Chem 107 - Hughbanks Exam 1 - Spring 2014

Name (Print) Key UIN # Section 504 Exam 1, Version # A On the last page of this exam, you’ve been given a periodic table and some physical constants. You’ll probably want to tear that page off to use during the exam – you don’t need to turn it in with the rest of the exam. The exam contains 9 problems, with 4 numbered pages. You have the full 50 minutes to complete the exam. Please show ALL your work as clearly as possible – this will help us award you partial credit if appropriate. Even correct answers without supporting work may not receive credit. You may use an approved calculator for the exam, one without extensive programmable capabilities or the ability to store alphanumeric information. Print your name above, provide your UIN number, and sign the honor code statement below.

On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam.

SIGNATURE:

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(1) (12 points, 3 pts each) For each of the following, enter the answer in the line provided. (No difficult calculations are necessary for these, so think first!)

(a) How many molecules are present in one mole of ozone (O3)?

Ans. (a) 6.022 × 1023

(b) How many moles of fluorine atoms are present in seven moles of CF4?

Ans. (b) 28 _

(c) How many moles of ammonium ions (NH4+) are present 3 L of 1.0 M

aqueous solution of (NH4)2SO4?

Ans. (c) 6 _

(d) What is the approximate mass (to the nearest gram) of 4 × 1023 F atoms?

Ans. (d) 13 _

(2) (12 pts, 4 pts each) Imagine that we had a balance that could compare the masses of small numbers of atoms of different elements. The figures below show the (hypothetical) results of 2 such experiments. The letters beside each balance pan identify the elements being weighed. (These are NOT actual elements! Your answer should be based ONLY on these pictures.)

Please circle your choice for each question below: (a) Of elements A and B, which has the greater molar mass?

element A element B neither, they’re equal

(b) Of elements C and D, which has more atoms per gram?

element C element D neither, they’re equal

(c) Of elements A and B, which has more atoms per mole?

element A element B neither, they’re equal

A BC

D

(For grading)

Scores 1 /12 2 /12 3 /10 4 /8 5 /10 6 /10 7 /15 8 /12 9 /12 Tot. /100

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(3) (10 points) In the first row of blanks beneath each of the following compounds, classify them as follows: weak acid (wa), strong acid (sa), weak base (wb), strong base (sb), or as a salt (s) that is neither acid or basic. In the second row of blanks, classify the same compounds as a weak electrolyte (we), strong electrolyte (se), or nonelectrolyte (ne).

HClO4 NH3 RbOH KCl CH3COOH

sa wb sb s wa

se we se se we

(4) (8 points) The following full sets of potential compounds are made up of common cations and anions. Which full sets are most likely to actually exist? (All 3 compounds in a set must be sensible to qualify.) There may be more than one correct choice; enter ALL the correct letters, in alphabetical order.

(a) Rb(SO4)2, CaCO3, ZnClO4 (b) (NH4)2PO4, NaCO3, KNO3 (c) Ca3(PO4)2, Ag2SO4, NH4NO3

(d) K2SO4, KClO4, Ca(OH)2 (e) (NH4)2PO4, K2CO3, Zn3(PO4)2

Ans. 4 c, d _

(5) (10 points) Calcium hydroxide can be dissolved in phosphoric acid to form calcium phosphate and water.

(i) (5 pts) Balance the equation below by filling in the blanks with coefficients:

_ 3 _ Ca(OH)2 + __ 2 __ H3PO4 → _1_ Ca3(PO4)2 + _6_ H2O

(ii) (5 pts) What volume (in mL) of 1.32 M H3PO4 is needed to react with 12.0 g of Ca(OH)2? (Write the letter of the correct answer below.)

(a) 0.184 mL (b) 81.8 mL (c) 184.1 mL (d) 122.7 mL (e) 0.082 mL

Ans. 5 (ii) __ b__

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Questions 6 - 9 – show all work (48 pts) (6) (10 pts) Adipic acid is used in the production of nylon, so it is manufactured in large quantities.

One method for the industrial synthesis of adipic acid is the reaction of cyclohexane with oxygen.

