Moles & Stoichiometry Answers

Chem Worksheet - Week 4 Moles & Stoichiometry Answers Key Questions & Exercises 1. The atomic weight of carbon is 12.0107 u, so a mole of carbon has a mass of 12.0107 g. Why doesn’t a mole of carbon weigh 12 g? The atomic weight refers to the weighted average of masses of the isotopes comprising a naturally occurring sample of carbon. A 12.0107-g sample of natural carbon would contain an Avogadro’s number of carbon atoms with all the naturally occurring mass numbers, present in their natural fractional abundance. 2. The atomic weight of oxygen is 16.00 u. What is the mass of a mole of O2(g)? How many O2 molecules does a mole of O2(g) contain? How many moles of oxygen atoms does a mole of O2(g) contain? The molecular weight of O2 is 32.00 u, so a mole of O2 would have a mass of 32.00 g and would contain 6.022 x 1023 O2 molecules. Each O2 molecule is composed of two oxygen atoms, so one mole of O2 contains two moles of oxygen atoms. 3. The mole is sometimes described as “the chemist’s dozen”. How is a mole like a dozen? A dozen is twelve things, and a mole is an Avogadro’s number (6.022 x 1023) of things. The difference is that a dozen things (12) is an exact number, but a mole of things (6.022 x 1023) is an inexact (measured) number. 4. Consider a 15.00-g sample of CO2 (m.w. = 44.01u). How many moles of CO2 are there in this sample? (15.00 g CO2)(1 mol CO2 /44.01 g CO2) = 0.3408 mol CO2 5. How many CO2 molecules are there in a 15.00-g sample of carbon dioxide? (0.3408 mol CO2)(6.022 x 1023 CO2 molecules/mol CO2) = 2.0525 x 1023 CO2 molecules 6. How many oxygen atoms are there in a 15.00-g sample of carbon dioxide. (2.0525 x 1023 CO2 molecules)(2 O atoms/CO2 molecule) = 4.105 x 1023 O atoms 7. Fluorine consists of a single isotope, 19F, with a mass of 19.00 u. What is the mass in grams of a single fluorine atom? 1 mol F = 19.00 g = 6.022 x 1023 atoms 8. If you have data for the percent composition of a compound, element by element, do you need to know the size of the sample in order to figure out the empirical formula? Why or why not? No, because the mass ratios of all the elements are the same, regardless of the sample size, as required by the Law of Constant Composition. 9. How is the molecular formula of a molecular compound related to its empirical formula? The molecular formula is always a whole-number multiple of the empirical formula of the compounds. Sometimes the whole-number multiple is 1, when the molecular and empirical formulas are the same. When there is a difference, the molecular formula is usually a small whole-number multiple of the empirical formula. For example, the molecular formula of benzene is C6H6, and its empirical formula is CH. The whole number multiplier in this case is the integer 6. Note that in going from the empirical formula to the molecular formula, only the subscripts (including implied 1's) are multiplied. It is not correct to write the molecular formula of benzene as 6 CH. 10. A compound is found to contain 54.52% C, 9.17% H, and 36.31% O. What is the empirical formula of the compound? If the compound is found to have molecular weight of 88.12 u, what is the molecular formula? Assume exactly 100 g of compound. Then the percentages are numerically equal to the numbers of grams of each element. C: (54.52 g C)(1 mol C/12.01 g C) = 4.540 mol C H: (9.17 g H)(1 mol H/1.01 g H) = 9.08 mol H O: (36.31 g O)(1 mol O/16.00 g O) = 2.269 mol O To find the simplest whole number ratio between these numbers of moles, take the smallest number and divide it into all the numbers. C: 4.540 mol/2.269 mol = 2.001 = 2 H: 9.08 mol/2.269 mol = 4.002 = 4 O: 2.269 mol/2.269 mol = 1 From this we obtain the empirical formula C2H4O. The formula weight for this is f.w. = 2(12.01 u) + 4(1.01 u) + 16.00 u = 44.06 u Dividing into the given molecular weight m.w./f.w. = 88.12/44.06 = 2 This shows that the molecular formula is twice the empirical formula; i.e., C4H8O2. 11. What is the empirical formula of an oxide of nitrogen whose composition is 25.94% nitrogen? The compound contains only nitrogen and oxygen. By subtraction, %O = 100% – 25.94% = 74.06% Assume 100 g of compound. N: (25.94 g N)(1 mol N/14.01 g N) = 1.852 mol N Y 1.852 mol/1.852 mol = 1 Y 2 O: (74.06 g O)(1 mol O/16.00 g O) = 4.629 mol O Y 4.629 mol/1.852 mol = 2.5 Y 5 Therefore, the empirical formula is N2O5. [Note that the originally found ratio N:O = 1:2.5 needed to be multiplied by 2 to make the integer ratio N:O = 2:5. Empirical formulas must have integer subscripts.] 12. A 2.554-g sample of a certain hydrocarbon is burned in excess oxygen, producing 8.635 g CO2(g) and 1.768 g H2O(l). If the molecular weight of the hydrocarbon is found to be 78.11 u, what is its molecular formula? [m.w. CO2 = 44.01 u; m.w. H2O = 18.02 u] A hydrocarbon only contains carbon and hydrogen. When burned, all of the carbon ends up in the CO2(g) produced, and all of the hydrogen ends up in the H2O(l) produced. First find the moles of CO2(g) and H2O(l), and then the moles of C and H in each of them. Then determine the empirical formula from the lowest whole-number ratio between the moles of C and moles of H. Finally, by dividing the given molecular weight by the formula weight, determine the factor by which the empirical formula must be multiplied to give the molecular formula. The empirical formula is CH, for which f.w. = 12.01 + 1.01 = 13.02. m.w./f.w. = 78.11/13.02 = 6 Therefore, the molecular formula is C6H6. 