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Exercises · 13.4

Q.The three stable isotopes of neon: 1020Ne^{20}_{10}\text{Ne}, 1021Ne^{21}_{10}\text{Ne} and 1022Ne^{22}_{10}\text{Ne} have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

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✓ Free question

Multiply each isotopic mass by its fractional abundance and sum the three terms. The average atomic mass of neon comes out to ≈20.18 u\approx 20.18\ \text{u}.

The weighted-average formula for three isotopes is:

M=f1m1+f2m2+f3m3M = f_1 m_1 + f_2 m_2 + f_3 m_3

Substituting the given abundances (as fractions) and masses:

M=(0.9051)(19.99 u)+(0.0027)(20.99 u)+(0.0922)(21.99 u)M = (0.9051)(19.99\ \text{u}) + (0.0027)(20.99\ \text{u}) + (0.0922)(21.99\ \text{u})

Computing each term:

(0.9051)(19.99)=18.0929 u(0.9051)(19.99) = 18.0929\ \text{u}

(0.0027)(20.99)=0.0567 u(0.0027)(20.99) = 0.0567\ \text{u}

(0.0922)(21.99)=2.0275 u(0.0922)(21.99) = 2.0275\ \text{u}

Adding:

M=18.0929+0.0567+2.0275=20.177 u≈20.18 uM = 18.0929 + 0.0567 + 2.0275 = 20.177\ \text{u} \approx 20.18\ \text{u}

This is very close to neon's accepted periodic-table atomic mass (20.18 u), confirming the calculation — the dominant isotope 20Ne^{20}\text{Ne} (90.5% abundant) pulls the average close to 20 u, with the heavier isotopes nudging it up slightly.

✓Final answer

M(Ne)≈20.18 uM(\text{Ne}) \approx \boxed{20.18\ \text{u}}

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