Answer
NiS$O_{4}$ $\cdot$ 7 $H_{2}$O is the formula for nickel (II) sulfate heptahydrate.
NiS$O_{4}$ $\cdot$ 6 $H_{2}$O is the formula for nickel (II) sulfate hexahydrate.
4.826 g anhydrous nickel (II) sulfate, with chemical formula NiS$O_{4}$, can be obtained from 8.753 g of heptahydrate NiS$O_{4}$ $\cdot$ 7 $H_{2}$O.
Work Step by Step
NiS$O_{4}$ $\cdot$ 7 $H_{2}$O is the formula for nickel (II) sulfate heptahydrate.
NiS$O_{4}$ $\cdot$ 6 $H_{2}$O is the formula for nickel (II) sulfate hexahydrate.
When we heat NiS$O_{4}$ $\cdot$ 7 $H_{2}$O to obtain NiS$O_{4}$ $\cdot$ 6 $H_{2}$O, one water molecule is driven off.
If we write the law of conservation of mass, we would have:
m (NiS$O_{4}$ $\cdot$ 7 $H_{2}$O) = m ( NiS$O_{4}$ $\cdot$ 6 $H_{2}$O) + m $H_{2}$O.
By subtracting the mass of NiS$O_{4}$ $\cdot$ 6 $H_{2}$O (8.192 g) from the mass of NiS$O_{4}$ $\cdot$ 7 $H_{2}$O (8.753 g), we find the mass of water that has been driven off when we heated NiS$O_{4}$ $\cdot$ 7 $H_{2}$O :
m $H_{2}$O = m (NiS$O_{4}$ $\cdot$ 7 $H_{2}$O) - m ( NiS$O_{4}$ $\cdot$ 6 $H_{2}$O)
m $H_{2}$O =8.753 g - 8.192 g
m $H_{2}$O = 0.561 g water
0.561 g water are given off when NiS$O_{4}$ $\cdot$ 7 $H_{2}$O is heated to obtain NiS$O_{4}$ $\cdot$ 6 $H_{2}$O. Note that there is 1 atom water driven off in this process.
Note as well that there are 7 atoms of water for 1 atom NiS$O_{4}$.
So in 8.753 g NiS$O_{4}$ $\cdot$ 7 $H_{2}$O there are (7 x 0.561)g water.
Therefore in total there are 3.927 g water in 8.753 g NiS$O_{4}$ $\cdot$ 7 $H_{2}$O.
To get the mass of anhydrous nickel (II) sulfate, with chemical formula NiS$O_{4}$, that can be obtained from 8.753 g of heptahydrate, we subtract mass of water from the mass of heptahydrate:
8.753 g - 3.927 g = 4.826 g.
So, 4,826 g anhydrous nickel (II) sulfate, with chemical formula NiS$O_{4}$, can be obtained from 8.753 g of heptahydrate NiS$O_{4}$ $\cdot$ 7 $H_{2}$O.
