Question

Carbon is found in nature as the mixture of C-12 and C-13 .If average atomic mass of Carbon is found as 12.011u then Find %abundance of C-12 and C-13?​

Answer 1

Answer :-

→ Abundance of C-12 isotope = 98.9 %

→ Abundance of C-13 isotope = 1.1 %

Explanation :-

We have :-

• Atomic mass of C-12 = 12 u

• Atomic mass of C-13 = 13 u

• Average atomic mass = 12.011 u

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Let the abundance of C-12 be x %

∴ Abundance of C-13 = (100 - x) %

Now putting the values in the formula of average atomic mass, we get :-

⇒ [12 × x/100] + [13(100 - x)/100] = 12.011

⇒ 0.12x + [1300 - 13x/100] = 12.011

⇒ [12x + 1300 - 13x]/100 = 12.011

⇒ 1300 - x = 12.011(100)

⇒ 1300 - x = 1201.1

⇒ - x = 1201.1 - 1300

⇒ - x = - 98.9

⇒ x = 98.9 %

∴ Abundance of C-12 = 98.9 %

∴ Abundance of C-13 :-

= (100 - 98.9) %

= 1.1 %

Answer 2

Answer:

Given :-

• Carbon is found in nature as the mixture of C-12 and C-13.
• Average atomic mass of Carbon is found as 12.011u.

To Find :-

• What is the abundance of C-12 and C-13.

Solution :-

Let,

$\mapsto$ Abundance of C-12 be x %

$\mapsto$ Abundance of C-13 be (100 - x)%

Given :

• Average Atomic mass of Carbon = 12.011 u
• Atomic mass of C-12 = 12 u
• Atomic mass of C-13 = 13 u

According to the question by using the formula we get,

$\implies \sf 12.011 =\: \dfrac{12 \times x}{100} + \dfrac{13(100 - x)}{100}$

$\implies \sf \dfrac{12011}{1000} =\: \dfrac{12x}{100} + \dfrac{1300 - 13x}{100}$

$\implies \sf \dfrac{12011}{1000} =\: \dfrac{12x + 1300 - 13x}{100}$

$\implies \sf \dfrac{12011}{1000} =\: \dfrac{1300 - x}{100}$

By doing cross multiplication we get,

$\implies \sf 1000(1300 - x) =\: 12011(100)$

$\implies \sf 1300000 - 1000x =\: 1201100$

$\implies \sf - 1000x =\: 1201100 - 1300000$

$\implies \sf {\cancel{-}} 1000x =\: {\cancel{-}} 98900$

$\implies \sf x =\: \dfrac{989\cancel{00}}{10\cancel{10}}$

$\implies \sf x =\: \dfrac{989}{10}$

$\implies \sf\bold{\purple{x =\: 98.9\: \%}}$

Hence, the required C-12 and C-13 are :

$\dashrightarrow$ Abundance of C-12 :

$\leadsto \sf x\%$

$\leadsto\sf\bold{\red{98.9\: \%}}$

And,

$\dashrightarrow$ Abundance of C-13 :

$\leadsto \sf (100 - x)\%$

$\leadsto \sf (100 - 98.9)\%$

$\leadsto \sf \bigg(100 - \dfrac{989}{10}\bigg)\%$

$\leadsto \sf \bigg(\dfrac{1000 - 989}{10}\bigg)\%$

$\leadsto \sf \bigg(\dfrac{11}{10}\bigg)\%$

$\leadsto \sf\bold{\red{1.1\: \%}}$

$\therefore$ The abundance of C-12 is 98.9 % and the abundance of C-13 is 1.1 %.