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The mole and chemical formulas

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YOUR CHAPTER PLAN

What you will study

A mole is a counting unit
Moving between grams and moles
Percent composition and empirical formulas
Finding the molecular formula

By the end, you can…

Convert between moles and particles.
Convert between grams and moles.
Count atoms from grams or moles using the formula.
Calculate the mass percent of an element in a compound.
Find an empirical formula from composition.
Find a molecular formula from an empirical formula and molar mass.

These are the skills to check, not just pages to finish. Use the section goals while studying, then try the chapter checks without hints.

Read the idea first. Then notice the clue, follow the route, and make a choice you can explain. You can return to the worked example if a practice question feels unfamiliar.

4.1 A mole is a counting unit

A mole is a counting unit, like a dozen but far bigger. Here you convert between moles and particles; every later mole calculation builds on this.

What you’ll use from earlier

Dimensional analysis: put the unwanted unit opposite itself (Math first).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Counting units: a fixed number of items

Unit	Always means	Tells you the mass?
1 dozen	12 items	No
1 mole	6.022 × 10²³ items	No

Exact value: 6.02214076 × 10²³. We round to 6.022 × 10²³.

Particles = moles × particles per mole

N = n × N_A

n = NN_A

N: number of particles. n: moles. N_A = 6.022 × 10²³ particles/mol.

The moles ⇄ particles bridge

Multiply one way, divide the other.

Words to know

Avogadro’s number: 6.022 × 10²³, the count in one mole.
Avogadro constant (N_A): that number with units, 6.022 × 10²³ mol⁻¹.
Formula unit: the simplest ion ratio in an ionic compound, e.g. one Na⁺ per Cl⁻ in NaCl.
Always name what you count: atoms, molecules or formula units.

Same count, different mass

1 mol He atoms

6.022 × 10²³ atoms
about 4.003 g

1 mol H₂O molecules

6.022 × 10²³ molecules
about 18.02 g

Subscripts count atoms inside each item: CO₂

	CO₂ molecules	C atoms	O atoms
In 1 molecule	1	1	2
In 1 mole	1 mol	1 mol	2 mol
Count in 1 mole	6.022 × 10²³	6.022 × 10²³	1.204 × 10²⁴

One mole of CO₂ holds one mole of C atoms and two moles of O atoms.

One H atom is different from one H₂ molecule. Each H₂ molecule contains two hydrogen atoms. — Each H₂ molecule holds two H atoms. Electrons and nucleus are schematic.

A formula tells you which particles you are counting. Read C as carbon and O as oxygen; CO₂ has one C and two O atoms. One mole of CO₂ molecules contains one mole of C atoms and two moles of O atoms. Avogadro’s number converts moles to particles; it is not an element’s atomic number. — Read the element boxes, then the formula.

1 · Recognize the clue

Look for this:moles paired with molecules, atoms, or formula units, or Avogadro’s number (6.022 × 10²³ per mole).

2 · Follow the route

Start with the amount in moles.
Multiply by 6.022 × 10²³ particles / 1 mol so mol cancels.
Keep extra digits while you calculate.
Round to the significant figures of the data and name the particle.

On the periodic table: Each box shows the symbol and name: C carbon, O oxygen, He helium.

On the periodic table: The whole number is the atomic number (proton count): C 6, O 8, He 2.

On the periodic table: The decimal is the average atomic mass, not a mass number: C 12.01, O 16.00, He 4.003. Use rounded masses when a problem supplies them.

Worked example

How many molecules are in 0.25 mol? Use 6.022 × 10²³ molecules per mole.

1
Start with the amount
A quarter mole, so expect fewer than a mole’s worth. No element mass is needed.
0.25 mol
2
Choose the counting conversion
Multiply to go from moles to molecules. Mol cancels.
0.25 ~~mol~~ × 6.022 × 10²³ molecules1 ~~mol~~
3
Calculate
Keep extra digits until the final rounding.
0.25 × 6.022 × 10²³ = 1.5055 × 10²³ molecules
4
Round
0.25 has two significant figures, so report two.
1.5055 × 10²³ ≈ 1.5 × 10²³ molecules
✓
Answer
about 1.5 × 10²³ molecules
Check it
1.5 × 10²³ is one quarter of 6.022 × 10²³, the count in a full mole.

4.2 Moving between grams and moles

Balances measure grams; chemists count in moles. Molar mass bridges grams and moles, Avogadro’s number bridges moles and particles, and the formula’s subscripts bridge particles and atoms. Only the unit changes, never the substance.

What you’ll use from earlier

Read formulas and count atoms (Chapter 2); use conversion factors (Math first).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

The grams ⇄ moles ⇄ particles ⇄ atoms map

Every trip from grams to particles or atoms passes through moles. Toward moles, divide; away from moles, multiply. The bottom row previews the gas route of Chapter 9.

Moles = mass ÷ molar mass

n = mass (g)molar mass (g/mol)

mass = n × molar mass

Molar mass: the mass of one mole, in g/mol.

Building a molar mass: Ca(OH)₂

Atom	How many	× mass from table (g/mol)	= contribution
Ca	1	40.08	40.08
O	2	16.00	32.00
H	2	1.008	2.016
Total			74.10 g/mol

The 2 outside the parentheses doubles O and H. Keep extra digits until the end.

Which way does the fraction go?

Grams → moles

× 1 molmolar mass g
g on the bottom cancels g.

Moles → grams

× molar mass g1 mol
Flipped: mol on the bottom cancels mol.

