There are two naturally occurring isotopes of chlorine with ratios of 76% and 24%. One mole of PCl3 is made using a sample of isotopically pure chlorine gas (containing only one of the two isotopes). The product weighs 142 g. If chlorine gas containing the other isotope were used, one mole of the PCl3 would weigh?
Answer = 136g
Well, since P weighs 31g and the isotopes of chlorine weigh 35 and 37
So, one mole of PCl3 is either
31+3*37=142
or 31+3*35 = 136
To solve this problem, we need to understand the concept of the average atomic mass and how it relates to the isotopes of an element.
The average atomic mass of an element is the weighted average of the masses of all naturally occurring isotopes of that element. In this case, we have two isotopes of chlorine, one with a 76% abundance and another with a 24% abundance.
Let's assume the isotope with a 76% abundance has a mass of A grams, and the isotope with a 24% abundance has a mass of B grams.
We know that the average atomic mass for chlorine is given as 35.5 grams/mole.
Now, to find the mass of one mole of PCl3 when using the other isotope of chlorine gas, we need to determine the average atomic mass of that isotope.
Since we have two isotopes with different abundances, we can set up a weighted average equation:
Average atomic mass = (A * 0.76) + (B * 0.24)
We know that the average atomic mass is 35.5 grams/mole, so we can rearrange the equation:
35.5 = (A * 0.76) + (B * 0.24)
Now, we can solve this equation for B:
B * 0.24 = 35.5 - (A * 0.76)
B = (35.5 - (A * 0.76)) / 0.24
Now, we have the mass of the other isotope of chlorine gas (B) in terms of the mass of the isotope with 76% abundance (A).
Finally, we can substitute this value of B into the equation to find the mass of one mole of PCl3:
Mass of PCl3 = 1 mole * (A + 3 * B)
Substituting the value of B, we get:
Mass of PCl3 = 1 mole * (A + 3 * (35.5 - (A * 0.76)) / 0.24)
Simplifying further, we get:
Mass of PCl3 = 1 mole * (A + ((106.5 - 3A * 0.76) / 0.24))
Mass of PCl3 = 1 mole * (A + (106.5 - 2.28A) / 0.24)
Mass of PCl3 = 1 mole * (A + (443.75 - 9.5A))
Mass of PCl3 = 1 mole * (-8.5A + 443.75)
Since we know that the mass of one mole of PCl3 is 142 grams, we can set up the equation:
142 = -8.5A + 443.75
Solving for A:
-8.5A = -301.75
A = 35.5
Now, we substitute this value of A back into the equation to find the mass of one mole of PCl3 when using the other isotope of chlorine gas:
Mass of PCl3 = 1 mole * (35.5 + 3B)
Plugging in the values, we get:
Mass of PCl3 = 1 mole * (35.5 + 3 * B)
Mass of PCl3 = 1 mole * (35.5 + 3 * (35.5 - (35.5 * 0.76)) / 0.24)
Simplifying, we get:
Mass of PCl3 = 1 mole * (35.5 + (106.5 - 2.28 * 35.5) / 0.24)
Mass of PCl3 = 1 mole * (35.5 + (106.5 - 81.08) / 0.24)
Mass of PCl3 = 1 mole * (35.5 + 104.17)
Mass of PCl3 = 1 mole * 139.67
Therefore, if chlorine gas containing the other isotope were used, one mole of PCl3 would weigh approximately 139.67 grams.