What was the mass of the anhydrous salt?

Question

A hydrate of aluminium chloride, AlCl3 X H2O was used.

Empty crucible - 15.807g

Crucible + hydrate 17.061g

Crucible + content

after 1st heat 16.503g

after 2nd heat 16.499g

a. What was the mass of the anhydrous salt? How many moles of anhydrous salt is this?

b. What was the mass of the water driven off? How many moles of water is this?

c. Find mole ratio of water to anhydrous salt?

d. What is the formula for this hydrate?

Answer 1

(Answers are written in bold)

A) you're determining the mass of the salt without water (anhydrous means without water). To do this, you subtract the mass of the salt after heating it by the mass of the empty crucible. Since the problem does not specifically say which mass recording to use, I am going to assume we are using the mass after the second heating.

16.499 grams (crucible + salt) - 15.807 grams (empty crucible) = 0.692 grams of anhydrous salt

Then you just need the molar mass of the anhydrous salt to determine the number of moles. The molar mass of aluminum chloride is 133.34 grams/mole. If you do not know how to find molar mass, just add the atomic weights of each element present in the formula.

Now we can convert to moles of anhydrous aluminum chloride from grams using a conversion factor:

0.629 grams AlCl₃ (1 mole / 133.34 grams) = 0.00519 moles of AlCl₃ (anhydrous)

B) To determine the amount of water evaporated from the salt, we have to first determine the original mass of the hydrated salt by subtracting the mass of the crucible and the hydrated salt by the mass of the empty crucible:

17.061 grams (crucible and hydrate) - 15.807 grams (empty crucible) = 1.254 grams of hydrated salt

From here, we have to determine what mass of water was evaporated from the salt. To do this, we will subtract the mass of the dehydrated salt that we calculated in part A from the mass of the hydrated salt that we just calculated. The difference in masses between these two quantities will equate to the amount of water you evaporated by heating the salt:

1.254 grams of hydrated salt - 0.692 grams of anhydrous salt = 0.562 grams of water evaporated

Now that we have the mass, we can write a conversion factor using the molar mass of water to convert our grams of water into moles of water. The molar mass of water is 18.02 grams/mole:

0.562 grams of water (1 mole / 18.02 grams) = 0.0312 moles of water were evaporated

C) Mole ratios are calculated by dividing the largest number of moles by the smallest number of moles. Since we have more moles of water than anhydrous salt, we will divide the moles of water by the moles of salt to get our mole ratio:

0.0312 moles / 0.00519 moles = 6.01

This tells us that in the formula for the original hydrated salt, there are 6 moles of water for every one mole of aluminum chloride. We can use this information to solve for part D. Since we know we need 6 moles of water in our chemical formula, we can now write the formula.

D) AlCl₃ x 6 H₂O (this compound would be called aluminum chloride hexahydrate, where hexahydrate means there are 6 water molecules in one formula unit for this compound).

Answer 2

1). Mass of crucible + hydrate = 17.061g

2). Mass of empty crucible = 15.807g

3). Mass of hydrate sample 1.254g subtract 1 from 2

Crucible + sample

4). after 1st heating = 16.503g

5). after 2nd heating = 16.499g

6). mass of water lost = 0.562g subtract 1 from 5

7). mass of anhydrous salt = 0.692g subtract 3 from 6

8). moles of anhydrous salt = 0.00518mol use 7 value and divide by the molar mass of AlCl₃

Al = 1 * 26.98 + Cl = 3 * 35.5

= 133.48g/mol

9). moles of water lost = 0.0312 mol use 6 value and divide by the molar mass of H₂O

H = 2 * 1 + O = 1 * 16

= 18g/mol

10). mole ratio of water to anhydrous = 6 9 answer/8 answer

11). Formula for hydrate = AlCl₃•6H_2O