CHEM 35

General Chemistry

EXAM #1

September 20, 2000

 

Name: Anne Serkey

 

SSN:

 

Lab T.A.:

 

INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, shall all of your work -- HOW you arrived at a particular answer is MORE important than the answer itself! Circle your final answer to numerical questions.

 

The entire exam is worth a total of 150 points. Attached are a periodic table and a formula sheet jam-packed with useful stuff!

 

Good Luck!

Page

Possible Points

Points Earned

2

15

 

3

26

 

4

26

 

5

17

 

6

24

 

7

24

 

8

18

 

TOTAL:

150

 

 


1.     (3 pts each) Carry out the following operations, and express the answer with the appropriate number of significant figures:

 

a.      1.24056 + 75.8 =

 

1.24056

+75.8

77.04056 -> Round to tenths place: 77.0

 

b.     320.55 (6104.5/2.3) =

 

320.55

-2654.1304

-2333.580 -> round to HUNDREDS place: -2300 or -2.3 x 103

 

c.      (0.0045 x 20,000.0) + (2813 x 12) =

 

90.000

+33756.000

33846.000 -> round to TEN THOUSANDS place: 34000 or 3.4 x 104

 

2.     (3 pts each) The all-time record lowest temperature ever recorded on this planet is 57.63 oC (recorded on July 21, 1983 in Vostok, the Russian Antarctic station).

 

a.     Express this temperature in oF.

 

oF = (9/5)oC + 32

= (9/5)(-57.63) + 32 = -71.734 oF -71.73 oF

 

b.     Express this temperature in Kelvins (K).

 

K = oC + 273.15

= -57.63 + 273.15 = 215.52 K 215.52 K

 


 

3.     (10 pts) A 26.27-g sample of a solid is placed in a flask. Toluene, in which the solid is insoluble, is added to the flask so that the volume of the solid and liquid together is 50.00 mL. The solid and liquid toluene together weigh 52.65 g. The density of toluene at the temperature of the experiment is 0.864 g/mL. What is the density of the solid?

 

 

solid + toluene = 52.65 g

solid = 26.27 g

Mass of toluene = 26.38 g toluene

 

 

26.38 g toluene x 1 ml toluene = 30.53241 mL toluene

0.864 g toluene

 

 

solid + toluene = 50.00 mL

toluene = 30.53241 mL

solid = 19.46759 mL

 

 

density of the solid = 26.27 g

19.46759 mL

 

= 1.34942 g/mL 1.35 g/mL

 

4.     (16 pts) Fill in the gaps in the following table:

 

Symbol

52Cr3+

107Ag+

75As3-

Protons

24

47

33

Neutrons

28

60

42

Electrons

21

46

36

Net Charge

3+

1+

3-

 


 

5.     (2 pts each) From this list of elements: Ar, H, Ga, Al, Ca, Br, Ge, K, O; pick the one that best fits each of the following descriptions. You may use each element only ONCE.

 

a.     an alkali metal: K

b.     an alkaline earth metal: Ca

c.     a noble gas: Ar

d.     a halogen: Br

e.     a metalloid: Ge

f.      a nonmetal listed in group IA: H

g.     a metal that forms a 3+ ion: Al

h.     a nonmetal that forms a 2- ion: O

i.       an element that resembles aluminum: Ga

 

6.     (1 pt each) Give the atomic symbol for the following elements:

 

a.     calcium: Ca

b.     sodium: Na

c.     mercury: Hg

d.     lead: Pb

 

 

7.     (1 pt each) Give the name of the element for the following atomic symbols:

 

a.     Cu: Copper

b.     K: Potassium

c.     H: Hydrogen

d.     Ag: Silver


 

8.     (3 pts each) Write a complete, balanced chemical equation for each of the following reactions:

 

a.     Solid calcium carbide (CaC2) reacts with water to form an aqueous solution of calcium hydroxide and acetylene gas (C2H2).

 

 

CaC2(s) + 2H2O(l) Ca(OH)2(aq) + C2H2(g)

 

 

b.     HBr + F2 HF + Br2

 

 

2HBr + F2 2HF + Br2

 

 

c.     CH4 + O2 CO2 + H2O

 

 

CH4 + 2O2 CO2 + 2H2O

 

 

9.     (8 pts) The element magnesium consists of three naturally occurring isotopes with masses 23.98504, 24.98584, and 25.98259 amu. The relative abundances of these three isotopes are 78.70, 10.13, and 11.17 percent, respectively. From these data, calculate the average atomic mass of magnesium.

 

23.98504(0.7870) + 24.98584(0.1013) + 25.98259(0.1117) =

 

18.87622648 + 2.531065592 + 2.9022553 = 24.3095474

 

= 24.31 amu

 

 


10. (6 pts) Show the Lewis structure for CH3Cl.

 

4 + 3 + 7 = 14 valence electrons

 

H

| _

H C Cl|

|

H

a.     (3 pts) What shape does VSEPR theory predict for this molecule?

 

Four bonding electron pairs = tetrahedral

 

b.     (3 pts) The electronegativity values for C, H and Cl are 2.55, 2.2 and 3.16, respectively. Which bond(s) is(are) the most polar?