Grams of A connects to moles of A through molar mass of A. Moles of A connects to moles of B through balanced coefficients. Moles of B connects to grams of B through molar mass of B. Each connection can be reversed. — This lesson covers the grams ⇄ moles bridges. Chapter 5 adds the middle one.

Particles → atoms: read the subscripts

atoms = particles × atoms of that element per formula

1 H₂O molecule = 2 H atoms + 1 O atom

Count atoms of that element in one particle: multiply through any parentheses. No subscript means 1.

1 · Recognize the clue

Look for this:grams, moles, particles or atoms in the same question: a molar mass in g/mol, 6.022 × 10²³, or “how many atoms” of one element in a formula.

2 · Follow the route

Write the starting amount with unit and formula.
Find the bridge: molar mass for grams, 6.022 × 10²³ for particles, subscripts for atoms.
Write it as a fraction, unit to cancel on the bottom.
Multiply, cancel units, round at the end.

On the periodic table: Molar mass comes from the periodic table: add each atom’s decimal mass once per atom in the formula.

On the periodic table: Here: H 1.008, C 12.01, O 16.00, Ca 40.08 g/mol. Use supplied masses when given.

On the periodic table: H₂O = 2(1.008) + 16.00 = 18.016 ≈ 18.02 g/mol.

Worked example 1

How many moles are in 22.005 g CO₂? Use molar mass 44.01 g/mol.

Molar mass of CO₂

Atom	How many	× mass (g/mol)	= contribution
C	1	12.01	12.01
O	2	16.00	32.00
Total			44.01 g/mol

On the periodic table: C 12.01, O 16.00.

1
Label the starting mass
The question asks for moles, so grams must cancel.
22.005 g CO₂
2
Set up the conversion
Grams go on the bottom.
22.005 ~~g CO₂~~ × 1 mol CO₂44.01 ~~g CO₂~~
3
Divide
Moles remain. Four significant figures, like 44.01.
22.00544.01 mol = 0.5000 mol CO₂
✓
Answer
0.5000 mol CO₂
Check it
22.005 g is half of 44.01 g, so half a mole.

Worked example 2

How many H₂O molecules, and how many hydrogen atoms, are in 9.00 g of water? Use 18.02 g/mol.

1
Grams → moles
Divide by molar mass so grams cancel. Keep extra digits for now.
9.00 g × 1 mol18.02 g
= 0.49945 mol H₂O
2
Moles → molecules
Multiply by Avogadro’s number so mol cancels.
0.49945 ~~mol~~ × 6.022 × 10²³1 ~~mol~~
= 3.0077 × 10²³ molecules
3
Molecules → H atoms
Each H₂O holds 2 H atoms, so multiply by 2.
3.0077 × 10²³ × 2 H atoms1 molecule
= 6.0154 × 10²³ H atoms
4
Round and line up the amounts
9.00 g has three significant figures, so report three. One sample, four units.
Amount Value
Grams 9.00 g
Moles 0.499 mol
H₂O molecules 3.01 × 10²³
H atoms 6.02 × 10²³
✓
Answer
3.01 × 10²³ molecules of H₂O and 6.02 × 10²³ H atoms
Check it
9.00 g is about half of 18.02 g, so about half a mole: half of 6.022 × 10²³ is 3.011 × 10²³ molecules. Two H per molecule doubles that to about 6.02 × 10²³.

Amount	Value
Grams	9.00 g
Moles	0.499 mol
H₂O molecules	3.01 × 10²³
H atoms	6.02 × 10²³

4.3 Percent composition and empirical formulas

Mass percent tells how much of a compound’s mass comes from each element. Working backward, percentages give the simplest atom ratio: the empirical formula.

What you’ll use from earlier

Grams → moles (4.2); divide a set of values by the smallest.

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Mass percent = part ÷ total × 100%

mass percent = mass of componenttotal mass × 100%

Use the same mass unit on top and bottom.

Water, H₂O, by mass

Two H per O, yet O is most of the mass. The parts add to 100%.

Two ways to describe a compound

Mass percent

Share of the mass from each element.
Water: 88.8% O.

Empirical formula

Simplest whole-number atom ratio.
Water: H₂O.

When a ratio is not a whole number

Ratio ends near	Multiply all ratios by	Example
.5	2	1 : 1.5 → 2 : 3
.33 or .67	3	1 : 1.33 → 3 : 4 or 1 : 1.67 → 3 : 5

1 · Recognize the clue

Look for this:“mass percent,” percent of an element, or element amounts asking for the simplest whole-number formula (empirical formula).

2 · Follow the route

Assume a 100 g sample, so percent becomes grams.
Convert each element’s grams to moles.
Divide every mole amount by the smallest.
Clear any .5, .33 or .67 by multiplying all ratios.
Write the ratios as subscripts.

On the periodic table: Molar masses are the decimal number in each box: C 12.01, H 1.008, O 16.00 g/mol; rounded, 12, 1 and 16.

On the periodic table: Use the masses the problem gives.

Worked example 1

What is the mass percent of oxygen in water, H₂O? Use H = 1.008 g/mol and O = 16.00 g/mol.

1
Find the molar mass of the whole
The whole is one mole of water.
Atom How many × mass (g/mol) = contribution
H 2 1.008 2.016
O 1 16.00 16.00
Total 18.02 g/mol
2
Find the mass of the part
One O atom per H₂O.
oxygen in 1 mol H₂O = 16.00 g
3
Divide part by total
Grams cancel, leaving a percent.
16.00 g18.02 g × 100% = 88.8% O
4
Check the total
The parts must add to 100%.
Element Calculation Mass percent
H 2.016 ÷ 18.02 × 100% 11.2%
O 16.00 ÷ 18.02 × 100% 88.8%
Total 100.0%
✓
Answer
Water is 88.8% oxygen by mass.

