CHEM 105 Problem Set 28 PDF

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Problem Set 28 – % Composition, Empirical & Molecular Formulas Chem 105 1. Give the percent composition (by mass) for each of the hydrocarbons below. (Unless otherwise specified, assume percent composition is always the mass %.) Which two have the same percent composition?

(b) (a )

(c) (d )

a) C6H6 molecular weight 78.11 g %C 72.06/78.11 x 100 = 92.38% %H 6.05/78.11x 100 = 7.75% b) C2H4 mw = 28.05 g % C 24.02/28.05 x 100 = 85.63 % H 4.03/28.05 x 100 = 14.37 c) C3H8 mw = 44.1 g % C 36.03/44.1 g x 100 = 81.7 % H 8.07/44.1 x 100 = 18.3 d) C6H12 mw 84.16 % C 72.06/86.16 x 100 = 85.62 % H 12.1/84/16 x 100 = 14.38 C2h4 and c6h12 have same percent composition 2. Ancient Egyptians used lead compounds including PbS, PbCO3, and Pb2Cl2CO3 as pigments in cosmetics, and many people suffered from chronic lead poisoning as a result. Calculate the percentage of lead in each of the compounds. Pbs: % Pb = (207.2/239.30) x 100 = 86.59% PbCO3 % Pb = 207.2/267/2 x 100 = 77.54% Pb2Cl2CO3 % Pb = 2 x 207.2 / 545.3 x 100 = 75.99% 3. The following ‘organic’ compounds have been detected in space. Which of them contains the greatest percentage of carbon by mass? a. naphthalene, C10H8 c. pentacene, C22H14 b. chrysene, C18H12 d. pyrene, C16H10 4. Unlike most metals, gold occurs in nature as the pure element. Miners in California in 1849 searched for gold nuggets and gold dust in streambeds, where the denser gold could be easily separated from sand and gravel. However, larger deposits of gold are found in veins of rock and can be separated chemically in a two-step process: (1) 4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) → 4 NaAu(CN)2(aq) + 4 NaOH(aq) (2) 2 NaAu(CN)2(aq) + Zn(s) → 2 Au(s) +Na2[Zn(CN)4](aq) If 23 kilograms of ore is 0.19% gold by mass, how much Zn is needed to react with the gold in the ore? Assume that both reactions are 100% efficient. 23 kg ore x 1000 g/ 1 kg x 0.0019 Au – 43.7 Au

43.7 g Au x 1 mol au/ 196.97 g x 4 mol NaAu(CN) / 4 mol Au x 2 mol Zn/ 4 mol Au x 65.38 g / 1 mol Zn = 7.3 g 5. Give the empirical and molecular formulas for the five oxides of nitrogen in (a) and the hydrocarbons in (b). Which molecules have the same empirical formulas? Do any have the same percent composition?

(a)

(b) a) N2O5 molecular and empirical formula N2O4 molecular formula, NO2 empirical formula NO molecular and empirical formula N2O3 molecular and empirical formula NO2 molecular and empirical formula N2O4 and NO2 have same empirical formulas and same percent composition b) C2H6 – molecular formula CH3 empirical C2H4 molecular CH2 empirical C2H2 molecular, CH empirical No same empirical formula or percent composition 6. (a) Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti or surgical-grade titanium. The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium. What is the empirical formula of surgical-grade titanium? Moles of titanium 64.39 / 47.87 = 1.345 mol Ti Moles of aluminum 24.19 / 26.98 = 0.8966 mol Al Moles of vanadium 11.42/ 50.94 = 0.2242 mol V Divide by smallest 0.2242 we get Ti:Al:V – 6:4:1 Empirical formula Ti6Al4V (b) Inhalation of asbestos fibers may lead to a lung disease known as asbestosis and to a form of lung cancer called mesothelioma. One form of asbestos, chrysotile, is 26.31% magnesium, 20.27% silicon, and 1.45% hydrogen by mass, with the remainder of the mass as oxygen. What is the empirical formula of chrysotile? %O = 100 – 26.31 + 20.27 + 1.45 = 51.97% Mg, 26.31/ 24.31 = 1.082 mol Mg Si, 20.27 / 28.09 = 0.7216 mol si H, 1.45 / 1.01 = 1.44 mol H O 51.97/ 16 = 3.248 mol O Divide by 0.7216 Mg:Si:H:O 1.5:1:2:4:5 Mg3Si2H4O9

