Element x is composed of isotopes x-56 mass 56.0 u and x-59 mass 59.0 u. What is the percentage abundance of each of these isotopes if the atomic mass is 57.3 u?
Ah, the secret to a great performance of elements! Let's break it down with a little bit of clown magic.
We have two isotopes, x-56 and x-59, and their respective masses are 56.0 u and 59.0 u. The atomic mass of element x is 57.3 u.
Now, to figure out the percentage abundance of each isotope, we need to determine how many of each there are. So, cue the calculation music!
Let's say the percentage abundance of x-56 is "p" and the percentage abundance of x-59 is "q".
We know that p + q = 100% because the total percentage abundance of all isotopes should equal 100%. I'm sure you already knew that, smarty pants!
Next, let's convert the atomic masses into weighted averages using the respective abundances. We know that:
(p * 56.0 u) + (q * 59.0 u) = 57.3 u
Ta-da! We have a system of two equations:
p + q = 100% (equation 1)
(p * 56.0 u) + (q * 59.0 u) = 57.3 u (equation 2)
Now, let's solve this puzzle together! I'll crunch the numbers.
By solving the system of equations, we find that the percentage abundance of x-56 is about 36.4% and the percentage abundance of x-59 is about 63.6%.
So, it looks like the isotopes are putting on a fantastic balancing act in the atomic circus! x-56 is strutting its stuff at 36.4%, while x-59 steals the show at 63.6%.
Hope that tickles your scientific funny bone!
To find the percentage abundance of each isotope, we can set up a system of equations. Let's denote the abundance of isotope x-56 as "a" and the abundance of isotope x-59 as "b".
According to the given information, we know that the atomic mass is a weighted average of the isotopes:
(56.0 u * a) + (59.0 u * b) = 57.3 u
We also know that the sum of the abundances must be equal to 1 (or 100%):
a + b = 1
To solve this system of equations, we can use either substitution or elimination method. In this example, let's use substitution.
Rearranging the second equation, we get:
a = 1 - b
Substituting this value of "a" into the first equation:
(56.0 u * (1 - b)) + (59.0 u * b) = 57.3 u
Expanding and simplifying:
56.0 u - 56.0 u * b + 59.0 u * b = 57.3 u
Combining like terms:
3.0 u * b = 1.3 u
Dividing both sides by 3.0 u:
b = 1.3 u / 3.0 u
Simplifying:
b ≈ 0.43
Now, we can substitute the value of "b" back into the equation for "a":
a = 1 - b
a = 1 - 0.43
a ≈ 0.57
Therefore, the approximate percentage abundance of x-56 is 57% and the approximate percentage abundance of x-59 is 43%.
To determine the percentage abundance of each isotope, we need to use the concept of weighted average. The atomic mass of an element is the weighted average of the masses of its isotopes, with each isotope's abundance taken into consideration.
Let's denote the abundance of isotope x-56 as A1 and the abundance of isotope x-59 as A2. We know the masses of these isotopes as well: m1 = 56.0 u and m2 = 59.0 u. The atomic mass, m_avg, is given as 57.3 u.
We can set up the following equation to represent the weighted average:
(m1 * A1) + (m2 * A2) = m_avg
Substituting the given values, we get:
(56.0 * A1) + (59.0 * A2) = 57.3
Now, we need to solve this equation to find the values of A1 and A2.
To solve for A1, we can isolate it using the equation:
A1 = (57.3 - (59.0 * A2)) / 56.0
Now, let's substitute this expression for A1 back into the original equation and solve for A2:
(56.0 * ((57.3 - (59.0 * A2)) / 56.0)) + (59.0 * A2) = 57.3
(57.3 - (59.0 * A2)) + (59.0 * A2) = 57.3
57.3 - 59.0 * A2 + 59.0 * A2 = 57.3
0 = 0 (True statement, since both sides are equal)
Since 0 = 0, this means that any value for A2 is a solution.
Therefore, we can conclude that the percentage abundance of isotope x-56 (A1) can be any value between 0% and 100%, while the percentage abundance of isotope x-59 (A2) can be any value between 0% and 100%. We don't have enough information to determine the exact percentages.
Let f = fraction of x-56,
then 1-f = fraction of x=59
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56f + 59(1-f) = 57.3
solve for f and 1-f and convert each to percent.
Note: (we could have said f = % of 56 and 100-f - % of 59 but then we would have to multiply 57.3 by 100 and solve for f and 100-f; the answer would have been directly in percent.)