Chemistry 171 Exam 1. January 13, Name. Periodic Table of the Elements

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1 Chemistry 171 Exam 1 January 13, 2011 Name N A = x kg = lb 1 in = 2.54 cm 1 L = qt 1 gal = 4 qt ( o F F) x (5 0 C/9 0 F) = 0 C Periodic Table of the Elements I A II A III B IV B V B VI B VII B VIII I B II B III A IV A V A VI A VII A inert gases 1 H Li Na K Rb Cs Fr (223) 4 Be Mg Ca Sr Ba Ra Sc Y * La * Ac (227) 22 Ti Zr Hf Rf (261) 23 V Nb Ta Db (262) 24 Cr Mo W Sg (263) 25 Mn Tc Re Bh (262) 26 Fe Ru Os Hs (265) 27 Co Rh Ir Mt (266) 28 Ni Pd Pt (269) 29 Cu Ag Au (272) 30 Zn Cd Hg (277) 5 B Al Ga In Tl ? 6 C Si Ge Sn Pb (289) 7 N P As Sb Bi O S Se Te Po (210) 9 F Cl Br I At (210) 2 He Ne Ar Kr Xe Rn (222) 58 Ce Th Pr Pa (231) 60 Nd U Pm (147) 93 Np (237) 62 Sm Pu (242) 63 Eu Am (243) 64 Gd Cm (247) 65 Tb Bk (247) 66 Dy Cf (251) 67 Ho Es (254) 68 Er Fm (253) 69 Tm Md (256) 70 Yb No (254) 71 Lu Lr (257)

2 Name Part 1: Short answer (43 points) 1) Give the missing IUPAC name, chemical formula, or element name, for each of the following: NiSO 3 nickel (II) sulfite potassium oxalate K 2 (C 2 O 4 ) Al(PO 4 ) _aluminum phosphate xenon tetrafluoride _XeF 4 CoP _cobalt (III) phosphide hydrobromic acid _HBr(aq) (NH 4 )(CH 3 COO)_ammonium acetate ruthenium (III) carbonate _Ru 2 (CO 3 ) 3 2) Balance the following chemical equation that describes the formation of nitrogen from ammonia. 4 NH O 2 _2 N H 2 O 3) Write the corresponding value (to 3 sig figs) for each of the following unit conversions. a) 1.20 mmol = _1.20 x 10 9 pmol b) 14.5 m/s 2 = 52.2 km/min mmol x 10-3 mol/mmol x pmol/10-12 mol = 14.5 m/s 2 x (60s/min) 2 x km/10 3 m = 4) Perform the following calculations and report the answer with the correct number of significant figures. a) ( )/ = b) (12.3 x 4.31) 1.1 = ) Provide the symbol of an element for each category below. a) transition metal many choices, non, main group metals. b) alkaline earth metal _Be, Mg, Ca, Sr, Ba, Ra c) forms an anion with -2 charge _O, S, Se d) exists as a diatomic molecule in elemental state _N, O, F, Cl, Br, I, H e) 5 th period element _Rb, Sr through I, Xe f) forms an ionic compound with formula XCl 2 Mg, Ca, Sr, Ba (where X is the element) 6) a) An ion of 55 Mn +4 is composed of _25 protons, 30 neutrons, and _21 electrons. b) The mass of one atom of manganese is amu c) The mass of one mole of manganese atoms is g 7) Give, as appropriate, the base SI unit name and symbol associated with each quantity below. a) Temperature Kelvin b) amount of substance mole c) Energy _Joule K mol J

3 8) Thyroxine, a thyroid hormone, has the chemical formula, I 4 C 15 H 10 O 4 N. What is the molar mass of this hormone? Let T be the abbreviation for thyroxine 1 mol T x 4 mol I/1 mol T x g/mol = g 1 mol T x 15 mol C/1 mol T x g/mol = g 1 mol T x 10 mol H/1 mol T x g/mol = g 1 mol T x 4 mol O/1 mol T x g/mol = g 1 mol T x 1 mol N/1 mol T x g/mol = g g/mol T 9) For each of the following, circle the best answer. Rubidium is a(n) ( element compound homogenous mixture ) Glyceraldehyde (C 3 H 6 O 3 ) is a(n) ( element compound homogenous mixture ) Tap water is a ( substance homogenous mixture heterogenous mixture ) 10) Consider the following equation showing the formation of a precipitate from combining solutions of potassium chromate and barium nitrate. In the space below, provide the ionic equation and the net ionic equation. K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) 2 KNO 3 (aq) + BaCrO 4 (s) ionic equation: 2 K + (aq) + CrO 4 2- (aq) + Ba 2+ (aq) + 2 NO 3 - (aq) 2 K + (aq) + 2 NO 3 - (aq) + BaCrO 4 (s) net ionic equation: CrO 4 2- (aq) + Ba 2+ (aq) BaCrO 4 (s) 11) In the space following each compound, indicate if it would be soluble (S) or insoluble (I) in water. a) KS 2 S b) MgS S c) CuSO 4 S d) Fe(OH) 3 I e) Co(ClO 3 ) 2 S f) AgCH 3 COO S Part 2: Multiple Choice (18 points) 1) Which one of the following represents the compounds which are ionic? i. Cl 2 S 3 ii. NaCl iii. PCl 3 iv. Ba(BrO 3 ) 2 v. TiO 4 vi XeO 2 vii. NH 4 NO 3 a. iv, vii, c. i, iii, v, vi e. i, ii, iii, iv, vii b. ii, iv, vii d. ii, iv,v, vii f. all are ionic 2) Which of the elements, Na, Al, In, Fe, Pb, Ag, form more than one type of cation? a) Fe, Ag b) Fe, Pb c) Al, In, Pb d) Fe, In, Pb e) Fe, Pb, In, Ag f) Na, Fe, Pb 3) Which element would have chemical properties most similar to Ca? a) Na b) Ba c) Mn d) Al e) Mo f) H

