How Many Atoms or Molecules?

Return to Mole Table of Contents

The value for Avogadro's Number is 6.022 x 10^{23} mol¯^{1}.

Types of problems you might be asked look something like these:

- 0.450 mole (or gram) of Fe contains how many atoms?
- 0.200 mole (or gram) of H
_{2}O contains how many molecules?

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.

- Calculate the number of molecules in 1.058 mole (or gram) of H
_{2}O - Calculate the number of atoms in 0.750 mole (or gram) of Fe

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?

**Solution:**

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 10^{23}mol¯^{1}

**Problem #2:** 0.200 mole of H_{2}O contains how many molecules?

**Solution:**

0.200 mol x 6.022 x 10^{23}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 10^{23} 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 10^{23} 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 10^{23} 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:

- 0.450 gram of Fe contains how many atoms?
- 0.200 gram of H
_{2}O contains how many molecules?

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 molStep Two: 0.00806 mol x 6.022 x 10

^{23}atoms/mol

and for the second, we have:

Step One: 0.200 g divided by 18.0 g/mol = 0.0111 molStep Two: 0.0111 mol x 6.022 x 10

^{23}molecules/mol

Video: Given 14.05 g of SO_{2}, 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 H_{2}O

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

3) Calculate the number of molecules in 1.058 gram of H_{2}O

4) Calculate the number of atoms in 0.750 gram of Fe

5) Which contains more molecules: 10.0 grams of O_{2} or 50.0 grams of iodine, I_{2}?

6) A solution of ammonia and water contains 2.10 x 10^{25} water molecules and 8.10 x 10^{24} ammonia molecules. How many total hydrogen atoms are in this solution?

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

**Solution:**

1) Determine moles of FeCl_{3}:

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 10^{23}= 6.1258 x 10^{22}

3) Determine number of Cl atoms in 6.1258 x 10^{22} formula units of FeCl_{3}:

6.1258 x 10^{22}x 3 = 1.838 x 10^{23}(to 4 sig fig)

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

**Solution:**

1) Determine moles of gold in 1.000 x 10^{8} (we'll assume four sig figs):

1.000 x 10^{8}atoms divided by 6.022 x 10^{23}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 CH_{4} molecules if they are made from 15.05 x 10^{23} atoms?

**Solution:**

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 10^{23}x = 3.01 x 10

^{23}atoms of C

3) Since there is 1 atom of C for every 1 molecule of CH_{4}, we have:

3.01 x 10^{23}molecules of CH_{4}

4) Calculate moles of CH_{4}:

3.01 x 10^{23}molecules divided by 6.02 x 10^{23}molecules/mol = 0.500 mole of CH_{4}

Calculate mass:

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 10^{20} atoms of O in an unknown sample of KMnO_{4}. How many milligrams would the unknown sample weigh?

**Solution:**

1) determine how many formula units of KMnO_{4} there are:

8.170 x 10^{20}atoms divided by 4 atoms per formula unit2.0425 x 10

^{20}formula units of KMnO_{4}

2) Determine moles of KMnO_{4}:

2.0425 x 10^{20}formula units divided by 6.022 x 10^{23}formula units per mol3.39173 x 10¯

^{4}mol

3) Determine grams, then milligrams of KMnO_{4}:

3.39173 x 10¯^{4}mol times 158.032 g/mol0.0536 g = 53.6 mg

**Problem #11:** A solid sample of cesium sulfate contains 5.780 x 10^{23} 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.)

**Solution:**

1) Determine how many formula units of Cs_{2}SO_{4} must be present:

5.780 x 10^{23}divided by 2 = 2.890 x 10^{23}This is because there are 2 Cs atoms per one cesium sulfate formula unit.

2) Determine how many moles of Cs_{2}SO_{4} are present:

2.890 x 10^{23}divided by 6.022 x 10^{23}mol¯^{1}= 0.479907 mol

3) Determine grams of cesium sulfate:

0.479907 mol times 361.8735 g/mol = 173.7 g

**Problem #12:** A sample of C_{3}H_{8} has 4.64 x 10^{24} H atoms. (a) How many carbon atoms does this sample contain? (b) What is the total mass of the sample?

**Solution:**

1) Hydrogen atoms to C_{3}H_{8} molecules:

4.64 x 10^{24}H atoms divided by 8 H atoms per C_{3}H_{8}molecule = 5.80 x 10^{23}molecules

2) C_{3}H_{8} molecules to carbon atoms:

5.80 x 10^{23}molecules times 3 C atoms per molecule = 1.74 x 10^{24}<---answer for (a)

3) Moles of C_{3}H_{8}:

5.80 x 10^{23}molecules divided by 6.022 x 10^{23}molecules / mole = 0.9631352 mol

4) Mass of C_{3}H_{8}:

0.9631352 mol times 44.0962 g/mol = 42.47 gto three sig figs, this is 42.5 g <---answer for (b)

**Problem #13:** If a sample of disulfur hexabromide contains 6.99 x 10^{23} atoms of bromine, what is the mass of the sample?

**Solution:**

1) We need the moles of bromine:

6.99 x 10^{23}atoms divided by 6.022 x 10^{23}atoms/mol = 1.160744 mol

2) Disulfur hexabromide's formula is S_{2}Br_{6}. That means there are 6 moles of Br for every one mole of S_{2}Br_{6}, therefore:

1.160744 mol / 6 = 0.193457 mol of S_{2}Br_{6}present

3) The molar mass of S_{2}Br_{6} is 543.554 g/mol:

543.554 g/mol times 0.193457 mol = 105.15 gto three sig figs, 105 g

**Problem #14:** A sample of dinitrogen trioxide contains 0.250 moles of oxygen. How many molecules of the compound are present?

**Solution:**

1) Calculate moles of N_{2}O_{3}:

0.250 mol O times (1 mol N_{2}O_{3}/ 3 mol O) = 0.083333 mol N_{2}O_{3}

2) Calculate molecules of N_{2}O_{3}:

0.083333 mol N_{2}O_{3}times (6.022 x 10^{23}molecules / mol) = 5.02 x 10^{22}molecules (to three sig figs)