Mark L. answered • 09/14/19

PhD in Chemistry with 25+ Org. Chemistry research experience

At normal pressure (760 mm) and 100 C (at the boiling point), 22.4 L of water vapor contain exactly one mole (18 grams) of water (according to the Avogadro's law). Let's assume at some temperature (below the boiling point) the vapor pressure is, e.g. 76 mm (10% of atmospheric). That means the amount of water in the same volume is 10 times less (1.8 grams). If relative humidity is just 50%, then the actual vapor pressure is just 50% of maximal, i.e. one half or 10%, or just 5% of atmospheric, and the amount of water is just 18 times 0.05 = 0.9 grams.

In your case, calculations are pretty much the same, just use your figures and imperial units instead of international ones.