Question

When a 1.69 g sample of AlCl3·xH2O is heated to remove all water, a mass of 0.933 g remains. Determine the value of x.

Answer 1

compare the moles of water to the moles of the anhydrate.

Answer 2

So to do that, I'd take the given g values then multiply by the molar masses of the compounds? (Ex. for AlCl3, 1.69g*(1mol AlCl/133.34g) Or am I way off on this logic?

Answer 3

you would divide, not multiply. Look at the formula, \(n=\dfrac{m}{M}\)

n=moles, m=mass, M=molar mass

Answer 4

also, you wouldn't use 1.69 g.

Answer 5

What would I use?

Answer 6

1.69 g is the mass of the hydrate it was heated it up and 0.933 g (of the anhydrate) were left.

Whats the mass of the water?

Answer 7

1.69-0.933?

Answer 8

\(\checkmark\)

Answer 9

So I calculated the moles of water to be 0.04213. From here, were do I go?

Answer 10

calculate the moles of the anhydrate

Answer 11

(0.993g AlCl3/133.34g/mol)=0.006997mol AlCl3

Answer 12

So from here, how would I compare the two?

Answer 13

divide the larger by the smaller

Answer 14

6.02! So 6?

Answer 15

yep !

Answer 16

Thank you VERY much!! So appreciated!!

Answer 17

no problem !