AP Chemistry 1: Structure of Matter
A.
Measurement (1.4 to 1.6)
1. science knowledge is advanced by observing patterns
(laws) and constructing explanations (theories), which
are supported by repeatable experimental evidence
a. theory lasts until disproven
b. theory is never 100 % certain
2. uncertainty in measurements
a. precision and accuracy
1. precise = consistent (even if incorrect)
2. accurate = correct (even if inconsistent)
b.
c.
precise
precise & accurate accurate
data analysis
1. accuracy is measured by percent difference
percent  = 100|mean – true|/true
2. precision is measured by percent deviation
%  = 100 |trial – mean|/N(mean)
(N is number of trials)

absolute  = |trail – mean|

average  = absolute )/N

%  = 100(average )/(mean)
significant figures (sf) indicate level of certainty
measurement includes all certain (numbered) plus
one estimated value  7.5 cm (2 sf)
d. rules for counting significant figures
1. all nonzero digits are significant
2. zero is sometimes significant, sometimes not
a. example: 0.00053000021000
never
always
?
b. (?) decimal vs. no decimal
1. significant with decimal: 120. (3 sf)
2. not significant w/o decimal: 120 (2 sf)
3. exact numbers (metric conversions, counting
or written numbers) have infinite number of sf
4. scientific notation: C x 10n
a. C contains only significant figures
b. 1200 with 3 sf: 1.20 x 103
e. rules for rounding off calculations
1. limited by least accurate measurement
2. x, : answer has the same number of sf as
the measurement with the fewest
3. +, –: answer has same end decimal position
as measurement with left most end position
3. SI measuring system
a. summary chart
Measurement
SI standards
Chemistry
mass
kilogram (kg)
gram (g)
volume
cubic meter (m3)
liter (L)
temperature
kelvin (K)
Celcius (oC)
time
second (s)
varies
Name __________________________
prefixes system (x 10X)
1. k3, c-2, m-3, µ-6, n-9
2. squared/cubed prefix: 1 cm2 = 1 x (10-2)2 m2
3. 1 mL = 1 cm3
4. 455 kg x 103 g x (10-2)3 m3 = 0.455 g
m3
1 kg
1 cm3
cm3
4. mass and volume measure amount of matter
a. density: d = m/V
1. units depend on units for m and V
2. dH2O = 1.00 g/mL = 1.00 g/cm3 = 1000 kg/m3
b. number of particles: mole = 6.022 x 1023 particles
1. periodic table mass equals formula mass in g
2. molar mass (MM)—sum of mass of atoms in
chemical formula (use 3 significant figures)
a. Al: 27.0 g/mol
b. H2O: 18.0 g/mol
c. conversions (dimensional analysis)
1. mass  moles (given formula or MM)
__ g x 1 mole/(MM) g = __ mole
2. volume  mass (given density–d)
__ mL x (d) g/1 mL = __ g
3. volume  mass  moles (given d and MM)
__ mL x (d) g/1 mL x 1 mole/(MM) g = __ mole
Atomic Nature of Matter (2.1 to 2.7)
1. historical perspective
a. Dalton's atomic theory (1805)
1. unique, indestructible atoms for each element
2. atoms are rearranging, not created during
chemical change
3. compounds are groups of atoms in fixed ratio
b. subatomic structure
1. J. J. Thomson (1897): measure charge-tomass ratio of electrons with cathode rays
2. Millikan (1909): measure electron charge with
oil drops in a vacuum chamber
3. Rutherford (1910): characterized dense, +
nucleus with alpha () radiation and gold foil
2. components of the atom
a. subatomic particles
Particle
Location
Charge
Mass
Symbol
1 p or 1 H
Proton
nucleus
+1
1.0
1
1
1 n
Neutron
nucleus
0
1.0
0
o e
Electron
outside
-1
.00055
-1
b. atomic number (Z)
1. number of protons
2. defines type of atom
c. mass number (A)
1. protons + neutrons
2. isotopes (same Z, different A)
3. nuclear symbol: AZX
d. ions are atoms where # electrons  # protons
1. e > p: (–) charged (anion): Xn2. e < p: (+) charged (cations): Xn+
e. unified atomic mass unit (u)
1. 1 u = 1/12 the mass of a C-12 atom
2. average atomic mass (periodic table mass)
a. isotopes have fixed % in natural sample
b. 100mav = %1m1 + %2m2 + ...
3. forms of matter
a. pure substance has a unique composition of
atoms unique formula and set of properties
1. elements—one type of atom
(diatomic: H2, N2, O2, F2, Cl2, Br2, I2)
2. compounds—two or more types of atoms
a. molecular—formula defines size
b. crystalline—formula shows ratio of atoms
b.
B.
b.
c.
mixture of pure substances in an object or container
1. variable composition (no set formula)
2. uniform: homogenous mixture = solution
3. non-uniform: heterogeneous
summary
Monatomic
Element
Molecular
Element
Molecular Homogeneous
Compound
Mixture
Crystalline Compound
Radioactivity (21.1 to 21.4)
1. forms of natural radiation
Mass #
Charge #
Stopping
Type
Symbol
(A)
(Z)
Shield
4 He
alpha
4
+2
paper