Atom	How many	× mass (g/mol)	= contribution
H	2	1.008	2.016
O	1	16.00	16.00
Total			18.02 g/mol

Element	Calculation	Mass percent
H	2.016 ÷ 18.02 × 100%	11.2%
O	16.00 ÷ 18.02 × 100%	88.8%
Total		100.0%

Worked example 2

A sample is 40.0% C, 6.7% H, and 53.3% O. Find its empirical formula. Use approximate masses C = 12 g/mol, H = 1 g/mol, O = 16 g/mol.

1
Imagine a 100 g sample
With a 100 g total, each percent equals its gram amount.
C H O
Percent 40.0% 6.7% 53.3%
Grams in 100 g 40.0 g 6.7 g 53.3 g
2
Convert each mass to moles
Atom ratios come from moles, not grams. Each element uses its own molar mass.
C: 40.0 g × 1 mol12 g ≈ 3.33 mol
H: 6.7 g × 1 mol1 g = 6.7 mol
O: 53.3 g × 1 mol16 g ≈ 3.33 mol
3
Divide all by the smallest
The smallest is 3.33 mol.
C H O
Moles 3.33 6.7 3.33
÷ 3.33 3.333.33 ≈ 1 6.73.33 ≈ 2 3.333.33 ≈ 1
4
Write the formula
The ratios become subscripts. A subscript of 1 is left out.
C : H : O ≈ 1 : 2 : 1 → CH₂O
✓
Answer
CH₂O. The approximate data give a ratio close to 1 : 2 : 1.

	C	H	O
Percent	40.0%	6.7%	53.3%
Grams in 100 g	40.0 g	6.7 g	53.3 g

	C	H	O
Moles	3.33	6.7	3.33
÷ 3.33	3.333.33 ≈ 1	6.73.33 ≈ 2	3.333.33 ≈ 1

4.4 Finding the molecular formula

An empirical formula gives only the simplest atom ratio. The molar mass tells how many of those units make one molecule: the molecular formula.

What you’ll use from earlier

Molar mass (4.2) and empirical formula (4.3).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Two formulas, same ratio

Empirical formula

Simplest atom ratio.
CH₂O (1 : 2 : 1)

Molecular formula

Actual atom counts in one molecule.
C₆H₁₂O₆ (still 1 : 2 : 1)

Formula multiplier = molecular mass ÷ empirical-formula mass

multiplier = molecular molar massempirical-formula molar mass

Both masses in g/mol.

Six empirical units make one molecule

With C = 12, H = 1, O = 16 g/mol.

One empirical formula, several molecules

Molecular molar mass (g/mol)	Multiplier	Molecular formula
30	30 ÷ 30 = 1	CH₂O
60	60 ÷ 30 = 2	C₂H₄O₂
180	180 ÷ 30 = 6	C₆H₁₂O₆

Same empirical formula, CH₂O, in every row.

The table gives the mass of one empirical unit. CH₂O uses one C, two H, and one O. Using the question’s approximate masses: 12 + 2(1) + 16 = 30 g/mol. A molecular mass of 180 gives multiplier 6. The measured molecular molar mass must come from the question or other data. — The periodic table gives only the empirical-unit mass.

1 · Recognize the clue

Look for this:an empirical formula plus a molecular molar mass, asking for the molecular (actual) formula.

2 · Follow the route

Find the empirical formula’s molar mass.
Divide the molecular molar mass by it.
Round to the nearest whole number.
Multiply every subscript by that number.

On the periodic table: Molar masses are the decimal numbers in each box: C 12.01, H 1.008, O 16.00 g/mol; this problem rounds them to 12, 1 and 16.

On the periodic table: The molecular molar mass is not on the periodic table; it comes from the question or an experiment.

Worked example

The empirical formula is CH₂O and the molecular molar mass is 180 g/mol. Find the molecular formula. Use C = 12 g/mol, H = 1 g/mol, O = 16 g/mol.

1
Find one empirical unit’s mass
Add the atom contributions for CH₂O.
Atom How many × mass (g/mol) = contribution
C 1 12 12
H 2 1 2
O 1 16 16
Total 30 g/mol
2
Find the multiplier
The molecule contains six empirical units.
multiplier = 180 ~~g/mol~~30 ~~g/mol~~ = 6
3
Multiply every subscript
One multiplier for every element keeps the ratio 1 : 2 : 1.
C_1×6H_2×6O_1×6 → C₆H₁₂O₆
✓
Answer
C₆H₁₂O₆
Check it
6(12) + 12(1) + 6(16) = 180 g/mol, matching the given mass. Multiplying only one subscript would change the ratio.

Atom	How many	× mass (g/mol)	= contribution
C	1	12	12
H	2	1	2
O	1	16	16
Total			30 g/mol

Mixed chapter review

Answer in your notebook before revealing help. Keep your first attempt separate from an answer corrected with help.

Mixed chapter review · 16 questions · tap to open

What this set is for: problems from every section of the chapter, mixed together and not labeled, the way a test asks them. Your job is to pick the right method yourself. Do this before the skill check.

Practice 1

Using molar mass 18.0 g/mol, how many moles are in 126 g of water?

Give me a hint

Mass and molar mass are given.

Check the answer

Answer: 7.00 mol

Explain the worked problem to me

What you know: 126 g of water, molar mass 18.0 g/mol. What you want: moles of water.

