Question

how to determine:
1) mass of BaCl2.XH2O
2) mass of barium chloride anhydrous
3) mass of water in BaCl2.XH2O
4) Percent composition of water in BaCl2.xH2O
5) Formula of barium chloride hydrate

Answer 1

THE INFORMATION
mass of empty crucible = 56.5622g
mass of crucible + hydrate before heating = 57.8310g
mass of crucible + hydrate after heating
(a) first reading = 57.6398g
(b) second reading = 57.6374g
(c) third reading = 57.6362g

Answer 2

@gazed @JoannaBlackwelder

Answer 3

Any ideas on how to find the mass of the hydrate?

Answer 4

i was thinking about "minus" the "before heating" and "the empty crucible."

Answer 5

Looks good to me :)

Answer 6

then, how to get the mass of barium chloride anhydrous?

Answer 7

I would average the three after heating masses and then "minus" again from the "empty crucible"

Answer 8

so i should find the "average first"?

Answer 9

Yeah, that's what I would do :)

Answer 10

so, for "mass of water" = "average after heat" - "before heat"
is it correct?

Answer 11

"before heat"-"average after heat"

Answer 12

oh okay.

Answer 13

i have the formula percentage composition of water
percentage of water = mass of water loss / mass of sample x 100

Answer 14

so is it...
mass of BaCl2.XH2O / mass of water in BaCl2.XH2O x 100 ??

Answer 15

I would do
percentage of water = mass of water/mass of hydrate x100

Answer 16

Okay I will do like that, so how can i find the formula for the last question?

Answer 17

I think we need to use the molar masses of BaCl2 and H2O to find the moles of each. Then use a ratio to find the composition.

Answer 18

i still cant understand the last part

Answer 19

Can you find the moles of BaCl2?

Answer 20

*Hint: use molar mass and mass of anhydrous

Answer 21

1)
57.8310g - 56.5622g = 1.2688 g hydrate
2)
57.6362g - 56.5622g = 1.07400 g anhydride
3)
57.8310g - 57.6362g = 0.1948 g water
4)
(0.1948 g water) / (1.2688 g hydrate) = 0.1535 = 15.35% water
5)
15.35% water = 84.65% BaCl2
(208.2330 g/mol BaCl2) / (0.8465) = 245.99 g/mol hydrate
((245.99 g/mol) - (208.2330 g/mol BaCl2)) / (18.01532 g H2O/mol) = 2.096
Round to the nearest whole number (2) to find the molar ratio of water to BaCl2:
BaCl2·2H2O