name: period: model 3 - proposed average atomic mass calculations marys method (78.99)(23.9850 amu) +…

name: period: 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 13. complete the three proposed calculations for the average atomic mass of magnesium in model 3. 14. 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? b. explain why the mathematical reasoning was incorrect for any method(s) in model 3 that did not give the correct answer for average atomic mass (the one on the periodic table). c. for the methods in model 3 that gave the correct answer for average atomic mass, explain why they give the same answer.

name: period: 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 13. complete the three proposed calculations for the average atomic mass of magnesium in model 3. 14. 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? b. explain why the mathematical reasoning was incorrect for any method(s) in model 3 that did not give the correct answer for average atomic mass (the one on the periodic table). c. for the methods in model 3 that gave the correct answer for average atomic mass, explain why they give the same answer.

Answer

Explanation:

Step1: Calculate Mary's Method

First, calculate the numerator: [ \begin{align*} &(78.99)(23.9850)+(10.00)(24.9858)+(11.01)(25.9826)\ =&78.99\times23.9850 + 10.00\times24.9858+11.01\times25.9826\ =&1899.98015+249.858 + 286.078426\ =&2435.916576 \end{align*} ] Then divide by 100: $\frac{2435.916576}{100}=24.35916576$ amu.

Step2: Calculate Jack's Method

[ \begin{align*} &(0.7899)(23.9850)+(0.1000)(24.9858)+(0.1101)(25.9826)\ =&0.7899\times23.9850+0.1000\times24.9858 + 0.1101\times25.9826\ =&18.9998015+2.49858+2.86078426\ =&24.35916576 \end{align*} ] amu.

Step3: Calculate Alan's Method

[ \begin{align*} &\frac{23.9850 + 24.9858+25.9826}{3}\ =&\frac{74.9534}{3}\ =&24.984466\cdots \end{align*} ] amu.

Step4: Answer part 14 - a

The average atomic - mass of magnesium on the periodic table is approximately 24.305 amu. Mary's and Jack's methods give the correct answer.

Step5: Answer part 14 - b

Alan's method is incorrect because it does not take into account the relative abundances of the isotopes of magnesium. The average atomic mass is a weighted - average based on the percentage abundances of each isotope, not a simple arithmetic mean of the masses of the isotopes.

Step6: Answer part 14 - c

Mary's and Jack's methods give the same answer because Mary's method has percentages out of 100 and Jack's method has the decimal - equivalent of those percentages. When you divide Mary's numerator (the sum of the products of percentage and mass) by 100, it is equivalent to multiplying by the decimal - form of the percentages as in Jack's method.

Answer:

  1. Mary's method: 24.35916576 amu; Jack's method: 24.35916576 amu; Alan's method: 24.984466… amu

a. Mary's and Jack's methods. b. Alan's method is incorrect because it does not consider relative abundances; it uses a simple arithmetic mean instead of a weighted - average. c. Mary's and Jack's methods are equivalent as Mary's method uses percentages out of 100 and Jack's uses the decimal - equivalent of those percentages.