How would I solve this:
The atomic mass of 6/3Li and 7/3Li are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of these 2 isotopes. The average atomic mass of Li is 6.941.
All I know is how to get the atomic mass and not the natural abundances.
Let a= the percent of 6/3 Li
let b= the percent of 7/3li
a + b = 100
6.0151a + 7.0160b = 6.941(100)
now solve like simultaneous equations in algebra
I figured it out
To calculate the natural abundances of the two isotopes given their atomic masses and the average atomic mass of Li, you can use the following steps:
Step 1: Assign variables:
Let x represent the natural abundance of 6/3Li (lithium-6).
Let y represent the natural abundance of 7/3Li (lithium-7).
Step 2: Write the equation for the average atomic mass:
The average atomic mass equation for a two-isotope element can be written as:
(Atomic mass of isotope A * Abundance of isotope A) + (Atomic mass of isotope B * Abundance of isotope B) = Average atomic mass
In this case, the equation becomes:
(6.0151 amu * x) + (7.0160 amu * y) = 6.941 amu
Step 3: Solve for one variable in terms of the other:
To solve for one variable in terms of the other, you can rearrange the equation above to isolate one variable. Let's solve for y in terms of x:
7.0160y = 6.941 - (6.0151x)
y = (6.941 - 6.0151x) / 7.0160
Step 4: Use the sum of abundances to solve for the remaining variable:
The sum of the natural abundances of the isotopes should always be equal to 1. Therefore, you can use this information to solve for the remaining variable:
x + y = 1
Substitute the expression for y from step 3 into this equation:
x + [(6.941 - 6.0151x) / 7.0160] = 1
Step 5: Solve for x:
Multiply the entire equation by 7.0160 to eliminate the fraction:
7.0160x + 6.941 - 6.0151x = 7.0160
Combine like terms:
0.0009x + 6.941 = 7.0160
Simplify:
0.0009x = 7.0160 - 6.941
0.0009x = 0.075
Divide both sides by 0.0009:
x = 0.075 / 0.0009
x = 83.33...
Therefore, the natural abundance of the isotope 6/3Li (lithium-6) is approximately 83.33%.
Step 6: Solve for y:
Substitute the value of x from step 5 into the equation from step 3:
y = (6.941 - 6.0151(83.33...)) / 7.0160
y = (6.941 - 6.461465) / 7.0160
y = 0.479535 / 7.0160
y = 0.06847...
Therefore, the natural abundance of the isotope 7/3Li (lithium-7) is approximately 6.85%.
Hence, the natural abundances of the two isotopes are approximately 83.33% and 6.85% for 6/3Li and 7/3Li, respectively.
To calculate the natural abundances of isotopes, you need to use the information provided and set up a system of equations. In this case, we are given the atomic masses of two isotopes of lithium (Li), along with the average atomic mass of Li.
Let's denote the natural abundance of the first isotope (6/3Li) as x and the natural abundance of the second isotope (7/3Li) as y. The sum of these abundances should be equal to 1 (since they represent all the possible proportions):
x + y = 1
Next, we need to set up an equation for the average atomic mass. The average atomic mass is the weighted average of the atomic masses of the isotopes, where the weights are the abundances. Using the given atomic masses and abundances, we can calculate the average atomic mass:
(x * 6.0151 amu) + (y * 7.0160 amu) = 6.941 amu
Now, we have a system of two equations:
x + y = 1
(x * 6.0151) + (y * 7.0160) = 6.941
You can solve this system of equations using various methods, such as substitution or elimination. One way to solve it is by substitution:
From the first equation, you can solve for x in terms of y:
x = 1 - y
Substitute this expression for x in the second equation:
((1 - y) * 6.0151) + (y * 7.0160) = 6.941
Simplify the equation further:
6.0151 - 6.0151y + 7.0160y = 6.941
Combine like terms:
1.0009y = 0.9259
Divide both sides by 1.0009:
y ≈ 0.9259
Now, substitute this value of y back into the first equation to solve for x:
x = 1 - 0.9259
x ≈ 0.0741
Therefore, the natural abundances of the two isotopes are approximately:
6/3Li: 0.0741 or 7.41%
7/3Li: 0.9259 or 92.59%