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HW 7 Answer Key

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Name _________KEY_____________ Date _____________

Class __________________________ Mr. D’Amico

Homework # 7

1) What is an isotope? How is it different from an ion?

Isotopes are atoms of a given element (same amount of protons) but have
differing amounts of neutrons. The varying amount of neutrons causes the
isotopes to have different masses.

2) A chemistry laboratory has analyzed the composition of isotopes of several


elements. Data for one of isotopes of the elements is given in the table below.

Element Atomic Mass Protons Electrons Neutrons Symbol Name


Number Number
22
neon 10 22 10 10 12 Ne Neon-22
46
calcium 20 46 20 20 26 Ca Calcium-46
17
oxygen 8 17 8 8 9 O Oxygen-17
57
iron 26 57 26 26 31 Fe Iron-57
64
zinc 30 64 30 30 34 Zn Zinc-64
204
mercury 80 204 80 80 124 Hg Mercury-
204

a) Determine the number of protons, neutrons, and electrons in each


isotope.

b) Name each isotope (including mass number) and give its symbol
(including mass number).

3) How do isotopes of the same element differ? Give specific examples.

Isotopes have different amounts of neutrons, and therefore different masses. For
example, argon-40 has 22 neutrons, while argon-42 has 24 neutrons.
4) Compare mass number and average atomic mass.

The average atomic mass is the weighted average of the masses of all of the
isotopes of an element. The mass number refers to the mass of one of those
elements, and is always a whole number.

5) Silver has two isotopes, 107Ag has a mass of 106.905 amu (52.00%) and 109Ag
has a mass of 108.905 amu (48.00%). What is the average atomic mass of
silver? Box your answer.

(106.905)(.5200) + (108.905)(.4800) = 107.86 amu

6) Chlorine, which has an average atomic mass of 35.453 amu, has two naturally
occurring isotopes, chlorine-35 and chlorine-37. Which isotope occurs in
greater abundance? Explain.

Chlorine-35 is in greater abundance due to the weighted average atomic mass of


35.453 amu.

7) A scientist investigating an element used for medical technology found that it


has 2 naturally occurring isotopes: one has a mass of 6.015 amu that occurs at
7.5% abundance; the other has a mass of 7.016 amu and occurs at 92.5%
abundance. Identify the element based on its average atomic mass. Box your
answer.

(.075)(6.015) + (.925)(7.016) = 6.94 amu; the element is lithium.

8) Titanium has 5 naturally occurring isotopes:


a) titanium-46 occurs at 8.0%
b) titanium-47 occurs at 7.30%
c) titanium-48 occurs at 73.80%
d) titanium-49 occurs at 5.50%
e) titanium-50 occurs at 5.40%
Calculate the average atomic mass of titanium. Box your answer.

(.08)(46) + (.0730)(47) + (.7380)(48) + (.0550)(49) + (.0540)(50) =

3.68 + 3.431 + 35.424 + 2.695 + 2.7 = 47.93 amu

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