Molar mass says 1 mol of water has a mass of 18.0 g. To turn grams into moles, divide by it so grams cancel: 126 g × (1 mol / 18.0 g).
Grams cancel, leaving mol. Do the arithmetic: 126 ÷ 18.0 = 7. Keep 3 significant figures: 7.00 mol.

Check: Reverse it: 7.00 mol × 18.0 g/mol = 126 g. That matches the starting mass.

Watch out: Do not multiply 126 × 18.0. That gives 2268, with units of g²/mol, which makes no sense.

Your route: Grams → Divide by molar mass.

On the periodic table: You can also build this from the table: hydrogen (H) is 1.008 and oxygen (O) is 16.00 g/mol, so water (H₂O) is 2 × 1.008 + 16.00 = 18.02 g/mol, which rounds to 18.0 g/mol.

That’s right: 7.00 mol

Practice 2

Using molar mass 18.0 g/mol, how many moles are in 144 g of water?

Give me a hint

Mass and molar mass are given.

Check the answer

Answer: 8.00 mol

Explain the worked problem to me

What you know: 144 g of water, molar mass 18.0 g/mol. What you want: moles of water.

Each mole of water weighs 18.0 g. Divide by molar mass so grams cancel: 144 g × (1 mol / 18.0 g).
Grams cancel, leaving mol: 144 ÷ 18.0 = 8. Both values have 3 significant figures, so the answer is 8.00 mol.

Check: 8.00 mol × 18.0 g/mol = 144 g. That is the mass you started with.

Watch out: Going from grams to moles always means dividing by molar mass. Multiplying would give a huge number with the wrong units.

Your route: Grams → Divide by molar mass.

That’s right: 8.00 mol

Practice 3

A substance has molar mass 40.0 g/mol. What mass is 0.900 mol?

Give me a hint

Moles and molar mass are given.

Check the answer

Answer: 36.0 g

Explain the worked problem to me

What you know: 0.900 mol of a substance, molar mass 40.0 g/mol. What you want: its mass in grams.

Molar mass says 1 mol has a mass of 40.0 g. To go from moles to grams, multiply so moles cancel: 0.900 mol × (40.0 g / 1 mol).
Mol cancels, leaving g: 0.900 × 40.0 = 36. Both values have 3 significant figures, so the answer is 36.0 g.

Check: Less than 1 mol should weigh less than 40 g. 36.0 g is a bit under 40 g, as expected.

Watch out: Dividing (0.900 ÷ 40.0 = 0.0225) is the grams → moles direction. Here you start with moles, so multiply.

Your route: Moles → Multiply by molar mass.

That’s right: 36.0 g

Practice 4

A substance has molar mass 40.0 g/mol. What mass is 1.05 mol?

Give me a hint

Moles and molar mass are given.

Check the answer

Answer: 42.0 g

Explain the worked problem to me

What you know: 1.05 mol of a substance, molar mass 40.0 g/mol. What you want: its mass in grams.

Each mole has a mass of 40.0 g. Multiply so moles cancel: 1.05 mol × (40.0 g / 1 mol).
Mol cancels, leaving g: 1.05 × 40.0 = 42. Both values have 3 significant figures, so the answer is 42.0 g.

Check: A little more than 1 mol should weigh a little more than 40 g. 42.0 g fits.

Watch out: Starting from moles means multiply by molar mass. Dividing would give about 0.026, which is far too small.

Your route: Moles → Multiply by molar mass.

That’s right: 42.0 g

Practice 5

Using 6.022 × 10²³ particles/mol, how many molecules are in 0.600 mol?

Give me a hint

A molecule count is requested from moles.

Check the answer

Answer: 3.61 × 10²³ molecules

Explain the worked problem to me

What you know: 0.600 mol, and 1 mol = 6.022 × 10²³ molecules. What you want: the number of molecules.

Avogadro’s number is a counting factor, like 12 per dozen. Multiply so moles cancel: 0.600 mol × (6.022 × 10²³ molecules / 1 mol).
Multiply the front numbers and keep the power of ten: 0.600 × 6.022 = 3.6132.
Round to 3 significant figures, because 0.600 mol has 3: 3.61 × 10²³ molecules.

Check: 0.6 mol is a bit more than half a mole, and 3.6 × 10²³ is a bit more than half of 6.0 × 10²³.

Watch out: Do not multiply the exponent by 0.6. Only the 6.022 part gets multiplied.

Your route: Moles → Multiply by Avogadro’s number.

That’s right: 3.61 × 10²³ molecules

Practice 6

Using 6.022 × 10²³ particles/mol, how many molecules are in 0.700 mol?

Give me a hint

A molecule count is requested from moles.

Check the answer

Answer: 4.22 × 10²³ molecules

Explain the worked problem to me

What you know: 0.700 mol, and 1 mol = 6.022 × 10²³ molecules. What you want: the number of molecules.

Avogadro’s number tells how many particles are in one mole. Multiply so moles cancel: 0.700 mol × (6.022 × 10²³ molecules / 1 mol).
Multiply the front numbers and keep 10²³: 0.700 × 6.022 = 4.2154.
Round to 3 significant figures, because 0.700 mol has 3: 4.22 × 10²³ molecules.

Check: 0.7 mol is less than 1 mol, so the count must be less than 6.022 × 10²³. 4.2 × 10²³ is.

Watch out: Dividing 0.7 by 6.022 × 10²³ gives a tiny number. Going from moles to particles always means multiplying.

Your route: Moles → Multiply by Avogadro’s number.

That’s right: 4.22 × 10²³ molecules

Practice 7

How many oxygen atoms are represented by 8 formula units of Ca(NO₃)₂?

Give me a hint

The formula contains parentheses.

