CHEM 121

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 à -71.73 ^oF

b. Express this temperature in Kelvins (K).

K = ^oC + 273.15

= -57.63 + 273.15 = 215.52 K à 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 à 1.35 g/mL

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

Symbol	⁵²Cr³⁺	¹⁰⁷Ag⁺	⁷⁵As^3-
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) à 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 – C – Cl|

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 à 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

We’ve 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, let’s 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 Å.

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

4pe_o 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

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