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Onkarabile Tiro (University of Witwatersrand, South Africa (ZA))

General question question in main group

When a 0.998g sample of CuSO4.xH20 was heated strongly,so that the waters of hydration were given off,the mass of the anhydrous salt remaining was found to be 0.638 g.What is the experimental value of the percent water of hydration?

April 21 2013, 10:15 pm

ALL ANSWERS
Leope Thobejane (University of Witwatersrand, South Africa (ZA))
First you need to calculate the amount of water lose (mass) that will be sample mass-anhydrous mass (0.998-0.638)g=m, Then calculate number of moles of water this will me your calculated mass of water lose "m" divided by the molar mass of water(18.02g/mol) {n(H2O)=m/18.02}. You can realize from your equation that the ratio Salt:water is 1:x therefore you can compare your calculated moles of water with the moles of the salt {n(salt)=0.638/Molar mass of salt}, then calculate your x.

Mass water losed= (0.998-0.638)g=0.36g
n(H2O)=0.36/18.02=0.01998 mol

Mass salt =0.638g
Molar Mass Salt=159.62g/mol
n(salt)=0.0039997 mol

x=0.01998mol/0.003997mol=5


April 22 2013, 10:52 am


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