Question

How would you identify the empirical formula of a compound CuCla bH2O.
You have at your disposal, an electronic balance and any other equipment you may need.

Answer 1

I need help on finding a.

Answer 2

Mass of water = 1.0 - 0.76 = 0.24 g = 0.013 moles, OK.

0.43 g Cu x (1 mole Cu / 63.5 g Cu) = 0.0068 moles Cu

0.33 g Cl x (1 mole Cl / 35.5 g Cl) = 0.0093 moles Cl.

Dividing by the smallest (0.0068), we get a Cu / Cl ratio of 1/ 1.36 which is 3:4 so your Cu3Cl4 isn't unreasonable (but I've never heard of Cu3Cl4). However, I think that it's MUCH more likely that the ratio is 1 / 1.5 and the correct formula is CuCl2. That compound has a molar mass of 134.4.

0.76 g CuCl2 x (1 mole CuCl2 / 134.4 g CuCl2) = 0.0057 moles CuCl2

0.013 moles H2O / 0.0057 moles CuCl2 = 2.3 moles H2O / mole CuCl2.

The likely formula is CuCl2 x 2H2O

Answer 3

You're supposed to conduct an experiment.

Answer 4

How would you know all of this?

Answer 5

by titration. this is an experiment ot find the mass% and hence the empirical formula..!

Answer 6

This is what I think please correct me if I am wrong:
Method:
1. Evaporate water off from the compound by heating it in a ceramic flask, over a bunsen burner and a tri-pod.
2. Weigh the anhydrous sample of CuCla.
3. React the CuCla whith K(s) wait until no further precipitate is formed.
So CuCla + K -> KCl + Cu
4. Weigh the sample of Cu.
5. Find the moles of Cu by deviding by molar mass.
Subtract mass of Cu from CuCla to get mass Cla, devide by its molar mass to get moles.
6 Ratio moles Cu with moles Cl.

Answer 7

@AnImEfReaK is this what u think is titration????

Answer 8

No I'm using metal displacement.

Answer 9

ohk well if u use metal displacement method then how would u find the emperical formula or the mass% ???

Answer 10

mass%?

Answer 11

idk

Answer 12

is my method right?

Answer 13

mass% is the % of that element in the whole compound..!

Answer 14

Ok i get that

Answer 15

i dont know how to find Emp. formula by metal displacemet.. but in titration i can help u!

Answer 16

I would ratio the moles of Cu and Cla
Cu:Cla
to get the mole ratios.

Answer 17

Okay can you explain step by step please?

Answer 18

yeah shure m sending u a link of a video in which titration is performed..!!
http://www.youtube.com/watch?v=sFpFCPTDv2w

Answer 19

alright

Answer 20

The video was helpful but lets say
you added 50mL of CuCla and 19mL of HCL at 2 mol/L reacted, even though how would you find moles of CuCLa?

Answer 21

ur talking about titration?

Answer 22

Yes

Answer 23

by molarity..!! @AnImEfReaK

Answer 24

?

Answer 25

molarity =moles / volume..

Answer 26

mol HCL = 38
mol CuCl = 50 mL * what concentration?

Answer 27

mol is never meseared in ml..!!

Answer 28

It is n = C times V

Answer 29

The molecular weight of a compound is simply the sum of all the individual components. In CuCl2 x 2H2O, there are simply two additional water molecules hydrating your CuCl2. Therefore,

MW (CuCl2 x 2H2O) = Cu + 2(Cl) + 2[2(H) + O]
= (63.54 g/mole) + 2(35.45 g/mole) + 2[2(1 g/mole) + 16 g/mole)]
= 170.44 grams/mole


If you have 0.502 g of this substance, then you will have:

0.502 g/(170.44 g/mole) = 0.00295 moles of CuCl2 x 2H2O

Answer 30

then u have the volume and now moles as well soo now can u find the molarity??

Answer 31

but how do you even know a is 2 and b is 2

Answer 32

i found it earlier..!!! scroll it..!!

Answer 33

Cu can take on +1 2 or 3 or 4 it has many oxidation states you can't assume it takes on that form.

Answer 34

I'm very confused...

Answer 35

there is a method of finding the empirical formula.. and by that i have found it..!! scroll these comments and u'll see the complete method i had done

Answer 36

Mass of water = 1.0 - 0.76 = 0.24 g = 0.013 moles, OK.

0.43 g Cu x (1 mole Cu / 63.5 g Cu) = 0.0068 moles Cu

0.33 g Cl x (1 mole Cl / 35.5 g Cl) = 0.0093 moles Cl.

Answer 37

I don't get this

Answer 38

Did you separate the compound?

Answer 39

yeah..!! i have foound moles of each seperately

Answer 40

But how? What did you do?

Answer 41

You have to write steps down so not figures.

Answer 42

I'll understand it better that way.

Answer 43

i am posting a generalised formula
\[\frac{ Vacid*Macid }{Vbase*Mbase }=\frac{ no. of moles of acid \in the balanced equation }{ no. of moles of base \in the balanced equation }\]