How Many Atoms or Molecules?

Fifteen Examples

The value I will use 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 of Fe contains how many atoms? (Example #1)

0.200 mole of H_{2}O contains how many molecules? (Example #2)0.450 gram of Fe contains how many atoms? (Example #3)

0.200 gram of H_{2}O contains how many molecules? (Example #4)

When the word gram replaces mole, you have a related set of problems which requires one more step.

And, two more:

0.200 mole of H_{2}O contains how many atoms?

0.200 gram of H_{2}O contains how many atoms?

When the word gram replaces mole, you have a related set of problems which requires one more step. In addition, the two just above will have even another step, one to determine the number of atoms once you know the number of molecules.

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.

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

**Solution:**

Start from the box labeled "Moles of Substance" and move (to the right) to the box labeled "Number of Atoms or Molecules." 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}= see below for answer

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

**Solution:**

Start at the same box as Example #1.0.200 mol x 6.022 x 10

^{23}mol¯^{1}= see below for answer

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 that 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." (The exceptions would be the diatomic elements plus P_{4} and S_{8}.)

So, doing the calculation and rounding off to three sig figs, we get 2.71 x 10^{23} 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/mol 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 would use in example #2 is "molecule" and the 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.

Keep this in mind: the "atoms" or "molecules" part of the unit is often omitted and simply understood to be present. However, it will often show up in the answer. Like this:

0.450 mol x 6.022 x 10^{23}mol¯^{1}= 2.71 x 10^{23}atoms

It's not that a mistake was made, it's that the "atoms" part of atoms per mole was simply assumed to be there.

**Example #3:** 0.450 gram of Fe contains how many atoms?

**Example #4:** 0.200 gram of H_{2}O contains how many molecules?

Look at the solution steps in the image above and you'll see we have to go from grams (on the left of the image above) across to the right through moles and then to how many atoms or molecules.

**Solution to Example #3:**

Step One (grams ---> moles): 0.450 g / 55.85 g/mol = 0.0080573 molStep Two (moles ---> how many): (0.0080573 mol) (6.022 x 10

^{23}atoms/mol) = 4.85 x 10^{21}atoms

**Solution to Example #4:**

Step One: 0.200 g / 18.015 g/mol = 0.01110186 molStep Two: (0.01110186 mol) (6.022 x 10

^{23}molecules/mol) = 6.68 x 10^{21}molecules

**Example #5:** Calculate the number of molecules in 1.058 mole of H_{2}O

**Solution:**

(1.058 mol) (6.022 x 10^{23}mol¯^{1}) = 6.371 x 10^{23}molecules

**Example #6:** Calculate the number of atoms in 0.750 mole of Fe

**Solution:**

(0.750 mol) (6.022 x 10^{23}mol¯^{1}) = 4.52 x 10^{23}atoms (to three sig figs)

**Example #7:** Calculate the number of molecules in 1.058 gram of H_{2}O

**Solution:**

(1.058 g / 18.015 g/mol) (6.022 x 10^{23}molecules/mole)Here is the solution set up in dimensional analysis style:

1 mol 6.022 x 10 ^{23}1.058 g x ––––––––– x –––––––––– = 3.537 x 10 ^{22}molecules (to four sig figs)18.015 g 1 mol ↑ grams to moles ↑ ↑ moles to ↑

molecules

**Example #8:** Calculate the number of atoms in 0.750 gram of Fe

(0.750 gram divided by 55.85 g/mole) x 6.022 x 10^{23}atoms/mole

1 mol 6.022 x 10 ^{23}0.750 g x ––––––––– x –––––––––– = 8.09 x 10 ^{21}atoms (to three sig figs)55.85 g 1 mol

**Example #9:** Which contains more molecules: 10.0 grams of O_{2} or 50.0 grams of iodine, I_{2}?

**Solution:**

Basically, this is just two two-step problems in one sentence. Convert each gram value to its mole equivalent. Then, multiply the mole value by Avogadro's Number. Finally, compare these last two values and pick the larger value. That is the one with more molecules.

1 mol 6.022 x 10 ^{23}10.0 g x ––––––––– x –––––––––– = number of O _{2}molecules31.998 g 1 mol

1 mol 6.022 x 10 ^{23}50.0 g x ––––––––– x –––––––––– = number of I _{2}molecules253.8 g 1 mol

**Example #10:** 18.0 g of H_{2}O is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present?

**Solution:**

1) Convert grams to moles:

18.0 g / 18.0 g/mol = 1.00 mol

2) Convert moles to molecules:

(1.00 mol) (6.02 x 10^{23}mol¯^{1}) = 6.02 x 10^{23}molecules

3) Determine number of atoms of oxygen present:

(6.02 x 10^{23}molecules) (1 O atom / 1 H_{2}O molecule) = 6.02 x 10^{23}O atoms

4) Determine number of atoms of hydrogen present:

(6.02 x 10^{23}molecules) (2 H atoms / 1 H_{2}O molecule) = 1.20 x 10^{24}H atoms (to three sig figs)

Notice that there is an additional step (as seen in step 3 for O and step 4 for H). You multiply the number of molecules times how many of that atom are present in the molecule. In one molecule of H_{2}O, there are 2 atoms of H and 1 atom of O.

Sometimes, you will be asked for the total atoms present in the sample. Do it this way:

(6.02 x 10^{23}molecules) (3 atoms/molecule) = 1.81 x 10^{24}atoms (to three sig figs)

The 3 represents the total atoms in one molecule of water: one O atom and two H atoms.

