calculate the atomic mass of element R

Question

Element R has three isotopes. The isotopes are present in 0.0398, 0.1614, and 0.7988 relative abundance. If their masses are 196, 178, and 151 respectively, calculate the atomic mass of element R. (No decimals)

Answer 1

To calculate the atomic mass of element R, we can use the following formula, which takes into account the masses and relative abundances of the isotopes:

Atomic Mass=(m₁×f₁)+(m₂×f₂)+(m₃×f₃)

Where:

1. m₁,m₂,m₃ are the masses of the isotopes,
2. f_1​,f₂​,f_3​ are their respective relative abundances.

Given values:

1. Masses: m₁​=196, m₂​=178, m₃=151
2. Relative abundances: f₁=0.0398, f₂=0.1614, f₃=0.7988

Let's calculate the atomic mass:

Atomic Mass=(196×0.0398)+(178×0.1614)+(151×0.7988)

Atomic Mass= 7.8008 + 28.7292+120.6188 = 157.1488 amu

The atomic mass of element R is approximately 157 (rounded to the nearest whole number)..

Answer 2

Hi Shannon!

to calculate the average atomic mass with given isotopes, all we need is each individual isotope's atomic mass and relative abundance.

For each individual isotope:

1. Convert the percentage of abundance to decimal form by dividing by 100 (In your situation, they've already done that work for us, so we can skip this step.
2. Multiply the atomic mass of that isotope by the abundance in decimal form

Once we have done those two steps, we add up all the values for each isotope, giving us our average atomic mass.

In this case, Isotope 1 has a mass of 196 and an abundance of 0.0398, isotope 2 has a mass of 178 and an abundance of 0.1614, and isotope 3 has a mass of 151 and an abundance of 0.7988.

Isotope 1: 196 * 0.0398 = 7.8008

Isotope 2: 178 * 0.1614 = 28.7292

Isotope 3: 151 * 0.7988 = 120.6188

Now we add up all 3 of those values:

7.8008 + 28.7292 + 120.6188 = 157.1488 amu

Here they do not want decimals, so we round this answer, giving us 157 amu as the average atomic mass of element R.

Note:

1. An average atomic mass will be a value closest to the isotope with the largest abundance. Since that was isotope 3 with an abundance of 0.7988 or 79.88%, we see that 157 is closest to 151.
2. If we were given abundance in percentage, we can either convert them into decimals or just keep them as percentages, but divide by 100 before I get my final answer.

Hope this helps!