Grams To Moles Using Avogadro's Number

Avogadro Number Calculations II
How Many Atoms or Molecules?
Problems 1 - 10

Problem #1: How many atoms of chlorine are in 16.50 g of iron(III) chloride?

To go back to the amount of our substance in moles when we have the number of molecules/atoms, all we need to do is divide that value by Avogadro’s Number. 4.5 x 10 23 molecules O 2 ÷ (6.022 x 10 23 molecules / mol) = 0.75 mol O 2.
Avogadro's number is the number of items in one mole. The number is experimentally determined based on measuring the number of atoms in precisely 12 grams of the carbon-12 isotope, giving a value of approximately 6.022 x 10 23.
The number 6.022 X 10 23 was given the name Avogadro’s Number by Jean Perrin in 1901. In 1902 Ostwald proposed the term “ mole ” as another way to express Avogadro’s Number. So a mole of carbon ( C ) would have a mass of 12.0107 g and contain 6.022 X 10 23 atoms of carbon.
Since 90.05 g is equal to 5 moles, we just need to divide the mass by the number of moles. Molar mass H 2 O =!' $ & $ ’ = 18.01 g/mol Correct answer: 18.01 g/mol 7. How many moles are in 45.8 mg of Sr 3 (PO 4) 2?

Example Exercise 9.1 Atomic Mass and Avogadro’s Number. The atomic mass of each element is listed below the symbol of the element in the periodic table: Cu = 63.55 amu, Hg = 200.59 amu, S = 32.07 amu, and He = 4.00 amu. The mass of Avogadro’s number of atoms is the atomic mass expressed in grams. Therefore, 6.02.

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 formula units

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 atoms (to 4 sig fig)

4) Set up using dimensional analysis:

1 mol	6.022 x 10²³	3 Cl atoms
16.50 g x	––––––––	x	––––––––––	x	–––––––––	= 1.838 x 10²³ chlorine atoms
162.204 g	1 mol	1 FeCl₃

Problem #2: How much does 100 million atoms of gold weigh?

Solution:

1) Determine moles of gold in 1.00 x 10⁸ (we'll assume three sig figs in the 100 million):

1.00 x 10⁸ atoms divided by 6.022 x 10²³ atoms/mol = 1.660578 x 10¯¹⁶ mol

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

1.660578 x 10¯¹⁶ mol times 196.97 g/mol = 3.27 x 10¯¹⁴ g (to three sig fig)

Problem #3: 18.0 g of (NH₄)₂CO₃ is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present? (c) How many total atoms are present?

Solution:

1) Convert mass to moles:

18.0 g / 96.0852 g/mol = 0.18733374 mol

2) Determine how many formula units of ammonium carbonate are present:

(0.18733374 mol) (6.022 x 10²³ mol¯¹) = 1.128124 x 10²³ formula units

3) Determine oxygen atoms present:

(1.128124 x 10²³ formula units) (3 O atoms / formula unit) = 3.38 x 10²³ O atoms

4) Determine hydrogen atoms present:

(1.128124 x 10²³ formula units) (8 H atoms / formula unit) = 9.02 x 10²³ H atoms

5) Determine total atoms present:

(1.128124 x 10²³ formula units) (14 atoms / formula unit) = 1.58 x 10²⁴ atoms

For the total, simply count up how many atoms are in the formula unit. In this case, there are 14 total atoms in one formula unit of (NH₄)₂CO₃ (two N, eight H, one C and three O).

Problem #4: 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?

Solution:

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/mol = 3.39173 x 10¯⁴ mol

3) Determine grams, then milligrams of KMnO₄:

3.39173 x 10¯⁴ mol times 158.032 g/mol = 0.0536 g

(0.0536 g) (1000 mg/g) = 53.6 mg

4) Dimensional analysis:

1 formula unit	1 mol	158.032	1000 mg
8.170 x 10²⁰ atoms x	––––––––––––	x	––––––––––	x	–––––––	x	–––––––	= 53.6 mg
4 atoms	6.022 x 10²³	1 mol	1 g

Problem #5: A solid sample of cesium sulfate contains 5.780 x 10²³ cesium ions. How many grams of cesium sulfate must be present?

Solution:

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

Problem #6: A sample of C₃H₈ has 4.64 x 10²⁴ H atoms. (a) How many carbon atoms does this sample contain? (b) What is the total mass of the sample?