13. In the complete combustion of propane, how many moles of CO2(g) are produced per mole of O2(g)? From the balanced equation, C3H8(g) + 5 O2(g) ÷ 3 CO2(g) + 4 H2O(l) we see that 3 moles of 3 CO2(g) are produced for every 5 moles of O2(g) consumed. Thus, for every mole of O2(g), 3/5 mole of CO2(g) is produced. 14. In the complete combustion of propane, how many moles of H2O(l) are produced per mole of O2(g)? From the balanced equation, C3H8(g) + 5 O2(g) ÷ 3 CO2(g) + 4 H2O(l) we see that 4 moles of 4 H2O(l) are produced for every 5 moles of O2(g) consumed. Thus, for every mole of O2(g), 4/5 mole of H2O(l) is produced. 15. In the complete combustion of propane, how many moles of H2O(l) are produced per mole of CO2(g)? From the balanced equation, C3H8(g) + 5 O2(g) ÷ 3 CO2(g) + 4 H2O(l) we see that 4 moles of 4 H2O(l) are produced for every 3 moles of CO2(g) that are produced. Thus, for every mole of CO2(g) produced, 1a mole of H2O(l) is produced. 16. A 1.638-g sample of propane is burned in excess oxygen. What are the theoretical yields (in grams) of CO2(g) and H2O(l) expected from the reaction? [m.w. C3H8 = 44.09 u, m.w. CO2 = 44.01 u, m.w. H2O = 18.02 u) The balanced equation is C3H8(g) + 2 O2(g) ÷ 3 CO2(g) + 4 H2O(l) Using the stoichiometric relationships, 17. If 4.750 g of CO2(g) was obtained from the combustion of 1.638 g of propane, what was the percent yield? 18. Define what is meant by the terms limiting reagent and excess reagent. The limiting agent is the reagent that is present in shortest supply, on the basis of the balanced chemical equation, and which will be completely consumed in a complete reaction. The excess reagent is present in more than a sufficient amount to react with the limiting reagent, and some of it will remain after a complete chemical reaction. The amount of excess reagent left can be determined by calculating the amount consumed on the basis of its stoichiometric relationship with the limiting reagent, and then subtracting that amount from the amount that was initially present. 19. In the reaction 2 A + 3 B ÷ products, if you have 0.500 mol A and 0.500 mol B, which is the limiting reagent? How much of the excess reagent will be left over, if compete reaction takes place? Dividing the amount of each reagent by its stoichiometric coefficient in the balanced equation, we see we have (0.500 mol A)/(2 mol/”set”) = 0.250 “set” of A and (0.500 mol B)/(3 mol/”set”) = 0.167 “set” of B Thus, B is the limiting reagent and will be completely consumed. Based on the balanced equation, 2 moles of A are consumed for every 3 mole of B, so the amount of A that is consumed will be mol A used = (0.500 mol B)(2 mol A/3 mol B) = 0.333 mol A Subtracting from the original0.500 mol A that was present, mol A left = (0.500 – 0.333) mol A = 0.167 mol A 20. What is the theoretical yield of Ca3(PO4)2(s) by the reaction 3 Ca(OH)2(s) + 2 H3PO4(l) 6 Ca3(PO4)2(s) + 3 H2O(l) when 10.00 g Ca(OH)2 and 10.00 g H3PO4 are mixed? [f.w. Ca(OH)2 = 74.10 u; m.w. H3PO4 = 97.99 u; f.w. Ca3(PO4)2 = 310.18 u] mol Ca(OH)2 = (10.00 g Ca(OH)2)(1 mol Ca(OH)2/74.10 g Ca(OH)2) = 0.134953 mol Ca(OH)2 mol H3PO4 = (10.00 g H3PO4)(1 mol H3PO4/97.99 g H3PO4) = 0.102051 mol H3PO4 To identify the limiting reagent divide each number of moles by its stoichiometric coefficient in the balanced equation, thereby determining the number of "sets" of each. "sets" Ca(OH)2 = (0.134953 mol Ca(OH)2)(1 "set" Ca(OH)2/3 mol Ca(OH)2) = 0.04498 "set" Ca(OH)2 "sets" H3PO4 = (0.102051 mol H3PO4)(1 "set" H3PO4/2 mol H3PO4) = 0.05103 "set" H3PO4 Y Ca(OH)2 limits, because it has the fewer "sets". Now, use the moles of Ca(OH)2 (not the number of "sets"!) in all subsequent calculations. g Ca3(PO4)2 = (0.134953 mol Ca(OH)2)(1 mol Ca3(PO4)2/3 mol Ca(OH)2) x (310.18 g Ca3(PO4)2/mol Ca3(PO4)2) = 13.95 g Ca3(PO4)2 Periodic Table of the Elements 1A 1 1 H 1.008 8A 18 2A 2 3A 13 4A 14 5A 15 6A 16 7A 17 2 He 4.003 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.941 9.012 10.81 12.01 14.01 16.00 19.00 20.18 13 14 15 16 17 18 Al Si P S Cl Ar 26.98 28.09 30.97 32.07 35.45 39.95 11 12 Na Mg 22.99 24.31 3B 3 4B 4 5B 5 6B 6 7B 7 +))Q 8 8B 9 ))), 10 1B 11 2B 12 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 [98] 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 [209] [210] [222] 87 88 89 104 105 106 107 108 109 110 111 112 Fr Ra Ac Rf Db Sg Bh Hs Mt Uun Uuu Uub [223] [226] [227] [261] [262] [263] [264] [265] [268] [269] [272] [277] 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 [145] 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 103 90 91 92 93 94 95 96 97 98 99 100 101 102 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 [237] [244] [243] [247] [247] [251] [252] [257] [258] [259] [262]