Chapter 2, Problem 2.25QP

Average Atomic Weight

Part 1: Consider the four identical spheres below, each with a mass of 2.00 g.

Calculate the average mass of a sphere in this sample.

Part 2: Now consider a sample that consists of four spheres, each with a different mass: blue mass is 2.00 g, red mass is 1.75 g, green mass is 3.00 g, and yellow mass is 1.25 g.

1. a Calculate the average mass of a sphere in this sample.
2. 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?

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.

1. a If you were to remove 50 blue spheres from the jar containing just the blue spheres, what would be the total mass of spheres left in the jar? (Note that the masses of the spheres are given in Part 2.)
2. b If you were to remove 50 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?
3. 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?
4. d If you had 80.0 grams of spheres from the blue sample, how many spheres would you have?
5. e If you had 60.0 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?

Part 4: Consider a sample that consists of three green spheres and one blue sphere. The green mass is 3.00 g, and the blue mass is 1.00 g.

1. a Calculate the fractional abundance of each sphere in the sample.
2. b Use the fractional abundance to calculate the average mass of the spheres in this sample.
3. c How are the ideas developed in this Concept Exploration related to the atomic weights of the elements?

Interpretation Introduction

Interpretation:

The average mass of a sphere in the given sample has to be calculated.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.

Answer to Problem 2.25QP

The average mass of sphere is $\text{2}\text{.00}\text{g}$.

Explanation of Solution

It is given that mass of single blue sphere is $\text{2}\text{.00}\text{g}$.  Totally, four spheres are given in the figure.  Therefore, the average mass can be calculated as shown below,

$\begin{array}{l}\text{Average mass = }\frac{\text{Sum of mass of all spheres}}{\text{Total number of spheres}}\\ =\frac{\text{2}\text{.00 g + 2}\text{.00 g + 2}\text{.00 g + 2}\text{.00 g}}{\text{4}}\\ =\frac{\text{8}\text{.00 g}}{\text{4}}=2.00\text{g}\end{array}$

Therefore, the average mass was calculated as shown above and found to be $\text{2}\text{.00}\text{g}$.

Conclusion

The average mass of the given spheres was calculated.

Interpretation Introduction

Interpretation:

The average mass of a sphere in the given sample has to be calculated.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.

Answer to Problem 2.25QP

The average mass of sphere is $\text{2}\text{.00}\text{g}$.

Explanation of Solution

It is given that mass of single blue sphere is $\text{2}\text{.00}\text{g}$.  Totally, four spheres are given in the figure.

Each sphere has different mass;

Blue sphere has mass of $\text{2}\text{.00}\text{g}$.

Red sphere has mass of $\text{1}\text{.75}\text{g}$.

Green sphere has mass of $\text{3}\text{.00}\text{g}$.

Orange sphere has mass of $\text{1}\text{.25}\text{g}$.

Therefore, the average mass can be calculated as shown below,

$\begin{array}{l}\text{Average mass = }\frac{\text{Sum of mass of all spheres}}{\text{Total number of spheres}}\\ =\frac{\text{2}\text{.00 g + 1}\text{.75 g + 3}\text{.00 g + 1}\text{.25 g}}{\text{4}}\\ =\frac{\text{8}\text{.00 g}}{\text{4}}=2.00\text{g}\end{array}$

Therefore, the average mass was calculated as shown above and found to be $\text{2}\text{.00}\text{g}$.

Conclusion

The average mass of the given spheres was calculated.

Interpretation Introduction

Interpretation:

How the average mass for sphere is same as that of in Part 1.  How such different samples can have same average mass has to be explained.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.

Explanation of Solution

The calculated mass is the average mass of spheres in both the samples.  It is not a mass of individual sphere.  As it does not represent the mass of individual spheres the average mass of the two samples that contain spheres is same.

Conclusion

The reason for the two samples having same average mass was explained.

Interpretation Introduction

Interpretation:

The total mass of spheres that is left in the jar after removing 50 blue spheres has to be calculated.

Concept Introduction:

The mass of the sample is the total of individual mass of all the contents in the sample under consideration.

Answer to Problem 2.25QP

The mass of spheres left in the jar is $\text{100}\text{.0}\text{g}$.

Explanation of Solution

It is given that the jar contains a total of 100 blue spheres.  Each blue sphere weighs $\text{2}\text{.00}\text{g}$.  If 50 blue spheres are removed from the jar, the total mass of spheres left in the jar can be calculated as,

$\frac{\text{50 spheres}}{\text{1}}\text{×}\frac{\text{2}\text{.00 g}}{\text{1 sphere}}\text{= 100}\text{.0}\text{g}$

The mass of spheres left out in the jar was calculated as shown above.

Conclusion

The mass of spheres left out in the jar was calculated.

Interpretation Introduction

Interpretation:

The total mass of spheres that is left in the jar after removing 50 spheres has to be calculated.

Concept Introduction:

The mass of the sample is the total of individual mass of all the contents in the sample under consideration.

Answer to Problem 2.25QP

The mass of spheres left in the jar is $\text{100}\text{.0}\text{g}$.