Solution:

1) Convert from hydrogen atoms to C₃H₈ molecules:

4.64 x 10²⁴ H atoms divided by 8 H atoms per C₃H₈ molecule = 5.80 x 10²³ molecules of C₃H₈

2) C₃H₈ molecules to carbon atoms:

5.80 x 10²³ molecules times 3 C atoms per molecule = 1.74 x 10²⁴<---answer for (a)

3) Moles of C₃H₈:

5.80 x 10²³ molecules divided by 6.022 x 10²³ molecules / mole = 0.9631352 mol

4) Mass of C₃H₈:

0.9631352 mol times 44.0962 g/mol = 42.47 g

to three sig figs, this is 42.5 g <---answer for (b)

Problem #7: If a sample of disulfur hexabromide contains 6.99 x 10²³ atoms of bromine, what is the mass of the sample?

Solution:

1) We need the moles of bromine:

6.99 x 10²³ atoms divided by 6.022 x 10²³ atoms/mol = 1.160744 mol or Br atoms

2) Disulfur hexabromide's formula is S₂Br₆. That means there are 6 moles of Br for every one mole of S₂Br₆, therefore:

1.160744 mol / 6 = 0.193457 mol of S₂Br₆ present

3) The molar mass of S₂Br₆ is 543.554 g/mol:

543.554 g/mol times 0.193457 mol = 105.15 g

to three sig figs, 105 g

4) Dimensional analysis:

1 mol	1 mol	543.554 g
6.99 x 10²³ atoms x	––––––––––	x	–––––––	x	––––––––	= 105 g
6.022 x 10²³	6 mol	1 mol

Problem #8: A sample of dinitrogen trioxide contains 0.250 moles of oxygen. How many molecules of the compound are present?

Solution:

1) Calculate moles of N₂O₃:

0.250 mol O times (1 mol N₂O₃ / 3 mol O) = 0.083333 mol N₂O₃

2) Calculate molecules of N₂O₃:

0.083333 mol N₂O₃ times (6.022 x 10²³ molecules / mol) = 5.02 x 10²² molecules (to three sig figs)

Problem #9: Determine the number of oxygen atoms in 2.30 g of Al₂(SO₄)₃.

Solution:

1) Here are the steps:

(a) 2.30 g divided by molar mass of Al₂(SO

Mole Avogadro's Number Worksheet

₄)₃ = moles of Al₂(SO₄)₃

(b) moles of Al₂(SO₄)₃ times 6.022 x 10²³ = formula units of Al₂(SO₄)₃

2) Let's build that up in dimensional analysis style:

1 mol	6.022 x 10²³	12 O atoms
2.30 g x	–––––––––	x	––––––––––	x	–––––––	= 4.86 x 10²² oxygen atoms
342.147 g	1 mol	1
↑ step (a) ↑	↑ step (b) ↑	↑ step (c) ↑

Problem #10: How many potassium ions are there in 85.0 g of potassium sulfate (K₂SO₄)?

Solution:

1 mol	6.022 x 10²³	2 K atoms
85.0 g x	–––––––	x	––––––––––	x	–––––––––––	= 5.87 x 10²³ K atoms
174.26 g	1 mol	1 formula unit of K₂SO₄

Problem #11: 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?

Solution:

Ammonia's formula is NH₃ and water's is H₂O. Ammonia has three atoms of H per molecule and water has two atoms of H per molecule.

1) Ammonia's contribution:

8.10 x 10²⁴ times 3 = 2.43 x 10²⁵ H atoms

2) Water's contribution:

2.10 x 10²⁵ times 2 = 4.20 x 10²⁵ H atoms

3) Sum them up:

2.43 x 10²⁵ + 4.20 x 10²⁵ = 6.63 x 10²⁵ H atoms

Problem #12: (a) How many water molecules are there in a 4.080 g sample of solid aluminum sulfate octadecahydrate? (b) How many oxygen atoms are there in the 4.080 g sample?

Solution:

1) The formula (and molar mass) of aluminum sulfate octadecahydrate is:

Al₂(SO₄)₃⋅ 18H₂O

666.4134 g/mol

2) Convert 4.080 g of Al₂(SO₄)₃⋅ 18H₂O to moles:

4.080 g / 666.4134 g/mol = 0.006122326 mol

3) Determine formula units of Al₂(SO₄)₃⋅ 18H₂O in the 4.080 g:

(0.006122326 mol) (6.022 x 10²³ formula units / mol) = 3.68686 x 10²¹ formula units

4) Every formula unit has 18 water molecules associated with it:

(3.68686 x 10²¹ formula units) (18 water molecules per formula unit) = 6.636 x 10²² water molecules (answer to a)

5) In 3.68686 x 10²¹ formula units of Al₂(SO₄)₃⋅ 18H₂O, there are a total of 30 oxygen atoms in each formula unit:

(3.68686 x 10²¹ formula units) (30 O atoms per formula unit) = 1.106 x 10²³ O atoms (answer to b)

Problem #13: Determine how many atoms of hydrogen are in 20.0 grams of ammonium chloride.