2
0 e
beta
0
-1
Al

-1
0 e
positron
0
+1
destroyed

1
0 
gamma
0
0
Pb

0
2. balancing nuclear reactions using nuclear symbols: AZX

balance A and Z values

determine symbol by Z number
238 U  4 He + 234 Th

92
2
90
3. nuclear instability
a. isotopes that are outside the "belt of stability" tend
to be radioactive
b. modes of decay
1. atomic number > 83— (alpha)
226 Ra  222 Rn + 4 He
88
86
2
2. Aisotope > Aaverage: 10n  11p + 0-1
14 C  14 N + 0 e
6
7
-1
3. Aisotope < Aaverage: 11p  10n + 01 (positron)
11 C  11 B + 0 e
6
5
1
C.
alpha decay
beta decay
4.
positron decay
transmutations
a. induced nuclear reactions by bombardment
b. 147N + 42He  178O + 11H
c. produce trans-uranium elements
5.
radioactive decay
a. rate of decay  number of radioactive atoms (Nt)
1. rate = kNt (k: rate constant)
2. time for half of remaining atoms to decay (t½)
is constant: k = (ln2)/t½
ln(No/Nt) = kt or Nt = Noe-kt
1. No = original amount
2. t and k must have same time units
Electron Structure—Bohr Model(6.3 to 6.4)
1. atomic spectrum
a. colors emitted by energized atoms (unique for
each element)
b.
D.
-25 J•m/
calculations: Ephoton
1.  = wavelength (m)
2. f = frequency (s-1) = c/  on AP test)
3. Ephoton = hf = hc/
hc = (6.63 x 10-34 J•s)(3.0 x 108 m/s) = 2.00 x 10-25 J•m
Bohr model—atoms with one electron only
a. energy levels (n)
1. Eelectron = -B/n2
2. for H: En = -2.18 x 10-18 J/n2
3. ground state (n = 1) electron has lowest
(most negative) energy
4. excited state (n > 1), electron energy
increases until ionized (E = 0 J)
5. Eelectron = En-final – En-initial
a. Eelectron > 0 when increasing n
b. Eelectron < 0 when decreasing n
b. |Eelectron| = Ephoton
b.
2.
434 nm 486 nm
656 nm
1.
2.
3.
A. Measurement
How many significant figures are there in?
0.008090 mL 1300.40 atm
13400 m
one liter
4
6
3

Express the answers to the correct number of sf.
(3.016)(4.23)
12.0 + 1.01 + 6
4100
0.19
0.0031
101.4
How much do you have when you double 12.28 g?
2 x 12.28 g = 24.56 g (4 sf because double is perfect #)
4.
A student measures the mass of an object to be 12.045 g.
The true mass is 12.000 g. What is the percent error?
|12.045 – 12.000|/12.000 = 0.38%
5.
Determine the % deviation for the following massings.
Mass
48.307 g
49.886 g
50.911 g
49.524 g
48.307 + 49.886 + 50.911 + 49.524 = 49.657 g
mean
4

