# Atomic Structure Review KEY

```Dr. G. Pernía - Atomic Structure Review – ANSWER KEY
ATOMIC PARTICLES: Define the following terms.
PROTON: IDENTIFIES an element
NEUTRON: Inside nucleus, stabilizes the nucleus, No charge
ELECTRON: negatively charged, outside nucleus
MASS NUMBER: number of particles inside nucleus: proton + neutron.
ATOMIC NUMBER: The number of PROTONS in an atom
ATOM: EQUAL number of electrons and protons
ION: A NEGATIVELY charged particle
ISOTOPE: Has a DIFFERENT number of electrons and protons therefore DIFFERENT atomic masses
What would happen to a neutral atom if
We changed the amount of PROTONS? ______CREATE A NEW ELEMENT__________________
We changed the amount of NEUTRONS? _____CREATE AN ISOTOPE______________________
We changed the amount of ELECTRONS? ____CREATE AN ION___________________________
AVERAGE ATOMIC MASS
1. What is the atomic mass of hafnium if, out of every 100 atoms, 5 have a mass of 176; 19 have a mass of
177; 27 have a mass of 178; 14 have a mass of 179; and 35 have a mass of 180.0?
(176 x 0.05) + (177 x 0.19) + (178 x 0.27) + (179 x 0.14) + (180 x 0.35)
= 178.55 amu
2. Calculate the average atomic mass of lithium, which occurs as two isotopes that have the following atomic
masses and abundances in nature: 6.017 amu with 7.30%; and 7.018 amu with 92.70%.
(6.017 x 0.073) + (7.018 x 0.927)
= 6.945 amu
3. Hydrogen is 99% 1H, 0.8% 2H, and 0.2% 3H. Calculate its average atomic mass.
(1 x 0.99) + (2 x 0.008) + (3 x 0.002)
= 1.012 amu
4. Calculate the average atomic mass of magnesium using the following data for three magnesium isotopes.
Isotope
mass (u)
relative abundance
Mg-24
23.985
0.7870
Mg-25
24.986
0.1013
Mg-26
25.983
0.1117
(23.985 x 0.7870) + (24.986 x 0.1013) + (25.983 x 0.1117)
= 24.310 amu
SCIENTISTS – fill in 4 scientists’ names and complete the table. (Only TWO of the scientists have
experiments!)
Nuclear Model
Gold Foil experiment
Discovered electrons
Atomic Theory
SCIENTIST
energy levels
positively charged nucleus
EXPERIMENT
Dalton
CONCEPTS/IDEAS
“Solar System”
“Billiard Ball”
Cathode Ray Tube
MODEL
Atomic Theory
“Billiard Ball”
Thomson
Cathode Ray Tube
Discovered electrons
Pudding”
Rutherford
Gold Foil experiment
positively charged
nucleus
Nuclear Model
energy levels
“Solar System”
Bohr
UNDERSTANDING ISOTOPIC NOTATION
DIRECTIONS: Fill in the missing information for the given element symbols. Round Atomic Masses to
the nearest whole number.
SYMBOL
K-40

2
Na+

Al3+
2
# OF
PROTONS
ELEMENT,
# OF
# OF
ION, or
NEUTRONS ELECTRONS ISOTOPE?
40
19
21
19
isotope
48
113
48
65
48
isotope
oxygen
8
18
8
10
10
ion
sodium
11
23
11
12
10
ion
carbon
6
14
6
8
6
isotope
aluminum
13
27
13
14
10
ion
sulfur
16
332
16
16
18
ion
barium
56
138
56
72
56
isotope
NAME
ATOMIC
NUMBER
potassium
19
ATOMIC
MASS
Ba-138
MOLAR MASS CONVERSIONS
How many ATOMS are in:
1.
6.15 moles of sulfur
6.15 mol S x 6.02 x 1023 atoms
1 mol S
2.
13 g of cadmium
13 g x
3.
= 3.70 x 1024 atoms
1 mol Cd
112.411 g
x 6.02 x 1023 atoms
1 mol Cd
= 6.96 x 1022 atoms
x 6.02 x 1023 atoms
1 mol Sn
= 8.93 x 1023 atoms
176 g of tin
176 g x
1 mol Sn
118.71 g
How many MOLES are in:
1. 891 g of sodium
891 g Na x
1 mol
= 38.8 mol
22.99 g
2. 2.03 x 1022 atoms of lead
2.03 x 1022 atoms Pb x
1 mol
= 0.0337 mol
23
6.02 x 10 atoms
Determine the MASS of:
1. 0.380 moles of potassium
0.380 moles K x 39.098 g
1 mol K
2. 9.64 x 1025 atoms of silver
9.64 x 1025 atoms x
= 14.86 g
1 mol Ag
6.02 x 1023 atoms
x
107.868 g
1 mol Ag
= 17,273 g
3. 46.99 x 1014 atoms of barium - report your answer in regular and scientific notation
46.99 x 1014 atoms x
1 mol Ba
x
137.327 g
= 1.071 x10-6 g =
6.02 x 1023 atoms
1 mol Ba
= 0.000001071 g
NUCLEAR REACTION EQUATIONS
1.
234
90ℎ
2.
228
89
4
2
_____0-1β___________ + 228
90ℎ
3. ___23290Th________
4.
5.
113
48
40
19
+ ____23088Ra__________
+ 10
+ −10
4
2
+
228
88
114
48Cd_________
40
20Ca_______
+
HALF-LIFE
1. How much of a 100.0g sample of 198Au is left after 8.10 days if its half-life is 2.70 days?
A = A0 x (½)n
8.1/2.7 = 3 half-lives.
A = 100 x (½)3 = 12.5 g
2. A 50.0g sample of 16N decays to 12.5g in 14.4 seconds. What is its half-life?
A = A0 x (½)n, so A/ A0 = (½)n
So 50/12.5 = (½)2 = 2 half-lives.
So 1 half-life = 14.4/2 = 7.2 s
3. The half-life of 42K is 12.4 hours. How much of a 750g sample is left after 62.0 hours?
62/12.4 = 5 half-lives.
A = A0 x (½)n
So A = 750 x (½)5 = 23.4375 g
4. What is the half-life of 99Tc if a 500g sample decays to 62.5g in 639,000 years?
A = A0 x (½)n
500/62.5 = (½)3 so 3 half lives
639,000/3 = 213,000 years = 1 half-life
5. There are 5.0g of 131I left after 40.35 days. How many grams were in the original sample if its half-life is
8.07 days?
A = A0 x (½)n, so A0 = A / (½)n
40.35/8.07 = 5 half-lives
So A = 5 / (½)5 = 160 g
```