Naturally occurring cobalt consists of only one isotope Co59 whose relative atomic mass is 58.9332. A synthetic radioactive isotope of cobalt Co60, relative atomic mass 59.9338, is used in radiation therapy for cancer. A 1.8516 sample of cobalt has an apparent "atomic mass" of 58.9901.Find the mass Co60 of in this sample
58.9332*(1-X)+59.9338*x=58.9901
solve for x, which is the decimal fraction of Co60, then multipy it by 1.8516 to get the mass of Co60
Let x = fraction of Co60; then
1-x = fraction of Co59.
mass Co59(1-x) + mass Co60(x) = 58.9901
Solve for x (and multiply by 100 if you want percent, then
fraction x 1.8516g = mass Co 60 in the 1.8616g sample.
When I worked it out I got .105 g, is that correct?
I have to get this question right, in order to make a B on my assignment
Yes.
To find the mass of Co60 in the given sample, we can use the concept of average atomic mass.
The average atomic mass is calculated by considering the relative abundance of each isotope and summing the products of their masses and abundances.
Let's assume that x is the mass of Co59 in the sample and y is the mass of Co60 in the sample.
Given:
Relative atomic mass of Co59 = 58.9332
Relative atomic mass of Co60 = 59.9338
Apparent "atomic mass" of the sample = 58.9901
Mass of the sample = 1.8516 g
By using the concept of average atomic mass, we know that the apparent "atomic mass" of the sample is the weighted average of Co59 and Co60.
So, we can set up the following equation:
(58.9332 * x + 59.9338 * y) / (x + y) = 58.9901
We also know that the total mass (x + y) of the sample is 1.8516 g.
Now, we can solve these equations to find the value of y (mass of Co60).
First, let's simplify the equation:
58.9332x + 59.9338y = 58.9901(x + y)
58.9332x + 59.9338y = 58.9901x + 58.9901y
(59.9338 - 58.9901)y = (58.9332 - 58.9901)x
0.9437y = -0.0569x
Now, we can rewrite the equation with x in terms of y:
x = (-0.9437 / 0.0569)y
Substituting this value of x in the equation (x + y = 1.8516) to solve for y:
(-0.9437 / 0.0569)y + y = 1.8516
(-0.9437 + 0.0569)y = 1.8516
-0.8868y = 1.8516
y = 1.8516 / -0.8868
y ≈ -2.0845 g
Since the mass cannot be negative, we discard this value.
This implies that there is no Co60 present in the given sample.