Average

%
6.
1.350 g
0.229 g
1.254 g
0.133 g
1.350 + 0.229 + 1.254 + 0.133 = 0.742
4
345 nm  m
350 mL  L
7.
Check the identity of the solid based on its density.
Al
Zn
Pb
d = 2.7 g/mL
d = 7.1 g/mL
d = 11 g/mL
c. Calculate the number of moles of solid.
Liquid: Mass a clean, dry 10 mL graduated cylinder (m1). Add
10.0 mL liquid to the cylinder. Mass the cylinder + liquid (m2).
d. Record the collected data. Calculate the change in
mass (m), change in volume (V) and density (d).
V
d
m2
–
m1
=
m
e.
345 nm = 345 x 10-9 m = 3.45 x 10-7 m
L
350 mL = 350 x 10-3 L = 0.35 L
155
cm3
= 155 mL = 155 x
10-3
L = .155 L
A student adds 7.76 g of pellets to a graduated cylinder
containing 5.00 mL. The total volume of the pellets and
water is 7.87 mL. What is the density of the pellets?
7.76 g/(7.87 mL – 5.00 mL) = 2.70 g/mL
8.
b.
10. mL
3640 cm2  m2 3640 cm2 = 3640 x (10-2)2 m2 = 3.64 m2
155
5.0 mL
0.742/49.657 x 100 = 1.49 %
Convert the following:
cm3
Density Lab
Measure the mass and volume of a solid, liquid and gas,
determine densities, and use the density to identify the
substance.
Solid: Add 5.0 mL (V1) water to a 10 mL graduated cylinder.
Mass the cylinder + water (m1). Add solid. Record the
volume to the nearest 0.1 mL (V2). Mass (m2).
a. Record the collected data. Calculate the change in
mass (m) and change in volume (V) and density (d).
d
m2 – m1 = m
V2 – V1 = V
A student measure the mass of an empty graduated cylinder
(10.076 g), then fills it with 10.0 mL of liquid. The total mass
of cylinder and liquid is 18.799 g. What is the density?
Circle the identity of the liquid based on its density.
C2H6O
H2 O
C3H8O3
d = 0.79 g/mL
d = 1.0 g/mL
d = 1.1 g/mL
f. Calculate the number of moles of liquid.
Gas: Add ½ scoop of baking soda (NaHCO3) to the flask and
½ fill the pipet with 6 M HCl (HANDLE WITH CARE).
Stopper the flask and mass the assembly (m1). Fill the gas
collecting bottle with water. Measure the volume of water
(V1). Insert the open tube into the gas collecting bottle. Add
the HCl to the flask, one drop at a time, until the gas
collecting bottle is nearly full. Mass the assembly (m2).
Measure the volume of water remaining in the bottle (V2).
g. Record the collected data. Calculate the change in
mass (m) and change in volume (V) and density (d).
d
m1 – m2 = m
V1 – V2 = V
(18.799 – 10.076)g/10.0 mL = 0.872 g/mL
9.
Calculate the molar masses to 3 significant figures.
NaCl
H2O
Cl2
58.5 g/mol
18.0 g/mol
70.9 g/mol
(NH4)2SO4
C4H7NO4
CuSO4•5 H2O
132 g/mol
133 g/mol
250. g/mol
10. Use dimensional analysis to determine the following
a. aluminum (MM = 27.0 g/mol, d = 2.70 g/cm3)
2.48 g x 1 mol = 0.0919 mol Al
2.48 g Al  mol
27.0 g
5.0 mL x 2.70 g = 13.5 g Al
3
5.00 cm Al  g
1 cm3
3
155 cm x 2.70 g x 1 mol = 15.5 mol Al
155 cm3 Al  mol
1 cm3
27 g
b. carbon dioxide (MM = 44.0 g/mol, d = 1.82 g/L)
85.0 g x 1 L = 46.7 L CO2
85.0 g CO2  L
1.82 g
3.15 mol x 44.0 g = 139 g CO2
3.15 mol CO2  g
1 mol
3.22 L x 1.82 g x 1 mol = .133 mol CO2
3.22 L CO2  mol
1L
44.0 g
h.
Circle the identity of the liquid based on its density.
H2
O2
CO2
d = 8.4 x 10-5 g/mL d = 1.3 x 10-3 g/mL d = 2 x 10-3 g/mL
i. Calculate the number of moles of gas.
B. Atomic Nature of Matter
11. How did Rutherford show that atoms have a nucleus?
Rutherford bombarded gold foil with  particles. Some
 particles bounced off of the foil, which could only
occur with a dense, positive nucleus.
12. Below is a modern view of an isotope of a sulfur atom.
16 protons
16 neutrons
18 electrons
Write the nuclear symbol for this ionized isotope.
32 S216
13. Complete the chart below.
Symbol
protons
27
56
13
14
13
+
19K
19
21
18
315P
15
16
18
3+
26
30
23

