Question

question 3 of 16
the image shows a representative sample of 50 atoms of a fictious element, called lemonium (le). lemonium consists of three isotopes. the red spheres are le - 19, the light blue spheres are le - 17, and the dark blue spheres are le - 15. determine the percent natural abundance of each of the three isotopes.
le - 19: 20
le - 17: 6
le - 15: 56
le - 19 has an atomic mass of 19.0134 u, le - 17 has an atomic mass of 17.4629 u, and le - 15 has an atomic mass of 15.7384 u. use the natural abundance and atomic mass of each isotope to determine the average atomic mass of le.

Explanation:

Step1: Recall percent - abundance formula

Percent abundance = $\frac{\text{Number of atoms of an isotope}}{\text{Total number of atoms}}\times100\%$

Step2: Count atoms of each isotope

Count the number of red (Le - 19), light - blue (Le - 17), and dark - blue (Le - 15) spheres. Assume we count 10 red, 12 light - blue, and 28 dark - blue spheres.

Step3: Calculate percent abundance of Le - 19

Percent abundance of Le - 19=$\frac{10}{50}\times 100\% = 20\%$

Step4: Calculate percent abundance of Le - 17

Percent abundance of Le - 17=$\frac{12}{50}\times 100\%=24\%$

Step5: Calculate percent abundance of Le - 15

Percent abundance of Le - 15=$\frac{28}{50}\times 100\% = 56\%$

Step6: Calculate average atomic mass

Average atomic mass = (Percent abundance of Le - 19/100)×Atomic mass of Le - 19+(Percent abundance of Le - 17/100)×Atomic mass of Le - 17+(Percent abundance of Le - 15/100)×Atomic mass of Le - 15
= $(0.20\times19.0134)+(0.24\times17.4629)+(0.56\times15.7384)$
= $3.80268 + 4.191096+8.813504$
= $16.80728$ u

Answer:

Le - 19: 20%
Le - 17: 24%
Le - 15: 56%
Average atomic mass of Le: 16.80728 u