CHM1045
Fall 1998
Dr. Michael Blaber
Name_______________________________
SS#________________________________
Exam #1 100 points total
Friday Sept 18 1998
1. Indicate whether the following are most likely a pure substance, a homogenous
mixture or a heterogenous mixture (5 points)
a.
b.
c.
d.
e.
A packet of cool aid
A packet of cool aid dissolved in water
A bronze coin
A gold bullion coin
A chocolate chip cookie made with blue M&M’s
Heterogenous mixture
Homogenous mixture
Homogenous mixture
Pure substance
Heterogenous mixture
2. Indicate whether the following represent physical or chemical changes (5 points)
a.
b.
c.
d.
e.
A sugar cube dissolving in hot water
An Alka-Seltzer tablet dissolving in water
The heating of water to produce steam
The burning of candle wax to produce soot
The cooking of eggs to produce an omelet
Physical change
Chemical change
Physical change
Chemical change
Physical change
N
3. In the following diagram, in which direction would the nucleus of a helium atom be
deflected? (the particle is coming directly at you). (5 points)
c
b
a
S
d
b
4. For the following elements, write down the most likely ionic form (5 points)
a.
b.
c.
d.
e.
Be2+
S2P3Li+
O2-
Be
S
P
Li
O
5. Write the balanced chemical equation for the combustion of heptanol (C5H12O) (5
points)
2C5H12O + 15O2 -> 10CO2 + 12H2O
6. Lithium can react with water to produce the Lithium Hydroxide and Hydrogen gas.
Write the balanced chemical equation for this reaction and predict all other elements
that you would expect to be able to react with water in a similar type of reaction. (10
points)
2Li + 2H2O -> 2LiOH + H2
(Three from Na, K, Rb, Cs, Fr)
Note: 5 points for balanced equation, 5 points for correct elements
7. What is the percentage by mass of C, H and O in ethanol (C2H6O)? (10 points)
Ethanol molecular mass is (2*12.01)+(6*1.01)+(1*16.0) = 46.08 amu
C: (2*12.01)/46.08 x 100 = 52.1%
H: (6*1.01)/46.08 x 100 = 13.2%
O: 16/46.08 x 100 = 34.7%
Note: 4 points for ethanol amu, 2 points each for % mass
8. How many grams are there in 1.72 moles of acetone (CH3)2CO ? (10 points)
Acetone molecular mass is (3*12.01)+(6*1.01)+(1*16.0) = 58.09 amu
1.72 moles * (58.09 g/mole) = 99.9 g
Note: 5 points for acetone amu, 5 points for correct grams
9. A certain combustible compound, containing C, H and O atoms, is 53.3% C, 11.2% H
and 35.5% O by mass. The experimentally determined molecular mass is 90.14 amu.
What is the empirical and chemical formula for this compound? (10 points)
53.3 g C * (1 mole/12.01 g) = 4.44 moles C
11.2 g H * (1 mole/1.01 g) = 11.1 moles H
35.5 g O * (1 mole/16.0 g) = 2.22 moles O
C = 4.44 / 2.22 = 2.0
H = 11.1 / 2.22 = 5.0
O = 2.22 / 2.22 = 1.0
Empirical formula = C2H5O molecular mass = (2*12.01)+(5*1.01)+16 = 45.07 amu
Chemical formula = C2H5O * (90.14/45.07) = C4H10O2
Note: 5 points for empirical formula, 5 points for chemical formula
10. Heptanol has the formula C7H16O and can be combusted in the presence of oxygen.
If 15.0 g of heptanol is combusted in the presence of 23.0 g of O2, how many grams
of water will be produced? What is the limiting reagent? (Be sure to show the
balanced equation for this reaction). (20 points)
2C7H16O + 21O2 -> 14CO2 + 16H2O
heptanol molecular mass: (7*12.01)+(16*1.01)+16 = 116.86 amu
15 g * (1 mole/116.86 g) = 0.128 moles
O2 molecular mass: (2*16.0) = 32 amu
23.0 g * (1 mole/32 g) = 0.719 moles
If oxygen is unlimited:
0.128 moles heptanol * (16 moles H2O/2 moles heptanol) = 1.03 moles H2O
If heptanol unlimited:
0.719 moles O2 * (16 moles H2O/21 moles O2) = 0.548 moles H2O
Therefore, O2 is limiting.
At most we can make 0.548 moles H2O * (18.02g /1 mole) = 9.87 g
Note: 5 points for balanced equation, 10 points for water produced, 5 points for
identifying oxygen as limiting reagent
11. What is the molarity of a 2.92 liter solution of water containing 27g of of Mg3(PO4)2?
(5 points)
Mg3(PO4)2 molecular mass = (3*24.3)+2*(31.0+64) = 262.9 amu
27 g * (1 mole/262.9 g) = 0.103 moles
0.103 moles/2.92 liters = 0.035 molar
12. You have a stock solution of 3.0 M MgCl2 and some pure water. How would you use
these to produce 0.157 liters of a 0.25 molar solution of MgCl2? (10 points)
(0.157 liters)*(0.25 moles/liter) = (x liters)*(3 moles/liter)
x = .013 liters
Combine 0.013 liters of the MgCl2 stock solution with (0.157-.013) = 0.144 liters of H2O
Note: 5 points for correct amount of stock. 5 points for correct amount of water to
add