Question

Element X has two natural isotopes. The isotope with a mass of 10.012 amu has a relative abundance of 19.91%. The isotope with a mass of 11.009 amu has a relative abundance of 80.09%. Calculate the atomic mass of this element

Answer 1

To find the atomic mass, multiply the individual atomic masses with their relative abundances and add them up.

10.012*0.1991 + 11.009*0.8009 = 10.810 amu (5 s.f.)

Element X is Boron.

Answer 2

The element with 2 isotopes, isotope 1 with an abundance of 19.91% (10.012 amu) and isotope 2 with an abundance of 80.09% (11.009 amu), has an atomic mass of 10.81 amu.

Element X has two natural isotopes. The isotope with a mass of 10.012 amu has a relative abundance of 19.91%. The isotope with a mass of 11.009 amu has a relative abundance of 80.09%. The atomic mass of element X is a weighted average that takes into account the mass ($m_i$) and the abundance ($ab_i$) of each isotope.

$am = \frac{\Sigma m_i \times ab_i }{100} = \frac{10.012 amu \times 19.91 + 11.009 \times 80.09}{100} = 10.81 amu$

The element with 2 isotopes, isotope 1 with an abundance of 19.91% (10.012 amu) and isotope 2 with an abundance of 80.09% (11.009 amu), has an atomic mass of 10.81 amu.

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