Ask your own question, for FREE!
Chemistry 8 Online
OpenStudy (anonymous):

How do I find the mass of the anhydrous salt that would remain if I heat 2.56 g of ZnSO4*7H20

OpenStudy (australopithecus):

use gravimetric factor \[Grams OfZnSO4*7H2O(\frac{Molecular Mass ofZnSO4}{MolecularMassofZnSO4*7H2O})\]

OpenStudy (anonymous):

Thanks! You're awesome :)

OpenStudy (australopithecus):

Notice how the units of ZnSO4*7H2O cancel out, and you are left with ZnSO4

OpenStudy (australopithecus):

I hope you understand how to use gravimetric factor and how it works from this, if you have any questions please ask

OpenStudy (anonymous):

I just realized that I never used the 2.56g in this... or is that just unnecessary information?

OpenStudy (australopithecus):

It is definitely needed to find the answer.

OpenStudy (australopithecus):

2.56g = the quantity of ZnSO4*7H2O you have, you simply want to find out how much of the 2.56g is ZnSO4. To do this you can simply multiply it by the ratio of ZnSO4/ZnSO4*7H20 molecular masses this will give you the the amount of ZnSO4 in ZnSO4*7H2O

OpenStudy (australopithecus):

So just say I had 1g of NaCl and I wanted to see how much Na was present in this sample. Molecular Mass Na = 23g/mol Molecular Mass Cl = 35g/mol Molecular Mass of NaCl = 58g/mol If I wanted to figure out the ratio of Na to NaCl I simply divide the molecular mass of sodium by sodium chloride like so, \[ \frac{23g/mol}{58g/mol} = 0.39\] so from this we know that 39% of sodium chloride is made up of sodium we can simply multiply this percent in the decimal form by any weight of sodium chloride to figure out how much sodium we have

OpenStudy (anonymous):

I understand! :) Thank you so much!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!