solve for fractional abundance

Question

nincompoop · Answer

The atomic weight of an element with varying isotopes can be calculated by the formula
AW=Σfm=Σf1m1+f2m2+f3m3...
 
where
AW : atomic weight
f : fractional abundance
m : mass

Suppose that the element Magnesium has an atomic weight of ≈24.31 amu and the following isotopes have the following mass (amu):
$$_{24}^{12}Mg=23.98504$$
$$_{25}^{12}Mg=24.98584$$
$$_{26}^{12}Mg=25.98259$$

solve for the fractional abundance of each isotopes.

anonymous · Answer

Data is insufficient to solve.

nincompoop · Answer

but how can you say that when we are able to solve if the unknown fractional abundance was 2?

abb0t · Answer

Hmm...is that all the information given?!

anonymous · Answer

if we take fractional abundace of one element as x, another as y then the third will be 1-x-y. now we have 2 variables but only one equation so it cannot be solved. If we had only two isotoped, we could use just x and 1-x and the equation would've been solvable.

nincompoop · Answer

I know that @rahul1995 and @abb0t ya that's all... I thought this was rather stupid.

nincompoop · Answer

maybe the math gods and goddesses can figure this out? lol where is that fibo chic?

abb0t · Answer

I think you're solving for "x". Set it equal to 100%.

anonymous · Answer

lol, noone can solve it without more information, you'll have infinite many solutions of this problem.

abb0t · Answer

rahul, ur so negative. u said that abt the integral I posted earlier. quit being such a debbie downer.

nincompoop · Answer

can one, just one, fractional abundance be solved?

anonymous · Answer

okey sorry, I was trying to help.

abb0t · Answer

Maybe if you had just two isotopes. Lol.

nincompoop · Answer

LOL !

abb0t · Answer

i actually don't know n e more. wat the funk? wat is lyf.