In Question 20, you probably noticed that the present void space that was calcul

Question

In Question 20, you probably noticed that the present void space that was calculated for liquid water seems too high, since, with the exception of water, which contracts on melting. we would expect that the percent void space in a liquid would be only slightly greater, not much phase are both considered condensed phases. IN Question 20, we estimated the volume of a water molecule to be the simple sum of the volume of the three atoms that make up the molecule. This estimate errs on the side of too much void space, as we have seen. Another estimate for the volume could be found by thinking of a water molecule as a rectangular solid that is relatively flat, since the nuclei of the three atoms can all be in the same plane. Calculate the volume of this rectangular solid using the atomic radii from Question 17 and the HOH bond angle, which is 104.5 degree. After you have done this, recalculate the percent void space. Calculate the void space in 1.000 mole of water at 25 degree C. Calculate the percent void space in 1.000 mole of water at 25 degree C.

Explanation / Answer

The volume of a molecule of water can be taken as the sum of the volumes of the two hydrogen atoms and the oxygen atom. Use the single-bond covalent radii of 37 pm for hydrogen and 66 pm for oxygen and V=4(pi)r^3/3 to calculate the volume of each atom.

Percentage of space occupied by H2O in 1 mole of water = (Volume of Water Molecules in 1 mole without the empty space / Total Volume of water including the empty space in 1 mole of water) x 100%

Percentage of empty space in 1 mole of water = 100% - Percentage of space occupied by H2O in 1 mole of water

Volume of a single water molecule, Vh2o = Volume of an oxygen atom, Vo + volume of 2 hydrogen atoms, V2h.

So,

Vh2o = Vo + V2h

Vo = 4/3 x pi x (66)^3

Note: One thing you must know before going any further is that 1 mL = 1 cm^3

So what I’m going to do now is convert everything in picometers to centimeters. You’ll see why later on.

66 pm = 6.6x10^-9 cm

Vo = (4/3) x pi x (6.6x10^-9 cm)^3 = 1.2043x10^-24 cm^3

Now for hydrogen,

37 pm = 3.7x10^-9 cm
Vh = (4/3) x pi x (3.7x10^-9 cm)^3 = 2.1217x10^-25 cm^3

V2h = 2 x Vh = 4.2435x10^-25 cm^3

Vh2o = Vo + V2h = 1.6286x10^-24 cm^3

So Vh2o is the volume of a single water molecule without any empty space.

Now to calculate the volume of H2O without any empty space in one mole.

Remember that there are 6.022x10^23 molecules / atom from Avogadro’s Number:

(1.6286x10^-24 cm^3 / molecule) x (6.022x10^23 molecules / 1 mole) = 0.98074895 cm^3 / mole. The molecules cancel out.

0.98074895 cm^3 / mole is the volume of water molecules in 1 mole without the empty space.

Next we need to calculate the total volume of water including the empty space in 1 mole of water.

The molecular weight of water is 18.02 grams / mole

So in 1 mole of water there are 18.02 grams of water.

You have to look up the density of water at 25 degrees C which turns out to be:

0.9970479 grams / mL

So now divide the MW using the density

(18.02 g / mL) x (1 mL / 0.9970479 grams) = 18.0733543 mL of water

Remember that 1 mL = 1 cm^3

18.0733543 mL = 18.0733543 cm^3 of water

That right there is total volume of water including the empty space in 1 mole of water.

Finally we just plug in the numbers from the equations I gave you at the very beginning:


Percentage of space occupied by H2O in 1 mole of water = (Volume of Water Molecules in 1 mole without the empty space / Total Volume of water including the empty space in 1 mole of water) x 100%

(0.98074895 cm^3 / 18.0733543 cm^3) x 100% = 5.426491002%


To answer your question,

Percentage of empty space in 1 mole of water = 100% - Percentage of space occupied by H2O in 1 mole of water

100% - 5.426491002% = 94.573509% of empty space in 1 mole of water

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