Boron has an atomic mass of 10.8 amu. It is known that naturally occurring boron is composed of two isotopes, boron-10 and boron-11. Determine the percent of each isotope of boron

Boron has an atomic mass of 108 amu It is known that naturally occurring boron is composed of two isotopes boron10 and boron11 Determine the percent of each iso class=

Respuesta :

B-10 = 19.9%
B-11 = 80.1%

Abundance of B10 = x
Abundance of B11= y
You know x+y = 1 because there are only the 2 isotopes.
Y= 1-x

10.01294x + 11.00931 (1-x) = 10.811
10.01294x + 11.00931 - 11.00931x = 10.811 - 0.99016x = -0.198
X = 0.200

Check:
10.01294(0.2) + 11.00931 (0.8) = 10.81

20% B10. 80% B11

Answer: Boron -10 : 20%

Boron -11 : 80%

Explanation:

Mass of isotope  boron -10= 10

% abundance of isotope 1 = x% = [tex]\frac{x}{100}[/tex]

Mass of isotope boron-11 = 11

% abundance of isotope 2 = (100-x)% = [tex]\frac{100-x}{100}[/tex]

Formula used for average atomic mass of an element :

[tex]\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})[/tex]

[tex]10.8=\sum[(10)\times \frac{x}{100})+(11)\times \frac{100-x}{100}]][/tex]

[tex]x=20[/tex]

Therefore, percent of boron -10 is 20% and isotope boron-11 is (100-20)= 80%.