Rite I just need explanation for this because i just cant get my head around it. its a basic AS level chemistry practical plan
barium bromide crystals contain water of crystallisation. heating the crystals drives off the water:
BaBr.2H2O ---> BaBr2(s) + 2H2O(g)
hydrated Barium approximately loses 11% of its mass when fully dehydrated. Confirm that there are 2 moles of water of crystallisation in one mole of the hydrated barium bromide BY CALCULATION Using: 25.00g of hydrated barium bromide, 2dp balance
My theory - well obviously the 11% lost from the barium bromide is the water right so how do i confirm there is two moles of water? ie BaBr.xH2O where x = 2. please explain as if you were explaining to a baby...i am dumb asap
Ok, these stoichiometric problems can be real mind benders, so ill take it real slow.
First, we need the relative atomic weights of the elements,
Ba = 137.34
Br = 79.90
H = 1.00
O = 16.00
Using the equation moles=mass/(relative atomic weight) we can find the mass of one mole of the Barium Bromide before and after the dehydration.
Mass = moles * [Ba + 2*Br + 2*(H2O)]
1*[137.34+(2*79.9)+2*((2*1.00)+16.0 0)] = 333.14 grams
Mass = moles * [Ba + 2*Br]
1*[137.34+(2*79.9)]= 297.14 grams
The confirmation that it is the loss of two moles of water can be seen, as i have included 2 moles of water in the 'before' calculation, and excluded them in the 'after' calculation.
Now, to show that this is equivalent to an 11% loss in mass
333.14 - 297.14 = 36
100*(36/333.14) = 10.8% ==> 11%