Check the answer

Answer: 48

Explain the worked problem to me

What you know: 8 formula units of Ca(NO₃)₂. What you want: the total number of oxygen atoms.

Read the inner subscript: each NO₃ group has 3 oxygen atoms.
Apply the parentheses: the 2 outside means two NO₃ groups, so one formula unit has 3 × 2 = 6 O atoms.
Multiply by the number of formula units: 8 × 6 = 48 oxygen atoms.

Check: Write it out: one unit is CaN₂O₆. Eight of those give 8 × 6 = 48 O.

Watch out: Forgetting the outside 2 gives 3 O per unit and 24 total. The subscript after a parenthesis multiplies everything inside.

Your route: Read inner subscript → Apply parentheses → Multiply formula units.

On the periodic table: Oxygen’s symbol is O, shown in its box, so count every O in the formula.

That’s right: 48

YOUR PERIODIC-TABLE REFERENCE

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 8

How many oxygen atoms are represented by 9 formula units of Ca(NO₃)₂?

Give me a hint

The formula contains parentheses.

Check the answer

Answer: 54

Explain the worked problem to me

What you know: 9 formula units of Ca(NO₃)₂. What you want: the total number of oxygen atoms.

Read the inner subscript: each NO₃ group has 3 O atoms.
Apply the parentheses: the outside 2 doubles the group, so each formula unit has 3 × 2 = 6 O atoms.
Multiply by the number of formula units: 9 × 6 = 54 oxygen atoms.

Check: 54 ÷ 9 = 6 O per unit, which matches CaN₂O₆.

Watch out: A common slip is 9 × 3 = 27, which ignores the 2 outside the parentheses.

Your route: Read inner subscript → Apply parentheses → Multiply formula units.

On the periodic table: Oxygen’s symbol is O, shown in its box, so count every O in the formula.

That’s right: 54

YOUR PERIODIC-TABLE REFERENCE

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 9

Using S = 32 g/mol and O = 16 g/mol, what is the molar mass of SO₃?

Give me a hint

A formula and atomic masses are supplied.

Check the answer

Answer: 80 g/mol

Explain the worked problem to me

What you know: The formula SO₃, with S = 32 g/mol and O = 16 g/mol. What you want: the molar mass of SO₃.

Count every atom: SO₃ has 1 S and 3 O.
Multiply each count by its atomic mass: S: 1 × 32 = 32 g/mol. O: 3 × 16 = 48 g/mol.
Add the contributions: 32 + 48 = 80 g/mol.

Check: With the table values, 32.06 + 3(16.00) = 80.06 g/mol, which rounds to 80 g/mol. It matches.

Watch out: The subscript 3 belongs to O only. Writing 3 × (32 + 16) = 144 is wrong.

Your route: Count every atom → Add atomic masses.

16S32.068O16.00On the periodic table: Sulfur (S) has an average atomic mass of 32.06 g/mol and oxygen (O) has 16.00 g/mol, the decimal numbers in their boxes; the problem rounds them to 32 and 16.

That’s right: 80 g/mol

Practice 10

Use C = 12 g/mol, H = 1 g/mol, O = 16 g/mol, S = 32 g/mol. What is sulfur’s mass percent in SO₂, nearest whole percent?

Give me a hint

Mass percent is requested.

Check the answer

Answer: 50%

Explain the worked problem to me

What you know: The formula SO₂, with S = 32 g/mol and O = 16 g/mol. What you want: sulfur’s mass percent.

Find the formula mass: 1 S + 2 O = 32 + 2(16) = 64 g/mol.
Sulfur’s part of that mass is 32 g/mol (one S atom).
Mass percent = part ÷ total × 100: (32 g/mol ÷ 64 g/mol) × 100 = 50. The units cancel. The answer is 50%.

Check: Oxygen’s share is 32 ÷ 64 × 100 = 50%. The two percents add to 100%.

Watch out: Do not count atoms instead of mass. S is 1 of 3 atoms, but 1/3 (33%) is not the mass percent.

Your route: Component mass ÷ Formula mass × 100.

On the periodic table: Sulfur (S) has an average atomic mass of 32.06 g/mol and oxygen (O) has 16.00 g/mol, the decimal numbers in their boxes; the problem rounds them to 32 and 16.

That’s right: 50%

YOUR PERIODIC-TABLE REFERENCE

16SSulfur32.06Group 16 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 11

An analysis gives 6 mol C, 6 mol H, and 3 mol O. Which is the simplest whole-number formula?

Give me a hint

Element amounts are already in moles.

Check the answer

Answer: C₂H₂O

Explain the worked problem to me

What you know: 6 mol C, 6 mol H, 3 mol O. What you want: the simplest whole-number formula (empirical formula).

The amounts are already in moles, so no gram conversion is needed. Find the smallest amount: 3 mol (O).
Divide every amount by 3: C: 6 ÷ 3 = 2. H: 6 ÷ 3 = 2. O: 3 ÷ 3 = 1.
All are whole numbers, so they become subscripts: C₂H₂O.

Check: 2 : 2 : 1 is the same ratio as 6 : 6 : 3, and no whole number larger than 1 divides all of 2, 2 and 1.

Watch out: C₆H₆O₃ keeps the right ratio but is not the simplest. Always divide by the smallest.

Your route: Mole amounts → Divide by smallest → Whole-number ratio.

That’s right: C₂H₂O

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 12

The mole ratio C:H:O is 1:1.5:0.5. Which empirical formula keeps that ratio using whole numbers?

Give me a hint

The ratio includes a half.

Check the answer

Answer: C₂H₃O

Explain the worked problem to me

What you know: The mole ratio C : H : O = 1 : 1.5 : 0.5. What you want: the empirical formula with whole numbers.