17
17
18
79Au
79
118
79
26Fe
34
electrons
13Al
40
31
neutrons
17Cl
197
14. Calculate the average atomic mass of Si, which consists of
three isotopes listed below.
Isotope
Si-28
Si-29
Si-30
Atomic Mass
27.98
28.98
29.97
Abundance
92.20%
4.70%
3.10%
100mav = m1(%1) + m2(%2) + m3(%3)
100mav = (27.98)(92.20) + (28.98)(4.70) + (29.97)(3.10)
mav = 28.09 u
15. Chlorine is primarily two isotopes Cl-35 and Cl-36 and has
an average atomic mass of 35.45 u.
a. Which isotope is more abundant? Explain
Cl-35 is more abundant because the average mass is
closer to 35 than 36.
b. Estimate the approximate abundances for the two
isotopes, Cl-35 and Cl-36 without using your calculator.
0.45 more than 35  Cl-36 = 45 % and Cl-35 = 55 %
16. Antimony has two isotopes: Sb-123 and Sb-121. Sb-121
has a mass of 120.9 u and an abundance of 57.25 %.
Antimony has an average atomic mass of 121.75.
a. What is the abundance of Sb-123?
100.00 = %Sb-123 + %Sb-121
100.00 = %Sb-123 + 57.25  %Sb-123 = 42.75 %
b. What is the atomic mass of Sb-123?
100mav = m1(%1) + m2(%2)
12175 = mSb-123(42.75) + (120.9)(57.25)
mSb-123 = 122.9 u
17. Consider the following molecules. Based on the model
below, write the unique formula and formula mass.
19. How is a molecular compound different from a nonmolecular compound?
A molecular compound exists as defined size equal to
the formula, whereas a non-molecular compound
exists as an aggregate of formula units.
Penny Isotope Lab
Mass 50 pennies, determine the average mass, use the
average to determine the percentages of heavy pennies
and light pennies, and compare to the actual numbers.
Count 50 pennies and mass the total.
a. Record the mass and calculate the average.
b.
Calculate the percentage of 2.5-g and 3.1-g penny.
Separate the pennies into pre-1982, 1982 and post-1982
piles. Mass each 1982 penny.
c. Record the number of pennies in each group and the
percentage of pennies that are 3.1 g and 2.5 g.
pre-1982
1982
post-1982
3.1 g
3.1 g
2.5 g
2.5 g
Total %:
d.
Total %:
Calculate the % difference between the actual % of 2.5
g penny (c) and the calculated % (b).
C. Radioactivity
20. Write nuclear equation for the radioactive process.
 42He + 22286Rn
Alpha emission of Ra-226
226
88Ra
Beta emission of I-131
131
53I
Positron emission of C-11
11
Th-231 decays to Pa-231
231
90Th
 23191Pa + 0-1e
Th-232 decays to Ra-228
232
90Th
 22888Ra + 42He
6C
 0 -1e + 13154Xe
 01e + 115B
21. What is the most likely mode of decay for the isotopes?
H-3
N-11
Co-60
Rn-222




22. Write a nuclear equation for the most likely mode of decay.
Formula
H2O
H2O2
CO
B-8
8
K-40
40
U-235
235
Co-60
60
CO2
18.0
34.0
28.0
44.0
Mass
18. Fill in the flow chart from the word bank (compound,
element, heterogeneous, homogeneous, matter, pure
substance, solution).
Matter

Is it uniform throughout?


Heterogeneous
Homogenous

Does it have a variable composition?