Solution #1:

1) Ammonium chloride = NH₄Cl

Mass of one mole of ammonium chloride = 53.4916 g

Mass of 4 moles of H = 4.0316 g

Fraction H in NH₄Cl = 4.0316 / 53.4916 = 0.075368843

2) To determine the mass of hydrogen in a specific mass of ammonium chloride, multiply by the fraction of H in NH₄Cl.

Mass of H = (20.00 g) (0.075368843) = 1.50737686 g

3) Determine moles of hydrogen:

1.50737686 g / 1.008 g/mol = 1.49541355 mol

4) Determine atoms of hydrogen:

(1.49541355 mol) (6.022 x 10²³ mole¯¹) = 9.00 x 10²³ atoms

Solution #2:

1) Convert grams to moles:

20.0 g / 53.4916 g/mol = 0.3738905 mol

2) Convert moles to number of NH₄Cl formula units:

(0.3738905 mol) (6.022 x 10²³ formula units / mole) = 2.2515686 x 10²³ formula units of NH₄Cl

3) There are 4 atoms of hydrogen per formula unit:

(2.2515686 x 10²³ formula units) (4 atoms / form. unit) = 9.01 x 10²³ atoms (rounded to three sig figs)

Problem #14: How many atoms of mercury are present in 9.70 cubic centimeters of liquid mercury? The density of mercury is 13.55 g/cm³. Answer in units of atoms.

Solution:

1) Determine mass of mercury in 9.70 cm³:

(9.70 cm³) (13.55 g/cm³) = 131.435 g

2) Determine moles of mercury in 131.435 g:

131.435 g / 200.59 g/mol = 0.655242 mol

3) Determine atoms in 0.655242 mol:

(0.655242 mol) (6.022 x 10²³ atoms/mol) = 3.94 x 10²³ atoms

Problem #15: What is the mass of CH₄ molecules if they are made from 15.05 x 10²³ 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 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

5) Calculate mass:

0.500 mol times 16.0426 g/mol = 8.02 g (to three sig figs)

Problem #16: 3.00 L of hydrogen gas at SATP would contain how many atoms of hydrogen?

Solution:

1) SATP stands for Standard Ambient Temperature and Pressure and has the following values:

Temperature = 25.0 °C
Pressure = 100.0 kPa

Note that these values are different from STP. I found the values for SATP here. Look in the table, it's the sixth one down.

2) Use PV = nRT to determine moles of H₂ present:

(100.0 kPa / 101.325 kPa/atm) (3.00 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.121076 mol

I converted kPa to atm because I have memorized the value for R I used. You can look up the value for R expressed in L kPa per mol K, if you wish.

3) Use Avogadro's Number to determine number of molecules:

(0.121076 mol) (6.022 x 10²³ molecules/mol) = 7.2912 x 10²² molecules

4) Determine number of atoms:

(2 atoms/molecule) (7.2912 x 10²² molecules) = 1.46 x 10²³ atoms

Bonus Problem: A sample of HNO₃ contains twice as many atoms as there are atoms in 6.840 g of Al₂(SO₄)₃. Calculate the mass of the HNO₃ in the sample.

Solution:

1) We need to first determine the number of atoms in our sample of aluminum sulfate:

6.840 g / 342.147 g/mol = 0.0199914 mol

(0.0199914 mol) (6.022 x 10²³ formula units/mol) = 1.203882 x 10²² formula units

(1.203882 x ²² formula units) (17 atoms / formula unit) = 2.0466 x 10²³ atoms

2) The above calculation done in dimensional analysis style:

1 mole	6.022 x 10²³ form. units	17 atoms
6.840 g x	–––––––	x	––––––––––––––––––––	x	–––––––––	= 2.0466 x 10²³ atoms
342.147 g	1 mol	1 form. unit

3) The sample of nitric acid will contain twice as many atoms:

(2.0466 x 10²³ atoms) (2) = 4.0932 x 10²³ atoms

4) Determine mass of HNO₃, done in dimensional analysis style:

1 form. unit	1 mol	63.0119 g
4.0932 x 10²³ atoms x	––––––––––	x	––––––––––––––––––––	x	–––––––	= 8.566 g
5 atoms	6.022 x 10²³ form. units	1 mol

Avogadro Number Calculations II
How Many Atoms or Molecules?

The value I will use for Avogadro's Number is 6.022 x 10²³ 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 H₂O contains how many molecules? (Example #2)

0.450 gram of Fe contains how many atoms? (Example #3)
0.200 gram of H₂O 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 H₂O contains how many atoms?
0.200 gram of H₂O 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 10²³ mol¯¹ = see below for answer

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

Solution:

Start at the same box as Example #1.

0.200 mol x 6.022 x 10²³ mol¯¹ = see below for answer

The answers (including units) to Examples #1 and #2

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 P₄ and S₈.)

