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Question
question
silver has two stable isotopes: $^{107}$ag with a fractional abundance on earth of 0.51839 and $^{109}$ag with a mass of 108.904752 amu. if the average atomic mass for silver on earth is 107.8682 amu, what is the mass of the most common isotope?
● your answer should have four significant figures (round to the nearest tenth)
Step1: Calculate the fractional abundance of \(^{109}Ag\)
1 - 0.51839 = 0.48161
Step2: Set up the average - atomic - mass equation
Let the mass of \(^{107}Ag\) be \(x\). The average - atomic - mass formula is \(Average\ atomic\ mass=(Mass\ of\ isotope_1\times Fractional\ abundance_1)+(Mass\ of\ isotope_2\times Fractional\ abundance_2)\). So, \(107.8682 = x\times0.51839+108.904752\times0.48161\).
Step3: Solve for \(x\)
First, calculate \(108.904752\times0.48161\approx52.4414\). Then the equation becomes \(107.8682 = 0.51839x+52.4414\). Rearrange to get \(0.51839x=107.8682 - 52.4414 = 55.4268\). Then \(x=\frac{55.4268}{0.51839}\approx106.92\). Rounding to four significant figures and to the nearest tenth gives 106.9 amu.
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106.9 amu