Question

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Math/chemistry question. Percent abundance.

Answer 1

@sweetburger @mathstudent55

Answer 2

this can help

http://www.chemteam.info/Mole/AvgAtomicWt-Reverse.html

Answer 3

I actually have that link open and it didn't help me, sorry

Answer 4

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element

Answer 5

I'm not sure if I have the percent abundance format down wrong or something... I got this:\[\frac{10.012937}{10.811}=92.6%\]

Answer 6

Oops, forgot the \(\times100\)

Answer 7

so we need the average atomic weight of boron

we already have the weights for the two isotopes

Answer 8

But yeah. I got 92.6%

Answer 9

10.811 u?

Answer 10

and the percentages add up to 100%

Answer 11

No... they don't.\[\frac{11.009305}{10.811}\ne100-92.6%\]

Answer 12

\[(10.012937)(x) + (11.009305)(1-x) = atomic~weight~of ~boron ~which~ is ~10.811\]

Answer 13

wait the x and the x-1 should be switched

Answer 14

??

Answer 15

BTW i got this from the information on the website

Answer 16

okay so \(^{10}B\) is 92.6%?

Answer 17

I feel like I calculated that wrong.

Answer 18

hold let me calculate my answer and see, if I don't understand u will need to tag someone that knows chemistry :(

Answer 19

Alright o:

Answer 20

I am getting 80% abundance for Boron 10

Answer 21

using the formula above

Answer 22

no sorry 20% abundance for Boron 10, and 80% for Boron 11

Answer 23

Alright

Answer 24

Wait, EXACTLY 80%?

Answer 25

80.09721% to be exact

Answer 26

Ah, I see
2 sig figs then, yes

Answer 27

and I am correct :)

https://en.wikipedia.org/wiki/Isotopes_of_boron
the first sentence reads,

"Boron (B) naturally occurs as isotopes, 10B and 11B, the latter of which makes up about 80% of natural boron. "

Answer 28

xD

Answer 29

Yeah, it's right. Thank you :)

Answer 30

sorry i don't know what you did in the beginning but the website i gave u first was pretty reliable ;)

Answer 31

np :)

Answer 32

OH, I know what I did... I divided the mass for \(^{10}B\) over that of \(B\)

Answer 33

lol

Answer 34

wait did u get that it is 20% for Boron 10 and not 80%,

80% is the percentage for Boron 11

Answer 35

apparently it wasnt that simple...

Answer 36

Yeah, I got it!

Answer 37

(if anyone wants to help w/ this) How many atoms are present in a 70.0 −cm3 sample of plumber's solder, a lead–tin alloy containing 67% Pb by mass and having a density of 9.4 g/cm3?
Express your answer using two significant figures.

Answer 38

Post this question again, sorry I have to go and I probably don't know this much chem anyway :(

Answer 39

Alright have a good day then :)