# Question b6ba3

Jan 22, 2015

I'm assuming you mean $\text{weight of}$, instead of $\text{water of}$, right? Here's how I think the data actually looks like:

The weight of the crucible is $\text{10.0 g}$.
The weight of the crucible + the weight of the copper (II) sulfate pentahydrate is $\text{14.98 g}$. Automatically, you know how much copper (II) sulfate pentahydrate you have

m_("hydrate") = "14.98 g" - "10.00 g" = "4.980 g"

The weight of the crucible + the weight of the residue is $\text{13.54 g}$. The "residue" actually represents the anhydrated ${\text{CuSO}}_{4}$, which means that you now know how much water has been evaporated

m_("water") = m_("hydrate") - "(13.54 g - 10.00 g)" = "4.980 g" - "3.540 g"

m_("water") = "1.440 g"

You know that water has a molar mass of $\text{18.0 g/mol}$. This will help you find the number of moles you have

"1.440 g" * ("1 mole")/("18.0 g") = 0.0800 $\text{moles}$

The number of molecules is

"0.0800 moles" * (6.022 * 10^(23) "molecules")/("1 mole") = 4.82 * 10^(22)# $\text{molecules}$