# c103exam1s09solutions

```Chem 103 Exam 1 Solutions
Spring, 2009
1. (3 pts) A person with a fever has a temperature of 101.0 oF. What is this temperature
in degrees Kelvin?
5
101.0  32.0  38.30 C
9
K  273.15  38.3  311.5
T
For the other version, the answers are 40.0 oC and 313.2K
2. (3 pts) Indicate whether each of the following is a physical change (P) or a chemical
change (C):
C
A metal is dropped into a beaker of water and bubbles appear
P
Water is heated over a bunsen burner until bubbles appear
P
3. (4 pts) A sample of metal that has a mass of 183.7 g is added to a graduated cylinder
filled with 231.0 mL of water. The level of the water rises to 240.5 mL. What is the
density of the metal?
Volume of water displaced = 340.5 – 231.0 = 9.5 mL
density 
183.7 g
 19 g / mL (2 sig fig)
9.5mL
210.5 mL – 161.0 mL = 49.5 mL
sig fig)
density = 173.4 g/49.5 mL = 3.50 g/mL (3
4. (4 pts) A jogger can run 1.2 miles in 15 minutes. What is this speed in km/hour? (1
mi = 1609 m)
km 1.2mi 60 min 1609m 1km

x
x
x
 7.7 km / hr
hr 15 min
1hr
1mi 10 3 m
km 1.8mi 60 min 1609m 1km

x
x
x
 10.86  11km / hr
hr 16 min
1hr
1mi 10 3 m
5. (3 pts) An atom of 134Xe contains 54 protons, 80 neutrons, and 54 electrons.
83
6. (2 pts) Give the symbol (
33
16
Kr contains 36 protons, 47 neutrons, and 36 electrons
) for an atom of sulfur with 17 neutrons.
or
S
34
16
S (depending on version)
7. (5 pts) A given element (designated as Z) has three naturally occurring isotopes: 31Z
(31.0065 amu), 32Z (32.2464 amu) and 33Z (33.1531 amu). The abundance of 31Z is
38.76%. The two other isotopes have equal abundance. Calculate the average atomic
mass of element Z.
0.3876 (31.0065 amu) + 0.3062 (32.2464) + 0.3062 (31.1531) = 32.04 amu
Other version:
0.5418 (31.0065 amu) + 0.2291 (32.2464) + 0.2291 (31.1531) = 31.78 am
8. (8 pts) For each element below, put an X in every box for each classification into
which the element falls. (i.e. if an element is a transition metal, you would mark both
transition element AND metal)
Li Co Se Xe
main group
element
X
transition element
metal
nonmetal
alkali metal
alkaline earth
metal
halogen
noble gas
X
X
X
X
X
X
main group
element
transition
element
metal
X nonmetal
alkali metal
alkaline earth
metal
halogen
X noble gas
X
Ar
Ni
X
P
Mg
X
X
X
X
X
X
X
X
X
9. (4 pts) What is the formula of the ionic compound that forms between the following
elements?
Ca and P
K and O
Ca3P2
K2O
Ba3P2
Ba and N
Li2O
Li and O
10. (10 pts) Name each of the following compounds.
Tetraphosphorous trisulfide
P4S3
Na2SO4
NH4ClO4
CaBr2
Fe(OH)3
Sodium sulfate
Ammonium perchlorate
Calcium bromide
Iron(III) hydroxide
Tetraphosphorous trisulfide
P4S3
Na2SO4
NH4ClO4
CaBr2
Fe(OH)3
Sodium sulfate
Ammonium perchlorate
Calcium bromide
Iron(III) hydroxide
11. (8 pts). Write the formula for each of the following compounds.
barium phosphate
copper(III) oxide
magnesium fluoride
lithium nitride
calcium phosphate
Ba3(PO4)2
Cu2O3
MgF2
Li3N
Ca3(PO4)2
Co2O3
cobalt(III) oxide
magnesium bromide
potassium nitride
MgBr2
K3N
12. (6 pts) One tablet of a common headache medication typically contains 324 mg of
aspirin (C9H8O4). How many molecules of aspirin are you consuming if you take one
tablet?
