View Full Version : Chemistry HW Problem

UCFinfan86

02-06-2007, 06:42 PM

Naturally occurring chlorine is 75.78 % 35Cl with atomic mass of 34.969 u and 24.22 % 37Cl. What is the atomic mass of 37Cl?

The answer is 36.966 u. But i have no idea how to figure it out, anyone know?

UCFinfan86

02-06-2007, 06:45 PM

actually i have a 2nd problem i dont understand(these are sample problems we should know for our quiz tomorrow)

Chalk is composed of calcium carbonate, CaCO3. How many atoms of Ca2+ are in a piece of chalk with a mass of 3.214 g?

The answer to this one is 1.934 x 10^23 atoms

Roman529

02-06-2007, 06:48 PM

I got a B+ in my Honors Chemistry class....but that was over 20 years ago. I just remember the girl next to me in class was hot. :lol:

dolphinfn3454

02-06-2007, 07:07 PM

I got a B+ in my Honors Chemistry class....but that was over 20 years ago. I just remember the girl next to me in class was hot. :lol:

Sitting next to the hot girl in class is all you should rememeber 20 years down the road.:wink:

King Felix

02-06-2007, 08:22 PM

i got a b in chemistry this first semester, i just have a tight teacher who gives a bunch of easy extra credit like cross words so i couldnt help you

Celtkin

02-06-2007, 08:36 PM

actually i have a 2nd problem i dont understand(these are sample problems we should know for our quiz tomorrow)

Chalk is composed of calcium carbonate, CaCO3. How many atoms of Ca2+ are in a piece of chalk with a mass of 3.214 g?

The answer to this one is 1.934 x 10^23 atoms

It's been a while for me too but:

3.214g CaCO3 x 1.00mol/100.0g CaCO3=0.03214 mol (check to make sure if your Professor wants you to round to 100 or use the actual grams per mole of CaCO3 which is 100.087...Just add up 1 calcium, 1 carbon, and 3 oxygen. 40.08 + 12.01 + 3(15.999) = 100.087 grams/ 1 mole of CaCO3.)

Now that you know that, you should be able to use Avogadro's number to solve for the number of Ca2+ atoms

Celtkin

02-06-2007, 08:41 PM

Naturally occurring chlorine is 75.78 % 35Cl with atomic mass of 34.969 u and 24.22 % 37Cl. What is the atomic mass of 37Cl?

The answer is 36.966 u. But i have no idea how to figure it out, anyone know?

Look on the periodic chart. Cl = 35.453 (with the combined isotopes). Once you know that, the problem is just algebraic.

UCFinfan86

02-06-2007, 08:51 PM

It's been a while for me too but:

3.214g CaCO3 x 1.00mol/100.0g CaCO3=0.03214 mol (check to make sure if your Professor wants you to round to 100 or use the actual grams per mole of CaCO3 which is 100.087...Just add up 1 calcium, 1 carbon, and 3 oxygen. 40.08 + 12.01 + 3(15.999) = 100.087 grams/ 1 mole of CaCO3.)

Now that you know that, you should be able to use Avogadro's number to solve for the number of Ca2+ atoms

When i do the first calculations i get 3.214/100= .03214. Then .03214 x 6.02 x 10^23(Avogadro's number) = 1.934 x 10^22 which is the answer im looking for almost. The correct answer is to the 23rd power. But thank you VERY much for helping

UCFinfan86

02-06-2007, 08:55 PM

Look on the periodic chart. Cl = 35.453 (with the combined isotopes). Once you know that, the problem is just algebraic.

Ya i know the Mass # for CL is 35.45 but im not sure the formula you plug it into. I thought the atomic mass was the same for CL, i didn't know it changed with the amount of atoms. I tried using (75.78)(35.45) + 24.22x but i couldn't figure it out that way.

Timmy54

02-07-2007, 11:02 AM

wow, i work as a chemist (medical tech) and i dont remember all that stuff.

GreenMonster

02-07-2007, 12:08 PM

Do what I did, drop Chemistry and take Astronomy or another science. Chemistry is very tough, you won't remember a thing when your done. A total waste of 2 sesmesters..

UCFinfan86

02-07-2007, 12:38 PM

Do what I did, drop Chemistry and take Astronomy or another science. Chemistry is very tough, you won't remember a thing when your done. A total waste of 2 sesmesters..

Chemsitry is a required for Engineering

Dors156

02-07-2007, 02:33 PM

Im in Chemistry now, but its high school chemistry so im not sure if i could help you.

retired opfinistic

02-07-2007, 04:11 PM

the answer is 4!

UCFinfan86

02-07-2007, 06:33 PM

Ehh to late, i just took the test and flunked, oh well

The Rev

02-07-2007, 10:11 PM

When i do the first calculations i get 3.214/100= .03214. Then .03214 x 6.02 x 10^23(Avogadro's number) = 1.934 x 10^22 which is the answer im looking for almost. The correct answer is to the 23rd power. But thank you VERY much for helping

My wife is a chemistry honors teacher in high school and she says that if you are doing the work right but getting the power wrong, it's probably a calculator error. That's what most of her students do. When you plug the 6.02 x 10^23 into the calculator, don't multiply by 10 then hit the exponent key for the 23. That just does an extra step and throws off your number. plug inthe 6.02 then the exponent symbol ( exp or ee depending on the calculator) then plug inthe 23 and that should do it.

If you are still stuck on the chlorine question, let me know and she says she can help out.

:wink:

UCFinfan86

02-07-2007, 10:43 PM

i tried .03214 x 6.02ee23 and still get the same answer. Its not a big deal though, but thanks for trying

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