Chemistry& 141
Clark College
Problem Solving Section
Please answer the following questions in the space provided. Place your answers in the boxes provided,
using correct units and significant figures! Show all of your work. No credit will be given to an
unsupported (no work shown) answer! Partial credit will be assigned when possible, and it is easier
to assign partial credit when work is shown.
1. Digoxin is a medication for the treatment of several heart conditions, and is prepared from extracts
of the plant ‘foxglove’ (Digitalis Purpuraea). Digoxins are also quite toxic, therefore dosages must
be highly controlled and are based on body weight, as even a modest overdose can be fatal (It was
recently a featured “poison” in the James Bond movie “Casino Royale”). The daily-maintenance
dosage requirements for digoxin is 11.0 µg digoxin/kg body mass (www.rxlist.com). How many
micrograms should a 225.0 lb patient receive? 1 kg = 2.205 lbs [10 pts]
1 kg
11.0 µ g
x
= 1122.4 µ g
2.205 lbs
1 kg
1 kg
11.0 µ g
Purple: 185.0 lbs x
x
= 922.9 µ g
2.205 lbs
1 kg
Green: 225.0 lbs x
Dosage = G 1120 µg P 923 µg
2. The vinegar that you purchase in a grocery store is a solution of acetic acid (HC2H3O2), with a
density of 1.01 g/mL. [12 pts]
a. What is the mass, in grams, of a 750.0 mL sample of vinegar?
750.0 mL x
1.01 g
= 757.5 g
mL
758 g vinegar solution
! 5.0 g acetic acid $
b. If the solution is 5.0 % acetic acid #
, how many grams of acetic acid are in the
" 100.0 g vinegar &%
sample?
758 g vinegar x
5.0 g acetic acid
= 37.9 g acetic acid
100.0 g vinegar
38 g acetic acid (HC2H3O2)
c. In that same sample, how many atoms of hydrogen are there?
38 g molecs HC2 H 3 O 2 x
1 mol
6.022 x 1023 molecules
4 atoms H
x
x
= 1.526 x 1024 atoms H
60.0 g
1 mole
1 molecule
1.5 x 1024 atoms H
Exam 1
Fall 2008
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Chemistry& 141
Clark College
3. Sodium benzoate is a common food preservative in acidic foods, such as sodas and salad dressings.
Elemental analysis of a solid sample of sodium benzoate gives the following results: 58.34 %C,
3.50% H, 22.21% O and 15.95% Na. What is the empirical formula for sodium benzoate? If the
molar mass is 144.1 amu, what is the molecular formula for sodium benzoate? [16 pts]
mol
= 4.86 mol
=7
0.693 mol
12.011 g
mol
H: 3.50 g x
= 3.47 mol
=5
0.693 mol
1.0079 g
mol
O: 22.21 g x
= 1.39 mol
=2
0.693 mol
15.9994 g
C: 58.34 g x
Na: 15.95 g x
mol
= 0.693 mol
=1
0.693 mol
23.0 g
Empirical Formula = C7 H 5 O 2 Na
Formula Weight = 144 g/mol
Since the formula weight is the same as the molar mass, the empirical formula and molecular formula are the same.
Empirical Formula = C7H5O2Na
Molecular Formula = C7H5O2Na
4. Sodium benzoate is an ionic compound, made of sodium ions and polyatomic benzoate ions. Using
your answer from the previous problem, and your knowledge of the charge on a sodium ion, what do
you think the charge on a benzoate ion is? [3 pts]
The charge of a sodium ion is +1. Since there is one sodium ion in the formula, the benzoate must
balance that charge.
-1
Exam 1
Fall 2008
Page 2 of 6
Chemistry& 141
Clark College
5. Write the full atomic symbols for the following species, based on the description provided. [9 pts]
a. An atom that contains 44 protons, 59 neutrons and 44 electrons
Green
Purple
44 + 59
44
Ru =
103
44
Ru
79+119
79
Au =
198
79
Au
b. An ion that contains 30 protons, 34 neutrons and 28 electrons.
Green
64
30
Zn +2
Purple
73
32
Ge +4
6. In the lab, a rectangular block of hematite (iron (III) oxide, Fe2O3) was analyzed to determine its
density. [12 pts]
a. Using vernier calipers, the block was measured and the following dimensions were recorded:
length = 8.72 cm, width = 3.46 cm, height = 2.03 cm. What is the volume of the block?
