Question

Question #0137a

Answer 1

Using your ratio gives the mass of `""_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,

`"0.506 90"m_79 + "0.493 10"m_81 = "79.904 u"`

Now,

`m_81/m_79 = 1.0356`

So,

`m_81 = 1.0356m_79`

∴ `"0.506 90"m_79 + "0.493 10" × 1.0356m_79 = "79.904 u"`

`"0.506 90"m_79 + "0.510 65"m_79 = "79.904 u"`

`"1.017 55"m_79 = "79.904 u"`

`m_79 = "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

∴ `"0.506 90"m_79 + "0.493 10" × 1.0253m_79 = "79.904 u"`

`"0.506 90"m_79 + "0.505 57"m_79 = "79.904 u"`

`"1.012 48"m_79 = "79.904 u"`

`m_79 = "79.904 u"/"1.012 48" = "78.919 u"`