Mole Concept and Molar Mass

Why Chemists Needed a Counting Unit

Eggs are sold by the dozen (12). A score is 20, a gross is 144. Each is a name for a fixed number. Atoms are so small that one drop of water holds more molecules than there are sand grains on Earth, so chemists needed a far bigger counting word: the mole.

Key Point: A dozen means 12, a score 20, a gross 144. A mole means a fixed, very large number of entities (atoms, molecules, ions, electrons, or any specified particle). It is a counting unit like a dozen, only far bigger.

The mole is an SI base quantity

The mole (symbol mol) is the seventh SI base quantity, for amount of substance (symbol $n$). It is not derived from the other six.

The exact definition (2019 SI)

Key Point (Definition): The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, in $\mathrm{mol^{-1}}$, and is called the Avogadro number.

1. "Exactly." Since 2019 the number is fixed by definition, not measured. Older books quote a measured $6.022 \times 10^{23}$ or $6.023 \times 10^{23}$. For calculations use $6.022 \times 10^{23}$.
2. "Elementary entities." Atom, molecule, ion, electron, or a specified group of particles. A mole of electrons or of $\mathrm{Na^+}$ ions is valid, but say which.
3. Same number for every substance. A mole of H atoms, of water molecules or of gold atoms holds $6.022 \times 10^{23}$ entities. Masses differ; counts do not.

How big is this number?

$$602{,}213{,}670{,}000{,}000{,}000{,}000{,}000$$

Twenty-four digits. If everyone on Earth (about $8 \times 10^{9}$) counted one atom per second, one mole would take over two million years. So we weigh atoms and let the mole count them.

[Board] "Define mole" is a regular two-mark question. Write exactly, the full number $6.02214076 \times 10^{23}$, and "seventh SI base quantity, amount of substance".

Where the Avogadro Number Came From

The carbon-12 link

One $^{12}\mathrm{C}$ atom is exactly 12 u (1 u is one-twelfth of it). The mole was chosen so that one mole of carbon-12 atoms has a mass of exactly 12 g: same number, grams instead of u.

A mass spectrometer gives the mass of one atom:

$$\text{mass of one } ^{12}\mathrm{C} \text{ atom} = 1.992648 \times 10^{-23} \ \mathrm{g}$$

Mass of a mole divided by mass of one atom gives the atoms in a mole:

$$N_A = \frac{12 \ \mathrm{g\ mol^{-1}}}{1.992648 \times 10^{-23} \ \mathrm{g\ atom^{-1}}} = 6.0221367 \times 10^{23} \ \mathrm{atoms\ mol^{-1}}$$

Check: $12 \div 1.992648 = 6.0221\ldots$, and $10^{0} \div 10^{-23} = 10^{23}$.

Key Point: The Avogadro number is the number of $^{12}\mathrm{C}$ atoms in exactly 12 g of carbon-12. It bridges the atomic mass unit and the gram.

The u-to-gram bridge

Since 1 u $= 1.66056 \times 10^{-24}$ g,

$$N_A \times 1 \ \mathrm{u} = 6.022 \times 10^{23} \times 1.66056 \times 10^{-24} \ \mathrm{g} = 1.000 \ \mathrm{g}$$

The Avogadro number of atomic mass units is one gram. So a mole of any entity has the same number of grams as one entity has in u. This is why molar mass in $\mathrm{g\ mol^{-1}}$ equals atomic or molecular mass in u.

Reciprocal: $1 \ \mathrm{u} = \dfrac{1}{N_A} \ \mathrm{g}$, so mass of one atom in grams is atomic mass divided by $N_A$.

"One mole of …" names the entity

NaCl is a lattice of ions with no discrete molecules, so we count formula units. One mole of NaCl holds $6.022 \times 10^{23}$ $\mathrm{Na^+}$ and $6.022 \times 10^{23}$ $\mathrm{Cl^-}$ ions.

[NEET] "1 mole of hydrogen" is ambiguous: hydrogen gas means $\mathrm{H_2}$ (2 g), hydrogen atoms means H (1 g).

Molar Mass

Key Point (Definition): The mass of one mole of a substance in grams is its molar mass (symbol $M$), unit $\mathrm{g\ mol^{-1}}$. Molar mass in grams per mole is numerically equal to the atomic, molecular or formula mass in u.

Compute the molecular mass (Section 5), keep the number, change the unit to $\mathrm{g\ mol^{-1}}$.

18.02 g of water, 58.5 g of salt, 32.00 g of oxygen gas and 12.011 g of carbon each contain one mole, $6.022 \times 10^{23}$ entities.

