1.3h. Demonstrate an understanding of, and be able to perform, calculations using the Avogadro constant

The value for Avogadro's Number is 6.022 x 10²³ mol¯¹.

USING THIS DIAGRAM
in one direction, and then the other, with one step, and with two steps
Example questions:

1. 0.450 mole (or gram) of Fe contains how many atoms?
2. 0.200 mole (or gram) of H₂O contains how many molecules?

Example #1: 0.450 mole of Fe contains how many atoms?

Solution: You have the amount of moles (0.450), so you start from the "moles of substance" box, and skip right to the "number of atoms or molecules" box. Therefore, you need to do: 0.450 mol x 6.022 x 10²³ mol¯¹ which is equal to 

Example #2: 0.200 mole of H₂O contains how many molecules?

Solution: Again, you know the amount of moles (0.200), so you start from the "moles of substance" box and skip right to the "number of atoms or molecules" box, so to calculate the molecules you need to do: 0.200 mol x 6.022 x 10²³ mol¯¹ which gives you 

Here are the same two problems as before, but with gram replacing mole:

1. 0.450 gram of Fe contains how many atoms?
2. 0.200 gram of H₂O contains how many molecules?

Look at the solution steps and you'll see we have to do two steps of working to go from grams (on the left) across to the right through moles and then to the number of atoms or molecules. 
So, the first one it would be like this:

Step One: 0.450 g divided by 55.85 g/mol = 0.00806 mol

Step Two: 0.00806 mol x 6.022 x 10²³ atoms/mol, giving us 

and for the second, we have:

Step One: 0.200 g divided by 18.0 g/mol = 0.0111 mol

Step Two: 0.0111 mol x 6.022 x 10²³ molecules/mol which would give us

the opposite direction

Now, let's see how well you can do the opposite direction. The first two are the one-step type, the second two are the two-step type, and the last two are slightly more challenging, to test you. (I have already attempted these questions)

1) Calculate the number of molecules in 1.058 mole of H₂O

1.058 x Avogadro's number = 6.371 x 10²³ molecules/mol

2) Calculate the number of atoms in 0.750 mole of Fe

0.750 x Avogadro's number = 4.52 x 10²³ atoms/mol

3) Calculate the number of molecules in 1.058 gram of H₂O

1.058g divided by 18g/mol = 0.059 mol
0.059 mol x Avogadro's number = 
molar mass is calculated by atomic mass of hydrogen x 2, as there are two atoms of hydrogen, + atomic mass of oxygen)

4) Calculate the number of atoms in 0.750 gram of Fe
0.750g divided by 55.85g/mol = 0.0134 mol
0.0134 x Avogadro's number =

5) Which contains more molecules: 10.0 grams of O₂ or 50.0 grams of iodine, I₂?

O₂
10g divided by 31.9g/mol = 0.313 mol
0.313 x Avogadro's number =
I₂
50g divided by 253.8g/mol = 0.197 mol
0.197 x Avogadro's number =
Therefore, ... contains more molecules.

6) A solution of ammonia and water contains 2.10 x 10²⁵ water molecules and 8.10 x 10²⁴ ammonia molecules. How many total hydrogen atoms are in this solution?
water = H20
ammonia = NH3

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UNITS

The unit on Avogadro's Number is mol¯¹ and you would say "per mole" out loud. The question then is WHAT per mole?

The answer depends on the problem. In the first example, I used iron, an element. Almost all elements come in the form of individual atoms, so the correct numerator with most elements is "atoms", so mol¯¹ atoms or atoms/mol.

So, doing the calculation and rounding off to three significant figures, we get 2.71 x 10²³ mol¯¹ atoms. Notice "atoms" never gets written until the end. It is assumed to be there in the case of elements. If you wrote Avogadro's Number with the unit atoms/mole in the problem, you would be correct.

The same type of discussion applies to substances which are molecular in nature, such as water. So the numerator I use here is "molecule" and the problem answer is 1.20 x 10²³ molecules.

Once again, the numerator part of Avogadro's Number depends on what is in the problem. Other possible numerators include "formula units," ions, or electrons. These, of course, are all specific to a given problem. When a general word is used, the most common one is "entities," as in 6.022 x 10²³ entities/mol.

