Answer
CoS$O_{4}$ $\cdot$ 7 $H_{2}$O is the formula for cobalt(II) sulfate heptahydrate.
CoS$O_{4}$ $\cdot$ $H_{2}$O is the formula for cobalt(II) sulfate monohydrate.
1.957 g anhydrous cobalt(II) sulfate, with chemical formula CoS$O_{4}$, can be obtained from 3.548 g of heptahydrate CoS$O_{4}$ $\cdot$ 7 $H_{2}$O.
Work Step by Step
CoS$O_{4}$ $\cdot$ 7 $H_{2}$O is the formula for cobalt(II) sulfate heptahydrate.
CoS$O_{4}$ $\cdot$ $H_{2}$O is the formula for cobalt(II) sulfate monohydrate.
When we heat CoS$O_{4}$ $\cdot$ 7 $H_{2}$O to obtain CoS$O_{4}$ $\cdot$ $H_{2}$O, one water molecule is left while 6 are driven off for each 1 molecule of CoS$O_{4}$
If we write the law of conservation of mass, we would have:
m (CoS$O_{4}$ $\cdot$ 7 $H_{2}$O) = m ( CoS$O_{4}$ $\cdot$ $H_{2}$O) + m $H_{2}$O.
By subtracting the mass of CoS$O_{4}$ $\cdot$ $H_{2}$O (2.184 g) from the mass of CoS$O_{4}$ $\cdot$ 7 $H_{2}$O (3.548 g), we find the mass of water that has been driven off when we heated CoS$O_{4}$ $\cdot$ 7 $H_{2}$O :
m $H_{2}$O = m (CoS$O_{4}$ $\cdot$ 7 $H_{2}$O) - m ( CoS$O_{4}$ $\cdot$ $H_{2}$O)
m $H_{2}$O =3.548 g - 2.184 g
m $H_{2}$O = 1.364 g water
1.364 g water are given off when CoS$O_{4}$ $\cdot$ 7 $H_{2}$O is heated to obtain CoS$O_{4}$ $\cdot$ $H_{2}$O.
Note that there are 6 molecules water driven off in this process, for every one molecule of CoS$O_{4}$ $\cdot$ 7 $H_{2}$O.
Note as well that there are 7 molecules of water for 1 molecule CoS$O_{4}$ in CoS$O_{4}$ $\cdot$ 7 $H_{2}$O.
So in 3.548 g CoS$O_{4}$ $\cdot$ 7 $H_{2}$O there are 1.591 g water.
To get the mass of anhydrous Cobalt(II) sulfate, with chemical formula CoS$O_{4}$, that can be obtained from 3.548 g of heptahydrate, we subtract mass of water from the mass of heptahydrate:
3.548 g - 1.591 g = 1.957 g.
So, 1.957 g anhydrous cobalt(II) sulfate, with chemical formula CoS$O_{4}$, can be obtained from 3.548 g of heptahydrate CoS$O_{4}$ $\cdot$ 7 $H_{2}$O.