Three cautions

1. Element versus molecule. O atoms: 16.00 $\mathrm{g\ mol^{-1}}$; $\mathrm{O_2}$ gas: 32.00 $\mathrm{g\ mol^{-1}}$. $\mathrm{H_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{F_2}$, $\mathrm{Cl_2}$ are diatomic. Noble gases (He, Ne, Ar) are monatomic: helium is 4.00 $\mathrm{g\ mol^{-1}}$.
2. Ionic compounds use formula mass: NaCl $= 58.5 \ \mathrm{g\ mol^{-1}}$.
3. Rounding. Boards accept H = 1, C = 12, O = 16, Na = 23, Cl = 35.5 unless precise values are given. Use the values supplied.

The central formula: $n = m / M$

$$n = \frac{m}{M} \qquad \Longleftrightarrow \qquad m = n \times M$$

$n$ in mol, $m$ in g, $M$ in $\mathrm{g\ mol^{-1}}$.

• Moles in 9.0 g of water: $n = \dfrac{9.0}{18.02} = 0.50$ mol.
• Mass of 0.20 mol NaCl: $m = 0.20 \times 58.5 = 11.7$ g.

Given a mass, first divide by molar mass. Everything else follows from moles.

The Mole Map: Mass, Moles, Particles and Gas Volume

Three roads out of "moles"

Key Point: Away from the centre, multiply; back to the centre, divide.

Mass to particles goes through moles:

$$N = \frac{m}{M} \times N_A$$

Molecules in 4.4 g of $\mathrm{CO_2}$: $4.4/44.01 = 0.100$ mol; $0.100 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$.

Molar volume of a gas

By Avogadro's law (Section 4), one mole of any gas has the same volume at the same temperature and pressure: the molar volume, $V_m$.

IUPAC moved standard pressure from 1 atm to 1 bar in 1982; at the lower pressure the gas expands slightly, so 22.4 L becomes 22.7 L. The textbook uses 22.7 L at 1 bar; competitive papers often still use 22.4 L at 1 atm.

[JEE Main] "At STP" with no pressure given: take 22.4 L unless the paper uses 1 bar. The two differ by about 1.3%; state the value used.

$$n_{\text{gas}} = \frac{V \ (\text{L at STP})}{22.7 \ \mathrm{L\ mol^{-1}}}$$

11.35 L at 1 bar, 273.15 K is 0.500 mol; 5.6 L at 1 atm, 273 K is 0.25 mol.

Key Point: Molar volume is for gases only. For liquids and solids use mass and density.

A full circuit: 8.0 g of oxygen gas

1. Moles: $n = 8.0/32.00 = 0.25$ mol.
2. Molecules: $N = 0.25 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$.
3. Volume at STP (1 bar): $V = 0.25 \times 22.7 = 5.68$ L (5.6 L at 1 atm).
4. Atoms: 2 per $\mathrm{O_2}$, so $3.01 \times 10^{23}$ O atoms.

Moles Inside Moles: Atoms in Molecules, Ions in Formula Units

The subscript is a mole ratio

One $\mathrm{C_2H_6}$ molecule has 2 C and 6 H atoms, so one mole of ethane has 2 mol C and 6 mol H atoms. Subscripts in a formula are mole ratios of atoms to molecules.

$$1 \ \mathrm{mol\ C_2H_6} \ \Longrightarrow \ 2 \ \mathrm{mol\ C\ atoms} + 6 \ \mathrm{mol\ H\ atoms} = 8 \ \mathrm{mol\ atoms\ in\ total}$$

Three moles of ethane: C atoms $= 3 \times 2 = 6$ mol; H atoms $= 3 \times 6 = 18$ mol; molecules $= 3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24}$.

(Electrons: $\mathrm{H_2O}$ has $2 + 8 = 10$, $\mathrm{CO_2}$ has $6 + 16 = 22$.)

Key Point: Moles of an atom $=$ (moles of compound) $\times$ (its subscript). Total moles of atoms $=$ (moles of compound) $\times$ (sum of subscripts).

Atoms in a given mass

$n = m / M$, times the subscript, times $N_A$. For 9.0 g of water: $n = 9.0/18.02 = 0.50$ mol; O atoms $= 0.50 \times 1 = 0.50$ mol $= 3.01 \times 10^{23}$; H atoms $= 0.50 \times 2 = 1.0$ mol $= 6.02 \times 10^{23}$.

Diatomic gases: the double-count trap

Atoms in 1 g of $\mathrm{Cl_2}$: molecules $= 1/71.0 = 0.0141$ mol; Cl atoms $= 2 \times 0.0141 = 0.0282$ mol $= 1.70 \times 10^{22}$. Drop the factor of 2 and you get half, which will be among the options.

