Question

boron has an average atomic mass of 10.81. one isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. the other isotope has a relative abundance of 80.20 percent. what is the mass of that isotope? report to two decimal places. amu done

Explanation:

Step1: Recall the formula for average atomic mass

The average atomic mass ($A$) of an element with isotopes is given by the sum of the product of each isotope's mass ($m_i$) and its relative abundance ($f_i$, in decimal form). For two isotopes, the formula is $A = m_1f_1 + m_2f_2$.
We know $A = 10.81$, $m_1 = 10.012938$, $f_1 = 0.1980$ (since 19.80% = 0.1980), and $f_2 = 0.8020$ (since 80.20% = 0.8020). We need to find $m_2$.

Step2: Rearrange the formula to solve for $m_2$

From $A = m_1f_1 + m_2f_2$, we can rearrange to get $m_2=\frac{A - m_1f_1}{f_2}$.

Step3: Substitute the known values into the formula

First, calculate $m_1f_1$: $10.012938\times0.1980 \approx 1.982562$.
Then, calculate $A - m_1f_1$: $10.81 - 1.982562 = 8.827438$.
Now, divide by $f_2$: $m_2=\frac{8.827438}{0.8020} \approx 11.01$.

Answer:

11.01