 14 mol As molasf 5 =14molAsF5 molas For the second calculation the conversion factor needed is the AsF 5 /F mole ratio from the balanced equation. Applying this yields: 9 mol F molasf 5 =1molAsF5 5molF Since the starting amount of F produces the smaller amount of AsF 5, fluorine is the limiting reactant and the maimum moles of AsF 5 is 1 moles.. When 6.5 g CO and.7 g H are allowed to react as shown below, CO(g) + H (g) CH OH(l) a) which is the limiting reactant? b) what is the theoretical yield in grams of CH OH? c) how much of the reactant in ecess remains? Answers: a) H b) 9. g c) 0.8 g

S.5.7 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS s a) What we know: g CO; g H ; balanced equation relating CO, H and CH OH which reactant, CO or H, is the limiting reactant Determine the limiting reactant by calculating how many moles of CH OH can form from each starting amount of reactant. The solution maps for these calculations are: g CO ö mol CO ö mol CH OH g H ö mol H ö mol CH OH For the first calculation the conversion factors needed are the molar mass of CO and the CH OH/CO mole ratio. 6.5 g CO 1molCO 1molCHOH = 0.946molCHOH 8.01 g CO 1molCO For the second calculation the conversion factors needed are the molar mass of H and the CH OH/H mole ratio..7 g H 1molH.016 g H 1molCHOH molh =0.918molCHOH Since the starting amount of H produces the smaller amount of CH OH, H is the limiting reactant. b) What we know: mol CH OH produced from limiting reactant theoretical yield of CH OH in grams Since the theoretical yield is the amount of product produced from the limiting reactant, the solution map for this calculation is: mol CH OH ö g CH OH The conversion factor needed for this calculation is the molar mass of CH OH.

S.5.8 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS Applying this yields: 0.918 molchoh.04g CH OH 1molCH OH =9.gCHOH c) What we know: mol CH OH produced from limiting reactant g CO remaining Use the theoretical yield of CH OH to calculate how much CO was used and subtract this from the starting quantity. The solution maps for these calculations are: mol CH OH ö mol CO ö g CO used g CO remaining = g CO initially - g CO used For the first calculation the conversion factors needed are the CO/CH OH mole ratio and the molar mass of CO. 0.918 molchoh 1molCO 1molCHOH 8.01g CO 1molCO =5.7gCOused Therefore, the mass of CO that remains equals 6.5 g - 5.7 g = 0.8 g. 4. How many kilograms of the ecess reactant remain when a miture of.50 kg of SiO and.50 kg of carbon react? SiO (s) + C(s) SiC(s) + CO(g) 1.00 kg What we know: kg SiO ; kg C; balanced equation relating SiO and C kg of ecess reactant remaining

S.5.9 CHAPTER 5 SOLUTIONS TO SUPPLEMENTARY CHECK FOR UNDERSTANDING PROBLEMS First, determine the limiting reactant by calculating how many moles of SiC can form from each starting amount of reactant. The solution maps for these calculations are: kg SiO ö g SiO ö mol SiO ö mol SiC kg C ö g C ö mol C ö mol SiC For the first calculation the conversion factors needed are that between kg and g, the molar mass of SiO and the SiC/SiO mole ratio from the balanced equation..50 kgsio 10 gsio 1kgSiO 1molSiO 1molSiC 1molSiO 60.09 gsio = 41.6molSiC For the second calculation the conversion factors needed are that between kg and g, the molar mass of C and the SiC/C mole ratio from the balanced equation..50 kg C 10 g C 1kgC 1molC 1.01 g C 1molSiC molc =08molSiC Since the starting amount of SiO produces the smaller amount of SiC, SiO is the limiting reactant and carbon is the reactant in ecess. Now use the theoretical yield of SiC (41.6 mol) to calculate how much C was used and subtract this from the starting quantity. The solution maps for these calculations are: mol SiC ö mol C used ö g CO used ö kg CO used kg C remaining = kg C initially - kg C used For the first calculation the conversion factors needed are the C/SiC mole ratio, the molar mass of C and the relationship between g and kg. 41.6 molsic molc 1molSiC 1.01 g C 1kgC =1.50kgCused 1molC 10 g C Therefore, the mass of C that remains equals.50 kg - 1.50 kg = 1.00 kg.