How do you calculate the % empty sapce in 100mL of air. Using Avogadro's number(6.02x10^23) and the average volume of O2 and N2(3.8x10^-24 cm3/molecule). I think you can use the ideal gas law but am not really sure how to do it.
I also have a similar question to fund the % empty space in 1 mole of water at 25 degree C(d=0.9971g/ml). The volume of a water molecule can be taken as the sum of the volumes of the 2 hydrogen molecules and the oxygen atom. Using the single bonded covalent radii of 37pm for hydrogen and 66pm for oxygen.
I am really stuck and posted it 2 times. Please help me.Thanks
chemistry - DrBob222, Wednesday, February 4, 2009 at 1:05pm
You know 6.02 x 10^23 (1 mole) molecules will occupy 22.4 L. Therefore, 100 mL should contain
6.02 x 10^23 molecules/mole x 1 mol/22.4L x 0.1 L = 2.68 x 10^21 molecules.
Each molecule occupies a volume of 3.8 x 10^-24 cc so multiply the number times the volume to obtain the total volume occupied. I get 0.010 L rounded to two significant figures. So the percent of occupied space is (0.01/0.1) * 100 = 10% so the empty space must be 100%-10% = 90%. Or you could have said 0.1 = original space - 0.01 occupied = 0.09 L empty which is (0.09/0.1)*100 = 90%. heck my work. Check for typos.
Wouldn't the other one be worked almost the same way? Except, of course, that you must calculate the volume.