Question #0137a

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Answer 1 · 2017-09-21T00:07:54

Using your ratio gives the mass of $_{35}^{79} Br$ $""_35^79"Br"$ as 78.525 u. Using the correct ratio gives a correct isotopic mass of 78.919 u.

Let's let $m_{79}$ $m_79$ and $m_{81}$ $m_81$ 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,

$0.506 90 m_{79} + 0.493 10 m_{81} = 79.904 u$ $"0.506 90"m_79 + "0.493 10"m_81 = "79.904 u"$

Now,

$\frac{m_{81}}{m_{79}} = 1.0356$ $m_81/m_79 = 1.0356$

So,

$m_{81} = 1.0356 m_{79}$ $m_81 = 1.0356m_79$

∴ $0.506 90 m_{79} + 0.493 10 \times 1.0356 m_{79} = 79.904 u$ $"0.506 90"m_79 + "0.493 10" × 1.0356m_79 = "79.904 u"$

$0.506 90 m_{79} + 0.510 65 m_{79} = 79.904 u$ $"0.506 90"m_79 + "0.510 65"m_79 = "79.904 u"$

$1.017 55 m_{79} = 79.904 u$ $"1.017 55"m_79 = "79.904 u"$

$m_{79} = \frac{79.904 u}{1.017 55} = 78.525 u$ $m_79 = "79.904 u"/"1.017 55" = "78.525 u"$

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

∴ $0.506 90 m_{79} + 0.493 10 \times 1.0253 m_{79} = 79.904 u$ $"0.506 90"m_79 + "0.493 10" × 1.0253m_79 = "79.904 u"$

$0.506 90 m_{79} + 0.505 57 m_{79} = 79.904 u$ $"0.506 90"m_79 + "0.505 57"m_79 = "79.904 u"$

$1.012 48 m_{79} = 79.904 u$ $"1.012 48"m_79 = "79.904 u"$

$m_{79} = \frac{79.904 u}{1.012 48} = 78.919 u$ $m_79 = "79.904 u"/"1.012 48" = "78.919 u"$