Question

The molar mass of boron atoms in a natural sample is 10.71 g/mol. The sample is known to consist of 10B (molar mass 10.013 g/mol) and 11B (molar mass 11.093 g/mol). What is the percentage abundance of 11B? Answer in units of %.

Answer 1

The percentage abundance of 11B is 8.37%.

Step-by-step explanation:

The percentage abundance of 11B can be calculated using the given molar masses. Let x represent the percentage abundance of 11B:

x(11.093 g/mol) + (100 - x)(10.013 g/mol) = 10.71 g/mol

Simplifying the equation:

11.093x + 1000.13 - 10.013x = 10.71

0.080x = 0.67

x = 8.37%

Therefore, the percentage abundance of 11B is 8.37%.

Answer 2

Answer: The percentage abundance of
`_(5)^(11)\textrm{B}` and 65.54 %

Step-by-step explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

`\text{Average atomic mass }=\sum_(i=1)^n\text{(Atomic mass of an isotopes)}_i* \text{(Fractional abundance})_i` .....(1)

Let the fractional abundance of
`_(5)^(10)\textrm{B}` isotope be 'x'. So, fractional abundance of
`_(5)^(11)\textrm{B}` isotope will be '1 - x'

• For
  `_(5)^(10)\textrm{B}` isotope:

Mass of
`_(5)^(10)\textrm{B}` isotope = 10.013 amu

Fractional abundance of
`_(5)^(10)\textrm{B}` isotope = x

• For
  `_(5)^(11)\textrm{B}` isotope:

Mass of
`_(5)^(11)\textrm{B}` isotope = 11.093 amu

Fractional abundance of
`_(5)^(11)\textrm{B}` isotope = 1 - x

Average atomic mass of boron = 10.71 amu

Putting values in equation 1, we get:

`10.71=[(10.013* x)+(11.093* (1-x))]\\\\x=0.3546`

Percentage abundance of
`_(5)^(10)\textrm{B}` isotope =
`0.3546* 100=35.46\%`

Percentage abundance of
`_(5)^(11)\textrm{B}` isotope =
`(1-0.3546)=0.6454* 100=64.54\%`

Hence, the percentage abundance of
`_(5)^(11)\textrm{B}` and 65.54 %