4 4) The formula of the compound formed between Mg 2+ and N 3- is a) Mg 2 N 3 b) MgN 1.5 c) Mg(N 2 ) 3 d) Mg 3 N 2 e) Mg 1.5 N 5) The mass of 1.5 mole Ca is a) 60.1 g b) 26.7 g c) 30.0 g d) 13.3 g e) 40.1 g f) 4.01 g 6) Which of the following, in water, would be strong electrolytes? i) HNO 3 ii) NaOH iii) HCH 3 COO iv) H 2 SO 4 v) H 3 PO 4 vi) NH 3 a. i, iv c. ii, vi e. vi b. i, iii, iv, v d. i, ii, iv f. ii, iv Part 3: Problems (39 points) 1) Consider the following compounds. The molar mass of each compound is parenthesis following the formula. 6 pts 1) NH 3 (17.03 g/mol) 2) N 2 O (44.02 g/mol) 3) CH(NH 2 ) 3 (61.1 g/mol) a) Consider 1.0 mole of each compound, which would have the largest mass of N? CH(NH 2 ) 3 1 mol NH 3 x 1 mol N/1 mol NH 3 x g/mol N = 14.00g 1 mol N 2 O x 2 mol N/1 mol N 2 O x g/mol N = 28.00g 1 mol CH(NH 2 ) 3 x 3 mol N/1 mol CH(NH 2 ) 3 x g/mol N = g b) Consider 1.0 mole of each compound, which would have the largest number of moles of N? _ CH(NH 2 ) 3 See above, CH(NH 2 ) 3 has 3 mol N/mol compound. c) Which compound has the largest mass percent N? NH 3 Using mass values of N in one mole from a and the given molar masses. % in NH 3 (14 g N/17.03 g) x 100 = 82.2 % % in N 2 O (28 g N/44.02 g) x 100 = 63.6 % % in CH(NH 2 ) 3 (42 g N/61.1 g) x 100 = 68.7 % 2) a) What is the mass, in grams, of 1.00 gallon of pentane (C 5 H 12 )? The density of pentane is g/ml 6 pts Convert volume units to ml and use density to obtain mass 1.00 gal x 4 qt/gal x 1 L/1.057 qt x ml/10-3 L x g/ml = 2,370 g b) The reaction of pentane with O 2 produces CO 2 and H 2 O. How many moles of O 2 are required to react stoichiometrically with 50.0 g pentane? (pentane molar mass = g/mol) 2C 5 H O 2 10 CO H 2 O path is mass pentane to moles pentane to moles O g pentane x mol pentane/72.15 mol x 16 mol O 2 /2 mol pentane = 5.54 mol O 2 c) If 50.0 g pentane reacts in the presence of excess O 2, how many molecules of CO 2 will be formed. path is mass pentane to moles pentane to moles CO 2 to molecules CO g pentane x mol pentane/72.15 mol x 10 mol CO 2 /2 mol pentane x x molecules/mol CO 2 = x molecules CO 2

