Question

You have a sample of Li2SiF6• xH20, which weighs 0.4813g. After heating the sample to get rid of the water of hydration, the weight of the anhydrous compound was found to be 0.3910 g. What is the empirical formula of the hydrate?

Show all steps.

1. Determine the mass of H2O

2. Convert mass to moles

3. Convert mass of anhydrate to moles

4. Find the ratio

5. Write the empirical formula of the hydrate.

Answer 1

1)
mass of = H2O = mass of hydrated salt - mass of anhydrous salt
mass of = H2O = 0.4813 g - 0.3910 g
mass of = H2O = 0.0903 g
Answer: 0.0903 g

2)

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass(H2O)= 0.0903 g

number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(0.0903 g)/(18.016 g/mol)
= 5.012*10^-3 mol
Answer: 5.012*10^-3 mol

3)
Molar mass of Li2SiF6,
MM = 2*MM(Li) + 1*MM(Si) + 6*MM(F)
= 2*6.968 + 1*28.09 + 6*19.0
= 156.026 g/mol


mass(Li2SiF6)= 0.3910 g

number of mol of Li2SiF6,
n = mass of Li2SiF6/molar mass of Li2SiF6
=(0.391 g)/(156.026 g/mol)
= 2.506*10^-3 mol
Answer: 2.506*10^-3 mol

4)
X = mol (H2O)/mol (Li2SiF6)
X = 5.012*10^-3 / 2.506*10^-3
X = 2
Answer: 2

5)
Empirical formula is Li2SiF6.2H2O