[NEET] "Atoms in X g of $\mathrm{N_2}$ / $\mathrm{O_2}$ / $\mathrm{H_2}$" always needs the factor of 2. Noble gases (He, Ne, Ar) are monatomic.

Hydrated salts: count the water. 0.1 mol of $\mathrm{Na_2SO_4 \cdot 10H_2O}$: $2 + 1 + 4 + 10(2 + 1) = 37$ atoms per formula unit, so $0.1 \times 37 = 3.7$ mol $= 2.23 \times 10^{24}$ atoms.

Comparing Samples and Single Molecules: Four Question Types

Type 1: Which sample has the most atoms?

Fixed mass: smaller molar mass, more moles, more particles. For atoms, also multiply by atoms per particle. Compare 1 g each of Au, Na, Li, $\mathrm{Cl_2}$:

Lithium wins: lightest atoms, most per gram. $\mathrm{Cl_2}$ passed Au only through the factor of 2.

Type 2: Moles, u, or grams?

Atoms in (i) 52 mol Ar, (ii) 52 u He, (iii) 52 g He:

• 52 mol Ar: $52 \times 6.022 \times 10^{23} = 3.131 \times 10^{25}$ atoms (monatomic).
• 52 u He: one He atom is 4 u, so $52 / 4 = 13$ atoms. A "u" is the mass of one atom, so this is a direct count.
• 52 g He: $n = 52 / 4.00 = 13$ mol; atoms $= 13 \times 6.022 \times 10^{23} = 7.829 \times 10^{24}$.

Key Point: Mass in u over atomic mass in u gives atoms. Mass in g over molar mass gives moles. They differ by $N_A$.

Type 3: Mass of one molecule or atom

$$\text{mass of one molecule} = \frac{M}{N_A}$$

One water molecule: $\dfrac{18.02}{6.022 \times 10^{23}} = 2.99 \times 10^{-23}$ g. One carbon-12 atom: $\dfrac{12}{6.022 \times 10^{23}} = 1.993 \times 10^{-23}$ g, the mass-spectrometer value.

Type 4: Molecules in a drop of water

Twenty drops make 1 mL, so one drop is 0.05 mL, or 0.05 g (density $\approx 1 \ \mathrm{g\ mL^{-1}}$).

$$n = \frac{0.05}{18.02} = 2.77 \times 10^{-3} \ \mathrm{mol}, \qquad N = 2.77 \times 10^{-3} \times 6.022 \times 10^{23} = 1.67 \times 10^{21} \ \text{molecules}$$

Common mistakes

Before calculating, underline the entity asked for and the quantity given, then walk the map between them.

Solved Examples

Question 1: Moles and molecules in a glass of water

How many moles of water and how many water molecules are present in 36.0 g of water? (H = 1.008, O = 16.00)

Answer: First the molar mass: $M(\mathrm{H_2O}) = 2(1.008) + 16.00 = 18.02 \ \mathrm{g\ mol^{-1}}$.

Moles: $n = \dfrac{36.0}{18.02} = 1.998 \approx 2.00$ mol.

Molecules: $N = 2.00 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$.

Ans: 2.00 mol of water, $1.20 \times 10^{24}$ molecules.

Question 2: From moles to mass and formula units

What is the mass of 0.25 mol of sodium chloride, and how many formula units does it contain? (Na = 23.0, Cl = 35.5)

Answer: $M(\mathrm{NaCl}) = 23.0 + 35.5 = 58.5 \ \mathrm{g\ mol^{-1}}$. NaCl is ionic, so I count formula units.

Mass: $m = 0.25 \times 58.5 = 14.6$ g.

Formula units: $0.25 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$.

That is also $1.51 \times 10^{23}$ $\mathrm{Na^+}$ and $1.51 \times 10^{23}$ $\mathrm{Cl^-}$, so $3.01 \times 10^{23}$ ions in total.

Ans: 14.6 g; $1.51 \times 10^{23}$ formula units.

Watch out: Each NaCl formula unit gives two ions, so "number of ions" is double the formula units.

Question 3: Three moles of ethane

In three moles of ethane ($\mathrm{C_2H_6}$), calculate (i) the number of moles of carbon atoms, (ii) the number of moles of hydrogen atoms, (iii) the number of molecules of ethane.

Answer: 1 mol $\mathrm{C_2H_6}$ has 2 mol C atoms and 6 mol H atoms.

(i) C atoms: $3 \times 2 = 6$ mol.

(ii) H atoms: $3 \times 6 = 18$ mol.

(iii) Molecules: $3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24}$.

