[pizza1512]

Can someone help me out with this question:

4.17g of hydrated barium bromide crystals (BaBr₂.nH₂O) gave 3.72g of anhydrouus barium bromide on heating to constant mass. Work out the relative molecular masss (M_r) of the hydrated barium bromide and the value of _n.

BaBr₂.nH₂O -> BaBr₄ + H₂O

Thanks

[enahs]

How much water was produced?

[pizza1512]

Well the water produced is 4.17g (hydrated barium bromide) - 3.72g (anhydrated barium bromide) which is 0.65g.  How does that help?

[enahs]

How many mols of water is that? Where does the water come from?


http://chem.lapeer.org/Chem1Docs/HydrateLab2.html

[Quantum07]

The mass of the anhydrous barium bromide subtracted from the hydrated barium bromide gives the mass of water that was lost.

Now find how many moles of water you just lost.

0.45 g of water lost is 0.025 moles water

Find how many moles of anhydrous Barium Bromide you have.

BaBr2= 297.16 g/mol

moles of anhydrous BaBr2= 0.0125

From here it's similar to finding %composition. Divide by lowest number of moles in this case BaBr2

0.0125 BaBr2 /0.0125 mol BaBr2 = 1

0.025 mol H20 / 0.0125 mol BaBr2= 2

The formula for the hydrate is

BaBr₂.2H₂O

The relative molecular mass is 297.16 g/mol BaBr₂ + 36.16 g/mol H₂O =

333.32 g/mol BaBr₂.2H₂O