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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 Li is 6.941 amu.

Natural abundance of $_{3}^{6} \mathrm{Li}=7.5 \%$ Natural abundance of $\frac{7}{3} \mathrm{Li}=92.5 \%$

Chemistry 101

Chapter 3

Mass Relationships in Chemical Reactions

Chemical reactions and Stoichiometry

University of Central Florida

University of Kentucky

University of Toronto

Lectures

04:02

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Chemical reactions can be either spontaneous, requiring no input of energy, or non-spontaneous, typically following the input of some type of energy, such as heat, light or electricity. Chemical reactions are usually characterized by a chemical change, and they yield one or more products after the reaction is complete. Chemical reactions are described with chemical equations, which symbolically present the starting materials, end products, and sometimes intermediate products and reaction conditions. Chemical reactions happen at a characteristic reaction rate at a given temperature and chemical concentration. Typically, reaction rates increase with increasing temperature because there is more thermal energy available to reach the activation energy necessary for breaking bonds between atoms.

08:02

In chemistry, a combination reaction is a chemical reaction in which two or more reactants combine to form more than one product. In a decomposition reaction, one reactant splits into two or more products.

04:42

The atomic masses of $^{6}…

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The atomic masses of 6 Li …

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The atomic masses of 6Li …

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The atomic masses of $^{20…

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Obtain the fractional abun…

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Use the following table of…

03:41

Lithium has two isotopes, …

04:13

The element rhenium (Re) h…

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Two isotopes of rubidium o…

03:03

in this problem into their goal. To find the natural abundance off both of the M six and M seven. However, they only give us the massive with them six math massively and seven and the total average atomic mess. So what we do know is that the average atomic mass equals the massive isotope one time financial abundance of isotope one plus the mass vices hope to times a natural abundance of isotope too. So we are going to use this formula to calculate the natural abundance of both of these isotopes. And first, we're going to plug all our numbers into this average atomic mass equation. So I'm just gonna breathing air around the masses to make it easier to write. So we have six times and one. The natural abundance is unknown at the time, plus seven times and to natural abundance is also unknown. That's going to equal our total 6.94 won a um you So right now we have one equation, this equation right here and two unknowns, and that's unsolvable. We need another equation before we can solve this problem. So what else we know. We know that there are only two isotopes of lithium. So the natural abundance must equal 100% or one if you're using fractions or decimals. So we know that and one plus in two equals 100% or one. So let's just right, use it with one that's a little easier so and one plus and two equals one. So now we have one equation here, another equation here, two equations and two unknowns. This is solvable. So we can do. First is rearrange this equation to solve for one of the unknown variables. So it's to end too equals one minus and one. So now what we're gonna do is we're gonna take this, we're gonna plug it in up here. So you have six times and won, plus seven times one minus. N one equals six point 941 So if we then solve for N one, what we end up getting is negative. 1.0 and one equals negative 0.75 and so and one equals 0.7 5/1 0.0 The native signs cancel out is gonna equal 0.75 Now he wants by this by 100 to put in a percent. We get 7.5%. Is the natural abundance four, lithium, six. So now, to find our abundance of lithium seven, what we can do is we can take this number and plug it back into this equation. So we would have 0.75 plus and two equals one rearranging and solving. For N two, we get in to equals 20.2 point nine. Sorry about that. You fixed sense point nine to five, multiplied by 100 equals 92.5 percent for L I seven.

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