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Gypsum, used for wallboard, has the formula CaSO_4 2H_2 O. What is the percent b

ID: 1024688 • Letter: G

Question

Gypsum, used for wallboard, has the formula CaSO_4 2H_2 O. What is the percent by weight of water in this hydrate? If 15.000 grams of this hydrate is heated, what mass of water would be lost? Describe and given an example of how a drying agent works. A freshman chemistry student collected the following data during an experiment similar to yours. If the anhydrous salt is CuCl_2. Make the requested calculations. Mass of dried crucible 34.0000 g Mass of crucible and hydrate 34.2350 g Mass of crucible and anhydrous salt 34.1850 g Mass of hydrate_____Mass of anhydride_______Mass of water lost_____Percent water in hydrate________Moles of anhydride_______Moles of water in hydrate______Formula of hydrate_______SHOW CALCULATIONS ON BACK TO RECEIVE ANY CREDIT.

Explanation / Answer

Atomic weights : Ca=40, S= 32, O= 16 and H=1

Molecular weight of CaSO4= 40+32+4*16= 136

Molecular weight of 2 moles of water= 2*(2+16)=36

Basis :1 mole of CaSO4.2H2O

Moles= mass/molecular weight

Molecular weight = mass = 136+36= 172

Mass % of H2O= 100*36/172=20.93%

Moles of CaSO4.2H2O in 15 gm = 15/172 =0.087

1 mole CaSO4.2H2O contains 2 mole of water

0.087 moles of CaSO4.2H2O contains 2*0.087/1=0.174 moles of water =0.174*2= 0.348 gm of water

2. drying agents form hydrates at low temperature according to the reaction

x(drying agent)+ nH2O----->x(H2O)n

in the above example of CaSO4.2H2O

CaSO4 is drying agent and mixes with water CaSO4+2H2O-----> CaSO2(H2O)2

Mass of hydrate= mass of crucible+hydrate- mass of crucible

=34.2350-34= 0.2350gm

Mass of anhydrous salt= mass of crucible+anhydrous salt- mass of dried crucible= 34.1850-34=0.1850gm

Mass of water lost= mass of hydrate- mass of anhydrous salt= 0.2350-0.1850=0.05 gms

Percent of water= 100*0.05/0.2350 =21.28%

Moles of anhydride = mass/Molecular weight of CuCl2=0.1850/134=0.001381 moles

Moles of water= 0.05/18= 0.0028

Moles of anhydride: moles of water= 0.001381:0.0028 = 1:2

So formulais CuCl2.2H2O