1) Deduce
the two masses and see the amount of water was driven off when heated:
5.03 g -
4.23 g = 0.8 g H2O given off
2) Change
mass from grams to moles of H2O:
0.8 g H2O
/ 18 g H2O in 1 mole = 0.044 mol H2O
3) Change
left over mass to moles of BaCl2 .
4.23 g
BaCl2 / 207 g BaCl2 in 1 mol = 0.021 mol BaCl2
4)Find
the ratio of mol H2O to mol BaCl2:
0.044 mol
H2O : 0.021 mol BaCl2
5) The
resulting ratio is 2:1 so two H2O for each BaCl2, thus, the hydrate was named:
Barium
chloride di-hydrate