7. (a) An unknown organic compound has the simplest formula CH2. If its molecular weight is 28 g/mol, what is the molecular formula? The molecular weight of CH is 12.011 g/ mol = 14.027 g/mol the actual molecular weight is 28 g/ mol = 2 28/ 14.027 = 2 Formula is C2H4 (b) Benzene, a common solvent, is a covalent molecular compound which contains only carbon and hydrogen. Its simplest (empirical) formula is CH, and its molecular weight to 2 significant digits is 78 g/mol. What is its molecular formula? N = 78/13 = 6 molecular formula is C6H6 8. The material often used to make artificial bones is the same material that gives natural bones their strength. Its common name is hydroxyapatite, and its formula is Ca5(PO4)3OH. a) Propose a systematic name for this compound. Calcium triphosphate hydroxide b) What is the mass percentage of calcium in it? mw of hydroxyapatite is 502.31 g/mol % ca = 5 x 40.08 / 502.31 x 100 = 39.89% c) When treated with hydrogen fluoride, hydroxyapatite becomes fluorapatite [Ca 5(PO4)3F], an even stronger substance. Does the percent mass of Ca increase or decrease as a result of this substitution? Mw of fluorapatite 504.3 g/mol % Ca 5 x 40.08 / 504.3 x100 = 39.74% the mass percent slightly decreases 9. Give brief answers to the following questions about combustion analysis (1-3 sentences each). a) Is it important for combustion analysis to be done in an excess of oxygen? Why or why not? Yes, more to react with will drive the reaction forward to make product. SO that all of the reagent is converted into the products b) Can the quantity of CO2 obtained in a combustion analysis be a direct measure of the oxygen content of the starting material? Why or why not? No, because oxygen can be from other reactant c) Can the results of a combustion analysis ever give the true molecular formula of a compound? It can determine the empirical formula d) What additional information is needed to determine a molecular formula if the empirical formula is known from combustion analysis? The molecular mass of the actual molecule 10. Charcoal (C) and propane (C3H8) are used as fuel in backyard grills. a) Write balanced chemical equations for the complete combustion reactions of C and C3H8. C + O2  CO2 C3H8 + 5O2  3CO2 + 4H2O b) How many grams of carbon dioxide are produced from burning 500.0 grams of each of the two fuels? 500.0 g C x 1 mol c/ 12.011 g x 1mol co2/ 1 mol c . 44.01 g co2/ 1 mol = 1832 g co2 500.0 g C3H8 x 1 mol C3H8 / 44.10 g x 3 mol CO2/ 1 mol C2H8 x 44.01 g CO2/ 1 mol = 1497 g CO2 11. (a) The compound geraniol is on the U.S. Food and Drug Administration’s GRAS (Generally Recognized as Safe) list and can be used in foods and personal care products. By itself, geraniol has a roselike odor, but it is frequently blended with other scents to produce the fruity fragrances of some personal care products. Complete combustion of 175 mg of geraniol produces 499 mg CO2 and 184 mg H2O. What is the empirical formula of geraniol? 0.498 g Co2 x 1 mole co2/ 44.01 g x 1 mole / 1 mol co2 = 0.0113 mol C 1.13 10 -2 mol c x 12.01 g C/ 1 mol = 0.13g C 0.184 g H2O x 1 mol H2o / 18.02 x 2 mol / 1 mol h2o = 0.0204 mol H 0.0204 mol H x 1.008 g h / 1 mol = 0.0206 gH 0.136 g + 0.0206 g = 0.1566 g Mass of O present = 0.175 g – 0.1566 g = 0.0184 O Moles of O in compound = 0.0184 O x 1 mol / 15.99 g = 1.15 x 10 -3 mol O

Divide by smallest molar amount gives ratio of 10C:18H:1O (b) Combustion analysis reveals that a certain organic compound with a molar mass of 62.1 g/mol is 38.7% C, 9.7% H, and 51.6% O by mass. What are its empirical and molecular formulas? 3.23 mol of c, 9.7 mol of H, 3.23 mol of O if we divide them by the smallest number (3.23) then the empirical formula is C1H3O1. Molar mass of ch3o is about 31 g/mol. It takes 2 empirical formula units to equal the molar mass, making the molecular formula C2H6O2 12. (a) 0.225 g of Caproic acid (an organic molecule that smells like dirty socks) is analyzed by combustion analysis. 0.512 g of CO2 and 0.209 g of H2O are produced. The molar mass of the compound is 116 g/mol. What are its empirical and molecular formulas? Convert masses to moles of C and H 0.512 g co2 / 1 mol/ 44.01 g x 1 mol c/ 1 mol co2 = 0.0116 mol C 0.209 g H20 1 mol/ 18.02 x 2 mol H / 1 mol H20 = 0.0232 mol H Calculate grams of C and H to find grams of O in original compound 0.0116 mol C 12.01/ 1 = 0.140 g C 0.0232 mol H x 1.008 g / 1 mol = 0.0234 g H 0.225g – 0.140 – o..o234 g = 0.062 g O 0.062 g O x 1/ 16.00 g = 0.0039 mol O Calculate the mole ratios mol c/ mol O = 0.0116 / 0.0039 = 3.0 mol c/ mol O Mol / mol O = 0.0232/ 0.0039 = 5.9 mol H/ mol O 1 mol O = 3 mol C = 6 mol H = C3H2O Compare mass of C3H6O to the molar mass: 116 g/mol x 1 / 58.08 g / mol = 2.00 2 unites = C6H12O2 (b) 7.70 The combustion of 40.5 mg of a compound extracted from the bark of the sassafras tree and known to contain C, H, and O produces 110.0 mg CO2 and 22.5 mg H2O. The molar mass of the compound is 162 g/mol. What are its empirical and molecular formulas? 0.1100 g co2 x 1 mol / 44.01 g x 1 mol c/ 1 mol co2 = 2.5 x 10 -3 mol C 2.50 x 10 -3 mol c x 12.01 g c / 1 mol c = 0.0300 g C 0.0225 g H20 x 1 mol h2o /18.02 g x 2 mol h / 1 mol h2o = 2.50 x 10-3 mol H 2.50 x 10 -3 mol h x 1.01 h / 1 mol – 0.00252 g h 0.0300 g + 0.00252 g = 0.0325 g Mass of O present = 0.0405 g – 0.0325 g = 0.0080 g O Moles of O in compound = 0.0080 g O x 1 mol / 16.00g = 5.0 x 10 -4 mol O 13. (a) 26.2 g of an unknown gas occupies 5.3 L at 40.ºC and 742 torr. What is the molar mass of this gas? Further studies indicate that this gas is monotomic – which element do you suspect this gas is? (b) Combustion analysis of an unknown gas gave an empirical formula of CH2. If 2.1 g of this gas occupies 3.1 L at 100. ºC and 0.50 atm, what is the molecular formula of this gas?...