 

For the C-H bonds: ΔEN = 2.55 2.2 = 0.35

 

For the C-Cl bond: ΔEN = 3.16 2.55 = 0.61

 

C-Cl bond has greatest ΔEN, so it is the most polar.

 

c.     (3 pts) Does this molecule have a net dipole? If so, indicate the positive and negative regions, on the Lewis structure, by the symbols d+ and d-, respectively.

 

Yes, there is a net dipole as the C-Cl bond is more polar than the C-H bonds. Since Cl is the most EN, the d- will be on the Cl. Since H is the least EN, the d+ will be on the three hydrogens.

 

 

11. (9 pts) Draw the Lewis and VSEPR structures for SF6.

 

S = 6 valence electrons

F = 7 x6 = 42 valence electrons

 

42 + 6 = 48 valence electrons

 

F F

| / 3 nonbonded electron pairs around

F - S F each F = 36 electrons

/ | +

F F 6 S-F bonds = 12 electrons = 48 e-

 

The VSEPR structure for SIX electron pairs is Octahedral

 

 

12. Ibuprofen, a potent headache remedy, has a molar mass of about 206 grams and has been determined to be 75.69 % C, 8.80 % H, and 15.51 % O by mass.

 

a.     (12 pts) Determine the empirical formula for ibuprofen.

 

75.69 g C x 1 mole C = 6.30188 mol C

12.0107 g C

 

8.80 g H x 1 mole H = 8.730678 mol H

1.00794 g H

 

15.51 g O x 1 mole O = 0.969411 mol O

15.9994 g O

 

C: 6.30188 mol = 6.50 x 2 = 13

0.969411 mol

 

H: 8.730678 mol = 9.00 x 2 = 18

0.969411 mol

 

O: 0.969411 mol = 1.00 x 2 = 2

0.969411 mol

 

So, C13H18O2

 

b.     (6 pts) Determine the molecular formula for ibuprofen.

 

13 C = 13 x 12 = 156

18 H = 18 x 1 = 18

2 O = 2 x 16 = 32

Empirical Mass = 206 g/mol = molar mass

 

Empirical Formula = Molecular Formula = C13H18O2

 

c.     (6 pts) Calculate the molar molecular mass of ibuprofen to the nearest mg/mol.

 

13 C = 13 x 12.0107 = 156.1391

18 H = 18 x 1.00794 = 18.1429

2 O = 2 x 15.9994 = 32.9988

206.2808 g/mol 206.281 g/mol

 

(206,281 mg/mol)

 

 

13. (9 pts each) A tablet of Advil contains 200. mg of ibuprofen.

 

a.     How many molecules of ibuprofen are in a single Advil tablet?

 

200. mg Ibu x 1 gram x 1 mol Ibu x 6.02214 x 1023 molecules =

1000 mg 206.281 g Ibu 1 mol Ibuprofen

 

= 5.8387733 x 1020 molecules

 

= 5.84 x 1020 molecules Ibuprofen

 

 

b.     How many moles of oxygen (from the ibuprofen) are there in a single Advil tablet?

 

200. mg Ibu x 1 gram x 1 mol Ibu x 2 moles O =

1000 mg 206.281 g Ibu mol Ibuprofen

 

= 1.93919249 x 10-3 mol O

 

= 1.94 x 10-3 mol O

 


EXTRA CREDIT! 10 points

 

Weve been claiming all semester that, for atomic- and molecular-sized systems, electromagnetic forces between and with atoms are much greater than gravitational forces between atoms and sub-atomic particles. Ok, lets demonstrate that with some calculations!

 

Calculate:

 

1)    The magnitude of the coulombic attractive force between the proton in the nucleus of a hydrogen atom and the electron whirling around it. For the purposes of this calculation, assume that the distance separating the proton and the electron is 5 .

2)    The magnitude of the gravitational attractive force between the same two particles in a hydrogen atom.

 

Based on your calculations, is our assumption justified?

 

1)    Coulombic attraction:

 

F = 1 x q1q2 e = 8.85 x 10-12 C2-N-1-m-2

4peo r2

 

= (1.60 x 10-19C)(-1.60 x 10-19C) =

4(3.14159)( 8.85 x 10-12 C2-N-1-m-2)(5 x 10-10 m)2

 

= -9.20761 x 10-10 N = -9.21 x 10-10 N

 

2)    Gravitational attraction:

 

F = - Gm1m2 G = 6.67 x 10-11 N-m2-kg-2

r2

 

mass electron:

5.4858 x 10-4 amu x 1.66054 x 10-24g x 1 kg = 9.10939 x 10-31 kg

1 amu 1000 g

 

mass proton:

1.007276 amu amu x 1.66054 x 10-24g x 1 kg = 1.67262 x 10-27 kg

1 amu 1000 g

 

 

F = -(6.67 x 10-11 N-m2-kg-2)( 9.10939 x 10-31 kg)( 1.67262 x 10-27 kg)

(5 x 10-10 m)2

 

F = -4.065116 x 10-49 N = -4.07 x 10-49 N