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Example · Example 3

Q.Naturally occurring chlorine consists of two isotopes: 35Cl^{35}\text{Cl} (mass 34.97 u34.97\ \text{u}, abundance 75.77%75.77\%) and 37Cl^{37}\text{Cl} (mass 36.97 u36.97\ \text{u}, abundance 24.23%24.23\%). Calculate the average atomic mass of chlorine.

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The average atomic mass is calculated as the sum of (fractional abundance ×\times isotopic mass) for every naturally occurring isotope.

For 35Cl^{35}\text{Cl}: fractional abundance =0.7577= 0.7577, mass =34.97 u= 34.97\ \text{u}

Contribution =0.7577×34.97=26.50 u= 0.7577 \times 34.97 = 26.50\ \text{u}

For 37Cl^{37}\text{Cl}: fractional abundance =0.2423= 0.2423, mass =36.97 u= 36.97\ \text{u}

Contribution =0.2423×36.97=8.96 u= 0.2423 \times 36.97 = 8.96\ \text{u}

Average atomic mass=26.50+8.96=35.46 u\text{Average atomic mass} = 26.50 + 8.96 = 35.46\ \text{u}

This matches the standard listed atomic mass of chlorine (35.45 u35.45\ \text{u}), confirming that the value on the periodic table is a weighted average over both naturally occurring isotopes, not the mass of either isotope alone.

[!ANSWER] The average atomic mass of chlorine is 35.46 u35.46\ \text{u}.

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