? molecules = 324 mg x
1g
1mol
6.022 x10 23 molecules
x
x
 1.08 x10 21 molecules
3
1mol
10 mg 180.16 g
1g
1mol
6.022 x10 23 molecules
? molecules = 415 mg x 3
x
x
 1.39 x10 21 molecules
1mol
10 mg 180.16 g
13. (6 pts) The molar mass of glucose (C6H12O6) is 180.2 g/mol. What is the mass of
one glucose molecule?
g
1molecule
x
1mol
180.2 g
x
 2.992 x10  22 g
23
6.022 x10 molecules 1mol
14. (10 pts) Peroxyacylnitrate (PAN) is one of the components of smog. It is composed
of C, H, N, and O. An analysis shows PAN to be composed of 19.8% C, 2.50% H,
11.6% N, and 66.1% O.
a. Determine the empirical formula for PAN
19.8 g C x 1 mol/12.011 g = 1.648 mol C
2.50 g H x 1 mol/1.008 g = 2.48 mol H
11.6 g N x 1 mol/14.008 g = 0.83 mol N
66.1 g O x 1 mol/15.9994 g = 4.13 mol O
C2H3NO5
b. What is the molecular formula for PAN if its molar mass is around 120 g?
Empirical weight of C2H3NO5 = 121 g/mol
Factor: 121/120 = 1
Molecular formula = C2H3NO5
15. (8 pts) A certain hydrate of potassium aluminum sulfate (alum) has the formula
KAl(SO4)2. xH2O. When a hydrate sample weighing 5.459 g is heated to drive off the
water, the mass of the remaining material is 2.583 g. How many waters of hydration are
there in alum?
Weight of water = 5.459 – 2.583 = 2.876 g
Moles of water = 2.87 g x 1 mol/18.01 g = 0.159 mol
Moles of alum = 2.583 g x 1 mol/258.195 g = 0.0100 mol KAl(SO4)2
Formula is KAl(SO4)2 . 16 H2O
16. (8 pts) 1.256 g of elemental sulfur is combined with excess fluorine to give 5.722 g
of a compound. Calculate the empirical formula for the resulting compound.
Mass of F = 5.722 total mass – 1.256 g Sulfur = 4.466 g F
Moles S = 1.256 g S x 1 mol S/32.0 g = 0.039 moles S
Moles F = 4.466 g F x 1 mol F/19 g = 0.235 moles F
Empirical formula is SF6
Other version:
Mass of F = 4.239 total mass – 1.256 g Sulfur = 2.983 g F
Moles S = 1.256 g S x 1 mol S/32.0 g = 0.039 moles S
Moles F = 2.983 g F x 1 mol F/19 g = 0.157 moles F
Empirical formula is SF4
17. (8 pts) Analysis of a metal chloride (XCl3) shows that it contains 67.2% Cl by mass.
Calculate the molar mass of the metal X and identify the metal.
67.2 g Cl x 1 mol/35.4 g = 1.9 mol Cl
moles X = 1.9 mol Cl x 1 mol X/3 mol Cl = 0.633 mol X
mass X = 100 g-67.2 g = 32.8 g
MM = 32.8 g/0.633 mol = 51.8 g/mol
Other version:
62.6 g Cl x 1 mol/35.4 g = 1.768 mol Cl
moles X = 1.768 mol Cl x 1 mol X/3 mol Cl = 0.589 mol X
mass X = 100 g-62.6 g = 37.4 g
MM = 37.4 g/0.589 mol = 63.5 g/mol
Extra Credit: The volume of a sphere is 4/3 r3. Calculate the density of the 64Zn
nucleus in g/cm3 given that the nuclear radius is 4.8 x 10-6nm and the mass of the atom is
1.06 x 10-22 g.
In this problem, you had to:
□ convert nm to cm correctly
□ find the atomic volume and the nuclear volume
□ subtract the nuclear volume from the atomic volume to find the volume of the
electrons
□ find the mass of the electrons by multiplying the number of electrons (30 fro Zn)
by the mass per electron.
All components had to be there and done correctly to get points for the extra credit
question. (Extra credit is almost always ‘all or nothing’)
The volume of the electrons is given as:
V 


4
  1.25 x10 8 cm 3  8.18 x10 24 cm 3
3
and the density is calculated as:
d


30 9.11x10 28 g
 3.34 x10 3 g / cm 3
8.18 x10 24 cm 3
```
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