Notice the halves (1.5 and 0.5). Subscripts must be whole numbers, and you cannot just round 1.5 or 0.5.
Multiply every part of the ratio by 2, the smallest number that clears a half: C: 1 × 2 = 2. H: 1.5 × 2 = 3. O: 0.5 × 2 = 1.
Use 2 : 3 : 1 as subscripts: C₂H₃O.

Check: Divide 2 : 3 : 1 by 2 and you get back 1 : 1.5 : 0.5.

Watch out: Multiply all three numbers, not just the fractions. Doubling only H and O would change the ratio.

Your route: Fractional ratio → Multiply all ratios to whole numbers.

That’s right: C₂H₃O

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 13

An empirical formula is CH₂O with mass 30 g/mol. Molecular molar mass is 210 g/mol. What is the molecular formula?

Give me a hint

An empirical formula and actual molar mass are given.

Check the answer

Answer: C₇H₁₄O₇

Explain the worked problem to me

What you know: Empirical formula CH₂O at 30 g/mol; molecular molar mass 210 g/mol. What you want: the molecular formula.

Find how many empirical units fit in one molecule: 210 g/mol ÷ 30 g/mol = 7. The g/mol units cancel, leaving a pure number.
Multiply every subscript in CH₂O (1 : 2 : 1) by 7: C: 1 × 7 = 7. H: 2 × 7 = 14. O: 1 × 7 = 7. The answer is C₇H₁₄O₇.

Check: 7(12) + 14(1) + 7(16) = 84 + 14 + 112 = 210 g/mol. It matches.

Watch out: Multiply every subscript, including the unwritten 1s on C and O. C₁H₁₄O₁ is a common slip.

Your route: Molecular mass ÷ Empirical mass → Multiply subscripts.

On the periodic table: You can also build the 30 g/mol from the table: carbon (C) is 12.01, hydrogen (H) is 1.008, and oxygen (O) is 16.00 g/mol, so CH₂O is 12.01 + 2 × 1.008 + 16.00 ≈ 30 g/mol.

That’s right: C₇H₁₄O₇

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 14

An empirical formula is CH₂O with mass 30 g/mol. Molecular molar mass is 240 g/mol. What is the molecular formula?

Give me a hint

An empirical formula and actual molar mass are given.

Check the answer

Answer: C₈H₁₆O₈

Explain the worked problem to me

What you know: Empirical formula CH₂O at 30 g/mol; molecular molar mass 240 g/mol. What you want: the molecular formula.

Find the multiplier: 240 g/mol ÷ 30 g/mol = 8. The molecule holds 8 empirical units.
Multiply every subscript in CH₂O by 8: C: 1 × 8 = 8. H: 2 × 8 = 16. O: 1 × 8 = 8. The answer is C₈H₁₆O₈.

Check: 8(12) + 16(1) + 8(16) = 96 + 16 + 128 = 240 g/mol. It matches.

Watch out: Divide molecular mass by empirical mass, not the other way. 30 ÷ 240 = 0.125 is not a useful multiplier.

Your route: Molecular mass ÷ Empirical mass → Multiply subscripts.

That’s right: C₈H₁₆O₈

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 15

How many oxygen atoms are in 0.500 mol of CO₂? Use 6.022 × 10²³ molecules/mol.

Give me a hint

Moles → molecules with Avogadro’s number, then multiply by the 2 O atoms in each CO₂ molecule.

Check the answer

Answer: 6.02 × 10²³ O atoms

Explain the worked problem to me

What you know: 0.500 mol CO₂, and 1 mol = 6.022 × 10²³ molecules. Each CO₂ molecule holds 1 C atom and 2 O atoms. What you want: the number of O atoms.

Moles to molecules: multiply by Avogadro’s number so mol cancels. 0.500 mol × (6.022 × 10²³ molecules / 1 mol) = 3.011 × 10²³ CO₂ molecules.
Molecules to atoms: the subscript 2 means 2 O atoms per molecule. 3.011 × 10²³ molecules × (2 O atoms / 1 molecule) = 6.022 × 10²³ O atoms.
0.500 has 3 significant figures, so round to 3. The answer is 6.02 × 10²³ O atoms.

Check: 0.500 mol CO₂ holds 2 × 0.500 = 1.00 mol of O atoms, and 1.00 mol is 6.022 × 10²³. It matches.

Watch out: Do not stop at 3.01 × 10²³. That counts CO₂ molecules, but the question asks for oxygen atoms.

That’s right: 6.02 × 10²³ O atoms

Practice 16

How many hydrogen atoms are in 3.60 g of H₂O? Use 18.0 g/mol and 6.022 × 10²³ molecules/mol.

Give me a hint

Go grams → moles → molecules, then multiply by the 2 H atoms in each H₂O molecule.

Check the answer

Answer: 2.41 × 10²³ H atoms

Explain the worked problem to me

What you know: 3.60 g of H₂O, molar mass 18.0 g/mol, and 1 mol = 6.022 × 10²³ molecules. Each H₂O molecule holds 2 H atoms. What you want: the number of H atoms.

Grams to moles: divide by molar mass so grams cancel. 3.60 g × (1 mol / 18.0 g) = 0.200 mol H₂O.
Moles to molecules: multiply by Avogadro’s number so mol cancels. 0.200 mol × (6.022 × 10²³ molecules / 1 mol) = 1.2044 × 10²³ molecules.
Molecules to atoms: 1.2044 × 10²³ molecules × (2 H atoms / 1 molecule) = 2.4088 × 10²³ H atoms.
The data have 3 significant figures (3.60 and 18.0), so round to 3. The answer is 2.41 × 10²³ H atoms.