Green: 8.72 cm x 3.46 cm x 2.03 cm = 61.2 cm 3
Purple: 4.08 cm x 5.64 cm x 3.92 cm = 90.2 cm 3
G = 61.2 cm3 P = 90.2 cm3
b. The mass of the block was also measured, and found to be 315.83 g. What is the density of the
block of hematite?
D=
m
V
G:
P:
315.83 g
3
61.2 cm
449.22 g
90.2 cm
3
= 5.16 g
cm 3
= 4.98 g
cm 3
G = 5.16 g/cm3 P = 4.98 g/cm3
c. The actual value of the density of hematite is 5.24 g/cm3. Calculate the percent difference from
the experimentally determined value and the actual value.
Watch for sig figs here! You must stop after the subtraction to get your sig figs, and then go on.
% difference =
G=
P=
actual - experimental
x 100
actual
5.24 g/cm 3 - 5.16 g/cm 3
5.24 g/cm 3
5.24 g/cm 3 - 4.98 g/cm 3
5.24 g/cm 3
x 100 =
x 100 =
0.08 g/cm 3
5.24 g/cm 3
0.26 g/cm 3
5.24 g/cm 3
x 100 = 2% (yes, one sig fig!)
x 100 = 5.0%
G = 2% P = 5.0%
Exam 1
Fall 2008
Page 3 of 6
Chemistry& 141
Clark College
7. Phosphorus and oxygen combine to give a binary compound. [8 pts]
a. Based on its position on the period table, what would be the value of the positive charge on a
phosphorus cation?
Phosphorus is in group 5, so it would be a +5 cation
+5
b. Write a balanced formula combining your phosphorus cation and an oxygen anion.
P 2 O5
8. Chlorine has two stable isotopes, 35Cl and 37Cl. The molar mass of chlorine is 35.453 amu. Which
isotope is more abundant, 35Cl or 37Cl? Explain or describe how you determined this. [6 pts]
35
Cl is the more abundant isotope. The molar mass is a weighted average of all isotopes of an
element. If the two isotopes were equally abundant (meaning there was a 50/50 mix), the molar
mass would be 36 amu. Since the molar mass is less than that, there must be a greater amount of
35
Cl to “pull” the average down.
Multiple Choice. Circle the choice that best answers the question statement. [3 pts each, 24 pts total]
9. What is the mass percent of oxygen in ibuprofen (C13H18O2)?
a. 6.06%
c. 15.5%
b. 7.75%
d. 20.4%
Mass Percent O =
mass of O
32.0 amu
x 100 =
x 100 = 15.5%
mass of C13 H18 O 2
206.3 amu
10. What is the correct name of HNO3(aq)?
a. Hydronitric acid
c. Nitrous acid
b. Hydrogen nitrate
d. Nitric acid
11. The correct formula for calcium nitride is:
a. CaN
c. Ca3N2
b. Ca2N3
d. CaN2
Ca is in group 2, so the charge is +2. N is in group 5, so its charge is 5-8 = -3.
Exam 1
Fall 2008
Page 4 of 6
Chemistry& 141
Clark College
+2
12. The number of electrons in a Ni ion is:
a. 26 e-
c. 30 e-
b. 28 e-
d. None of these.
13. How would you classify sugar water?
a. Pure substance – compound
c. Mixture – homogeneous
b. Pure substance – element
d. Mixture – heterogeneous
14. The radius of a palladium atom is 0.137 nm. What is this in centimeters?
a. 1.37 x 10-10 m
c. 1.37 x 10-7 m
b. 1.37 x 10-8 m
d. 1.37 x 10-5 m
0.137 nm x
1 x 10-9 m 100 cm
x
= 1.37 x 10-8 cm
1 nm
1m
15. Which of the following properties does not describe a substance in the gas phase?
a. There is a large amount of molecular movement.
b. The substance has a defined volume.
c. The substance has an undefined shape.
d. The substance has high compressibility.
16. A block of wood has a volume of 0.2587 ft3. What is its volume in cm3? (1 ft = 12 in, 1 in = 2.54 cm
exactly)
a. 7326 cm3
c. 7.33 x 103 cm3
b. 7.89 cm3
d. 7.885 cm3
Don’t forget to cube all unit conversions!!!
0.2587 ft 3 x
Prob 1
/10
Exam 1
2
3
/12
4
/16
123 in 3
1 ft
3
x
2.543 cm 3
1 in
5
/3
3
6
/9
Fall 2008
= 7326 cm 3
8
7
/12
/8
MC
/6
/24
Total
/100
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