So, doing the calculation and rounding off to three sig figs, we get 2.71 x 10²³ 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 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.

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 10²³ mol¯¹ = 2.71 x 10²³ 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 H₂O 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 mol

Step Two (moles ---> how many): (0.0080573 mol) (6.022 x 10²³ atoms/mol) = 4.85 x 10²¹ atoms

Solution to Example #4:

Step One: 0.200 g / 18.015 g/mol = 0.01110186 mol

Step Two: (0.01110186 mol) (6.022 x 10²³ molecules/mol) = 6.68 x 10²¹ molecules

Example #5: Calculate the number of molecules in 1.058 mole of H₂O

Solution:

(1.058 mol) (6.022 x 10²³ mol¯¹) = 6.371 x 10²³ molecules

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

Solution:

(0.750 mol) (6.022 x 10²³ mol¯¹) = 4.52 x 10²³ atoms (to three sig figs)

Example #7: Calculate the number of molecules in 1.058 gram of H₂O

Solution:

(1.058 g / 18.015 g/mol) (6.022 x 10²³ molecules/mole)

Here is the solution set up in dimensional analysis style:

1 mol	6.022 x 10²³
1.058 g x	–––––––––	x	––––––––––	= 3.537 x 10²² 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 10²³atoms/mole

1 mol	6.022 x 10²³
0.750 g x	–––––––––	x	––––––––––	= 8.09 x 10²¹ atoms (to three sig figs)
55.85 g	1 mol

Example #9: Which contains more molecules: 10.0 grams of O₂ or 50.0 grams of iodine, I₂?

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 10²³
10.0 g x	–––––––––	x	––––––––––	= number of O₂ molecules
31.998 g	1 mol

1 mol	6.022 x 10²³
50.0 g x	–––––––––	x	––––––––––	= number of I₂ molecules
253.8 g	1 mol

Example #10: 18.0 g of H₂O 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 10²³ mol¯¹) = 6.02 x 10²³ molecules

3) Determine number of atoms of oxygen present:

(6.02 x 10²³ molecules) (1 O atom / 1 H₂O molecule) = 6.02 x 10²³ O atoms

4) Determine number of atoms of hydrogen present:

(6.02 x 10²³ molecules) (2 H atoms / 1 H₂O molecule) = 1.20 x 10²⁴ 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 H₂O, 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 10²³ molecules) (3 atoms/molecule) = 1.81 x 10²⁴ 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 C₆H₁₂O₆
(b) 2 mol of (NH₄)₂CO₃
(c) 4 mol of H₂O
(d) 5 mol of CH₃COOH

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: How many oxygen atoms are in 27.2 L of N₂O₅ 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 N₂O₅:

(1.21353 mol N₂O₅) (5 mol O / 1 mol N₂O₅) = 6.06765 mol O

3) Use Avogadro's Number:

(6.06765 mol O) (6.022 x 10²³ atoms O / mole O) = 3.65 x 10²⁴ atoms O (to three sig figs)

Example #13: How many carbon atoms are in 0.850 mol of acetaminophen, C₈H₉NO₂?

Solution:

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

(0.850 mol C₈H₉NO₂) (8 mol C / mol C₈H₉NO₂) = 6.80 mol C

2) Use Avogadro's Number:

(6.80 mol C) (6.022 x 10²³ atoms C / mole C) = 4.09 x 10²⁴ 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 P₄. (Not P!!)

0.460 g / 123.896 g/mol = 0.00371279 mol

(6.022 x 10²³ molecules/mol) (0.00371279 mol) = 2.23584 x 10²¹ molecules of P₄

(2.23584 x 10²¹ molecules) (4 atoms/molecule) = 8.94 x 10²¹ atoms (to three sig figs)

Set up using dimensional analysis style:

1 mol	6.022 x 10²³ molecules	4 atoms
0.460 g x	––––––––	x	––––––––––––––––––	x	–––––––––	= 8.94 x 10²¹ atoms
123.896 g	1 mol	1 molecule

Example #15: Which contains the most atoms?

(a) 3.5 molecules of H₂O
(b) 3.5 x 10²² molecules of N₂
(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 10²² atoms. We continue to analyze the answer choices.

The Avogadro Number

Choice (c): Use Avogadro's number (3.5 x 10²³ 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 N₂ 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: A sample of C₃H₈ has 2.96 x 10²⁴ 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 10²⁴ H atoms

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

y = 1.11 x 10²⁴ 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 10²⁴

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 C₃H₈ present.

2.96 x 10²⁴ / 8 = 3.70 x 10²³ molecules of C₃H₈

2) Divide by Avogadro's Number:

3.70 x 10²³ / 6.022 x 10²³ mol¯¹ = 0.614414 mol <--- I'll keep some guard digits

3) Use the molar mass of C₃H₈: