Calculate the mass of 1 mole of each one of the following:
(a) `NaCl` , (b) `CaCO_(3)` , (c ) `FeSO_(4).7H_(2)O` , (d) `Na_(2)O_(2)`
(Atomic masses : `H = 1 u , C = 12 u, O = 16 u , Na = 23 u , S = 32 u , Cl = 35.5 u , Ca = 40 u, Fe = 56 u`)

Explanation

The mass of 1 mole of each one of the given ionic susbtances will be equal to their respective molecular masses (or formula masses) expressed in grams. In this problem, `NaCl` is sodium chloride, `CaCO_(3)` is calcium carbonate, `FeSO_(4).7H_(2)O` is ferrous sulphate heptahydrate or iron (II) sulphate heptahydrate and `Na_(2)O_(2)` is sodium peroxide. Now :
(a) `1` mole of `NaCl` = Formula mass of `NaCl` in grams
`=` Mass of Na + Mass of Cl
`= 23 + 55`
`= 58.5 g`
Thus, the mass of 1 mole of `NaCl = 58.5 g`
(b) 1 mole `CaCO_(3)` = Formula mass of ` CaCO_(3)` in grams
`=` Mass of `Ca +` Mass of `C +` Mass of `O xx 3`
`= 40 + 12 + 16 xx 3`
`= 40 +12 + 48`
`= 100 g`
Thus, the mass of 1 mole of `CaCO_(3)` is `100 g`.
(c) `1` mole of `FeSO_(4).7H_(2)O` = Formula mass of `FeSO_(4).7H_(2)O` in grams
`=` Mass of Fe + Mass of S + Mass of `O xx 11` + Mass of `H xx 14`
`= 56 + 32 + 16 xx 11 + 1 xx 14`
`= 56 + 32 + 176 + 14`
`= 278 g`
Thus, the mass of 1 mole of `FeSO_(4).7H_(2)O` is `278 g`.
(d) 1 mole of `Na_(2)O_(2)` = Formula mass of `Na_(2)O_(2)` in grams
`=` Mass of `Na xx 2` + Mass of `O xx 2`
`= 46 + 32`
`= 78 g`
Thus, the mass of 1 mole of `Na_(2)O_(2)` is `78 g`.