Explanation of Solution

It is given that the jar contains a total of 100 blue spheres.  If 50 spheres are removed from the jar, then 50 spheres are left in the jar.  The average mass obtained in Part 1 and Part 2a is $\text{2}\text{.00 g}$ for a single sphere.  The total mass of spheres left in the jar can be calculated as,

$\frac{\text{50 spheres}}{\text{1}}\text{×}\frac{\text{2}\text{.00 g}}{\text{1 sphere}}\text{= 100}\text{.0}\text{g}$

The mass of spheres left out in the jar was calculated as shown above.

Conclusion

The mass of spheres left out in the jar was calculated.

Interpretation Introduction

Interpretation:

In case of mixture of spheres, is the average mass represent the mass of individual sphere has to be explained.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.

Explanation of Solution

The calculated mass is the average mass of spheres in the sample.  It is not a mass of individual sphere.  As it does not represent the mass of individual spheres, the average mass of the sphere cannot be mass of individual sphere.

Interpretation Introduction

Interpretation:

The total number of blue sphere in the given sample which weighs $\text{80}\text{.0}\text{g}$ has to be calculated.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.  From this we can calculate the total number of elements also.

Answer to Problem 2.25QP

The total number of blue spheres is 40.0

Explanation of Solution

It is given that mass of single blue sphere is $\text{2}\text{.00}\text{g}$.  Totally, the mass of the sample is given as $\text{80}\text{.0}\text{g}$.

Therefore, the total number of blue sphere present can be calculated as shown below,

$\begin{array}{l}\text{Total number of spheres = }\frac{\text{Sum of mass of all spheres}}{\text{Average mass}}\\ =\frac{\text{80}\text{.0 g}}{\text{2}\text{.00}}\\ =40.0\end{array}$

Therefore, the total number of blue spheres was calculated as shown above and found to be 40.0.

Conclusion

The total number of blue spheres was calculated.

Interpretation Introduction

Interpretation:

The total number of sphere in the given sample which weighs $\text{60}\text{.0}\text{g}$ has to be calculated.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.  From this we can calculate the total number of elements also.

Answer to Problem 2.25QP

The total number of spheres is 30.0

Explanation of Solution

It is given that average mass of sphere is $\text{2}\text{.00}\text{g}$.  Totally, the mass of the sample is given as $\text{60}\text{.0}\text{g}$.

Therefore, the total number of sphere present can be calculated as shown below,

$\begin{array}{l}\text{Total number of spheres = }\frac{\text{Sum of mass of all spheres}}{\text{Average mass}}\\ =\frac{\text{60}\text{.0 g}}{\text{2}\text{.00}}\\ =30.0\end{array}$

Therefore, the total number of spheres was calculated as shown above and found to be 30.0.

We can assume that the sample of the spheres was very well mixed.

Conclusion

The total number of spheres was calculated.

Interpretation Introduction

Interpretation:

The fractional abundances for the each sphere in the given sample has to be obtained.

Concept Introduction:

If the element under consideration has isotope means then fractional abundance has to be considered for the isotope that naturally occurs on the planet.  This is usually expressed in percentage of the isotopes for the element under consideration in nature.  The total fractional abundance of the isotopes of a particular element is always equal to 1.

Answer to Problem 2.25QP

The fractional abundance of green sphere is 0.750

The fractional abundance of blue sphere is 0.250

Explanation of Solution

The total of fractional abundances of spheres is always equal to 1.

The fractional abundance of green spheres can be calculated as,

$X=\frac{\text{3 green}}{\text{3 green + 1 blue}}=0.750$

The fractional abundance of green sphere was found as 0.750

The fractional abundance of blue sphere can be calculated as,

$X=\frac{\text{1 blue}}{\text{3 green + 1 blue}}=0.250$

The fractional abundance of blue sphere was found as 0.250

Conclusion

The fractional abundance of the two spheres were calculated.

Interpretation Introduction

Interpretation:

The average mass of a sphere in the given sample has to be calculated using fractional abundance.

Concept Introduction:

Average mass is the sum of mass of all the elements present divided by the total number of elements.

Answer to Problem 2.25QP

The average mass of sphere is $\text{2}\text{.50}\text{g}$.

Explanation of Solution

The fractional abundance of green sphere is 0.750 and blue sphere is 0.250.

From the fractional abundance, the average mass can be calculated as shown below,

$\begin{array}{l}\text{Average mass = (0}\text{.750)}\times \frac{\text{3}\text{.00 g}}{\text{1 green sphere}}+\text{(0}\text{.250)}\times \frac{\text{1}\text{.00 g}}{\text{1 blue sphere}}\\ =\frac{\text{2}\text{.250 g + 0}\text{.250 g}}{1}\\ =2.50\text{g}\end{array}$

Therefore, the average mass was calculated as shown above and found to be $\text{2}\text{.50}\text{g}$.

Conclusion

The average mass of the given spheres was calculated using fractional abundance.

Interpretation Introduction

Interpretation:

How the ideas are developed for this concept explorations related to atomic weight has to be given.

Explanation of Solution

Atomic weight of an element is the average atomic weight of all the isotope present in it.  This gives a universal weight.  The fractional abundances of the individual isotope of the element can be used to calculate atomic weight of element.