Ans: (i) 6 mol C atoms; (ii) 18 mol H atoms; (iii) $1.807 \times 10^{24}$ molecules.

Question 4: Mass of a single molecule and a single atom

Calculate (i) the mass of one molecule of water and (ii) the mass of one atom of carbon-12, in grams.

Answer: One mole is $M$ g and holds $N_A$ particles, so one particle is $M / N_A$.

(i) Water: $\dfrac{18.02}{6.022 \times 10^{23}} = 2.992 \times 10^{-23}$ g.

(ii) Carbon-12: $\dfrac{12.00}{6.022 \times 10^{23}} = 1.993 \times 10^{-23}$ g.

The mass-spectrometer value for $^{12}\mathrm{C}$ is $1.992648 \times 10^{-23}$ g, matching to four figures.

Ans: (i) $2.99 \times 10^{-23}$ g; (ii) $1.99 \times 10^{-23}$ g.

Watch out: A single-molecule mass should be around $10^{-22}$ to $10^{-24}$ g.

Question 5: Molecules in a drop of water

Assuming 20 drops of water make 1.0 mL and the density of water is $1.0 \ \mathrm{g\ mL^{-1}}$, find the number of water molecules and the total number of atoms in one drop.

Answer: One drop $= 1.0/20 = 0.050$ mL, so mass $= 0.050 \times 1.0 = 0.050$ g.

Moles: $n = \dfrac{0.050}{18.02} = 2.775 \times 10^{-3}$ mol.

Molecules: $2.775 \times 10^{-3} \times 6.022 \times 10^{23} = 1.671 \times 10^{21}$.

Each $\mathrm{H_2O}$ has 3 atoms: $3 \times 1.671 \times 10^{21} = 5.01 \times 10^{21}$ atoms.

Ans: About $1.67 \times 10^{21}$ molecules and $5.0 \times 10^{21}$ atoms.

Question 6: Volume of a gas at STP — both conventions

Find the volume occupied by 8.0 g of oxygen gas (i) at STP (1 bar, 273.15 K) and (ii) at 1 atm and 273.15 K. Also state the number of molecules present.

Answer: $M(\mathrm{O_2}) = 2 \times 16.00 = 32.00 \ \mathrm{g\ mol^{-1}}$, so $n = 8.0 / 32.00 = 0.25$ mol.

(i) At 1 bar: $V = 0.25 \times 22.7 = 5.68$ L.

(ii) At 1 atm: $V = 0.25 \times 22.4 = 5.6$ L.

Molecules: $0.25 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$ ($3.01 \times 10^{23}$ O atoms).

Ans: (i) 5.68 L; (ii) 5.6 L; $1.51 \times 10^{23}$ molecules.

Watch out: Match the molar volume to the pressure given.

Question 7: Which 1 g sample has the most atoms?

Which of the following will have the largest number of atoms: (i) 1 g Au(s), (ii) 1 g Na(s), (iii) 1 g Li(s), (iv) 1 g $\mathrm{Cl_2}$(g)? (Au = 197.0, Na = 23.0, Li = 6.94, Cl = 35.5)

Answer: Au: $1/197.0 = 5.08 \times 10^{-3}$ mol $= 3.06 \times 10^{21}$ atoms.

Na: $1/23.0 = 4.35 \times 10^{-2}$ mol $= 2.62 \times 10^{22}$ atoms.

Li: $1/6.94 = 1.44 \times 10^{-1}$ mol $= 8.68 \times 10^{22}$ atoms.

$\mathrm{Cl_2}$ ($M = 71.0$): $1/71.0 = 1.41 \times 10^{-2}$ mol molecules; atoms $= 2 \times 1.41 \times 10^{-2} = 2.82 \times 10^{-2}$ mol $= 1.70 \times 10^{22}$.

Li $>$ Na $>$ $\mathrm{Cl_2}$ $>$ Au.

Ans: (iii) 1 g of Li, about $8.7 \times 10^{22}$ atoms.

Watch out: Each $\mathrm{Cl_2}$ molecule has two atoms.

Question 8: Moles, u and grams

Calculate the number of atoms in (i) 52 moles of Ar, (ii) 52 u of He, (iii) 52 g of He. (He = 4.00 u)

Answer: (i) Argon is monatomic: $52 \times 6.022 \times 10^{23} = 3.131 \times 10^{25}$ atoms.

(ii) One He atom is 4 u, so $52/4 = 13$ atoms. No $N_A$ needed.

(iii) $n = 52/4.00 = 13$ mol; atoms $= 13 \times 6.022 \times 10^{23} = 7.829 \times 10^{24}$, exactly $N_A$ times (ii).

