Cooper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given Cooper's atomic weight of 63.546. what is the percent abundance of each isotope?
Let's assume that the percent abundance of Cu-63 is x, and the percent abundance of Cu-65 is (1-x) since the total percent abundance is 100%.
The atomic weight can be calculated using the formula:
Atomic weight = (mass of isotope 1 * percent abundance of isotope 1) + (mass of isotope 2 * percent abundance of isotope 2)
63.546 = (62.9296 * x) + (64.9278 * (1-x))
Now, let's solve for x:
63.546 = 62.9296x + 64.9278 - 64.9278x
63.546 - 64.9278 = -1.9982x
-1.3818 = -1.9982x
x = -1.3818 / -1.9982
x ≈ 0.691 or 69.1% (rounded to one decimal place)
Therefore, the percent abundance of Cu-63 is approximately 69.1% and the percent abundance of Cu-65 is approximately 30.9%.
To find the percent abundance of each isotope, we can let x represent the percent abundance of Cu-63 and (100 - x) represent the percent abundance of Cu-65.
The atomic weight of an element is the weighted average of the masses of its isotopes, taking into account their respective abundance.
We can set up the equation as follows:
(x/100) * (62.9296 amu) + ((100 - x)/100) * (64.9278 amu) = 63.546 amu
Simplifying the equation:
(62.9296x + (100 - x) * 64.9278) / 100 = 63.546
Now, solve for x:
62.9296x + 64.9278(100 - x) = 63.546 * 100
62.9296x + 6492.78 - 64.9278x = 6354.6
-1.9982x = -138.18
x = -138.18 / -1.9982
x ≈ 69.15
So, the percent abundance of Cu-63 is approximately 69.15%, and the percent abundance of Cu-65 is approximately (100 - 69.15) = 30.85%.
To determine the percent abundance of each isotope, we need to use the atomic weight and the masses of the isotopes.
Let's assume the abundance of Cu-63 is x and the abundance of Cu-65 is y. The sum of their percentages should be 100%, so we can write the equation:
x + y = 100
Next, we need to consider the atomic weights and masses of the isotopes. The atomic weight is the weighted average of the masses of each isotope, taking into account their relative abundance. We can express this as:
(0.01 * Cu-63 mass) + (0.99 * Cu-65 mass) = atomic weight
(0.01 * 62.9296) + (0.99 * 64.9278) = 63.546
0.629296 + 64.378522 = 63.546
64.007818 = 63.546
Now we can solve the system of equations:
x + y = 100 (Eq. 1)
0.01x + 0.99y = 0.546 (Eq. 2)
To simplify Eq. 2, we can multiply all terms by 100 to remove decimals:
x + 99y = 54.6
Now we can solve this system of equations. Typically, there are various methods to solve it (such as substitution or elimination), but as an AI explanation bot, I will solve it using substitution:
From Eq. 1, we can express x in terms of y:
x = 100 - y
Substituting this into Eq. 2:
100 - y + 99y = 54.6
Combining like terms:
98y = 45.4
Dividing both sides by 98:
y = 0.4636735
Now we can substitute the value of y back into Eq. 1 to find x:
x + 0.4636735 = 100
x = 99.5363265
Therefore, the percent abundance of Cu-63 is approximately 99.54%, and the percent abundance of Cu-65 is approximately 0.46%.