Yes 
pure substance
solution

Can it be separated into simpler substances?
No
Yes
Element
Compound
5B
 84Be + 01e
19K
 4020Ca + 0-1e
92U
27Co
 23190Th + 42He
 6028Ni + 0-1e
23. Fill in the missing part of the nuclear transmutations.
+ 10n  2411Na + 42He
27
13Al
14
7N
+ 42He  178O + 11H
16
8O
+ 11H  137N + 42He
58
26Fe
+ 10n  5927Co + 0-1e
24. Pu-239 undergoes nuclear fission when bombarded by a
neutron. Determine the missing (?) product.
27. C-14 (t½ = 5715 yrs) decays to N-14. As a result, the 14C/12C
ratio in organic material decreases upon death.
a. What is the rate constant k for C-14?
k = ln2/t½
k = ln2/5715 = 1.213 x 10-4 yr-1
b.
A = 1 + 239 – 95 – 2(1) = 143
Z = 0 + 94 – 40 – 1(0) = 54  14354Xe
25. The graph shown below illustrates the decay of
which decays via positron emission.
How old is an organic artifact whose 14C/12C ratio is
65.4 % of a living plant?
ln(No/Nt) = kt
ln(100/65.4) = (1.213 x 10-4)t  t = 3510 yr
28.
88
42Mo,
238U
a.
(t½ = 4.5 x 109 yr) naturally decays to 206Pb.
What is the rate constant k for U-238?
k = ln2/t½
k = ln2/4.5 x 109 = 1.5 x 10-10 yr-1
b.
What is the age of the rock whose
206Pb/238U
is 1.32/1?
ln(No/Nt) = kt
ln(2.32/1) = (1.5 x 10-10)t  t = 5.6 x 109 yr
Radioactive Decay Lab
Calculate the number of radioactive atoms that remain
after given periods of time, graph the data and compare
the graph to a ½-life graph.
a. Calculate the rate constant k given t½ = 2.77 minutes.
b.
t
Calculate the number of radioactive atoms that remain
after each minute.
0
1
2
3
4
5
6
7
8
9 10
Nt
c.
a.
88
Write a nuclear equation for the decay of
42Mo
b.
 01e + 8841Nb
80
What is the half-life of the decay?
The half-life is 7 minutes.
c.
60
What is the rate constant for the decay?
k = ln2/t½ = ln2/7 min = 0.1 min-1
d.
Graph Nt vs. t.
88 Mo.
42
Nt
40
What percent of the original sample remains after 12
minutes?
20
0.3/1.0 x 100 = 30 %
e.
How many minutes does it take the sample to go from
0.8 g to 0.5 g?
7 min – 2 min = 5 min
26. The half-life of radioactive S-35 is 88 days. Determine
a. the rate constant.
0
d.
1
3
5
7
9
t (minutes)
The slope of the line represents the rate of decay.
How does the rate of decay change with time?
k = ln2/t½ = ln2/88 = 7.9 x 10-3 d-1
b.
the number of days for the sample to be ¼ as
radioactive (without a calculator).
2 x 88 days = 176 days
the number of days for the sample to be ¼ as
radioactive (with a calculator).
ln(No/Nt) = kt
ln(100/25) = (7.9 x 10-3 d-1)t  t = 175 days
D. Electron Structure—Bohr Model
29. Calculate the missing value.
Photon energy
Wavelength
3.25 x 10-18 J
6.15 x 10-8 m
1.64 x 10-18 J
1.216 x 10-7 m
6.60 x 10-19 J
3.03 x 10-7 m
5.80 x 10-19 J
345 nm
c.
d. the percent that remain radioactive after 290 day.
ln(No/Nt) = kt
ln(100/Nt) = (7.9 x 10-3 d-1)(290 d)  Nt = 10 %
30. A hydrogen electron transitions from n = 1 to n = 4.
a. What is the electron's energy at n = 1?
d.
Match the results from (a) and (c) and record the
transition that produced each spectral line.
4.10 x 10-7 m 4.34 x 10-7 m 4.86 x 10-7 m 6.56 x 10-7 m
Eelectron = 2.18 x 10-18 J/n2
Eelectron = -2.18 x 10-18 J/12 = -2.18 x 10-18 J
b.
What is the electron's energy at n = 4?
e.
What do these transitions have in common?
Eelectron = 2.18 x 10-18 J/n2
Eelectron = -2.18 x 10-18 J/42 = -1.36 x 10-19 J
c.
The electron gains energy during the transition.
d.
What is the change in energy of the electron?
E = E4 – E1
E = -1.36 x 10-19 J – (-2.18 x 10-18 J) = 2.04 x 10-18 J
31. A hydrogen electron transitions from n = 9 to n = 7.
a. What is the energy of the electron when n = 9.
Eelectron = 2.18 x 10-18 J/n2
Eelectron = -2.18 x 10-18 J/92 = -2.69 x 10-20 J
b.
Multiple Choice (No calculator)
Briefly explain why the answer is correct in the space provided.
1
2
3
4
5
6
7
8
9
10 11 12 13
D D D B D B B B
A
C
D
C
D
1. Based on the data, the density of the solid in g/mL is
Mass of metal
19.611 g
Volume of water
12.4 mL
Volume of water + metal
14.9 mL
(A) 7.8444 (B) 7.844 (C) 7.84
(D) 7.8
V = 14.9 mL – 12.4 mL = 2.5 mL
d = m/V = (19.611 g)/(2.5 mL) = 7.8 g/mL
