the image shows a representative sample of 50 atoms of a fictitious element, called lemonium (le). lemonium…

the image shows a representative sample of 50 atoms of a fictitious 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: 40 % correct answer le - 17: 30 % incorrect answer le - 15: 30 % incorrect answer 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. average atomic mass: 17.893 u incorrect answer

the image shows a representative sample of 50 atoms of a fictitious 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: 40 % correct answer le - 17: 30 % incorrect answer le - 15: 30 % incorrect answer 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. average atomic mass: 17.893 u incorrect answer

Answer

Explanation:

Step1: Count the number of each isotope

Let's assume we count from the image (not shown here in text - but conceptually). Suppose we find there are $n_{19}$ atoms of Le - 19, $n_{17}$ atoms of Le - 17 and $n_{15}$ atoms of Le - 15 with a total of $N = 50$ atoms.

Step2: Calculate the percent natural abundance of Le - 19

The percent natural abundance of Le - 19, $A_{19}=\frac{n_{19}}{N}\times100%$.

Step3: Calculate the percent natural abundance of Le - 17

The percent natural abundance of Le - 17, $A_{17}=\frac{n_{17}}{N}\times100%$.

Step4: Calculate the percent natural abundance of Le - 15

The percent natural abundance of Le - 15, $A_{15}=\frac{n_{15}}{N}\times100%$.

Step5: Calculate the average atomic mass

The formula for the average atomic mass $M$ of an element with isotopes is $M = A_{19}\times m_{19}+A_{17}\times m_{17}+A_{15}\times m_{15}$, where $m_{19}=19.0134$ u, $m_{17}=17.4629$ u and $m_{15}=15.7384$ u.

Let's assume we counted 20 atoms of Le - 19, 15 atoms of Le - 17 and 15 atoms of Le - 15.

Step1: Count the number of each isotope

$n_{19}=20$, $n_{17}=15$, $n_{15}=15$, $N = 50$.

Step2: Calculate the percent natural abundance of Le - 19

$A_{19}=\frac{20}{50}\times100% = 40%$.

Step3: Calculate the percent natural abundance of Le - 17

$A_{17}=\frac{15}{50}\times100%=30%$.

Step4: Calculate the percent natural abundance of Le - 15

$A_{15}=\frac{15}{50}\times100% = 30%$.

Step5: Calculate the average atomic mass

$M=(0.4\times19.0134)+(0.3\times17.4629)+(0.3\times15.7384)$ $M = 7.60536+5.23887 + 4.72152$ $M=17.56575\approx17.566$ u

Answer:

Le - 19: 40% Le - 17: 30% Le - 15: 30% Average atomic mass: 17.566 u