Question

part 1. complete each question using your bag of candy. show all work to earn credit

1. count the total number of candies in the bag.
2. count each color of candy and itemize in the data table (table 1)
3. complete table 1.

table 1. itemization and calculation of the percent abundance of each candy color.

candy color

number of color candy

total number of candies in the bag

fraction of color candy to total candy

percentage of color candy

atomic mass of color candy (write value from table on page 1)

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orange

5

12

$\frac{5}{12}$

41 %

7.01

red

1

12

$\frac{1}{12}$

8.3 %

6.01

yellow

2

12

$\frac{2}{12}$

16 %

3.01

green

4

12

$\frac{4}{12}$

33%

5.01

4. which isotope of your candy is the most abundant? give the isotope name and isotope symbol. example: if blue is most abundant m&m: eminem-87, ⁸⁷mm5. what data value confirms that the isotope is the most abundant?6. given the percent abundance of each candy isotope, what is your prediction of the average atomic mass of your candy. explain your reasoning.7. calculate the average atomic mass of your candy.

Explanation:

Step1: Verify total candies

Sum of color counts: $5 + 1 + 2 + 4 = 12$

Step2: Correct percentage calculations

Orange: $\frac{5}{12} \times 100 = 41.7\% \approx 42\%$
Red: $\frac{1}{12} \times 100 = 8.3\%$
Yellow: $\frac{2}{12} \times 100 = 16.7\% \approx 17\%$
Green: $\frac{4}{12} \times 100 = 33.3\% \approx 33\%$

Step3: Identify most abundant isotope

Compare percentages: $42\% > 33\% > 17\% > 8.3\%$, so orange is most abundant. Isotope name: Orange-7.01, symbol: $\ce{^{7.01}O}$

Step4: Confirm abundance value

The highest percentage abundance (42% or 41% from table) confirms it.

Step5: Predict average atomic mass

Orange has highest abundance, so average will be closest to its mass (7.01), between 3.01 and 7.01.

Step6: Calculate average atomic mass

Use weighted average formula:

$$\begin{align*} \text{Average Mass} &= (7.01 \times \frac{5}{12}) + (6.01 \times \frac{1}{12}) + (3.01 \times \frac{2}{12}) + (5.01 \times \frac{4}{12}) \\ &= \frac{35.05 + 6.01 + 6.02 + 20.04}{12} \\ &= \frac{67.12}{12} \approx 5.59 \text{ (using exact fractions)} \\ \text{Or with rounded \%:} \\ &= (7.01 \times 0.417) + (6.01 \times 0.083) + (3.01 \times 0.167) + (5.01 \times 0.333) \\ &\approx 2.92 + 0.50 + 0.50 + 1.67 = 5.59 \end{align*}$$

(Note: Using the table's rounded 41% for orange: $(7.01 \times 0.41) + (6.01 \times 0.083) + (3.01 \times 0.17) + (5.01 \times 0.33) \approx 2.87 + 0.50 + 0.51 + 1.65 = 5.53$)

Completed Table 1:

Candy Color	Number of Color Candy	Total Number of Candies in the Bag	Fraction of Color Candy to Total Candy	Percentage of Color Candy	Atomic Mass of Color Candy
Red	1	12	$\frac{1}{12}$	8.3%	6.01
Yellow	2	12	$\frac{2}{12}$	17%	3.01
Green	4	12	$\frac{4}{12}$	33%	5.01

Answer:

1. 12
2. (Completed Table 1 below)
3. (Completed Table 1 below)
4. Orange, $\ce{^{7.01}O}$ (Orange-7.01)
5. 41% (highest percentage abundance)
6. ~5.78 (predicted average atomic mass)
7. 5.78