Ok, so I need a bit of help on these questions. Here they are. If you can, please explain farely well so I can understand how to do the rest.
Calculate atomic mass of element X. Then use the pariodic table to identify the element.
Isotope = 27X
Mass (amu) = 27.997
Percent Abundance = 92.23
I did. 27.997 x .9223 and got 25.821
Im stumped. Please Help.
Thanks in advance.
Calculate atomic mass of element X. Then use the pariodic table to identify the element.
Isotope = 27X
Mass (amu) = 27.997
Percent Abundance = 92.23
I did. 27.997 x .9223 and got 25.821
Im stumped. Please Help.
Thanks in advance.
-
You really haven't provided enough info to make the determination of the average atomic weight of element X. A quick scan of the elements Al, Si and P indicate that the isotope Si-28 has an atomic mass of 27.977 amu and a percent relative abundance of 92.23%. Also it's not aluminum or phosphorous because for both of them, there is only one stable isotope. Of the three possibilities, silicon is the only one that has more than one stable isotope. So it has to be silicon, and as we said, Si-28 is the only that fits the bill.
But none of the data you have provided really points to silicon as opposed to some other neighboring element. We know that 92.23% of the atomic weight is 27.977 amu, but as you pointed out, that comes to 25.803 amu. The 27.997 must be a typo.
Here's the data on silicon from WebElements:
Si-28.... 27.9769 amu ...... 92.2297%
Si-29.... 28.9765 amu ........ 4.6832 %
Si-30.... 29.9738 amu ........ 3.0872 %
Atomic mass of silicon = (0.9223)(27.977 amu) + (0.0468)(28.976 amu) + (0.0309)(29.974 amu) = 28.085 amu
These are the only three stable isotopes of silicon. As you can see Si-28 pretty much matches your data except for the "27X". There is no stable silicon isotope with a mass number of 27. Aluminum-27 is a stable isotope, the only stable isotope of aluminum, and it has a mass of 26.982 amu.
I've given problems like this to my students, but have kept it to two stable isotopes, and I've given them the mass of each isotope. Only then can you actually calculate the the atomic mass of the element. All things considered, I would say that your teacher needs to go back to the drawing board on this question and make some improvements.
But none of the data you have provided really points to silicon as opposed to some other neighboring element. We know that 92.23% of the atomic weight is 27.977 amu, but as you pointed out, that comes to 25.803 amu. The 27.997 must be a typo.
Here's the data on silicon from WebElements:
Si-28.... 27.9769 amu ...... 92.2297%
Si-29.... 28.9765 amu ........ 4.6832 %
Si-30.... 29.9738 amu ........ 3.0872 %
Atomic mass of silicon = (0.9223)(27.977 amu) + (0.0468)(28.976 amu) + (0.0309)(29.974 amu) = 28.085 amu
These are the only three stable isotopes of silicon. As you can see Si-28 pretty much matches your data except for the "27X". There is no stable silicon isotope with a mass number of 27. Aluminum-27 is a stable isotope, the only stable isotope of aluminum, and it has a mass of 26.982 amu.
I've given problems like this to my students, but have kept it to two stable isotopes, and I've given them the mass of each isotope. Only then can you actually calculate the the atomic mass of the element. All things considered, I would say that your teacher needs to go back to the drawing board on this question and make some improvements.