PS 28 2022 key~

docx

School

Brigham Young University *

*We aren’t endorsed by this school

Course

105

Subject

Chemistry

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by samuelsbox

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? 1. A) C 6 H 6 , M.W-78.11 g Mass%- (mass of the element/ Total mass) X 100 %C - 72.06/78.11 X 100 = 92.38 %H - 6.05/78.11 X 100 = 7.75 b) C 2 H 4 , M.W- 28.05 g %C - 24.02/28.05 X 100 = 85.63 %H - 4.03/28.05 X 100 = 14.37 c) C 3 H 8 , M.W- 44.1 g %C- 36.03/44.1 X 100 = 81.7 %H- 8.07/ 44.1 X 100 = 18.3 d) C 6 H 12 , M.W- 84.16 %C - 72.06/84.16 X 100 =85.62 %H - 12.1/84.16 X 100 = 14.38 C 2 H 4 and C 6 H 12 have same percent composition 2. Ancient Egyptians used lead compounds including PbS, PbCO 3 , and Pb 2 Cl 2 CO 3 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% PbCO 3 : %Pb= (207.2/267.2) X 100= 77.54% Pb 2 Cl 2 CO 3 %Pb= (2X207.2)/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, C 10 H 8 %C = (12.01X10)/128.2 X100= 93.69% C c. pentacene, C 22 H 14

%C = (12.01X22)/278.4 X100= 94.91% C b. chrysene, C 18 H 12 %C= (12.01X18)/228.3 X100= 94.70% C d. pyrene, C 16 H 10 %C= (12.01X16)/202.3 X100= 95.00% C Pyrene has the greatest % of c 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 ) + O 2 ( g ) + 2 H 2 O(ℓ) → 4 NaAu(CN) 2 ( aq ) + 4 NaOH( aq ) (2) 2 NaAu(CN) 2 ( aq ) + Zn( s ) → 2 Au( s ) +Na 2 [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 g Au 43.7 g Au X 1 mole Au 196.97 g X 4 NaAu ( CN ) 2 4 Au × 1 mol Zn 2 mol NaAu ( CN ) 2 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) N 2 O 5 molecular and empirical formula N 2 O 4 molecular formula, NO 2 empirical formula NO- molecular and empirical formula N 2 O 3 – molecular and empirical formula NO 2- molecular and empirical formula N 2 O 4 and NO 2 have same empirical formulas and same percent composition. (b) C 2 H 6 – molecular formula, CH 3 – empirical formula C 2 H 4 – molecular formula, CH 2 - empirical formula C 2 H 2 – molecular formula, CH- empirical formula No same empirical formulas 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 aluminium- 24.19 / 26.98= 0.8966 mol Al Moles of vanadium- 11.42/ 50.94 = 0.2242 mol V Dividing them all by the smallest, 0.2242, we get Ti:Al:V – 6:4:1 Empirical formula- Ti 6 Al 4 V (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 Dividing these by smallest 0.7216, we get Mg:Si:H:O-1.5:1:2:4.5. Multiplying these by two gives a whole number. The empirical formula is Mg 3 Si 2 H 4 O 9 7. (a) An unknown organic compound has the simplest formula CH 2 . If its molecular weight is 28 g/mol, what is the molecular formula? The molecular weight of CH is 12.011 g/mol + 1.008 g/mol = 14.027 g/mol. The actual molecular weight is 28 g/mol. actual molecular weight weight of empirical formula = 28 14.027  2 The molecular weight is 2 times the empirical molecular weight, so the molecular formula must be C 2 H 4 . (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, The molecular formula is C 6 H 6 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 Ca 5 (PO 4 ) 3 OH. a) Propose a systematic name for this compound. b) What is the mass percentage of calcium in it? c) When treated with hydrogen fluoride, hydroxyapatite becomes fluorapatite [Ca 5 (PO 4 ) 3 F], an even stronger substance. Does the percent mass of Ca increase or decrease as a result of this substitution? a) Calcium triphosphate hydroxide b) M.W of hydroxyapatite is 502.31 g/mol %Ca= (5X40.08)/502.31 X 100= 39.89% C) M.W of fluorapatite- 504.3 g/mol %Ca=(5X40.08)/504.3 X100= 39.74%

