# Chemistry 205 Chapter 2 Concept Explorations

Topics: Atom, Isotope, Neutron Pages: 5 (1110 words) Published: May 27, 2013

Assignment Chapter 2
Concept Explorations
2.25. Average Atomic Mass
Part 1:
Consider the four identical spheres below, each with a mass of 4.00 g.
a. Calculate the average mass of a sphere in this sample.  (4.00 + 4.00 + 4.00 + 4.00)/4= 16.00/4= 4.00g

Part 2:
Now consider a sample that consists of four spheres, each with a different mass: blue mass is 4.00 g, red mass is 3.75 g, green mass is 3.00 g, and yellow mass is 1.25 g.
* a. Calculate the average mass of a sphere in this sample.  (4.00 + 3.75 + 3.00 + 1.25)/4= 3.00g

* b. How does the average mass for a sphere in this sample compare with the average mass of the sample that consisted just of the blue spheres? How can such different samples have their averages turn out the way they did?

The sizes of these spheres are different. This can be caused by many factors such as the temperature could have changed the composition or size of the spheres in part two or the spheres could be different sizes. These factors are unknown, but what is known is that the masses are different, so the average mass changed for varying masses as compared to a constant mass. Part 3:

Consider two jars. One jar contains 100 blue spheres, and the other jar contains 25 each of red, blue, green, and yellow colors mixed together.
* a. If you were to remove 30 blue spheres from the jar containing just the blue spheres, what would be total mass of spheres left in the jar? (Note that the masses of the spheres are given in Part 2.)  70 X 4.00= 280.00g

* b. If you were to remove 30 spheres from the jar containing the mixture (assume you get a representative distribution of colors), what would be the total mass of spheres left in the jar?

70 X 3.00= 210.00g
* c. In the case of the mixture of spheres, does the average mass of the spheres necessarily represent the mass of an individual sphere in the sample? No, because the average is not the exact mass for each sphere. It is an estimated number that represents the all the spheres. * d. If you had 60.0 grams of spheres from the blue sample, how many spheres would you have?

x X 4.00=60.0, 4.00x=60.0, x=60.0/4.00, x=15spheres
* e. If you had 50.0 grams of spheres from the mixed-color sample, how many spheres would you have? What assumption did you make about your sample when performing this calculation?  x X 3.00=50.0, 3.00x=50.0, x=50.0/3.00, x=16.67spheres

Part 4:
Consider a sample that consists of three green spheres and one blue sphere. The green mass is 2.00 g, and the blue mass is 3.00 g.  * a. Calculate the fractional abundance of each sphere in the sample.  There are 4 total spheres, but three spheres are green and one is blue. The fractional abundance of green spheres is ¾=.75

The fractional abundance of blue spheres is ¼=.25

* b. Use the fractional abundance to calculate the average mass of the spheres in this sample. Average atomic mass= (.75 X 2.00) + (.25 X 3.00)= 1.50 + .75= 2.25
* c. How are the ideas developed in this Concept Exploration related to the atomic masses of the elements? The average atomic mass is equal to the sum of the fractional abundance of the isotope. This is calculated using the same concept that you would use in calculating this for elements. However, you can’t go to check with element it is on the periodic table and to make sure the atomic mass is near the mass of the isotope.

2.26 Model of the Atom
Consider the following depictions of two atoms, which have been greatly enlarged so you can see the subatomic particles.
* a. How many protons are present in atom A? 3 protons

* b. What is the significance of the number of protons depicted in atom A or any atom?  Protons give the nucleus a positive charge. It also identifies which element it is. The number of protons in the nucleus is also the atomic number.

* c. Can you identify the real element...