Ans: (i) $3.131 \times 10^{25}$ atoms; (ii) 13 atoms; (iii) $7.829 \times 10^{24}$ atoms.

Watch out: "u" gives a direct count of atoms; "g" gives moles.

Question 9: Atoms of each kind in a mass of carbon dioxide

For 4.4 g of $\mathrm{CO_2}$ find (i) the number of molecules, (ii) the total number of atoms, (iii) the number of oxygen atoms. (C = 12.011, O = 16.00)

Answer: $M(\mathrm{CO_2}) = 12.011 + 2(16.00) = 44.01 \ \mathrm{g\ mol^{-1}}$.

Moles: $n = 4.4/44.01 = 0.100$ mol.

(i) Molecules: $0.100 \times 6.022 \times 10^{23} = 6.02 \times 10^{22}$.

(ii) Each molecule has $1 + 2 = 3$ atoms: $3 \times 6.02 \times 10^{22} = 1.81 \times 10^{23}$ atoms.

(iii) O atoms, 2 per molecule: $2 \times 6.02 \times 10^{22} = 1.20 \times 10^{23}$.

Ans: (i) $6.02 \times 10^{22}$ molecules; (ii) $1.81 \times 10^{23}$ atoms; (iii) $1.20 \times 10^{23}$ O atoms.

Question 10: Ions and electrons in a pinch of salt

A 5.85 g sample of NaCl is taken. Calculate (i) the total number of ions present, (ii) the total number of electrons present. (Atomic numbers: Na = 11, Cl = 17)

Answer: Moles of NaCl: $5.85/58.5 = 0.100$ mol formula units.

(i) Each formula unit gives one $\mathrm{Na^+}$ and one $\mathrm{Cl^-}$, so 2 ions. Moles of ions $= 0.200$; number $= 0.200 \times 6.022 \times 10^{23} = 1.20 \times 10^{23}$.

(ii) Electrons per formula unit: $\mathrm{Na^+}$ has $11 - 1 = 10$, $\mathrm{Cl^-}$ has $17 + 1 = 18$, total 28. (Neutral Na + Cl also gives $11 + 17 = 28$.)

Electrons: $0.100 \times 28 = 2.8$ mol $= 2.8 \times 6.022 \times 10^{23} = 1.69 \times 10^{24}$.

Ans: (i) $1.20 \times 10^{23}$ ions; (ii) $1.69 \times 10^{24}$ electrons.

Question 11: Identifying a gas from its particle count and mass

A sample of a gas contains $1.505 \times 10^{23}$ molecules and has a mass of 11.0 g. (i) How many moles of gas are present? (ii) What volume will it occupy at STP (1 bar, 273.15 K)? (iii) What is its molar mass, and suggest what the gas could be.

Answer: (i) $n = \dfrac{1.505 \times 10^{23}}{6.022 \times 10^{23}} = 0.250$ mol.

(ii) $V = 0.250 \times 22.7 = 5.68$ L at 1 bar.

(iii) $M = \dfrac{11.0}{0.250} = 44.0 \ \mathrm{g\ mol^{-1}}$.

Molar mass 44 fits $\mathrm{CO_2}$ (44.01), also $\mathrm{N_2O}$ (44.01) or propane $\mathrm{C_3H_8}$ (44.1). Without more data, $\mathrm{CO_2}$ is the usual answer.

Ans: (i) 0.250 mol; (ii) 5.68 L; (iii) 44 $\mathrm{g\ mol^{-1}}$, consistent with $\mathrm{CO_2}$.

Question 12: Just how big is a mole?

(i) How many moles and molecules of water are in 1.00 L of water (density $1.00 \ \mathrm{g\ mL^{-1}}$)? (ii) If a machine could count $10^{7}$ molecules every second, how many years would it take to count one mole?

Answer: (i) $1.00 \ \mathrm{L} = 1000$ mL, so mass $= 1000$ g.

Moles: $1000/18.02 = 55.5$ mol (pure water is 55.5 mol per litre).

Molecules: $55.5 \times 6.022 \times 10^{23} = 3.34 \times 10^{25}$.

(ii) Seconds: $\dfrac{6.022 \times 10^{23}}{10^{7}} = 6.022 \times 10^{16}$ s.

One year $\approx 365 \times 24 \times 3600 = 3.154 \times 10^{7}$ s, so $\dfrac{6.022 \times 10^{16}}{3.154 \times 10^{7}} = 1.91 \times 10^{9}$ years.

Ans: (i) 55.5 mol, $3.34 \times 10^{25}$ molecules; (ii) about 1.9 billion years.