Check: 3.60 g is one fifth of 18.0 g, so 0.200 mol of water. That holds 2 × 0.200 = 0.400 mol of H atoms, and 0.400 × 6.022 × 10²³ = 2.4088 × 10²³. It matches.

Watch out: Do not double the molar mass for the two H atoms. 18.0 g/mol already covers the whole molecule; the 2 is used only in the last step.

That’s right: 2.41 × 10²³ H atoms

On the periodic table: H is 1.008 and O is 16.00, so H₂O is 2(1.008) + 16.00 = 18.016 g/mol, rounded here to 18.0 g/mol.

Skill check: the chapter’s key skills

Try each fresh question without the worked examples. Record your first answer and whether you used a hint. Explain the route as well as the answer.

Skill check · 6 questions · tap to open

What this set is for: one question on each key skill from this chapter. Try each on your own, without hints, then mark how it went in the tracker below. A miss tells you which section to review.

Skill 1: Convert between moles and particles.

Find molecules in 2.0 mol, using 6.022 × 10²³ molecules/mol.

Give me a hint

Multiply moles by Avogadro’s number in molecules per mole. The mole units cancel.

Check the answer

1.2 × 10²⁴ molecules (two significant figures).

Explain the worked problem to me

What you know: 2.0 mol, and 1 mol = 6.022 × 10²³ molecules. What you want: the number of molecules.

Multiply by Avogadro’s number so moles cancel: 2.0 mol × (6.022 × 10²³ molecules / 1 mol) = 12.044 × 10²³ molecules.
Rewrite in scientific notation: 12.044 × 10²³ = 1.2044 × 10²⁴. Moving the decimal one place left raises the exponent by 1.
2.0 has 2 significant figures, so round to 2. The answer is 1.2 × 10²⁴ molecules.

Check: Two moles should be about twice 6 × 10²³, which is 12 × 10²³ = 1.2 × 10²⁴.

Watch out: Keeping 10²³ after moving the decimal gives 1.2 × 10²³, which is ten times too small.

That’s right: 1.2 × 10²⁴ molecules (two significant figures).

Skill 2: Convert between grams and moles.

Find moles in 8.0 g using 16 g/mol.

Give me a hint

Divide grams by grams per mole. The gram units cancel, leaving moles.

Check the answer

8.0/16 = 0.50 mol.

Explain the worked problem to me

What you know: 8.0 g of a substance, molar mass 16 g/mol. What you want: moles.

Molar mass says 1 mol weighs 16 g. Divide so grams cancel: 8.0 g × (1 mol / 16 g).
Grams cancel, leaving mol: 8.0 ÷ 16 = 0.50. Both values have 2 significant figures, so the answer is 0.50 mol.

Check: 8.0 g is half of 16 g, so it should be half a mole. Reverse it: 0.50 mol × 16 g/mol = 8.0 g.

Watch out: Multiplying gives 128 g²/mol, which is not a mole amount. Grams to moles means divide.

That’s right: 8.0/16 = 0.50 mol.

Skill 3: Find an empirical formula from composition.

A sample contains 2 mol C and 6 mol H. Empirical formula?

Give me a hint

Divide each mole amount by the smallest to find the simplest whole-number atom ratio.

Check the answer

CH₃, the simplest 1:3 ratio.

Explain the worked problem to me

What you know: 2 mol C and 6 mol H. What you want: the empirical formula.

The amounts are already in moles. Find the smallest: 2 mol (C).
Divide each amount by 2: C: 2 ÷ 2 = 1. H: 6 ÷ 2 = 3.
Use 1 : 3 as subscripts (a 1 is not written): CH₃.

Check: 1 : 3 is the same ratio as 2 : 6, and it cannot be reduced further.

Watch out: C₂H₆ has the right ratio but is not the simplest. The empirical formula always uses the smallest whole numbers.

That’s right: CH₃, the simplest 1:3 ratio.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 4: Find a molecular formula from an empirical formula and molar mass.

Empirical formula CH₂ has mass 14 g/mol; molecular mass is 42 g/mol. Formula?

Give me a hint

Divide the molecular molar mass by the empirical-formula molar mass, then multiply every subscript by that whole number.

Check the answer

42/14 = 3; C₃H₆.

Explain the worked problem to me

What you know: Empirical formula CH₂ at 14 g/mol; molecular molar mass 42 g/mol. What you want: the molecular formula.

Find the multiplier: 42 g/mol ÷ 14 g/mol = 3. The units cancel, so one molecule holds 3 empirical units.
Multiply every subscript in CH₂ by 3: C: 1 × 3 = 3. H: 2 × 3 = 6. The answer is C₃H₆.

Check: 3(12) + 6(1) = 36 + 6 = 42 g/mol. It matches.

Watch out: Remember C has an unwritten subscript of 1. It must be multiplied too, giving C₃, not C₁.

That’s right: 42/14 = 3; C₃H₆.

On the periodic table: You can also build the 14 g/mol from the table: carbon (C) is 12.01 and hydrogen (H) is 1.008 g/mol, so CH₂ is 12.01 + 2 × 1.008 ≈ 14 g/mol.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

6CCarbon12.01Group 14 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 5: Count atoms from grams or moles using the formula.

How many hydrogen atoms are in 8.00 g of CH₄? Use 16.0 g/mol and 6.022 × 10²³ molecules/mol.

Give me a hint

Grams → moles (divide by 16.0 g/mol), moles → molecules (multiply by Avogadro’s number), then multiply by the 4 H atoms in each CH₄.

