Delta Hvap equation

Question

How much energy is required to vaporize 48.7 g of dichloromethane (CH2Cl2) at its boiling point, if its ΔHvap is 31.6 kJ/mol?

I just need to know the formula on how to solve this equation

Answer 1

If the ΔHvap for dichloromethane (CH2Cl2) is 31.6 kJ/mol, this means that it takes 31.6 kJ of energy to vaporize 1 mole of dichloromethane (CH2Cl2) at its boiling point.

 

But you aren't given 1 mole of dichloromethane, you are trying to vaporize 48.7 g of it.  Since we know the heat of vaporization "per mole", we need to first convert grams to moles and then find the heat.  To do that we need the molar mass of CH2Cl2, which is 12.01 + (2 x 1.01) + (2 x 35.45) = 84.93 grams per mole of CH2Cl2

 

We can do this in one string of calculations, using the definitions of molar mass and molar heat of vaporization.

 

We know that 31.6 kJ = 1 mol CH2Cl2 that can be vaporized

We know that 1 mol CH2Cl2 = 84.93 g CH2Cl2

 

Start with your given quantity and convert it to the energy required to vaporize it by using conversion factors based on these relationships:

 

48.7 g CH2Cl2 x (1 mol CH2Cl2 / 89.3 g CH2Cl2) x (31.6 kJ / 1 mol CH2Cl2) = 17.2  kJ   

 

You should always be able to check to see if the answer makes sense.  Since it takes 31.6 kJ to vaporize one mole, and we had less than a mole, it should take less than 31.6 kJ, right?