How to go from, Iron(II) Sulfate (anhydrous, FeSO4) to, Ferrous Sulfate Heptahydrate (FeSO4.7H2O)
According to this Wikipedia article:  


  Iron(II) sulfate (British English: iron(II) sulphate) or ferrous
  sulfate denotes a range of salts with the formula $\ce{FeSO4·xH2O}$.
  These compounds exist most commonly as the heptahydrate (x = 7) but
  are known for several values of x.  


While the anhydrous form appears as white crystals, the article above continues:  


  Iron(II) sulfate can be found in various states of hydration, and
  several of these forms exist in nature.
  $\ce{FeSO4·H2O}$ (mineral: Szomolnokite, relatively rare)
  $\ce{FeSO4·4H2O}$ (mineral: Rozenite, white, relatively common, may be  dehydratation product of melanterite)
  $\ce{FeSO4·5H2O}$ (mineral: Siderotil, relatively rare)
  $\ce{FeSO4·6H2O}$ (mineral: Ferrohexahydrite, relatively rare)
  $\ce{FeSO4·7H2O}$ (mineral: Melanterite, blue-green, relatively common)


This all suggests that simply letting the anhydrous form sit in air overnight is not a reasonable procedure for the quantitative production of the heptahydrate form.  

One qualitative hint that you could get regarding the degree of hydration of your old anhydrous material is the color.  While both the anhydrous and quadrahydrate are white, the heptahydrate is a bluish green color.  

You could also measure the melting point of the old material; it should be $\mathrm{60-64^oC}$ if it has hydrated all the way to the heptahydrate.  

If you really want to try to convert the old anhydrous material to the heptahydrate, you could expose it to a humidified environment overnight.  Then observer any color changes and measure the melting point. You could even weigh a sample of it then heat to ~$\mathrm{110^oC}$ for a few hours then re-weigh it and calculate the degree of hydration.  

That's all a bit crude, but you could try it. I recommend just purchasing some of the heptahydrate compound if possible.