G chem problems

[deleted31120]

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i was wondering if any body could help me deal with conversions

like finding the molality of 14 g of NAF in 400 mg of solvent.

an how many atoms of carbon in 4.00 x 10^-8 grams of C3H8

i know the the conversion ie

10^-6 micro
10^-3 milli
10^-9 nano

but does anyone know how to do the actual calculation without all the zeros

Has anyone seen alot of conversions on the DAT

 

[croco]

#1 -- convert NAF from grams to moles: 14g*1mol/42g=.33 moles
convert 400mg solvent to kg: 400mg*1KG/1000000mg and cancelling the zeros you end up with 4/10000 or 4*10^-4
so if .33 is also 1/3 then 1/3/4 is equal to 1/12 which I believe is around .08. so .08/10^-4 or 800 molals or close to it.
#2 -- 4*10^-8 g of C3H8*36g of C/44g of C3H8 where 36/44 can be reduced to 9/11. So 4*9=36/11 which is close to 3.25.
So now we have 3.25*10^-8 g of C*1mol/12gC or something close to (.28*10^-8 mol of C*6.022*10^23 atoms of C/1 molC) and it seems something close to 1.7*10^15 atoms
--Note these numbers may not be exact but I think they should be close.

 

[deleted31120]

hey thanks that realy helps,

the second answer is dead on,
but the first answer is 0.83 m in barrons test 1 Q58?

barrons could be wrong i know they have alot of errors especially in the chem and org chem section

 

[croco]

Im pretty sure it should be close to 800 molals or something close to it, however someone can correct me if they find an error in my calculation

 

[croco]

I think where Barrons screwed up is that they calculated by using grams instaed of milligrams.

400g*1KG/1000g=4/10KG
so you have 1/3/4/10 or 10/12 which reduces to 5/6 which is .83 molals. As you can see Barrons gives excellent problems but has calculation problems.

 

[jjflynn2]

croco had

400mg*1KG/1000000mg

but there are only 1000mg in a KG, so change his answer accordingly, and it matches the book. His math and logic help should be very clear though.

 

[croco]

No, there are 1000g in one kg? there are 1000mg in 1 g so 1000*1000=1000000