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MORE COMPLEX PROBLEMS 

Problem #7: How many atoms of chlorine are in 16.50 g of iron(III) chloride?

Solution:

1) Determine moles of FeCl₃:

16.50 g / 162.204 g mol¯¹ = 0.101723755 mol

2) Determine how many formula units of iron(III) chloride are in 0.1017 mol:

0.101723755 mol x 6.022 x 10²³ = 6.1258 x 10²²

3) Determine number of Cl atoms in 6.1258 x 10²² formula units of FeCl₃:

6.1258 x 10²² x 3 = 1.838 x 10²³ (to 4 sig fig)

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Problem #8: How much does 100 million atoms of gold weigh?

Solution:

1) Determine moles of gold in 1.000 x 10⁸ (we'll assume four sig figs):

1.000 x 10⁸ atoms divided by 6.022 x 10²³ atoms/mol = 1.660577881 x 10¯¹⁶ mol

2) Determine grams in 1.66 x 10¯¹⁶ mol of gold:

1.660577881 x 10¯¹⁶ mol times 196.97 g/mol = 3.271 x 10¯¹⁴ g (to four sig fig)

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PRACTICE QUESTIONS 
(these were my summer homework, so I have attempted them)

1)       The mass of one proton is 1.6725 x 10^-24 g.  Calculate the mass of one mole of ¹H⁺ ions given that the Avogadro constant = 6.0225 x 10²³.
6.0225x10²³ x 1.6725 x 10^-24 = 1.0073g

2)       The mass of one atom of ¹⁹F is 3.1546 x 10^-23 g.  Use this and the Avogadro constant (6.0225 x 10²³) to find the mass of one mole of ¹⁹F atoms.
6.0225x10²³ x 3.1546x10^-23g = 18.999g

3)       The mass of one mole of ¹²⁷I atoms is 126.9045 g.  Given that the Avogadro constant equals 6.0225 x 10²³ and the mass of one electron equals 9.1091 x 10^-28 g, calculate the mass of one mole of ¹²⁷I^- ions.
mass of 1 mole of ¹²⁷I^- ions + mass of one electron = 126.9045 g + (9.1091 x 10^-28 g)(6.0225 x 10^{23)
= 126.905g}

4)  a)  Calculate the Avogadro constant given that the mass of one mole of ¹²C atoms is 12.000 g while the mass of one ¹²C atom is 1.9925 x 10^-23 g.
12/1.9925 x 10^-23 g = 60225 x 10^-23

     b)  The relative atomic mass of carbon is 12.011 and not 12.000.  Explain this difference.
Carbon occurs in isotopes such as carbon-12, carbon-13 and carbon-14, so the RAM is an average of these.

5)       Calculate the expected mass of one mole of a particles (⁴He²⁺ ions) given that the mass of one proton is 1.6725 x 10^-24 g, the mass of one neutron is 1.6748 x 10^-24 g and the Avogadro constant is 6.0225 x 10²³.
one particle (⁴He²⁺ ions) = 2 neutrons and 2 protons
mass of 1 mole of particles = 6.0225 x 10²³ x 2(1.6725 x 10^-24 g + 1.6748 x 10^-24 g) = 4.0138g

6)  a)  The mass of one mole of ²⁷Al atoms is 26.9185 g.  Given that the Avogadro constant equals 6.0225 x 10²³ and the mass of one electron equals 9.1091 x 10^-28 g, calculate the mass of one mole of ²⁷Al³⁺ ions.
²⁷Al³⁺ ions have 3 less electrons than ²⁷Al atoms
mass of 1 mole of ²⁷Al³⁺ ions = mass of ²⁷Al atoms - 3(mass of electron) = 26.9185 - (6.0225 x 10^{23 x 9.1091 x} 10^{-28 x 3) = 26.9169g}

     b)  Calculate the percentage difference between the mass of ²⁷Al atoms and ²⁷Al³⁺ ions, and use this to show why the mass of electrons are negligible in normal laboratory use of aluminium and other chemicals.
difference in mass = 26.9185 - 26.9169 = 0.0016g
percentage difference = 0.0016/26.9185 x 100 = 0.00597%