csus Department of Chemistry Data Part Il Chem. 1A Experiment 2 Mixture number M
ID: 807013 • Letter: C
Question
csus Department of Chemistry Data Part Il Chem. 1A Experiment 2 Mixture number Mass of the clean dry test tube: Mass of test tube and hydrated mixture: Mass of test tube and anhydrous residue after 1t heating: Mass of test tube and anhydrous residue after 2nd heating: 12 Mass after 1st heating-mass after 2nd heating: (must be 0.010g) Mass of test tube and anhydrous residue after 3rd heating: lif necessary) Mass of water lost: (using the last mass) o, 00 mo Calculations and Results Part II Moles of H2O in lost by the hydrate: (Show calculation for credit) 18.020-0.0bs Moles of BaCl2·2H2O in your sample: (Show calculation for c redit) mol /n Grams of BaCl2·2H2O in your sample: (Show calculation for credit) (244.26 g/mol)l Percent by mass BaCl2·2H2O in your original sample: (Show calculation)Explanation / Answer
Moles of BaCl2.2H2O= 0.0044 mol
n=Moles of H2O/2 [As each mole pf BaCl2.2H2O contains 2 moles water]
=0.0088/2=0.0044mol
_________
Grams of BaCl2.H2O=1.074
Mass of BaCl2.2H2O=Moles *Mol Mass
=0.0044*244.26=1.074g
________
Percent of Mass=53.98%
%Mass=(Mass of BaCl2.2H2O/Total Mass)*100
=(1.074/(13.5487-11.5574))*100=53.98%
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