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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 (Le Chatelier’s principle). So that all of the reagent is converted into the products. b) Can the quantity of CO 2 obtained in a combustion analysis be a direct measure of the oxygen content of the starting material? Why or why not? No, because the oxygen in the CO 2 comes from both the O 2 and any oxygen in the 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 (C 3 H 8 ) are used as fuel in backyard grills. a) Write balanced chemical equations for the complete combustion reactions of C and C 3 H 8 . C + O 2  CO 2 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O b) How many grams of carbon dioxide are produced from burning 500.0 grams of each of the two fuels? C 3 H 8 (g) + 5O 2 (g)  3CO 2 +4H 2 O 500.0 gC × 1 molC 12.011 g × 1 molCO 2 1 molC × 44.01 gCO 2 1 mol = 1832 gCO 2 500.0 gC 3 H 8 × 1 molC 3 H 8 44.10 g × 3 molCO 2 1 molC 3 H 8 × 44.01 gCO 2 1 mol = 1497 gCO 2 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 CO 2 and 184 mg H 2 O. What is the empirical formula of geraniol? 0.498gCO 2 x 1molCO 2 /44.01g x 1mol C/1mol CO 2 = 1.13x10 -2 mol C 1.13x10 -2 mol C x 12.01gC/1mol = 0.136g C 0.184g H 2 Ox1mol H 2 O/18.02g x 2mol H/1mol H 2 O = 0.0204 mol H 0.0204 mol H x 1.008g H/1mol = 0.0206 gH Total mass of C and H = 0.136g + 0.0206g = 0.1566g Mass of O present = 0.175g – 0.1566g = 0.0184 O Moles of O in compound = 0.0184 O x 1mol/15.999g = 1.15x10-3 mol O Dividing the moles of C,H,O by the smallest molar amount (1.15x10-3mol) gives a 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? Say there are 100 gof thiscompound ,thenthe massof eachelement is : 100 g× 0.387 C = 38.7 gC 100 g× 0.097 H = 9.7 gH 100 g× 0.516 O = 51.6 gO Thismeansthat thereare 3.23 moleof C , 9.7 moleof H , ∧ 3.23 moleof O .If we dividethem up by the smallest number (3.23), then the empirical formula is C 1 H 3 O 1

The molar mass of C 1 H 3 O 1 is about 31g/mol. Therefore it takes 2 empirical formula units to equal the molar mass, making the molecular formula C 2 H 6 O 2 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 CO 2 and 0.209 g of H 2 O are produced. The molar mass of the compound is 116 g/mol. What are its empirical and molecular formulas? Convert the masses to moles of C and H 0.512 gC O 2 1 mol 44.01 g 1 molC 1 molC O 2 = 0.0116 molC 0.209 g H 2 O 1 mol 18.02 g 2 mol H 1 mol H 2 O = 0.0232 mol H Calculate grams of C and H to find grams of O in original compound: 0.0116 molC 12.01 g 1 mol = 0.140 gC 0.0232 mol H 1.008 g 1 mol = 0.0234 g H 0.225g – 0.140g – 0.0234g = 0.062g O 0.062 gO 1 mol 16.00 g = 0.0039 molO Calculate the mole ratios: molC molO = 0.0116 0.0039 = 3.0 molC molO mol H molO = 0.0232 0.0039 = 5.9 mol H molO 1 mol O = 3 mol C = 6 mol H = C 3 H 6 O Compare the mass of C 3 H 6 O to the molar mass: 116 g / mol 1 unit 58.08 g / mol = 2.00 units 2 units = C 6 H 12 O 2 (b) 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 CO 2 and 22.5 mg H 2 O. The molar mass of the compound is 162 g/mol. What are its empirical and molecular formulas? 0.1100g CO 2 x 1mol Co 2 /44.01g x 1mol C/1mol CO 2 = 2.5 x 10 -3 mol C 2.50x10 -3 mol Cx12.01gC/1 mol C = 0.0300g C 0.0225g H 2 O x 1mol H 2 O/18.02g x 2 mol H/1 mol H 2 O = 2.50 x 10 -3 mol H 2.50x 10 -3 mol H x 1.01 H/1mol = 0.00252g H total mass of C and H = 0.0300g + 0.00252g = 0.0325g Mass of O present = 0.0405g – 0.0325g = 0.0080g O Moles of O in compound = 0.0080g O x 1mol/16.00g = 5.0x10 -4 mol / 5.0x10 -4 mol = 1 Moles of C in compound = 2.5 x 10 -3 mol / 5.0x10 -4 mol = 5 Moles of H in compound = 2.5 x 10 -3 mol / 5.0x10 -4 mol = 5 Empirical Formula = C 5 H 5 O = 81 g/mol  x2  Molecular Formula = C 10 H 10 O 2

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 monoatomic – which element do you suspect this gas is? Use the gas laws to determine the number of moles in the sample: n = PV RT = ( 742 torr× 1 atm 760 torr ) ( 5.3 L ) 0.0821 L∙atm mol∙K × 313 K = 0.20 mol Use the mass and the moles to calculate the molar mass: (26.2 g)/(0.20 mol)=130 g/mol This looks like xenon (131 g/mol) (b) Combustion analysis of an unknown gas gave an empirical formula of CH 2 . 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? First, we need to find the molar mass of the gas: n = PV RT = ( 0.50 atm )( 3.1 L ) ( 0.08206 Latm mol K ) ( 373 K ) = 0.051 mol M = m n = 2.1 g 0.051 mol = 41 g / mol Now, we can find the molecular formula: 41 g mol × 1 unit 14 g mol = 2.9 units = 3 units C 3 H 6 is the molecular formula.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version