Chemistry
OpenStudy (anonymous):

We did a lab with Copper Chloride (II) where we dehydrated a sample for 30 minutes, weighed it, heated it up for another 10 minutes, and then put it over boiling water and rehydrated the sample, Here's out numbers: First weigh (before dehydration):.589 g Second weigh (30 min in the oven): 17.501 g third (after an extra 10 min in the oven): 17.499 and after re-hyradtion: 1.256 g I need help finding the ratio of water to copper chloride in the sample.

OpenStudy (anonymous):

and it's the mole ratio I need help with

OpenStudy (atlas):

Is the after rehydration weight 1.256kg

OpenStudy (anonymous):

yeah that was are final number

OpenStudy (anonymous):

My partner got back to me saying it was a 243.9 : 1 ratio, but no idea how he would get that number

OpenStudy (atlas):

k ...u wrote 1.256g in your question :P

OpenStudy (atlas):

you can safely assume that the weight of given amount of CuCl2 (completely dehydrated) is around 17.500g

OpenStudy (anonymous):

yes

OpenStudy (atlas):

you can see this because after putting the sample in the oven for additional time - its weight does not reduce right??

OpenStudy (atlas):

Increase in weight after rehydration is -> (1256 - 17.5)g So this is the weight of the water that combines with the CuCl2

OpenStudy (anonymous):

it reduced very slightly...taking the excess water that was left out

OpenStudy (atlas):

yeah it reduced very slightly because all the water was (almost) already taken out :P

OpenStudy (atlas):

1238.5 g water (H20) for 17.5 g of CuCl2

OpenStudy (atlas):

Can you find how many moles of water will be there in 1238.5g of H20?

OpenStudy (atlas):

1 mole of water = 18g

OpenStudy (anonymous):

68.8 mol?

OpenStudy (anonymous):

what do I do with this number?

OpenStudy (atlas):

right!!.......sry I was away

OpenStudy (atlas):

now find the number of CuCl2 moles in 17.5g of CuCl2

OpenStudy (anonymous):

what is the number of CuCl2? Look up the atomic mass?

OpenStudy (atlas):

yeah u can do that atomic mass of Cu + 2* atomic mass of Cl

OpenStudy (anonymous):

so 63.546 + 2 =127.092* 35.45 = 4,505.4114? right?

OpenStudy (atlas):

Nope

OpenStudy (anonymous):

it seemed wronged

OpenStudy (atlas):

M(Cu) = 63.546 M(Cl) = 35.45 M(CuCl2) = M(Cu) + 2M(Cl)

OpenStudy (anonymous):

so 63.546 + 70.9 =134.446?

OpenStudy (atlas):

right!!

OpenStudy (atlas):

go on find the number of moles in 17.5g of CuCl2

OpenStudy (anonymous):

is it 7.682 /18? so .426? and thanks for all the help, by the way.

OpenStudy (atlas):

Nope 1 mole of CuCl2 is equal to the atomic mass of CuCl2 you calculate -> 134.446g So use simple unitary method to find how many moles will be there in 17.5g of CuCl2

OpenStudy (atlas):

134.46g of CuCl2 === 1mole -> 1g of CuCl2 === (1/134.46)mole so 17.5 g of CuCl2 === ??? moles

OpenStudy (anonymous):

2,353.5 g?

OpenStudy (anonymous):

134.46 * 17.5

OpenStudy (atlas):

nope

OpenStudy (atlas):

it is 17.5 *(1/134.46)

OpenStudy (anonymous):

.00743

OpenStudy (anonymous):

where do I go from here?

OpenStudy (atlas):

you still have your calculations wrong there!

OpenStudy (atlas):

anyway if you find the number of moles of water in the sample and then the number of moles of CuCl2 -> you can divide them both to get the mole ratio

OpenStudy (anonymous):

hmm I keep getting that .000743 or .130

OpenStudy (atlas):

0.130 is right!!

OpenStudy (anonymous):

Oh great! so divide .130 and 134.46?

OpenStudy (atlas):

I just observed that you have to find the mole ration of water and CuCl2 in the original sample..............SO wt of the sample b4 dehydration was - 589g and after dehydration was 17,5 In dehydration all water was lost = 589g -17.5g =571.5g so number of moles of water in ORIGINAL SAMPLE = 571.5/18 = 31.75

OpenStudy (atlas):

so divide 31.75 and 0.13

OpenStudy (atlas):

OpenStudy (anonymous):

244.2:1

OpenStudy (anonymous):

Thanks for all the help!

OpenStudy (atlas):

right!!

OpenStudy (atlas):

you are welcome..........i hope u understood the concept

OpenStudy (anonymous):

Yeah, itsdefinitely made more sense as I worked through it