What is the energy of the electron when n = 7
Eelectron = 2.18 x 10-18 J/n2
Eelectron = -2.18 x 10-18 J/72 = -4.45 x 10-20 J
c.
Practice Quiz
Does the electron gain or lose energy in the transition?
2.
What is the change in energy for the transition?
E = E7 – E9
E =– -4.45 x 10-20 J – (-2.69 x 10-20 J) = -1.76 x 10-20 J
d.
e.
Thomson: mass/charge ratio, Millikan: electron charge,
Bohr: atomic model with quantum numbers.
Is energy absorbed or released during the transition?
Energy is released during the transition.
3.
What is the wavelength of the light emission?
|Ephoton| = hc/
1.76 x 10-20 J = (2.00 x 10-25 J•m)/ = 1.14 x 10-5 m
Hydrogen Spectrum Lab
Use hydrogen spectrum data to determine the electron
transition which generates each color.
a. Calculate Ephoton for each wavelength.
4.
Calculate En for each value of n.
4
2
5
3
6
c.
Copper has two isotopes, 63Cu and 65Cu. What is the
abundance of 63Cu if the average atomic mass is 63.5?
(A) 90%
(B) 75%
(C) 50%
(D) 20%
mav(100) = (Mass1 x %1) + (Mass2 x %2)
6350 = (63)( x) + (65)(100 – x)  x = 75 %
5.
1
Which represents a pair of isotopes?
(A) 146C and 147N
(B) 189F and 3517Cl
56
2+
56
3+
(C) 26Fe and 26Fe
(D) 3517Cl and 3617Cl
Isotopes: atoms of element with same # protons, but
different # neutrons  different A.
4.10 x 10-7 m 4.34 x 10-7 m 4.86 x 10-7 m 6.56 x 10-7 m
b.
Which scientist is correctly matched with the discovery?
(A) Millikan discovered the electron charge-to-mass ratio.
(B) Thomson discovered the charge of an electron.
(C) Bohr discovered the four quantum numbers.
(D) Rutherford discovered the nucleus.
Which of the following is correct about beta particles?
I. mass number of 4 and a charge of +2
II. more penetrating than alpha particles
III. electron
(A) I only
(B) III only (C) I and II (D) II and III
 particle, 0-1e,electron, with a mass 0 and charge -1,
more penetrating than  because of its smaller mass.
6.
Calculate E for each transition listed below.
6 to 5
5 to 4
4 to 2
6 to 4
5 to 3
4 to 1
For the types of radiation given, which is the correct order
of increasing ability to penetrate a piece of lead?
(A)  <  < 
(B)  <  < 
(C)  <  < 
(D)  <  < 
The ability to penetrate depends on mass. The more
mass the less penetration  <  < .
7.
249
96Cm is radioactive and decays by the loss of one beta
particle. The other product is
(A) 24594Pu (B) 24997Bk (C) 24896Cm (D) 25096Cm
249
6 to 3
5 to 2
3 to 2
6 to 2
5 to 1
3 to 1
6 to 1
4 to 3
2 to 1
8.
96Cm
 0-1 + 24997Bk
251 Cf  2 1 n + 131 Xe + . . .
98
0
54
What is the missing product in the nuclear reaction?
(A) 11842Mo (B) 11844Ru (C) 12042Mo (D) 12044Ru
A: 252 = 2(1) + 131 + x  x = 118
Z: 98 = 2(0) + 54 + x x = 44  11844Ru
9.
The radioactive decay of C-14 to N-14 occurs by
(A) beta particle emission (B) alpha particle emission
(C) positron emission
(D) electron capture
14
6C
Mass the butane lighter before and after bubbling
butane into the gas collecting bottle.
d.
 147N + 0-1
How would you determine the density of butane?
Divide the mass of butane by the volume of butane.
10. What is the resulting nucleus after 21484Po emits 2  and 2
 particles?
(A) 20683Bi (B) 21083Bi (C) 20682Pb (D) 20882Pb
6.
A: 214 – 4 – 2(0) – 4 = 206
Z: 84 – 2 – 2(-1) – 2 = 82  20682Pb
11.
238 g + 6(19 g) = 352 g
b.
+ 10n  14155Cs + 3 10n + X
Neutron bombardment of uranium can induce the reaction
represented above. Nuclide X is which of the following?
(A) 9235Br
(B) 9435Br
(C) 9137Rb (D) 9237Rb
235
92U
A: 235 + 1 = 141 + 3(1) + x  x = 92
Z: 92 + 0 = 55 + 3(0) + x  x = 37  9237Rb
c.
If 87.5 % decay then 12.5 % remains, which is 1/8
(1  ½  ¼  1/8). 3 half-lives = 36 days  t½ = 12 d
13. The half-life of isotope Y is 12 minutes. What mass of Y was
originally present if 1 g is left after 60 minutes?
(A) 8 g
(B) 16 g
(C) 24 g
(D) 32 g
60 minutes is 5 half-lives, after which 1/32 of the original
should remain. (1/32)x = 1  x = 32
8.
Determine the % deviation for the following massings.
Mass
17.215 g
17.135 g
16.988 g
17.455 g
17.215 + 17.135 + 16.988 + 17.455 = 17.198 g
mean
4

