Element X has three istotopes, X⁹⁵, X⁹⁶, and X⁹⁸. Element X has an atomic mass of 96.344u. If X-98 has a mass of 98.0001u and an abundance of 19.8%, what are the abundances of X-96 (mass of 95.9988u) and X-95 (mass of 95.0011u)?
See your other two posts above.
To find the abundances of X-96 and X-95, we need to use the atomic mass and the abundance of X-98.
Let's assume the abundances of X-96 and X-95 are represented by 'x' and 'y' respectively.
First, write down the equation based on the total atomic mass:
98.0001u * 0.198 + 95.9988u * x + 95.0011u * y = 96.344u
Rearrange the equation:
19.4198198u + 95.9988u * x + 95.0011u * y = 96.344u
Combine the terms:
95.9988u * x + 95.0011u * y = 96.344u - 19.4198198u
95.9988u * x + 95.0011u * y = 76.9241802u
Now we have a system of equations:
1) 95.9988u * x + 95.0011u * y = 76.9241802u
2) x + y = 1 (since the sum of the abundances should be 100%)
To solve this system, we can use one of several methods, such as substitution or elimination. Since the second equation is simpler, let's solve it for one variable and substitute it in the first equation.
From equation 2, we have:
x = 1 - y
Substituting this in equation 1:
95.9988u * (1 - y) + 95.0011u * y = 76.9241802u
Simplify the equation:
95.9988u - 95.9988u * y + 95.0011u * y = 76.9241802u
Combine like terms:
190u * y = 76.9241802u - 95.9988u
190u * y = -19.0746198u
Now divide both sides by 190u:
y = (-19.0746198u) / 190u
Simplify:
y = -0.100393786u
Since the abundance of an isotope cannot be negative, we can conclude that the abundance of X-95 (y) is 0.
Now substitute the value of y into equation 2:
x + 0 = 1
x = 1
So, the abundance of X-96 (x) is 1 or 100%.
To summarize:
- The abundance of X-95 (mass of 95.0011u) is 0%.
- The abundance of X-96 (mass of 95.9988u) is 100%.