Question

1. the periodic table does not show the atomic mass of every isotope for an element.

a. explain why this would be an impractical goal for the periodic table.
b. is it important to the average scientist to have information about a particular isotope of an element? explain.
10.what could be practical ways of showing the mass of magnesium atoms on the periodic table given that most elements occur as a mixture of isotopes?
11.propose a possible way to calculate the average atomic mass of 100 magnesium atoms. your answer may include a mathematical equation, but it is not required.
model 3: proposed average atomic mass calculations
marys method
(78.99)(23.9850 amu)+(10.00)(24.9858 amu)+(11.01)(25.9826 amu)/100
jacks method
(0.7899)(23.9850 amu)+(0.1000)(24.9858 amu)+(0.1101)(25.9826 amu)
alans method
(23.9850 amu + 24.9858 amu + 25.9826 amu)/3
12.complete the three proposed calculations for the average atomic mass of magnesium in model 3.
13.consider the calculations in model 3.
a. which methods shown in model 3 give an answer for average atomic mass that matches the mass of magnesium on the periodic table?

Explanation:

Step1: Calculate Mary's Method

\[

$$\begin{align*} &\frac{(78.99)(23.9850\ amu)+(10.00)(24.9858\ amu)+(11.01)(25.9826\ amu)}{100}\\ =&\frac{78.99\times23.9850 + 10.00\times24.9858+11.01\times25.9826}{100}\\ =&\frac{1904.56015+249.858 + 286.078426}{100}\\ =&\frac{2440.496576}{100}\\ =&24.40496576\ amu \end{align*}$$

\]

Step2: Calculate Jack's Method

\[

$$\begin{align*} &(0.7899)(23.9850\ amu)+(0.1000)(24.9858\ amu)+(0.1101)(25.9826\ amu)\\ =&0.7899\times23.9850+0.1000\times24.9858 + 0.1101\times25.9826\\ =&18.9557015+2.49858+2.86078426\\ =&24.31506576\ amu \end{align*}$$

\]

Step3: Calculate Alan's Method

\[

$$\begin{align*} &\frac{23.9850\ amu + 24.9858\ amu+25.9826\ amu}{3}\\ =&\frac{74.9534\ amu}{3}\\ =&24.98446667\ amu \end{align*}$$

\]

Step4: Answer question 13a

The method that gives an answer that matches the mass of magnesium on the periodic - table is Jack's method. The average atomic mass of magnesium on the periodic table is approximately \(24.31\ amu\) and Jack's method gives \(24.31506576\ amu\) which is closest.

Answer:

9a. There are a large number of isotopes for many elements. Listing the atomic mass of every isotope would make the periodic table extremely cluttered and difficult to read and use.
9b. Yes, it can be important. For example, in nuclear chemistry and some medical applications (like radioactive tracers), the properties of specific isotopes are crucial.

1. One practical way could be to show the average atomic mass of the element, which takes into account the relative abundances of its isotopes.
2. One way is to use the formula \(\text{Average Atomic Mass}=\sum_{i = 1}^{n}x_iM_i\), where \(x_i\) is the relative abundance of the \(i -\)th isotope and \(M_i\) is the atomic mass of the \(i -\)th isotope.
3. Mary's Method: \(24.40496576\ amu\); Jack's Method: \(24.31506576\ amu\); Alan's Method: \(24.98446667\ amu\)

13a. Jack's method.