Can someone help me on a chemistry problem, please?

Question

Ok, I'm completely stuck on this one! Any help would be appreciated!

The problem is...

"When a mole of CuSO4 X 5H2O (copper sulfate pentahydrate) is heated to 110 degrees C, it loses four moles of H2O to form CuSO4 X H2O (copper sulfate hydrate). When it is heated to temperatures above 150 degrees C, the other mole of H2O is lost.

a. How many grams of CuSO4 X H2O could be obtained by heating 675 g of CuSO4 X 5H2O to 110 degrees C?

b. How many grams of anhydrous (no water, dry) CuSO4 could be obtained by heating 584 g of CuSO4 X 5H20 to 180 degrees C?

The teacher provided the answer (but no work) and it's...

a. 480 g CuSO4 X H2O

b. 373 g CuSO4

Please show all work and thanks for the help!

Answer 1

To answer these you need to write a balanced equation and work out the number of moles of each starting material.

a)

CuSO4.5H2O --------110 deg----> CuSO4.H2O + 4 H2O

moles = mass / molar mass

molar mass CuSO4.5H2O = 63.55 + 32.07 + (9 x 16.00) + (10 x 1.008) = 249.7 g/mol

Therefore moles CuSO4.5H2O = 675 g / 249.7 g/mol

= 2.70 moles of CuSO4.5H2O

Froom the balanced equation, 1 mole of CuSO4.5H2O produces 1 mole of CuSO4.H2O, therefore if you have 2.70 moles of CuSO4.5H2O you will produce 2.70 moles of CuSO4.H2O.

Now, if moles = mass / molar mass

then mass = moles x molar mass

molar mass CuSO4.H2O = 63.55 + 32.07 + (5 x 16.00) + (2 x 1.008) = 177.636 g/mol

So mass CuSO4.H2O = 2.70 moles x 177.636 g/mol

= 479.62 g

= 480 g (3 sig figs)

b)

CuSO4.5H20 -----180 deg -----> CuSO4 + 5H20

moles CuSO4.5H20 = 584 g / 249.7 g/mol

= 2.34 moles of CuSO4.5H2O

1 moles CuSO4.5H2O gives 1 moles CuSO4, therefore from 2.34 moles of CuSO4.5H2O  you will get 2.34 moles of CuSO4

molar mass CuSO4 = 63.55 + 32.07 + (4 x 16.00) = 159.62 g/mol

mass CuSO4 =  moles x molar mass

= 2.34 mol x 159.62 g/mol

= 373.51 g

= 374 g (3 sig figs)

Answer 2

First of all calculate the molar mass of CuSO4 that is 159,62 g/mol and then the one of H2O which is 18,016 g/mol. Now calculate the number of moles of CuSO4 x 5H2O that correspond to 675 g: 675 g / (159,62 + 18,016 x 5) g/mol = 2,7 mol. Now find out how much (in percentage) CuSO4 is contained in CuSO4 x 5H2O: 249,7 : 100 = 159,62 : X --> X = 63,9% so the 5H2O will be the remaining 36,1% with 1H2O being 36,1/5 = 7,22%. So the percentage of CuSO4 x H2O contained in CuSO4 x 5H2O will be 63,9+7,22= 71,12%. So 675 : 100 = X : 71,12 --> X = 480,06 g.

Now that you have all the data, do the same thing for the other one:

584 : 100 = X : 63,9 --> X = 373,176 g