Question

independent practice: working silently and independently, answer the following questions on isotopes and average atomic mass. be sure to write out all calculations, write out all steps and remember to put the correct units!!

1. a 100.00 - gram sample of naturally occurring boron contains 19.78 grams of boron - 10 (atomic mass 10.01 atomic mass units) and 80.22 grams of boron - 11 (atomic mass = 11.01 atomic mass units). calculate the average atomic mass of boron.
2. a 100.0 gram sample of element x contains 72.0g of an isotope of element x with an isotopic mass of 84.9 amu and 28g of the isotope of an element x with an isotopic mass of 87.0 amu. calculate the average atomic mass of element x.
3. calculate the average atomic mass of neon using the following data.
4. show the numerical setup for calculating the average atomic mass of the following atom that has 3 naturally occurring isotopes. show all your working out.

Explanation:

Question 1

Step1: Calculate the percentage abundance of each isotope

The percentage abundance of boron - 10: \(\frac{19.78}{100}\times100\% = 19.78\%\)
The percentage abundance of boron - 11: \(\frac{80.22}{100}\times100\%=80.22\%\)

Step2: Use the formula for average atomic mass \(M=\sum_{i}M_i\times x_i\) (where \(M_i\) is the atomic mass of isotope \(i\) and \(x_i\) is its percentage abundance in decimal form)

For boron, \(M=(10.01\times0.1978)+(11.01\times0.8022)\)
First term: \(10.01\times0.1978 = 1.979978\)
Second term: \(11.01\times0.8022=8.832222\)
Sum: \(1.979978 + 8.832222=10.8122\approx10.81\) atomic mass units

Step1: Calculate the percentage abundance of each isotope

The percentage abundance of the first isotope of \(X\): \(\frac{72.0}{100}\times 100\%=72.0\%\)
The percentage abundance of the second isotope of \(X\): \(\frac{28}{100}\times100\% = 28\%\)

Step2: Use the formula \(M=\sum_{i}M_i\times x_i\)

\(M=(84.9\times0.72)+(87.0\times0.28)\)
First term: \(84.9\times0.72 = 61.128\)
Second term: \(87.0\times0.28 = 24.36\)
Sum: \(61.128+24.36=85.488\approx85.5\) amu

Step1: Use the formula \(M=\sum_{i}M_i\times x_i\)

For neon, \(M=(19.99\times0.90)+(20.99\times0.003)+(21.99\times0.088)\)
First term: \(19.99\times0.90=17.991\)
Second term: \(20.99\times0.003 = 0.06297\)
Third term: \(21.99\times0.088=1.93512\)
Sum: \(17.991+0.06297 + 1.93512=19.98909\approx20.0\)

Answer:

The average atomic mass of boron is \(10.81\) atomic mass units.

Question 2