Chemistry 112 Sample Exam (1st Hour) Harwood KEY

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1 1. Phenylalanine is one of the compounds found in low calorie sweeteners. It contains C, H, N, and O. Analysis of a g sample of phenylalanine gives g C, g N, and g O. Calculate the empirical formula of phenylalanine. mass of H = g g g g = g H moles C = ( g)(1 mol / g) = mol C moles N = ( g)(1 mol / g) = mol N moles O = ( g)(1 mol / g) = mol O moles H = ( g)(1 mol / g) = mol H C ==> mol / mol = 9 N ==> mol / mol = 1 O ==> mol / mol = 2 H ==> mol / mol = 11 C 9 H 11 NO 2 2. Give the chemical formula for each of the following: a. xenon difluoride ef 2 b. calcium hydrogen sulfite Ca(HSO 3 ) 2 c. dinitrogen tetraoxide N 2 O 4 3. Consider the compound, BaCrO 4, a. Name this compound barium chromate b. Determine its molar mass g / mol c. Determine the %Ba in this compound. %Ba = ( / ) 100% = % d. How many oxygen atoms are there in 0.50 mol of BaCrO 4 molecules # O atoms = (0.50 mol)( molecules / 1 mol)(4 O atoms / 1 molecule) = O atoms e. How many grams of BaCrO 4 are in 0.50 mol? mass = (0.50 mol)( g / 1 mol) = g 4. How many protons, neutrons, and electrons are in each of the following atoms? 15 a. N 7 protons, 8 neutrons, 7 electrons b. c Co I 27 protons, 33 neutrons, 27 electrons 53 protons, 78 neutrons, 53 electrons 5. Identify the following elements: a. Mg 24

2 12 b. c Ni Pt 6. a. What is the maximum theoretical yield of CO 2 (in moles) which can be obtained from a reaction mixture of g CH 3 CH 2 OH ( g/mol) and g of O 2 ( g/mol) in the following reaction? CH 3 CH 2 OH + 3 O 2 2 CO H 2 O mol CH 3 CH 2 OH = (52.45 g)(1 mol / g) = mol mol O 2 = (28.73 g)(1 mol / g) = mol mol CO 2 from CH 3 CH 2 OH = ( mol)(2 mol CO 2 /1 mol CH 3 CH 2 OH) = mol CO 2 mol CO 2 from O 2 = ( mol)(2 mol CO 2 /3 mol O 2 ) = mol CO 2 O 2 is limiting reactant mol CO 2 = mol CO 2 b. Identify the reactant that is in excess and calculate the number of grams of that reactant left over at the end of the reaction. CH 3 CH 2 OH is in excess. mol CH 3 CH 2 OH reacted = ( mol O 2 )(1 mol CH 3 CH 2 OH / 3 mol O 2 ) = mol CH 3 CH 2 OH reacted mol CH 3 CH 2 OH left over = mol mol = mol mass CH 3 CH 2 OH left over = ( mol)( g / 1 mol) = g 7. Element react with element Y to give a product containing 3+ ions and Y 2 ions. a. Is element likely to be a metal or a nonmetal? Explain. metal because it forms a cation b. Is element Y likely to be a metal or a nonmetal? Explain. nonmetal because it forms an anion c. What is the formula of the product? 2 Y 3 d. What groups of the periodic table are elements and Y likely to be in? - Group 3a, Y - Group 6a 8. Name the following compounds: a. Na 2 Cr 2 O 7 sodium dichromate b. Cl 2 O 7 dichlorine heptoxide

3 c. NaNO 2 sodium nitrite d. LiHSO 3 lithium hydrogen sulfite e. HClO 3 chloric acid f. HBr hydrobromic acid 9. Balance the following equations: a. NaHCO 3 + SrCl 2 + NaOH SrCO NaCl + H 2 O b. 3 Ca(OH) H 3 PO 4 6 H 2 O + Ca 3 (PO 4 ) 2 c. 2 AgNO 3 + H 2 SO 4 Ag 2 SO HNO 3 d. 16 Cr + 3 S 8 8 Cr 2 S 3 10.Indicate which of the following compounds are likely to exist, based on your understanding of the periodic table and the formulas for polyatomic ions: Na 3 SO 4 no, must be Na 2 SO 4 to be a neutral compound (Na forms +1 ion and sulfate is 2 ion) BaO KF 3 CaI 2 yes, Ba forms +2 ion and O forms 2 ion no, must be KF to be a neutral compound (K forms +1 ion and F forms 1 ion) yes, Ca forms +2 ion and I forms 1 ion 11.Ethylene gas, C 2 H 4, reacts with water at high temperature to yield ethyl alcohol, C 2 H 6 O. a. How many grams of ethylene are needed to react with mol of water? How many grams of ethyl alcohol result? C 2 H 4 + H 2 O C 2 H 6 O mass C 2 H 4 = (0.133 mol H 2 O)(1 mol C 2 H 4 /1 mol H 2 O)( g/1 mol) = 3.73 g C 2 H 4 mass C 2 H 6 O = (0.133 mol H 2 O)(1 mol C 2 H 6 O/1 mol H 2 O)( g/1 mol) = 6.13 g C 2 H 6 O b. How many grams of water are needed to react with mol of ethylene? How many grams of ethyl alcohol result? mass H 2 O = (0.371 mol C 2 H 4 )(1 mol H 2 O / 1 mol C 2 H 4 )( g/1 mol) = 6.68 g C 2 H 4 mass C 2 H 6 O = (0.371 mol C 2 H 4 )(1 C 2 H 6 O/1 mol C 2 H 4 )( g/1 mol) = 17.1 g C 2 H 6 O 12.Acetic acid (CH 3 CO 2 H) reacts with isopentyl alcohol (C 5 H 12 O) to yield water and isopentyl acetate (C 7 H 14 O 2 ), a fragrant substance with the odor of bananas. If the yield from the reaction of acetic acid with isopentyl alcohol is 45%, how many grams of isopentyl acetate are actually formed from 3.58 g of acetic acid and 4.75 g of isopentyl alcohol?

