How to find a formula of Hydrated salt.

PURPOSE

The aim is to find the formula of a hydrated salt. This might, for example, involve finding the value of x in the formula of salts such as hydrated barium sulfate, BaSO4xH2O or hydrated iron (II) sulfate, FeSO4xH2O. This can be done by heating a measured sample of the salt to dryness and finding the loss in mass.

METHOD

 A lid is usually necessary in the early stages of heating to prevent the loss of crystal fragments while the bulk of the water is being driven off.

Step1: Weigh a crucible and lid. Place the planned quality of hydrated salt in the crucible and reweigh the hydrated salt, lid and crucible.

 Step2: Heat the crucible and its contents for about 10 minutes. Start heating gently and then more strongly. During the last minute of heating remove the cover so that any moisture which has collected on the underside of the cover can evaporate.

Step3: Allow the crucible to cool. Then reweigh the crucible, lid and contents.

Step4: Replace the cover and heat the crucible for 5 more minute, removing the cover during the last minute of heating as before cool and reweigh the crucible, lid and contents. This last mass should agree with the previous mass to within the accuracy of balance. If it does not, repeat this heating until a constant mass is reached.

Steps3 and 4 are repeated until the results shown that are all the water of crystallization has been driven off. Once this has happened the combined mass of the crucible, lid and residue ceases to fail. Chemist call this 'heating to constant mass'

NOTE: check safety before carrying out any practical procedure.

Results of crucible Plus Lid= 18.55 g

Results of crucible, lid and hydrated salts crystal= 23.95 g

Mass of crucible, lid and residue after heating to constant weight= 21.49 g

CALCULATIONS

Mass of hydrated iron (II) sulfate before heating=5.40 g

Mass of anhydrous FeSO4 after heating= 2.94 g

Loss in mass of water=2.46 g

                                                                         FeSO4                                                                H2O

Masses combining/g                                      2.94                                                                        2.46

Molar mass/g ^mol                                         152                                                                         18

Amounts combining/mol                     2.94/152 =0.0193                                                       2.46/18= 0.137

Amounts Mol/ (Smaller amount/mol) 0.0193/0.0193= 1.00                                           0.137/0.0193 =7.10

Simplest whole numbers ratio                                   1                                                                              7

Value of x in FeSO4xH2O= 7

Formula of hydrated salt= FeSO4.7H20