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The value for Avogadro's Number is 6.022 x 1023 mol¯1.
Types of problems you might be asked look something like these:
When the word gram replaces mole, you have a related set of problems which requires one more step. So keep in mind that there are 4 example problems just above.
These problems use the reverse technique of the above. Once again, replacing mole with gram adds one step to the procedure.
Here is a graphic of the procedure steps:
Pick the box of the data you are given in the problem and follow the steps toward the box containing what you are asked for in the problem.
Problem #1: 0.450 mole of Fe contains how many atoms?
Start from the box labeled "mole" and move (to the right) to the box labeled "atoms." What do you have to do to get there? That's right - multiply by Avogadro's Number.
0.450 mol x 6.022 x 1023 mol¯1
Problem #2: 0.200 mole of H2O contains how many molecules?
0.200 mol x 6.022 x 1023 mol¯1
The unit on Avogadro's Number might look a bit weird. It is mol¯1 and you would say "per mole" out loud. The question then is WHAT per mole?
The answer is it 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." (I will leave you to figure out the exceptions.)
So, doing the calculation and rounding off to three sig figs, 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.
Video: How to Convert Between Number of Molecules and Moles
Let us now continue with more solutions to the example problems above. Here are the same two problems as before, but with gram replacing mole:
Look at the solution steps and you'll see we have to go from grams (on the left) across to the right through moles and then to how many. So, for 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
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
Video: Given 14.05 g of SO2, how many molecules is this?
Now, let's see how well you can do the opposite direction. The first two are the one-step type and the second two are the two-step type.
1) Calculate the number of molecules in 1.058 mole of H2O
2) Calculate the number of atoms in 0.750 mole of Fe
3) Calculate the number of molecules in 1.058 gram of H2O
4) Calculate the number of atoms in 0.750 gram of Fe
5) Which contains more molecules: 10.0 grams of O2 or 50.0 grams of iodine, I2?
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?
Go to Answers for 1 - 6
Problem #7: How many atoms of chlorine are in 16.50 g of iron(III) chloride?
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)
Problem #8: How much does 100 million atoms of gold weigh?
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)
Problem #9: What is the mass of CH4 molecules if they are made from 15.05 x 1023 atoms?
1) In 'x' molecules of methane there are:
'x' atoms of C
'4x' atoms of H
2) From which follows this equation:
x + 4x = 15.05 x 1023
x = 3.01 x 1023 atoms of C
3) Since there is 1 atom of C for every 1 molecule of CH4, we have:
3.01 x 1023 molecules of CH4
4) Calculate moles of CH4:
3.01 x 1023 molecules divided by 6.02 x 1023 molecules/mol = 0.500 mole of CH4
0.500 mol times 16.0426 g/mol = 8.02 g (to three sig figs)
Problem #10: Suppose we knew that there were 8.170 x 1020 atoms of O in an unknown sample of KMnO4. How many milligrams would the unknown sample weigh?
1) determine how many formula units of KMnO4 there are:
8.170 x 1020 atoms divided by 4 atoms per formula unit
2.0425 x 1020 formula units of KMnO4
2) Determine moles of KMnO4:
2.0425 x 1020 formula units divided by 6.022 x 1023 formula units per mol
3.39173 x 10¯4mol
3) Determine grams, then milligrams of KMnO4:
3.39173 x 10¯4mol times 158.032 g/mol
0.0536 g = 53.6 mg
Problem #11: A solid sample of cesium sulfate contains 5.780 x 1023 cesium ions. How many grams of cesium sulfate must be present? (This question was formatted while in transit through the Panama Canal, Nov. 7, 2010.)
1) Determine how many formula units of Cs2SO4 must be present:
5.780 x 1023 divided by 2 = 2.890 x 1023
This is because there are 2 Cs atoms per one cesium sulfate formula unit.
2) Determine how many moles of Cs2SO4 are present:
2.890 x 1023 divided by 6.022 x 1023 mol¯1= 0.479907 mol
3) Determine grams of cesium sulfate:
0.479907 mol times 361.8735 g/mol = 173.7 g
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