# Question #0137a

Sep 21, 2017

Using your ratio gives the mass of $\text{_35^79"Br}$ as 78.525 u. Using the correct ratio gives a correct isotopic mass of 78.919 u.

#### Explanation:

Let's let ${m}_{79}$ and ${m}_{81}$ represent the masses of the two isotopes.

The average atomic mass is the weighted average of the isotopic masses.

You multiply each isotopic mass by its relative abundance (percentage as a decimal fraction) in the mixture.

Thus,

$\text{0.506 90"m_79 + "0.493 10"m_81 = "79.904 u}$

Now,

${m}_{81} / {m}_{79} = 1.0356$

So,

${m}_{81} = 1.0356 {m}_{79}$

$\text{0.506 90"m_79 + "0.493 10" × 1.0356m_79 = "79.904 u}$

$\text{0.506 90"m_79 + "0.510 65"m_79 = "79.904 u}$

$\text{1.017 55"m_79 = "79.904 u}$

${m}_{79} = \text{79.904 u"/"1.017 55" = "78.525 u}$

Using the correct mass ratio

Your ratio of the two atomic masses is incorrect.

The correct ratio is 1.0253.

Using this number, we get

$\text{0.506 90"m_79 + "0.493 10" × 1.0253m_79 = "79.904 u}$

$\text{0.506 90"m_79 + "0.505 57"m_79 = "79.904 u}$

$\text{1.012 48"m_79 = "79.904 u}$

${m}_{79} = \text{79.904 u"/"1.012 48" = "78.919 u}$