The atomic masses of 3^6 Li and 3^7 Li are 6.0151 amu and 7.0160 amu, respectively. Calculate

The atomic masses of 3^6 Li and 3^7 Li are 6.0151 amu and...

The atomic masses of ${ }_{3}^{6} \mathrm{Li}$ and ${ }_{3}^{7} \mathrm{Li}$ are $6.0151 \mathrm{amu}$ and 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of $\mathrm{Li}$ is $6.941 \mathrm{amu}$.

Key Concepts

Natural Abundance

Natural abundance is the relative proportion of each isotope of an element as it occurs in nature. These proportions are expressed as fractions or percentages and are critical in determining the average atomic mass of an element based on its constituent isotopes.

Weighted Average

The weighted average is a calculation that takes into account the different contributions of each isotope’s mass according to its natural abundance. By multiplying each isotopic mass by its abundance and summing the results, one obtains the average atomic mass of the element.

Isotopic Mass

Isotopic mass refers to the mass of a specific isotope of an element, usually measured in atomic mass units (amu). It reflects the mass contribution of protons, neutrons, and electrons for that isotope, and it is used in calculations of the element's average atomic mass when combined with natural abundances.

System of Linear Equations

Solving for natural isotopic abundances often involves setting up a system of linear equations, where one equation represents the sum of the abundances equalling unity and another represents the weighted average of the isotopic masses. This method is fundamental in quantitative chemical analysis and understanding the composition of elements.

Transcript

00:01 In this problem, it is our goal to find the natural abundance of both lithium -6 and lithium -7.

00:07 However, they only give us the mass of lithium -6, mass of lithium -7, and the total average atomic mass.

00:16 So what we do know is that the average atomic mass equals the mass of isotope 1 times the natural abundance of isotope 1, plus the mass of isotope 2 times the natural abundance of isotope 2.

00:28 So we are going to use this formula to calculate the natural abundance of both of these isotopes.

00:36 And first, we are going to plug all our numbers into this average atomic mass equation.

00:41 So i'm just going to round the masses to make it easier to write.

00:47 So we have six times n1.

00:52 The natural abundance is unknown at the time, plus seven times n2.

00:59 Natural abundance is also unknown.

01:01 That's going to equal our total 6 .941amu.

01:09 So right now we have one equation, this equation right here, and two unknowns, and that's unsolvable.

01:18 We need another equation before we can solve this problem.

01:23 So what else do we know? we know that there are only two isotopes of lithium.

01:29 So the natural abundance must, equal 100 % or 1 if you're using fractions or decimals.

01:36 So we know that n1 plus n2 equals 100 % or 1.

01:47 So let's just use it with 1 because that's a little easier.

01:51 So n1 plus n2 equals 1.

01:54 So now we have one equation here, another equation here, two equations and two unknowns...

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