CH141, Fall 2017 Exam 1 10/3/2017. Section (please circle): A(Rice) B(McKinney)

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1 C141, Fall 2017 Exam 1 10/3/2017 Name: Section (please circle): A(Rice) B(McKinney) Part I. Short Answer: Answer 8 of the following 9 questions. Clearly indicate which question you do NOT want counted, otherwise the first 8 answered questions will be graded. 4 points each (no partial credit given for this section). 1. If identical quantities (by moles) of barium chloride and potassium sulfate are allowed to react forming barium sulfate and potassium chloride, which reactant is the limiting reagent? A) barium chloride B) potassium sulfate C) barium sulfate D) potassium chloride E) the reactants are at stoichiometric equivalence Complete the following table (you must do both parts a and b of problems 2-4: NAME FORMULA 2a. pentaphosphorus hexoxide P 5O 6 2b. sodium hydrogen phosphate Na 2PO 4 3a. hydriodic acid I 3b. mercury (II) oxide go 4a. cobalt (II) nitride Co 3N 2 4b. sulfuric acid 2SO 4 5. Which of the following is a derived unit? A) kilogram B) liter C) Kelvin D) mole E) second 6. Which of the following is a specific illustration of the Law of Constant Composition (also known as the Law of Definite Proportions)? A) The number of oxygen atoms for each carbon atom in CO 2 is exactly twice that in CO. B) CO and CO 2 are both gases at room temperature. C) The percent oxygen in CO 2 is 72.7%, regardless of the size of the sample. D) The mass of 2 mol of CO is exactly twice the mass of 1 mol of CO. E) The density of CO 2 is 1.96 g/l, regardless of the size of the sample.

2 C141, Fall 2017 Exam 1 10/3/ Complete the following table: Symbol # of protons # of neutrons # of electrons atomic number mass number 89 Sr I If 2 mol of C (s) reacted according to the following equation, what quantity of TiCl 4 would result? 2 TiO 2 (s) + 3 C (s) + 4 Cl 2 (g) à 2 TiCl 4 (g) + CO 2 (g) + 2 CO (g) A) 2 mol B) 6 mol C) 0.67 mol D) 3 mol E) 1.33 mol 9. Complete the following arithmetic operation, giving the answer with the proper significant figures. ( ) ( ) = 2.37 x 10 5 Part II. Short Problems: Answer all 3 questions. 10 points each (you must show your work partial credit is available). 10. Acetone, the primary component of nail polish remover, has the following structural formula: O C C C (A) What is the molecular formula for acetone? C36O (B) What is the formula weight for acetone? (please round your answer to 4 significant figures) FW = g mol g mol g mol = g mol

3 C141, Fall 2017 Exam 1 10/3/ Iron has a density of 7.9 g/cm 3. What is the length in inches of one side of a cube of iron with a mass of 1.31 kg? (Note: 1.00 in = 2.54 cm) 1.31 kg cm! 7.9 g 1000 g 1 kg 1! in! 2.54! cm! = in!! in! = 2.18 in 12. Copper has two isotopes, 63 Cu and 65 Cu with exact masses of amu and amu, respectively. The atomic weight of copper is amu. What are the percent abundances of the two isotopes? (please round your answer to 4 significant figures) Fraction 63 Cu à x Fraction 65 Cu à 1 x AW Cu = x amu + 1 x amu = amu x = x = % 63 Cu 30.85% 65 Cu Part III. Problems (begins next page): Answer 2 of the following 3 questions. Clearly indicate which question you do NOT want counted for a grade, otherwise the first two answered questions will be graded. 14 points each (you must show your work partial credit is available).

4 C141, Fall 2017 Exam 1 10/3/ Oxalic acid, a natural compound used commercially in cosmetics and ceramics, contains only carbon, hydrogen, and oxygen. If g of oxalic acid is combusted in air, g carbon dioxide ( mol) and g water ( mol) result. If the formula weight of oxalic acid is g/mol, what is its molecular formula? C x yo z + a O 2 à b CO 2 + c 2O g? g g g mass O 2 = (0.501g g) 0.513g = 0.091g moles O 2 consumed = 0.091g / g/mol = mol O 2 a, b, & c: (divide mol qtys by ) à a=1, b 4, c 2 moles C on right (using CO 2 coefficient) = 4 mol C moles on right (using 2O coefficient) = 2 2 = 4 mol (since C and are only in oxalic acid on left, x=4 & y=4) moles O on right (using 2O & CO 2 coefficients) = (2 4) + (2) = 10 mol O moles O in O 2 (using coefficients) = 2 mol O z = moles of O on right moles O in O 2 = 8 Formula = C 4 4O 8 à so empirical formula = CO 2 FW of empirical formula = = Actual formula weight = = 2 FW(empirical) Therefore, C22O4 14. ydrazine (N 2 4), an important precursor molecule for industrial and pharmaceutical processes, decomposes into ammonia and nitrogen gas according to the following unbalanced reaction. 3 N 2 4 (l) à 4 N 3 (g) + 5 N 2 (g) What volume (in L) of hydrazine (FW = g/mol; d = g/ml) would be needed to yield kg of ammonia (FW = g/mol)? kg N! 1000 g N! kg N! mol N! g N! 3 mol hyd 4 mol N! g hyd mol hyd ml hyd g hyd L hyd 1000 ml hyd = L hydrazine

5 C141, Fall 2017 Exam 1 10/3/ Mercury and bromine will react with each other to produce mercury (II) bromide: g (l) + Br 2 (l) à gbr 2 (s) (A) If 10.0 g of g and 9.00 g Br 2 are reacted, what mass of which reagent (in g) is left after the reaction is complete? 10.0 g g mol g g g = mol g 9.00 g Br! mol Br! g Br! = mol Br! 1:1 stoichiometry, therefore g is limiting: 0 g g remaining Br 2: = mol remaining mol Br! g Br! mol Br! = 1.0 g Br! remaining (B) If the yield of the reaction is 86.3% of the theoretical yield, what mass of gbr 2 (in g) is produced? mol gbr! g gbr! mol gbr! = 15.5 g gbr! yielded Extra Credit (2 points each)!! What was the subject of the 2017 Nobel Prize in Physiology / Medicine? circadian rhythms What WILL BE the specific subject of the 2017 Nobel Prize in Chemistry? electron microscopy

Counting by mass: The Mole. Unit 8: Quantification of Chemical Reactions. Calculating molar mass. Particles. moles and mass. moles and particles

Counting by mass: The Mole. Unit 8: Quantification of Chemical Reactions. Calculating molar mass. Particles. moles and mass. moles and particles Unit 8: Quantification of Chemical Reactions Chapter 10: The mole Chapter 12: Stoichiometry Counting by mass: The Mole Chemists can t count individual atoms Use moles to determine amounts instead mole