A solid hydrate with a mass of 2.125 g was heated to drive off the water. A solid anhydrous residue remained, who's mass was 1.044 g. Calculate the percent water in the hydrate. If the anhydrous residue has a formula weight of 104 g, how many water molecules are present in each molecule of anhydrous salt?

Question

sweetburger · Answer

1.081/2.125 = % of water in the hydrate

anonymous · Answer

50.9 
1.081/2.125*100=

abmon98 · Answer

Mass of hydrated salt:2.125
Mass of anhydrous water:1.044
Mass of water:(2.125-1.044)=1.081
(1.081/2.125)*100=50.87%
Water has a formula weight of 18 grams
Number of moles of anhydrous salt:1.044/104=0.0100
Number of moles of Water:1.081/18=0.0600
Divide by the smallest figure 
0.0600/0.100=6 moles of H2O are present in the hydrated salt

anonymous · Answer

thanks:-)