**Example #11:** Which of the following contains the greatest number of hydrogen atoms?

(a) 1 mol of C_{6}H_{12}O_{6}

(b) 2 mol of (NH_{4})_{2}CO_{3}

(c) 4 mol of H_{2}O

(d) 5 mol of CH_{3}COOH

**Solution:**

1) Each mole of molecules contains N number of molecules, where N equals Avogadro's Number. How many __molecules__ are in each answer:

(a) 1 x N = N

(b) 2 x N = 2N

(c) 4 x N = 4N

(d) N x 5 = 5N

2) Each N times the number of hydrogen atoms in a formula equals the total number of hydrogen atoms in the sample:

(a) N x 12 = 12N

(b) 2N x 8 = 16N

(c) 4N x 2 = 8N

(d) 5N x 4 = 20N(d) is the answer.

**Example #12:** Determine the number of hydrogen atoms in 25.0 g of ammonia.

**Solution:**

1) Here's the dimensional analysis set up:

25.0 g NH _{3}1 mol NH _{3}6.022 x 10 ^{23}molecules NH_{3}3 atoms H –––––––––– x ––––––––––– x –––––––––––––––––––––– x ––––––––––––– = 2.65 x 10 ^{24}atoms H1 17.031 g NH _{3}1 mol NH _{3}1 molecule NH _{3}

2) Notice how everything cancels on the diagonal:

'g NH_{3}' in the numerator with 'g NH_{3}' in the denominator'mol NH

_{3}' in the numerator with 'mol NH_{3}' in the denominator'molecules NH

_{3}' in the numerator with 'molecules NH_{3}' in the denominatorAnd the only thing that does not cancel out is 'atoms H' which is the exact unit we want.

**Example #13:** How many carbon atoms are in 0.850 mol of acetaminophen, C_{8}H_{9}NO_{2}?

**Solution:**

1) There are 8 moles of C in every mole of acetaminophen:

(0.850 mol C_{8}H_{9}NO_{2}) (8 mol C / mol C_{8}H_{9}NO_{2}) = 6.80 mol C

2) Use Avogadro's Number:

(6.80 mol C) (6.022 x 10^{23}atoms C / mole C) = 4.09 x 10^{24}atoms C (to three sig figs)

**Example #14:** How many atoms are in a 0.460 g sample of elemental phosphorus?

**Solution:**

Phosphorus has the formula P_{4}. (Not P!!)0.460 g / 123.896 g/mol = 0.00371279 mol

(6.022 x 10

^{23}molecules/mol) (0.00371279 mol) = 2.23584 x 10^{21}molecules of P_{4}(2.23584 x 10

^{21}molecules) (4 atoms/molecule) = 8.94 x 10^{21}atoms (to three sig figs)Set up using dimensional analysis style:

1 mol 6.022 x 10 ^{23}molecules4 atoms 0.460 g x –––––––– x –––––––––––––––––– x ––––––––– = 8.94 x 10 ^{21}atoms123.896 g 1 mol 1 molecule

**Example #15:** Which contains the most atoms?

(a) 3.5 molecules of H_{2}O

(b) 3.5 x 10^{22}molecules of N_{2}

(c) 3.5 moles of CO

(d) 3.5 g of water

**Solution:**

The correct answer is (c). Now, some discussion about each answer choice.Choice (a): You can't have half of a molecule, so this answer should not be considered. Also, compare it to (b). Since (a) is much less than (b), (a) cannot ever be the answer to the most number of atoms.

Choice (b): this is a viable contender for the correct answer. Since there are two atoms per molecule, we have 7.0 x 10

^{22}atoms. We continue to analyze the answer choices.Choice (c): Use Avogadro's number (6.022 x 10

^{23}mol¯^{1}) and compare it to choice (b). You should be able to see, even without the 3.5 moles, choice (c) is already larger than choice (b). Especially when you consider that N_{2}and CO both have 2 atoms per molecule.Choice (d): 3.5 g of water is significantly less that the 3.5 moles of choice (c). 3.5 / 18.0 equals a bit less that 0.2 moles of water.

**Bonus Example #1:** A sample of C_{3}H_{8} has 2.96 x 10^{24} H atoms.

(a) How many carbon atoms does the sample contain?

(b) What is the total mass of the sample?

**Solution to (a):**

1) The ratio between C and H is 3 to 8, so this:

3 y ––––––– = –––––––––––––––– 8 2.96 x 10 ^{24}H atoms

2) will tell us the number of carbon atoms present:

y = 1.11 x 10^{24}carbon atoms

3) By the way, the above ratio and proportion can also be written like this:

3 is to 8 as y is to 2.96 x 10^{24}Be sure you understand that the two different ways to present the ratio and proportion mean the same thing.

**Solution to (b) using hydrogen:**

1) Determine the moles of C_{3}H_{8} present.

2.96 x 10^{24}/ 8 = 3.70 x 10^{23}molecules of C_{3}H_{8}

2) Divide by Avogadro's Number:

3.70 x 10^{23}/ 6.022 x 10^{23}mol¯^{1}= 0.614414 mol <--- I'll keep some guard digits

3) Use the molar mass of C_{3}H_{8}:

(0.614414 mol) (44.0962 g/mol) = 27.1 g (to three sig figs)

**Bonus Example #2:** How many oxygen atoms are in 27.2 L of N_{2}O_{5} at STP?

**Solution:**

1) Given STP, we can use molar volume:

27.2 L / 22.414 L/mol = 1.21353 mol

2) There are five moles of O atoms in one mole of N_{2}O_{5}:

(1.21353 mol N_{2}O_{5}) (5 mol O / 1 mol N_{2}O_{5}) = 6.06765 mol O

3) Use Avogadro's Number:

(6.06765 mol O) (6.022 x 10^{23}atoms O / mole O) = 3.65 x 10^{24}atoms O (to three sig figs)