here is a really retarded chem. question

[ManBearPig]

uh...wow. i knew how to do this before but i cant do it now...any ideas? its not really hugely important, but im just going over stuff and for some reason i cant figure it out.

If for the element B you determine the abundance of 10B to be 19.9 % and the abundance of 11B to be 80.1 %, what is the atomic weight for boron

i thought i would do .199 * 10.81 + .801 * 10.81 and then divide that by 1.6606e-24

thanks lol

 

[thesurge]

10*.199 + 11*.801

Don't know what you are doing.

 

[TecHNooB]

I don't think dividing by avagadro's # is necessary. I don't think you can just take the atomic weight of boron and multiply by the percentages because the atomic weights on the periodic table already take abundance into account. You might have to add up the weight of the individual protons and neutrons. In this case, 11B would have an extra neutron. The sum should get you the answer.

 

[soydios]

Originally posted by: thesurge
10*.199 + 11*.801

Don't know what you are doing.

/agree

(I took AP Chemistry last year).

 

[ManBearPig]

oh, the answer is in amu haha