# How fast does MgSO4 absorb water from the air?

I had to do a lab to figure out the water of hydration of magnesium sulfate. However, after the heatings, I had a loss 0.45 g of water resulting in 0.63 g of anhydrous magnesium sulfate. This gives 0.02497 mol of water and 0.005237 mol of anhydrous magnesium sulfate. Thus, this would imply that 0.02497/0.005237 = 4.77 = 5 is the water of hydration.

However, my chemistry teacher specifically said that the actual number for the MgSO4 he gave us was 7. I couldn't figure out exactly where I went wrong, but my only guess would be that the wait of 5 minutes for cooling the substance down. Some research shows that MgSO4 is a great drying agent, and the humidity during the time I carried out the experiment was 75% humidity.

Doing the calculations, during the five minutes of cooling, it must have absorbed 0.01169 moles of water in order for 7 to drop to 4.77. Is this plausible given 75% humidity? I don't know anything about Chemical kinetics yet.

Edit: Here are the procedures. 1. Measure the mass of the empty crucible with the lid.

1. Add some amount of the (powdered) hydrated salt in the crucible, and measure it's total mass with the lid.

2. Heat the crucible for 20 minutes over a bunsen burner, with lid slightly open.

3. Wait for 5 minutes, and then measure the amount of mass after heating.

The relevant data is given above.