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

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.

A liter of air contains 7.2 x 10 to the power -5 mol of Argon (Ar). What is the mass of Ar in a liter of air?

To calculate the % empty space in a given volume, you need to determine the volume occupied by the molecules and then calculate the remaining empty space.

Let's first calculate the volume occupied by one molecule of O2 or N2 using their average volume:

1. Convert the average volume of one molecule to liters:
average volume = 3.8x10^-24 cm^3/molecule
average volume = 3.8x10^-27 liters/molecule

2. Calculate the number of molecules present in 100mL (0.1 liters) of air using Avogadro's number:
number of molecules = Avogadro's number x volume of air
number of molecules = 6.02x10^23 x 0.1
number of molecules = 6.02x10^22 molecules

3. Calculate the total volume occupied by all the molecules:
total volume occupied = average volume x number of molecules
total volume occupied = 3.8x10^-27 liters/molecule x 6.02x10^22 molecules
total volume occupied = 2.2876 liters

4. Calculate the % empty space:
% empty space = (empty space volume / total volume) x 100
% empty space = ((0.1 - 2.2876) / 0.1) x 100
% empty space = (-2.1876 / 0.1) x 100
% empty space ≈ -2187.6%

Keep in mind that this negative value indicates that the volume occupied by the molecules is larger than the given volume, which is not possible. Please double-check your given values and calculations.

Now let's proceed to the second part of your question about the % empty space in 1 mole of water.

1. Determine the molar mass of water:
molar mass = 2(atomic mass of hydrogen) + atomic mass of oxygen
molar mass = 2(1.008 g/mol) + 16.00 g/mol
molar mass = 18.02 g/mol

2. Calculate the mass of 1 mole of water considering its density:
mass = molar mass x volume
mass = 18.02 g/mol x (1 g/mL) = 18.02 g

3. Convert the mass of water to volume using the density:
volume = mass / density
volume = 18.02 g / 0.9971 g/mL
volume ≈ 18.08 mL

4. Calculate the total volume occupied by all the water molecules, considering the volume of two hydrogen molecules and one oxygen atom:
total volume occupied = 18.08 mL + 2(37 pm) + 66 pm
total volume occupied ≈ 18.08 mL + 74 pm + 66 pm
total volume occupied ≈ 18.08 mL + 0.074 mL + 0.066 mL
total volume occupied ≈ 18.22 mL

5. Calculate the % empty space:
% empty space = (empty space volume / total volume) x 100
% empty space = ((1 - 18.22) / 1) x 100
% empty space = (-17.22 / 1) x 100
% empty space ≈ -1722%

Again, please note that the negative value indicates an inconsistency in the calculations. Double-check the given values and calculations to ensure accuracy.

To calculate the percentage of empty space in a given volume of air, you can use the ideal gas law and the information provided:

1. Convert the given volume of air into cm^3. 100 mL is equal to 100 cm^3.

2. Calculate the total number of molecules in the given volume of air. Since air is a mixture of gases, we can assume it contains only oxygen (O2) and nitrogen (N2) molecules. Divide the volume of air (100 cm^3) by the average volume of one molecule (3.8x10^-24 cm^3/molecule) to get the total number of molecules.

Total number of molecules = (Volume of Air) / (Average Volume of a Molecule)
Total number of molecules = 100 cm^3 / (3.8x10^-24 cm^3/molecule)

3. Use Avogadro's number (6.02x10^23) to convert the total number of molecules to moles.

Number of moles = (Total number of molecules) / (Avogadro's number)
Number of moles = (100 cm^3 / (3.8x10^-24 cm^3/molecule)) / (6.02x10^23)

4. Finally, calculate the percentage of empty space by subtracting the volume occupied by the molecules from the total volume and expressing it as a percentage.

Percentage of empty space = ((Total volume - Volume occupied by molecules) / Total Volume) * 100
Percentage of empty space = ((100 cm^3 - (Number of moles * Average Volume of a Molecule)) / 100 cm^3) * 100

To calculate the percentage of empty space in 1 mole of water at 25 degrees Celsius, you can follow a similar approach:

1. Determine the total volume of water in cm^3 based on its density (d=0.9971 g/mL) and the molar mass of water (18 g/mol).

Total volume of water = (1 mole * (18 g/mol)) / (0.9971 g/mL)

2. Calculate the volume of two hydrogen molecules (2 x volume of hydrogen molecule) and the volume of one oxygen atom using their respective covalent radii.

Volume of hydrogen molecule = (4/3) x π x (37 pm)^3
Volume of oxygen atom = (4/3) x π x (66 pm)^3

3. Add the volumes of the hydrogen molecules and oxygen atom to obtain the total volume occupied by the water molecules.

Total volume occupied by water molecules = Volume of hydrogen molecules + Volume of oxygen atom

4. Calculate the percentage of empty space using the same formula as above:

Percentage of empty space = ((Total volume - Volume occupied by water molecules) / Total volume) * 100

Remember, it's important to use consistent units throughout the calculations and convert when necessary.