AS Chemistry Calculations

work out number of moles of BaCl2 you have in 8.3g
work out number of moles of h20 you have in 1.7g

simplify ratio to 1:x, and thus you have your answer...

Reply 2

Moles of BaCl anhydrous = mass/moles = 8.53/208 = 0.041
1:1 ratio of Hydrous to Anhydrous
Therefore 0.041 Moles of Hydrous

Mr of Hydrous = mass/moles
Mr = 10/0.041 = 244

Mr of Hydrous = 244

Mr of xH2O = 244 - 208 = 36

Mr of one H2O molecule is 18

36/18 ... therefore x=2

Reply 3

OP

mc_watson87

Moles of BaCl anhydrous = mass/moles = 8.53/208 = 0.041
1:1 ratio of Hydrous to Anhydrous
Therefore 0.041 Moles of Hydrous

Mr of Hydrous = mass/moles
Mr = 10/0.041 = 244

Mr of Hydrous = 244

Mr of xH2O = 244 - 208 = 36

Mr of one H2O molecule is 18

36/18 ... therefore x=2

Yay, I got that x=2, but I worked it out differently I think:

BaCl2
8.53/208
=0.041
0.041/0.041
= 1

H20
1.47/18
= 0.0817
0.0817/0.041
=2 (almost)

That works, right?

Reply 4

kieshaxxx

Yay, I got that x=2, but I worked it out differently I think:

BaCl2 H20
8.53/208.3 1.47/18
=0.041 =0.0817
0.041/0.041 0.0817/0.041
=1 =2 (well almost)

That works, right?

Yep

...twas Stevo's method

Reply 5

OP

mc_watson87

Yep

...twas Stevo's method

Ah good good. I didn't realise that's what he meant :redface:

Reply 6

*makes note to self, explain things better*

still, you got there in the end... :smile:

Reply 7

OP

Ok, I'm stuck on this one now :frown:

:

When 585g of the salt UO(C204).6H20 was left in a vacuum desciccator for 48hrs, the mass changed to 535mg. What formula would you predict for the resulting substance?

Anyone help, please?

Would all of the H20 be removed by the desiccator, just leaving UO(C204)?

Reply 8

kieshaxxx

Ok, I'm stuck on this one now :frown:

:

When 585g of the salt UO(C204).6H20 was left in a vacuum desciccator for 48hrs, the mass changed to 535mg. What formula would you predict for the resulting substance?

Anyone help, please?

Would all of the H20 be removed by the desiccator, just leaving UO(C204)?

Find Moles of initial Salt... Moles = Mass/Mr = 585/450 = 1.3

1:1 Moles Ratio with substance after being in desiccator.

1.3 Moles of Salt after dessicating.

Mr= Final mass/moles= 535/1.3 = 412.

Mr of xH2O = 412 - 342 = 70

70/18 = 3.9

So me thinx x=4
...UO(C204).4H20

Reply 9

*thinks before writes*

work out how many moles of UO(C204).6H20 there were in 585g. i'd do it but i don't have my relative atomic masses with me.

however, if there was 2 moles of the salt, there would be 12 moles of water. work out how many moles of water were lost, and then divide the remaining number by your inital number of moles, so if there were two, divide it by 2. you would use this number as the number of h20's with the UO(C2O4)...

edit: yeah, he explained it better than me but it was the same thing...

Reply 10

OP

Mr of UO(C204) = 238+16+24+64 = 342

Mol = 585/342 = 1.71mol

I'm sorry, I have no idea where to go from here.

Reply 11

you need to add the 6h2o's on as well for the mr of your salt...

238+16+24+64+12+96=450

number of moles = 585/450 = 1.3

*looks at question again*

are the masses meant to be 585g and then 535mg. thats one helluva drop...

Reply 12

OP

El Stevo

you need to add the 6h2o's on as well for the mr of your salt...

238+16+24+64+12+96=450

number of moles = 585/450 = 1.3

*looks at question again*

are the masses meant to be 585g and then 535mg. thats one helluva drop...

Oops, typo. Both meant to be mg.

Reply 13

kieshaxxx

Mr of UO(C204) = 238+16+24+64 = 342

Mol = 585/342 = 1.71mol

I'm sorry, I have no idea where to go from here.

Mr of UO(C204).6H20 = 238+16+24+64+6*18 = 450
Moles = mass/mr = 585/450 =1.3
Therefore 1.3 Moles of UO(C204).xH20
Mr=mass/moles
Mr= 535/1.3 = 412

Mr of xH2O = 412-342=70
70/18= 3.9

so x = 4 i think
...UO(C204).4H20

Reply 14

okay...

(0.585)/450 = 0.0013 moles of salt beforehand...

dessicator removes at least some water... but 0.0013moles of pure salt, ie, anhydrous salt remains...

weight of salt = 342*0.0013
weight of salt = 0.4446

weight of water after dessication = weight after dessication - weight of 0.0013 moles of anhydrous salt = 0.535-0.4446 = 0.0904g

g=mr
0.0904=m18
m=0.005

initial ratio was 0.0013:0.0013 = 1:1
after ratio is 0.0013:0.005 = 1:4 ish

so 4 lots of water...

Reply 15

El Stevo

okay...

(0.585)/450 = 0.0013 moles of salt beforehand...

dessicator removes at least some water... but 0.0013moles of pure salt, ie, anhydrous salt remains...

weight of salt = 342*0.0013
weight of salt = 0.4446

weight of water after dessication = weight after dessication - weight of 0.0013 moles of anhydrous salt = 0.535-0.4446 = 0.0904g

g=mr
0.0904=m18
m=0.005

initial ratio was 0.0013:0.0013 = 1:1
after ratio is 0.0013:0.005 = 1:4 ish

so 4 lots of water...

Phew... we got the same! :biggrin:

Reply 16

tis always good when that happens :smile:

Reply 17

El Stevo

tis always good when that happens :smile:

Btw do you know... Wat an organic compound reacting with sodium hydrogencarbonate solution releasing CO2 shows?
Wat can u determine about the compound from that?

Reply 18

acid + carbonate gives *mumbles* + *mumbles* + co2

the organic compound is a carboxylic acid...

Reply 19