Sodium only has one naturally occuring isotope, Na23
, with a relative atomic mass of 22.9898 u.
A synthetic, radioactive isotope of sodium, Na22
, is used in positron emission tomography. Na22
has a relative atomic mass of 21.9944 u.
A 1.6421 g
sample of sodium containing a mixture of Na23
and Na22
has an apparent "atomic mass" of 22.9441 u
. Find the mass of Na22
contained in this sample.
Let X = fraction of Na atoms with a mass of 22
Then 1 - X = fraction of Na with a mass of 23
===========================
21.9944X + 22.9898(1-X) = 22.9441
Solve for X and 1- X
21.9944 X + 22.9898 - 22.9898X = 22.9441
Finish for X
Then Na22 = 1.6421 g sample x fraction of Na22 = g Na22.
Post your work if you get stuck.
Oh, Na22, the radioactive rebel of the sodium family! Let's solve this atomic mystery, shall we?
First, we know the relative atomic mass of Na23 is 22.9898 u, and Na22 is 21.9944 u. Our sample apparently has an atomic mass of 22.9441 u.
To find the mass of Na22, we first need to figure out the mass contribution of Na23 in the sample. We can do this by subtracting the atomic mass of Na22 from the atomic mass of the sample:
Mass of Na23 = Atomic mass of sample - Atomic mass of Na22
= 22.9441 u - 21.9944 u
Now, let's calculate that:
Mass of Na23 = 0.9497 u
Since we have the total mass of the sample given as 1.6421 g, we can find the mass of Na22 by subtracting the mass of Na23 from the total sample mass:
Mass of Na22 = Total sample mass - Mass of Na23
= 1.6421 g - 0.9497 g
And now, the moment of truth:
Mass of Na22 = 0.6924 g
So, in this 1.6421 g sodium sample, there is approximately 0.6924 g of the mischievous Na22. Keep an eye on that radioactivity, folks!
To find the mass of Na22 contained in the sample, we can set up a system of equations.
Let x be the mass of Na22 in the sample.
Then, the mass of Na23 in the sample can be calculated as (1.6421 - x) g.
Using the relative atomic masses and the mass of each isotope, we can set up the following equation:
[(mass of Na22 * mass of Na22 in the sample) + (mass of Na23 * mass of Na23 in the sample)] / total mass of the sample = apparent "atomic mass" of the sample.
Substituting the given values:
[(21.9944 * x) + (22.9898 * (1.6421 - x))] / 1.6421 = 22.9441.
Now, let's solve for x.
(21.9944 * x) + (22.9898 * (1.6421 - x)) = 22.9441 * 1.6421.
21.9944x + 22.9898 * 1.6421 - 22.9898x = 37.78013061.
21.9944x - 22.9898x = 37.78013061 - 22.9898 * 1.6421.
(-0.9954x) = 2.651848.
x = 2.651848 / -0.9954.
x ≈ -2.6616 g.
Since the mass cannot be negative, we therefore conclude that there is no Na22 in the sample.
To find the mass of Na22 contained in the sample, we can use the concept of weighted averages.
The "apparent atomic mass" of the sample (22.9441 u) can be thought of as a weighted average of the atomic masses of Na23 and Na22, where the weights are the relative abundance of each isotope.
Let's denote the mass of Na23 as m23 and the mass of Na22 as m22. We know that the relative atomic mass of Na23 is 22.9898 u, while the mass of Na22 is 21.9944 u.
Now, let's set up two equations based on the information given:
1. Mass equation: m23 + m22 = 1.6421 g (total mass of the sample)
2. Atomic mass equation: (m23 * 22.9898) + (m22 * 21.9944) = 22.9441 * 1.6421 (weighted average atomic mass of the sample)
We can now solve these equations simultaneously to find the values of m23 and m22.
Rearranging equation 1, we get: m23 = 1.6421 g - m22
Substituting this value in equation 2, we get:
(1.6421 g - m22) * 22.9898 + (m22 * 21.9944) = 22.9441 * 1.6421
Simplifying, we have:
37.65057 g - 22.9898 m22 + 21.9944 m22 = 37.69600461
Combining like terms:
-0.9954 m22 = 0.04540461
Solving for m22:
m22 = 0.04540461 / -0.9954
m22 ≈ -0.0456 g
Since mass cannot be negative, there seems to be an error in the calculations. Please make sure that the given values are accurate and the equations have been set up correctly.