Answer:
Co(NO₃)₂·6H₂O
Explanation:
Assume that you have 100 g of the hydrate.
Then you have 62.9 g Co(NO₃)₂ and 37.1 g H₂O.
1. You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.
[tex]M_{r}[/tex]: 182.94 18.02
Co(NO₃)₂·xH₂O ⟶ Co(NO₃)₂ + xH₂O
Mass/g: 62.9 37.1
2. Use the molar masses of each compound to calculate its number of moles.
[tex]\text{Moles of Co(NO}_{3})_{2} = \text{62.9 g} \times \frac{\text{1 mol} }{\text{182.94 g}} = \text{0.3438 mol}[/tex]
[tex]\text{Moles of H}_{2}\text{O} = \text{37.1 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 2.059 mol}[/tex]
3. Calculate the molar ratio of the two products.
Divide each number by the smaller number of moles (0.3438 mol).
[tex]\text{Moles of Co(NO}_{3})_{2}:\text{Moles of H}_{2}\text{O} = \frac{\text{0.3438 mol}}{\text{0.3438 mol}}:\frac{\text{2.059 mol}}{\text{0.3438 mol}} = 1: 5.988[/tex]
4. Round off each number to the closest integer.
[tex]\text{Moles of Co(NO}_{3})_{2}:\text{Moles of H}_{2}\text{O} \approx 1:6[/tex]
1 mol of Co(NO₃)₂ combines with 6 mol H₂O, so x = 6.
The formula of the hydrate is Co(NO₃)₂·6H₂O.
