Please Help- Random Questions?

1. If cobalt (II) sulfate is heated too strongly, the following reaction will occur: CoSO4 (s) --%26gt; CoO (s) + SO3 (g)





If you're heating a sample of CoSO4 · 6H2O %26amp; this reaction occurs along with dehydration, what will happen to the experimental percent water? Please Explain.





2. A 6.000 g sample of CaCl2 · xH2O after heating gives an anhydrous residue of 4.531 g. What is the value of x?





3. A hydrate has the following percent composition: Pb= 54.61%, C= 12.66%, H= 1.61%, O= 16.88% %26amp; H2O= 14.24%. What is the formula of the hydrate?


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*Do I just divide #2?


Please help explain these to me. Thank you so much!

Please Help- Random Questions?
1) If "dehydration" occurs, that means that you're losing water from the cobolt sulfate in the form of water vapor, so the experimental percent water will go down.





2) For this part, first calculate the mass of water released. Since all you're doing here is heating the sample, and the only change is water loss, then the difference in weight is equivalent to the mass of water released.





mass H2O = mass comp'd pre heat - mass anhydrous comp'd


mass H2O = 6g - 4.531g


mass H2O = 1.469 g





You can then calculate the number of moles water released from the mass water released calculated above and the molecular weight of water:





H= 1g/mole


O= 16g/mole





Mol wt H2O = (2*1) + (1*16)= 18g/mol





Moles H2O = mass of H2O released/ Molecular wt H2O


Moles H2O = 1.469g/ 18g/mol


Moles H2O = 0.08 moles





Now use the same process to calculate the moles of anhydrous CaCl2 from the mass of anhydrous compound (given = 4.531 g) and the molecular weight of CaCl2 (MW = 110.98g/mole):





Moles Anhydrous CaCl2 = mass anhydrous CaCl2/ MW CaCl2


Moles Anhydrous CaCl2 = 4.531 g / 110.98 g/mole


Moles Anhydrous CaCl2 = 0.04 moles





If you look at your original formula for hydrated calcium cloride, you'll notice that for every 1 mole of CaCl2 there are x moles of H2O. So the amount of hydration of this compound is a multiple of CaCl2. To calculate exactly what that number is, just divide the moles of water calculated above, by the moles of anyhydrous CaCl2 (also calculated above):





x = moles H2O released/ moles anhydrous CaCl2


x = 0.08 / 0.04


x = 2











3) For this problem, the percent of each compound relates directly to the mass of the entire compound and the molecular weight of each element.








Divide the percent composition of element by it's molecular weight:





Pb= 54.61/ 207.2 = 0.2636


C = 12.66/12 = 1.055


H = 1.61/ 1 = 1.61


O = 16.88/ 16 = 1.055


H2O = 14.24/18 = 0.7911





To change these into whole numbers, just divide all by the smallest value (ie: Pb = 0.2636):





Pb= 0.2636/0.2636 = 1


C = 1.055/ 0.2636 = 4


H = 1.61/ 0.2636 = 6


O = 1.055/0.2636 = 4


H2O = 0.7911/0.2636 = 3





You know that lead either has to be +2 or +3, and since C, H, and O are all multiples of 2, then you're dealing with lead II. So the formula for the hydrate has to be:





Pb(C2H3O2)2 * 3 H2O





I hope this is clear.