Name ______________________
You should be able to complete this worksheet after studying Sections 4.1 Ð 4.3 in your textbook. This will let you see if you understand the terms we use to describe the structure of an atom.
1) What are the three major subatomic particlesÕ masses (amuÕs and grams), charge (elementary charges), and location (general vicinity)?
|
Name of Particle |
Charge on particle (el ch) |
Location |
|
Proton |
+1 |
In the nucleus. |
|
Neutron |
0 |
In the nucleus |
|
Electron |
-1 |
Outside the nucleus. |
2) Fill in the cells.
|
Symbol |
Atomic Number (Z) |
Mass Number (A) |
# of Protons |
# of Neutrons |
# of Electrons |
Atomic Mass (amu) |
|
1) 37Li |
3 |
7 |
3 |
4 |
3 |
6.941 |
|
2) 11H |
1 |
1 |
1 |
0 |
1 |
1.00794 |
|
3) 1939K1+ |
19 |
39 |
19 |
20 |
18 |
39.0983 |
|
4) 15O2- |
8 |
15 |
8 |
7 |
10 |
15.9994 |
|
5) 1327Al 3+ |
13 |
27 |
13 |
14 |
10 |
26.981538 |
|
6) 3065 Zn 2+ |
30 |
65 |
30 |
35 |
28 |
65.39 |
|
7) 24 He |
2 |
4 |
2 |
2 |
2 |
4.002602 |
|
8) 1735 Cl 1- |
17 |
35 |
17 |
18 |
18 |
35.453 |
|
9) 1428 Si 4+ |
14 |
28 |
14 |
14 |
10 |
28.0855 |
|
10) 92238U |
92 |
238 |
92 |
146 |
92 |
238.02891 |
Is #1 the most common isotope of Li? _yes Is #4 the most common isotope of O? no Briefly explain your reasoning.
I am assuming there is one dominant isotope that is responsible for most of the abundance of the element. If this is true, then rounding off the Atomic Mass to the nearest whole number will give the mass number of the most abundant isotope. Li rounds to 7, O rounds to 16.
3) Use the data in the table below to find the atomic mass of each element. Compare your calculation to the atomic mass listed on your periodic table.
* Natural Percent Abundance of Some Elements
|
Name |
Symbol |
Natural Percent Abundance |
Mass (AMU) |
Calculated Average Atomic Mass |
Accepted Average Atomic Mass |
|
Hydrogen |
1H 2H 3H |
99.985 .015 negligible |
1.0078 2.0141 3.0160 |
1.0079 |
1.00794 |
|
Oxygen |
16O 17O 18O |
99.759 .037 .204 |
15.995 16.995 17.999 |
15.995 |
15.9994 |
|
Chlorine |
35Cl 37Cl |
75.77 24.23 |
34.969 36.966 |
35.453 |
35.453 |
* Source: Prentice Hall Chemistry (P. 114)
Show your work in the space provided below.
Hydrogen
Average Atomic Mass
(.99985 x 1.0078) + ( .00015 x
2.0141) = 1.0079
Oxygen
Average Atomic Mass
(.99759 x 15.995) + ( .00037 x (16.995)
+ (.00204 x 17.999) = 15.995
Chlorine
Average Atomic Mass
(.7577 x 34.969) + ( .2423 x
36.966) = 35.453