Average

%
3.
0.017 g
0.063 g
0.210 g
0.257 g
0.017 g + 0.063 g + 0.210 g + 0.257 g = 0.137 g
4
0.137 g/17.198 g x 100 = 0.797 %
A student adds 33.552 g of pellets to a graduated cylinder
containing 5.00 mL. The total volume of the pellets and
water is 8.05 mL. What is the density of the pellets?
33.552 g/(8.05 mL – 5.00 mL) = 11.0 g/mL
4.
An empty graduated cylinder (25.044 g) is filled with 50.0 mL
of liquid. The total mass of cylinder and liquid is 69.886 g.
What is the density of the liquid?
D = m/V = (69.886 g – 25.044 g)/50.0 mL = 0.897 g/mL
5.
You have a butane lighter for the purpose of measuring
butane's density at room temperature and pressure.
a. How would you collect a sample of butane?
Release the butane underwater so that it bubbles up
into the water filled gas collecting bottle.
b.
How would you determine the volume collected?
Measure the volume of water remaining and subtract
from the bottle's capacity.
c.
Write a brief description of the scientists' contribution.
Scientist
Contribution
J. J. Thomson
Measured electron charge-to-mass ratio
using cathode rays.
Millikan
Measured electron charge using oil
drops.
Rutherford
Discovered the dense positive nucleus
using  radiation and gold foil.
Bohr
Proposed a quantum model of hydrogen
atoms using atomic spectrum data.
Write the symbol for an atom that contains 24 protons, 28
neutrons and 21 electrons.
52
[(77.500 g – 75.011 g)/77.500 g] x 100 = 3.21 %
2.
How would you determine the mass collected?
What is the volume of 25.0 g of UF6?
25.0 g x 1 L/15.7 g = 1.59 L
7.
Free Response (Calculator)
A student measures the mass of an object to be 75.011 g.
The true mass is 77.500 g. What is the percent error?
How many moles of UF6 have a volume of 1.00 L?
1.00 L x 15.7 g/1 L x 1 mol/353 g = 0.0446 mol
12. If 87.5 % of a sample of pure Pb-210 decays in 36 days,
what is the half-life of Pb-210?
(A) 6 days (B) 8 days (C) 12 days (D) 14 days
1.
Consider the gas, UF6, which has a density of 15.7 g/L.
a. What is the molar mass of UF6?
9.
3+
24Cr
Consider two variations of 2311 Na; 2411Na and 2311Na+. How
is each different from Na-23 and what is it called?
How is it different?
What is it called?
24
11Na
23
11Na
+
different # of neutrons
isotope
different # of electrons
ion (cation)
10. Calculate the average atomic mass of Pb given the atomic
masses and abundances of its stable isotopes.
Isotope
Pb-204
Pb-206
Pb-207
Pb-208
Atomic Mass
204 u
206 u
207 u
208 u
Abundance
1.4%
24.1%
22.1%
52.4%
100mav = (204)(1.4) + (206)(24.1) + (207)(22.1) + (208)(52.4)
100mav = 286 + 4965 + 4575 + 10900  mav = 207
11. Copper has two isotopes, Cu-63 and Cu-65. What is the %
Cu-65 if the average mass of copper is 63.5?
100(mav) = (Mass1 x %1) + (Mass2 x %2)
6350 = (63)(100 – x) + (65)(x)  x = 25 %
12. Determine the molar mass of the following pure substances.
NaCl
O2
C6H12O6
NH3
58.5 g
32.0 g
H+,
13. Place the forms of H (H,
Atom
Cation
H
H+
180. g
H-,
17.0 g
H2) in the table.
Anion
Molecule
H-
H2
14. Predict the mode of decay for the following nuclei and then
write a balanced nuclear equation for the process.
Nuclei
Decay Mode Balanced Nuclear Equation
N-13
+
13
7N
 01e + 136C
Cu-68
-
68
29Cu
Np-241

