Question 2c98c

Feb 21, 2015

The mole ratio between the water and the anhydrous salt is usually determined using the mass of the hydrate, the mass of the anhydrous salt, and the mass of the evaporated water.

Basically, you would prepare a sample of the hydrate and heat it until all the water has evaporated. Once that happens, the hydrate sample will be reduced to the andyrous salt, which means you can weigh it again and see how much water the sample contained.

Here's an example using nichel (II) sulfate. Assume you have a sample of $N i S {O}_{4}$ hydrate that weighs $\text{40.00 g}$. After heating the sample until the water is evaporated, its weight is now $\text{20.74 g}$. Determine the formula of the hydrate.

The first thing you need to do is determine how much water was evaporated; this is done by subtracting the mass of the anhydrate from the mass of the hydrate

${m}_{\text{water") = m_("hydrate") - m_("anhydrate}}$

m_("water") = "40.00 g" - "20.74 g" = "19.26 g"#

Next, determine the number of moles of water and anhydrous salt you have by using their respective molar masses

$\text{19.26 g water" * ("1 mole")/("18.0 g") = "1.070 moles water}$, and

$\text{20.74 g anhydrate" * ("1 mole")/("154.8 g") = "0.1340 moles anhydrate}$

Now divide both these numbers to find the mole ratio water has with the andhydrous salt

$\text{1.070 moles water"/"0.1340 moles anhydrate} = 7.985 \cong 8.00$

This means that you get 8 moles of water for every mole of $N i S {O}_{4}$; as a result, the formula will be

$N i S {O}_{4} \cdot 8 {H}_{2} O$

One mole of $N i S {O}_{4}$, 8 moles of ${H}_{2} O$.