CHM1045
Fall 1999
Dr. Michael Blaber

Name_______________________________

SS#________________________________

Exam #1 100 points total

Wednesday September 22 1999

  1. Indicate whether the following are most likely a pure substance, a homogenous mixture or a heterogenous mixture (5 points)

    2. a. Potting soil heterogenous mixture

    b. Non-iodized table salt pure substance

    c. A paperclip homogenous mixture

    d. Gatoraid homogenous mixture

    e. The gold in Fort Knox pure substance

     

  2. Indicate whether the following represent physical or chemical changes (5 points)

    3. a. Toasting a bagel until it is nice and brown Chemical

    b. Melting butter on a hot bagel Physical

    c. Burning gasoline in your car engine Chemical

    d. Recycling glass Physical

    e. Making wine Chemical

      

  3. In the following diagram, the particles from a cathode ray tube are coming directly towards you. In which direction would you expect the particles to be deflected in response to the magnetic field? (write the appropriate letter). (5 points)

    4. C

     

  4. For the following elements, write down the most likely ionic form (5 points)

    5. a. Br Br-

    b. Al Al3+

    c. Se Se2-

    d. K K+

    e. Ca Ca2+

     

  5. Write the balanced chemical equation for the combustion of hexanol (C6H14O) (5 points)

    6.  

    C6H14O + 9O2 -> 6CO2 + 7H2O
6C	       6C
14H	       14H
19O	       19O

     

  6. Magnesium can burn in air to produce Magnesium oxide. Write the balanced chemical equation for this reaction and predict two other elements which might be expected to participate in a similar type of reaction with air (10 points)

    7. 2Mg + O2 -> 2MgO

    Other group 2A elements (Be, Ca, Sr, Ba, Ra) might participate in similar reaction with O2

     

     

  7. What is the percentage by mass of C, H and O in acetic acid (i.e. vinegar: C2H4O2)? (10 points)

    8. 2 C = (2 * 12.01) = 24.02

    4 H = (4 * 1.008) = 4.032

    2 O = (2 * 16.00) = 32.00

    total = 60.05 amu

    % C = 24.02/60.05 = 0.400 or 40%

    % H = 4.032/60.05 = 0.067 or 6.7%

    % O = 32.00/60.05 = 0.533 or 53.3%

     

  8. How many grams are there in 3.98 moles of the following compound: (10 points)

2 C = (2 * 12.01) = 24.02

6 H = (6 * 1.008) = 6.048

1 O = (1 * 16.00) = 16.00

total = 46.07 amu OR 46.07 g/mole

46.07g/mole * 3.98 mole = 183g

 

  1. A certain combustible compound, containing C, H and O atoms, is 65.7% C, 12.4% H and 21.9% O by mass. The experimentally determined formula weight is 146 amu. What is the empirical and chemical formula for this compound? (10 points)

    2. 65.7g * (1mole C/12.01g) = 5.47 mole C

    12.4g * (1mole H/1.008g) = 12.3 mole H

    21.9g * (1mole O/16.0g) = 1.37 mole O

    5.47/1.37 = 3.99 mole C, which is ~ 4 mole C

    12.3/1.37 = 8.98 mole H, which is ~ 9 mole H

    1.37/1.37 = 1 mole O

    Therefore, the empirical formula is: C4H9O

    This has a formula weight of:

    4 C = (4 * 12.01) = 48.04

    9 H = (9 * 1.008) = 9.072

    1 O = (1 * 16.00) = 16.00

    (48.04 + 9.072 + 16.00) = 73.1 amu

    This is one-half of the experimentally derived formula weight of 146 amu.

    Therefore, the chemical formula is: C8H18O2

  2. Hexane has the formula C6H14 and can be combusted in the presence of oxygen. If 8.0 g of hexane is combusted in the presence of 5.2 g of O2, what is the limiting reagent? How many grams of water will be produced? (Be sure to show the balanced equation for this reaction). (20 points)

    3. 2C6H14 + 19O2 -> 12CO2 + 14H2O

    C6H14 = (6 * 12.01) + (14 * 1.008) = 86.2 amu or 86.2 g/mole

    O2 = (2 * 16.00) = 32.0 amu or 32.0 g/mole

    8.0 g C6H14 * (1 mole/86.2g) = 0.0928 moles C6H14

    5.2 g O2 * (1 mole/32.0g) = 0.163 moles O2

    0.0928 moles C6H14 * (14H2O/2C6H14) = 0.650 moles H2O if oxygen unlimited

    0.163 moles O2 * (14H2O/19O2) = 0.120 moles of H2O if hexane unlimited

    Therefore, O2 is the limiting reagent, and the most H2O we can make is 0.120 moles

    H2O = (2 * 1.008) + (1 * 16.0) = 18.01 amu or 18.01 g/mole

    18.01 g/mole * 0.120 moles H2O = 2.16 g H2O

     

  3. What is the molarity of a 3.5 liter solution of water containing 84.5 g of Ammonium Sulfate (NH4)2SO4? (5 points)

    4. 2N = (2 * 14.01) = 28.02 amu

    8H = (8 * 1.008) = 8.064 amu

    1S = (1 * 32.07) = 32.07 amu

    4O = (4 * 16.0) = 64.00 amu

    (NH4)2SO4 = (28.02 + 8.064 + 32.07 + 64.0) = 132 amu or 132.0 g/mole

    84.5 g * (1mole / 132 g) = 0.64 moles

    0.64moles / 3.5 L = 0.183 molar solution

     

     

  4. You have a stock solution of 5.0 M NaCl and some pure water. How would you use these to produce 4.2 liters of a 0.15 molar solution of NaCl (state exactly how much volume of each you would use)? (10 points)

(4.2 L * 0.15 M) = (X L * 5.0 M)

X L = (4.2 L * 0.15 M)/(5.0 M) = 0.126 L

4.2 L - 0.126 L = 4.074 L

Therefore, take 0.126 L of the 5.0 M stock NaCl solution and add to it 4.074 L of H2O.