4 CH 3 CO 2 H + C 5 H 12 O H 2 O + C 7 H 14 O 2 mol acetic acid = (3.58 g)(1 mol / g) = mol acetic acid mol isopentyl alcohol = (4.78 g)(1 mol / g) = mol alcohol isopentyl alcohol is limiting reactant mass acetate = ( mol)( g / 1 mol) = g isopentyl acetate actual yield = (0.45)( g) = 3.2 g isopentyl acetate 13.Element, a member of group 5A, forms two chlorides, Cl 3 and Cl 5. Reaction of an excess of Cl 2 with g of Cl 3 yields g of Cl 5. What is the atomic mass and the identity of the element? (Possible bonus-type question.) Cl 3 + Cl 2 Cl 5 mass of Cl 2 = g g = g Cl 2 mol Cl 2 = (4.504 g)(1 mol / g) = mol Cl 2 from balanced equation, 1 mol Cl 2 reacts with 1 mol Cl 3 to give 1 mol Cl 5 so if mol Cl 2 reacts then mol of Cl 3 must react with it molar mass of Cl 3 = g / mol = g/mol atomic mass of = g/mol 3( g/mol) = g/mol = Phosphorus 14.Predict whether a precipitation reaction will occur when aqueous solutions of the following substances are mixed: a. MnCl 2 + Na 2 S yes, MnS will precipitate b. HNO 3 + CuSO 4 no c. (NH 4 ) 2 SO 4 + NiCl 2 no 15.Write balanced equations for the following reactions: a. Aqueous perchloric acid is neutralized by aqueous calcium hydroxide. 2 HClO 4(aq) + Ca(OH) 2(aq) Ca(ClO 4 ) 2(aq) + 2 H 2 O (l) b. Aqueous hydrochloric acid reacts with magnesium metal. 2 HCl (aq) + Mg (s) MgCl 2(aq) + H 2(g) 16.In the following reaction, which element is oxidized and which is reduced? Cl 2(g) + 2 NaBr (aq) Br 2(aq) + 2 NaCl (aq) Cl is reduced (0 1); Br is oxidized ( 1 0) 17.A sample of 1.50 g of lead(ii) nitrate is mixed with 125 ml of M sodium sulfate solution. a. Write a balanced equation for the reaction. Pb(NO 3 ) 2(s) + Na 2 SO 4 (aq) PbSO 4(s) + 2 NaNO 3(aq) b. What is the limiting reactant in the reaction? mol Pb(NO 3 ) 2 = (1.50 g)(1 mol / g) = mol Pb(NO 3 ) 2

5 mol Na 2 SO 4 = (0.125 L)(0.100 mol / 1 L) = mol Na 2 SO 4 mol PbSO 4 from Pb(NO 3 ) 2 = ( mol Pb(NO 3 ) 2 )(1 mol PbSO 4 /1 mol Pb(NO 3 ) 2 ) = mol PbSO 4 mol PbSO 4 from Na 2 SO 4 = ( mol Na 2 SO 4 )(1 mol PbSO 4 /1 mol Na 2 SO 4 ) = mol PbSO 4 lead(ii) nitrate is the limiting reactant b. Calculate the concentrations of all ions that remain in the solution after reaction. no Pb 2+ is left in solution mol SO 4 2 removed from solution in the form of solid PbSO4 SO 4 2 left in solution = mol mol = mol SO4 2 [SO 4 2 ] = mol / L = M [Na+] = [( mol Na 2 SO 4 )(2 mol Na+/1 mol Na 2 SO 4 )] / L = M [NO 3 ]=[( mol Pb(NO3 ) 2 )(2 mol NO 3 /1 mol Pb(NO3 ) 2 )]/0.125 L = M

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