Question #0137a

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Answer 1 · 2017-09-21T00:07:54

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.

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"#