Question #3037d

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Question #3037d

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Answer 1 · 2017-06-17T22:27:59

$^{6} Li$ $""^6"Li"$ : $7.59 %$ $7.59%$

$^{7} Li$ $""^7"Li"$ : $92.4 %$ $92.4%$

Explanation:

We're asked to find the relative abundance (expressed commonly as a percentage abundance of the total number of isotopes) of the two naturally-occurring isotopes of $Li$ $"Li"$ .

The relative atomic mass of $Li$ $"Li"$ is $6.94$ $6.94$ $amu$ $"amu"$ (from periodic table).

We can set up an equation representing the fractional abundance of each isotope:

$overbrace((x))^(% "abundace lithium-"6)(6.01512"amu") + overbrace((1-x))^("remaining is lithium-"7)(7.01601"amu") = 6.94"amu"$

Let's use algebra to solve this equation:

$(6.01512"amu")x + 7.01601"amu" - (7.01601"amu")x = 6.94$

Combining like terms:

$(-1.00089)x = -0.07601$

$x = color(red)(0.07594$

This represents the fractional abundance of $""^6"Li"$ . The remaining quantity ( $1$ minus this value) represents that of $""^7"Li"$ :

$1- color(red)(0.07594) = 0.92406$

Therefore, the percentage abundance of the two isotopes is

$%""^6"Li" = (color(red)(0.07594))(100) = color(red)(7.59%$

$%""^7"Li" = (color(red)(0.92406))(100) = color(red)(92.4%$

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Question #3037d

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