5 3) Consider strontium phosphate (Sr 3 (PO 4 ) 2 ), molar mass = g/mol). 5 pts a) How many moles of O are there in 3 moles of strontium phosphate? 3 mol strontium phosphate x 8 mol O/mol strontium phosphate = 24 mol O b) What is the mass percent O in strontium phosphate? 1 mol strontium phosphate x 8 mol O/mol strontium phosphate x g/mol O = 126 g O/mol Sr 3 (PO 4 ) 2 (126 g O/452.8 g Sr 3 (PO 4 ) 2 x 100 = 28.3 % c) What is the mass of P in 15.0 g strontium phosphate? path is mass Sr 3 (PO 4 ) 2 to mol Sr 3 (PO 4 ) 2 to mol P to mass P 15.0 g x mol Sr 3 (PO 4 ) 2 /452.8 g x 2 mol P/1 mol Sr 3 (PO 4 ) 2 x g/mol P = 2.05 g 4) Magnesium has three isotopes with 12, 13, and 14 neutrons respectively. 5 pts a) Which of the following best describes the number of protons in magnesium? i) all isotopes have the same number, 12, of protons ii) the number of protons in each isotope equals the number of neutrons iii) the number of protons cannot be determined from the information given. Magnesium Isotope (neutrons) b) Complete the columns for mass number and atomic number in the table below. c) Using the information in the 4 th and 5 th columms below and the fact that the average atomic mass of magnesium is amu, calculate the mass of the magnesium atom that has 14 neutrons. Mass number Atomic number Isotopic abundance (%) Atomic mass (amu) (0.7899)(24.05) + (0.1000)(24.91) + (0.1101)(x) = (0.1101)(x) = (0.1101)(x) = x = ) Sodium chloride is a very soluble salt. 5 pts a) What volume of a M solution would contain moles? mol x mol/l = L = 78.2 ml b) What mass of NaCl is required to prepare 150 ml of a M solution? path is vol solution to mol NaCl to mass NaCl Convert ml to L L x mo/l x g/mol = 5.04 g c) What would be the final concentration if 50 ml of a M solution of NaCl was added to 250 ml H 2 O? This is a dilution problem and C 1 V 1 = C 2 V 2. Convert volumes to L or note that units will cancel. (0.575 M)(0.050 L) = (0.300 L)( x M) x = M

6 6) Hopamoxin is a naturally occurring compound containing C, H, and S. A sample weighing 0.35 g was burned in oxygen giving mol CO 2. A second sample also weighing 0.35 g was subjected to a series of reactions that transformed all of the sulfur in the compound to sulfate and was found as BaSO 4. The amount of BaSO 4 formed from 0.35 g hopamoxin was g. The molecular mass of hopamoxin is g/mol. 6 pts a) What is the empirical formula of hopamoxin C 2 H 5 S with a mass of 61.1 g Determine C from CO 2 ; Determine S from BaSO 4 ; Determine H from mass sum b) What is the molecular formula of hopamoxin? Determine n and multiply subscripts g CO 2 x mol CO 2 /44.0 g x 1 mol C/1 mol CO 2 x g /mol C = g C g BaSO 4 x mol BaSO 4 /233.4 g x 1 mol S/1 mol BaSO 4 x g/mol S = g 0.35 g sample = mass C + mass S + mass H = g g + x x = g Convert all mass values to moles: g C x mol/12.00 g = mol C / = g S x mol/32.07 g = mol S / = g H x mol/1.008 = mol H / = 5 Divide by smallest number of moles, which in this case is Use these numbers as molecule subscripts to get the empirical formula To get the molecular formula, divide the molecular mass by the empirical mass to get n. Multiply subscripts by n /61.1 = n Molecular formula is C 4 H 10 S 2 7) Nylon (H 2 C 6 H 8 O 4, molar mass g/mol) is produced by a controlled reaction (see below) between cyclohexane (C 6 H 12, molar mass g/mol) and oxygen. 2 C 6 H 12 (l) + 5 O 2 (g) 2 H 2 C 6 H 8 O 4 (l) + 2 H 2 O (l) a) If 25.0 g of cyclohexane and 35.0 g of oxygen are reacted, what mass (in g) of nylon can be formed? Determine how much product can be formed from each reagent if the other is present in excess g C 6 H 12 x mol C 6 H 12 /84.12g x 2 mol nylon/2 mol C 6 H 12 = 0.29 mol nylon 35.0 g O 2 x mol O 2 /32.00 g x 2 mol nylon/5 mol O 2 = mol nylon 0.29 mol nylon is lesser and C 6 H 12 is the limiting reagent mol x g/mol nylon = 43.4 g nylon can form b) What mass of the excess reagent is left after the reaction is complete. Prepare a reaction table The table must be in moles. Convert 25.0 g C 6 H 12 to moles and 35.0 g O 2 to moles, using g and g as molar masses respectively. C 6 H 12 O 2 nylon H 2 O initial change final Information above indicates C 6 H 12 is the limiting reagent and will be used up. Use reaction stoichiometry to complete table mol C 6 H 12 x 5 mol O 2 /2 mol C 6 H 12 = mol O 2 used 0.29 mol C 6 H 12 x 2 mol nylon/2 mol C 6 H 12 = 0.29 mol nylon made c) What is the percent yield if 31.7 g nylon are formed? (actual yield/theoretical yield) x 100 (31.7 g/ 43.4 g) x 100 = 73.0 %