Question

question bromine has two stable isotopes: ⁷⁹br and ⁸¹br with masses of 78.9183371 amu and 80.9162906 amu. if the average atomic mass for bromine on earth is 79.904 amu, what is the mass of ⁷⁹br in a 10.0 g sample? hint: set the abundance of the ⁷⁹br isotope equal to x. then the abundance of the other isotope will equal 1 - x. your answer should have three significant figures, so round to the nearest hundredth.

Explanation:

Step1: Set up the equation for average atomic mass

Let the abundance of $^{79}$Br be $x$, then the abundance of $^{81}$Br is $1 - x$. The formula for average atomic mass $A$ is $A=m_1x + m_2(1 - x)$, where $m_1$ and $m_2$ are the masses of the isotopes. So, $79.904=78.9183371x+80.9162906(1 - x)$.

Step2: Expand and solve for $x$

Expand the right - hand side: $79.904=78.9183371x + 80.9162906-80.9162906x$. Combine like terms: $79.904 - 80.9162906=(78.9183371 - 80.9162906)x$. So, $-1.0122906=-1.9979535x$. Then $x=\frac{1.0122906}{1.9979535}\approx0.506$.

Step3: Calculate the mass of $^{79}$Br in the sample

The mass of $^{79}$Br in a 10.0 g sample is $m = 10.0\ g\times0.506 = 5.06\ g$.

Answer:

5.06 g