Check the answer

1.20 × 10²⁴ H atoms

Explain the worked problem to me

What you know: 8.00 g of CH₄, molar mass 16.0 g/mol, 1 mol = 6.022 × 10²³ molecules. Each CH₄ molecule holds 1 C atom and 4 H atoms. What you want: the number of H atoms.

Grams to moles: 8.00 g × (1 mol / 16.0 g) = 0.500 mol CH₄.
Moles to molecules: 0.500 mol × (6.022 × 10²³ molecules / 1 mol) = 3.011 × 10²³ molecules.
Molecules to atoms: 3.011 × 10²³ molecules × (4 H atoms / 1 molecule) = 12.044 × 10²³ = 1.2044 × 10²⁴ H atoms.
Keep 3 significant figures, matching 8.00 g and 16.0 g/mol: 1.20 × 10²⁴ H atoms.

Check: 0.500 mol CH₄ holds 4 × 0.500 = 2.00 mol of H atoms, and 2.00 × 6.022 × 10²³ = 1.2044 × 10²⁴. It matches.

Watch out: Use the subscript of the element you are asked about. Multiplying by 1 (for C) would count carbon atoms, 3.01 × 10²³, not hydrogen atoms.

That’s right: 1.20 × 10²⁴ H atoms

On the periodic table: C is 12.01 and H is 1.008, so CH₄ is 12.01 + 4(1.008) = 16.042 g/mol, rounded here to 16.0 g/mol.

YOUR PERIODIC-TABLE REFERENCE

6CCarbon12.01Group 14 · Period 2

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 6: Calculate the mass percent of an element in a compound.

What is the mass percent of carbon in carbon dioxide, CO₂? Use C = 12.01 g/mol and O = 16.00 g/mol. Give the answer to the nearest 0.1%.

Give me a hint

First add up the molar mass of CO₂ (one C and two O). Then divide the mass of the one C by that total and multiply by 100%.

Check the answer

27.3% C

Explain the worked problem to me

What you know: CO₂ has 1 C and 2 O; C = 12.01 g/mol, O = 16.00 g/mol. What you want: the mass percent of C.

Find the molar mass of CO₂: 12.01 + 2 × 16.00 = 12.01 + 32.00 = 44.01 g/mol.
Mass of C in one mole: 1 × 12.01 = 12.01 g.
Divide part by total: 12.01 g ÷ 44.01 g = 0.2729.
Multiply by 100%: 0.2729 × 100% = 27.29%, which rounds to 27.3%.

Check: Oxygen should be the rest: 32.00 ÷ 44.01 × 100% = 72.7%, and 27.3% + 72.7% = 100.0%.

Watch out: Count both O atoms in the total. Using 12.01 + 16.00 = 28.01 g/mol would give 42.9%, which is wrong.

That’s right: 27.3% C

On the periodic table: carbon (C) is 12.01 g/mol and oxygen (O) is 16.00 g/mol, the same values given in the question.

YOUR PERIODIC-TABLE REFERENCE

6CCarbon12.01Group 14 · Period 2

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

What to do next

Mark each skill: “I can do it independently,” “I needed help,” or “I need another explanation.” Revisit the section with the same goal for any skill needing help, then try a different question. A successful first attempt is evidence of progress; one answer alone does not establish lasting mastery.

Practice with instant feedback Next: Chapter 5

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Original ChemGem questions.

The mole and chemical formulas

What you will study

By the end, you can…

4.1 A mole is a counting unit

How many molecules are in 0.25 mol? Use 6.022 × 1023 molecules per mole.

4.2 Moving between grams and moles

How many moles are in 22.005 g CO2? Use molar mass 44.01 g/mol.

How many H2O molecules, and how many hydrogen atoms, are in 9.00 g of water? Use 18.02 g/mol.

4.3 Percent composition and empirical formulas

What is the mass percent of oxygen in water, H2O? Use H = 1.008 g/mol and O = 16.00 g/mol.

A sample is 40.0% C, 6.7% H, and 53.3% O. Find its empirical formula. Use approximate masses C = 12 g/mol, H = 1 g/mol, O = 16 g/mol.

4.4 Finding the molecular formula

The empirical formula is CH2O and the molecular molar mass is 180 g/mol. Find the molecular formula. Use C = 12 g/mol, H = 1 g/mol, O = 16 g/mol.

Mixed chapter review

Practice 1

Practice 2

Practice 3

Practice 4

Practice 5

Practice 6

Practice 7

Practice 8

Practice 9

Practice 10

Practice 11

Practice 12

Practice 13

Practice 14

Practice 15

Practice 16

Skill check: the chapter’s key skills

Skill 1: Convert between moles and particles.

Skill 2: Convert between grams and moles.

Skill 3: Find an empirical formula from composition.

Skill 4: Find a molecular formula from an empirical formula and molar mass.

Skill 5: Count atoms from grams or moles using the formula.

Skill 6: Calculate the mass percent of an element in a compound.

What to do next

Take the 70-question practice final

How many molecules are in 0.25 mol? Use 6.022 × 10²³ molecules per mole.

How many moles are in 22.005 g CO₂? Use molar mass 44.01 g/mol.

How many H₂O molecules, and how many hydrogen atoms, are in 9.00 g of water? Use 18.02 g/mol.

What is the mass percent of oxygen in water, H₂O? Use H = 1.008 g/mol and O = 16.00 g/mol.

The empirical formula is CH₂O and the molecular molar mass is 180 g/mol. Find the molecular formula. Use C = 12 g/mol, H = 1 g/mol, O = 16 g/mol.