241
 0-1e + 6830Zn
93Np
 42He + 23791Pa
15. Complete and balance the following nuclear equations.
32
7
16S
4Be
+ 10n  11H + 3215P
+ 0-1e  73Li
187
75Re
235
92U
 18776Os + 0-1e
+ 10n  13554Xe + 2 10n + 9938Sr
16. What is produced after 21484Po produces 2  and 2 ?
A = 214 – 2(4) – 2(0) = 206
Z = 84 – 2(2) – 2(-1) = 82  20682Pb
17. The rate constant for tritium (H-3) is 0.0564 yr-1.
a. What is the half-life?
t½ = ln2/k = ln2/0.0564 = 12.3 yr
b.
How long will it take 75% of the sample to decay?
2(12.3 yr) = 24.6 yr
18. The half-life of Cr-51 is 27.8 days.
a. Calculate the rate constant of Cr-51.
k = ln2/t½ = ln2/27.8 days = 0.0249 d-1
b.
What percent of the sample will remain after 100 days?
ln(No/Nt) = kt
ln(100/Nt) = (0.0249)(100)  Nt = 8.29 %
19. If 87.5 % of a sample of pure 131I decays in 24 days, what
is the half-life of 131I? (without calculator)
87.5 % decay = 12.5 % remains, which is 1/8
(1  ½  ¼  1/8)  3 half-lives = 24 d (t½ = 8 d)
20. The half-life of Y is 30 s. What mass of Y was originally
present if 1 g is left after 60 s?
60 minutes is two half-lives, after which ¼ of the
original should remain. ¼ x = 1  x = 4 g
21. How old is a wooden artifact, which has a C-14 (t½ = 5715 yr)
activity of 10.4 counts/min•g compared to living wood that has
a C-14 activity of 13.6 counts/min•g?
k = ln2/T½= ln2/5715 = 1.213 x 10-4 yr-1
ln(No/Nt) = kt  ln(13.6/10.4) = (1.213 x 10-4)t  t = 2200 yr
22. A line in the spectrum of hydrogen is associated with the
electronic transition from n = 6 to n = 2.
a. Find the energy that is given off in the transition from
n = 6 to n = 2 for hydrogen (En = -2.18 x 10-18/n2 J).
E = E2 – E6 = -2.18 x 10-18 J/22 – -2.18 x 10-18 J/62
E = -5.45 E-19 J – (-6.05 E-20 J) = -4.85 x 10-19 J
b.
Calculate the wavelength of the radiation associated
with the spectral line. (Ephoton = (2.00 x 10-25 J•m)/)
Ephoton = (2.00 x 10-25 J•m)/
4.85 x 10-19 J = (2.00 x 10-25 J•m)/  = 4.12 x 10-7 m