(a) Give the molecular formulas for (2 pts) cyclohexane: ____C6H12____ (3 pts) adipic acid: ___ C6H10O4___

(b) (5 pts) Balance the skeleton equation above by putting in the smallest possible whole number coefficients in the four boxes provided.

(7) (14 points) Shown below is the molecular structure of nepetalactone, which is a cat attractant first

isolated by steam distillation from catnip and also present in the wood of tartarian honeysuckle, shavings of which are often used in cat toys.

nepetalactone

(a) (5 pts) Give the molecular formula for nepetalactone.

C10H14O2

(b) (4 pts) Assuming your answer is correct; give the molar mass of nepetalactone.

166.21 g/mol

(c) (5 pts) How many carbon atoms are there in 1.0 g of nepetalactone?

(1 g/166.21 mol/g)(10 C atoms/mol) = 3.62 × 1022 C atoms/g

O

OHO

OH+ O2

cyclohexane adipic acid

+ H2O2 5 22

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(8) (12 points) A mixture of methane (CH4) and propane (C3H8) has a total mass of 59.68 g. When a mixture is burned completely in excess oxygen, the CO2 and H2O products have combined mass of 285.94 g. Calculate the mass of methane in the original mixture.

The key to the problem is to realize that it is the ratio of product mass and reactant mass that matters. (This was a study problem.)

Let x = mole fraction of CH4 , then (1 – x) = mole fraction of C3H8 x CH4 + excess O2 → x CO2 + 2x H2O

(1 – x) C3H8 + excess O2 → 3(1 – x) CO2 + 4(1 – x) H2O ––––––––––––––––––––––––––––––––––––––––––––––––

x CH4 + (1 – x) C3H8 + excess O2 → (3 – 2x) CO2 + (4 – 2x) H2O

285.94 g

59.68 g= 4.791 =

total product mass

methane + propane mass=44.01 g/mol( ) 3 − 2x( ) + 18.02 g/mol( ) 4 − 2x( )

16.04 g/mol( ) x + 44.09 g/mol( ) 1 − x( )

solving for x, x = 0.690. Now you need to multiply the mass fraction of methane by the total original mass,

original methane mass =16.04 g/mol( ) 0.69 mol( )

16.04 g/mol( ) 0.69 mol( ) + 44.09 g/mol( ) 0.31 mol( ) 59.68 g( ) = 26.70 g

(9) (12 points) There are many isotopes of uranium, but only two are present in greater than trace quantities on earth: 235U and 238U. The data in the table below given can be used to calculate the molar mass of uranium, which is 238.0289 g/mol. The molar mass of uranium on earth hasn’t always had this value, however, because both of these isotopes have been very slowly decaying since the earth was formed about 4.5 billion years ago. The % abundances of the two isotopes have therefore changed since the earth was formed. Measurements of the rates of decay of the two isotopes allows us to calculate the how much of each isotope has decayed: 50.25% of the 238U and 98.81% of the 235U that were present 4.5 billion years have decayed.

Isotope Isotopic Molar Mass (g/mol) Current Abundance (%)

235U 235.043993 0.727 238U 238.050788 99.273

Compute the molar mass of uranium on earth 4.5 billion years ago.

If today you have 0.727 mol of 235U , 4.5 billion years ago you would have had

0.727 mol( ) 100%

100% − 98.81%

⎛⎝⎜

⎞⎠⎟= 61.09 mol

If today you have 99.273 mol of 238U , 4.5 billion years ago you would have had

99.273 mol( ) 100%

100% − 50.25%

⎛⎝⎜

⎞⎠⎟= 199.54 mol

The abundances 4.5 billion years ago were

235U : 61.09

61.09 + 199.54⎛⎝⎜

⎞⎠⎟= 0.2344 (23.44%) 238U :

199.54

61.09 + 199.54⎛⎝⎜

⎞⎠⎟= 0.7656 (76.56%)

Molar mass 4.5 billion years ago = 235.043993( ) 0.2344( ) + 238.050788( ) 0.7656( ) = 237.35 g/mol