# Question 9b036

Aug 23, 2017

Natural abundance: 50.69%

Mass: $79.919$ $\text{amu}$

#### Explanation:

The natural abundance of $\text{Br-81}$ is 49.31%.

This means that 49.31% of all of the world's bromine $\text{Br}$ exists in the form of $\text{Br-81}$.

As there is only one other isotope of bromine, 100% - 49.31% = 50.69% of it exists in the form of $\text{Br-79}$.

Therefore, the natural abundance of $\text{Br-79}$ is 50.69%

The atomic mass of bromine is $79.904$ $\text{amu}$.

$R i g h t a r r o w 79.904$ $\text{amu}$ = 50.69% times x + 49.31% times 80.9163# $\text{amu}$

We need to solve for $x$, which in this case is the mass of $\text{Br-79}$:

$R i g h t a r r o w 79.904$ $\text{amu}$ $= 0.5069 x + 0.4931 \times 80.9163$ $\text{amu}$

$R i g h t a r r o w 79.904$ $\text{amu}$ $= 0.5069 x + 39.89982753$ $\text{amu}$

$R i g h t a r r o w 40.00417247$ $\text{amu}$ $= 0.5069 x$

$R i g h t a r r o w 78.919259164$ $\text{amu}$ $= x$

$\therefore x = 78.919$ $\text{amu}$

Therefore, the mass of $\text{Br-79}$ is $79.919$ $\text{amu}$.