Mole Concept and Molar Mass
The Mole — A Chemical Counting Unit
Atoms are extremely small particles whose individual masses are impossible to measure directly on a balance. To work with meaningful quantities, chemists use a unit called the mole. Just as a dozen means 12 items regardless of what they are, a mole is a fixed number of particles. The concept bridges the gap between the atomic scale and the macroscopic (laboratory) scale.
Gram Atom (Mole of an Element): The atomic mass of an element expressed in grams equals one gram atom, also called one mole of that element. For example, 1 mole of carbon = 12 g, 1 mole of magnesium = 24 g, and 1 mole of uranium = 238 g.
Gram Molecule (Mole of a Compound): The molecular mass of a molecular substance expressed in grams equals one gram molecule, or one mole of that substance. For example, 1 mole of water ($H_2O$) = 18 g, and 1 mole of sucrose ($C_{12}H_{22}O_{11}$) = 342 g.
Gram Formula (Mole of an Ionic Compound): Ionic compounds do not exist as discrete molecules. The sum of atomic masses of the ions in the formula unit, expressed in grams, is called one gram formula or one mole of the ionic substance. For example, 1 mole of NaCl = 58.50 g, and 1 mole of $Na_2CO_3$ = 106 g.
Gram Ion (Mole of an Ion): The ionic mass of a charged species expressed in grams is called one gram ion or one mole of ions. For example, 1 mole of $OH^-$ = 17 g, and 1 mole of $SO_4^{2-}$ = 96 g.
Mole Equivalents — Quick Reference
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1 gram atom of hydrogen = 1.008 g of H atoms
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1 gram atom of sodium = 23 g of Na atoms
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1 gram atom of carbon = 12.000 g of C atoms
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1 gram molecule of $H_2SO_4$ = 98.0 g
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1 gram molecule of water = 18.0 g
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1 gram formula of NaCl = 58.50 g
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1 gram formula of $AgNO_3$ = 170 g
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1 gram ion of $OH^-$ = 17 g
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1 gram ion of $CO_3^{2-}$ = 60 g
The number of moles of any substance is calculated by dividing the given mass by its molar mass. This fundamental relationship applies to elements, molecular compounds, and ionic compounds alike.
$$n = \frac{m}{M}$$
Calculates the number of moles from the mass and molar mass of a substance
$n$=Number of moles (dimensionless)(mol)
$m$=Mass of the substance(g)
$M$=Molar mass (atomic, molecular, or formula mass as appropriate)(g mol$^{-1}$)
Avogadro's Number ($N_A$)
Avogadro's number ($N_A$) is the number of particles (atoms, molecules, or ions) present in one mole of any substance. Its value is $6.02 \times 10^{23}$ particles per mole. A key insight is that one mole of any substance always contains the same number of particles, even though different substances have vastly different masses per mole.
Same Count, Different Masses: 1 mole of hydrogen (1.008 g) and 1 mole of uranium (238 g) both contain exactly $6.02 \times 10^{23}$ atoms. The heavier element has heavier atoms, so you need more mass to reach the same count.
Elements: 23 g of Na = 1 mole of Na = $6.02 \times 10^{23}$ atoms of Na. Different masses of different elements contain the same number of atoms because their atoms differ in mass.
Compounds: 18 g of $H_2O$ = 1 mole = $6.02 \times 10^{23}$ molecules. 180 g of glucose ($C_6H_{12}O_6$) = 1 mole = $6.02 \times 10^{23}$ molecules. Different masses, same molecule count.
Ions: 96 g of $SO_4^{2-}$ = 1 mole = $6.02 \times 10^{23}$ ions of $SO_4^{2-}$. 62 g of $NO_3^-$ = 1 mole = $6.02 \times 10^{23}$ ions of $NO_3^-$.
Three related formulas connect mass to the number of particles (atoms, molecules, or ions) using Avogadro's number. These are derived by combining $n = m/M$ with $N = n \times N_A$.
$$N = \frac{m \times N_A}{M}$$
Calculates the number of particles from mass, Avogadro's number, and the mass of one particle (atomic, molecular, or ionic mass)
$N$=Number of particles (atoms, molecules, or ions)(particles (dimensionless count))
$m$=Given mass of the substance(g)
$N_A$=Avogadro's number ($6.02 \times 10^{23}$)(mol$^{-1}$)
$M$=Atomic mass, molecular mass, or ionic mass(g mol$^{-1}$)
$M$ = atomic mass
→$N$ gives the number of atoms of an element
$M$ = molecular mass
→$N$ gives the number of molecules of a compound
$M$ = ionic mass
→$N$ gives the number of ions of an ionic species
Number of Atoms: To find atoms of an element, use atomic mass as $M$. For example, 10 g of Mg (atomic mass = 24) gives $\frac{10 \times 6.02 \times 10^{23}}{24} = 2.51 \times 10^{23}$ atoms.
Number of Molecules: To find molecules of a compound, use molecular mass as $M$. For example, 10 g of $H_2O$ (molecular mass = 18) gives $\frac{10 \times 6.02 \times 10^{23}}{18} = 3.34 \times 10^{23}$ molecules.
Atoms Within Molecules: Once you know the number of molecules, multiply by the subscript of each element. In $H_2O$, each molecule has 2 H atoms and 1 O atom, so 3.34 × 10²³ molecules contain 6.68 × 10²³ H atoms and 3.34 × 10²³ O atoms.
Molar Mass — Unified Mass Measure
Molar mass is the mass of one mole of a substance expressed in grams per mole (g mol$^{-1}$). It is numerically equal to the atomic mass for elements, the molecular mass for molecular compounds, or the formula mass for ionic compounds. The concept unifies the three types — gram atom, gram molecule, and gram formula — under one term.
For Elements (Gram Atom Mass): The molar mass of an element equals its atomic mass in g mol$^{-1}$. Carbon = 12 g mol$^{-1}$, sodium = 23 g mol$^{-1}$, magnesium = 24.305 g mol$^{-1}$.
For Molecular Compounds: The molar mass equals the sum of atomic masses of all atoms in the molecule. Water ($H_2O$) = $2(1) + 16 = 18$ g mol$^{-1}$. Sulphuric acid ($H_2SO_4$) = $2(1) + 32 + 4(16) = 98$ g mol$^{-1}$.
For Ionic Compounds: The molar mass equals the formula mass — the sum of atomic masses of all ions in the formula unit. NaCl = $23 + 35.5 = 58.5$ g mol$^{-1}$. $Na_2CO_3$ = $2(23) + 12 + 3(16) = 106$ g mol$^{-1}$.
Molar Mass Calculation Examples
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$H_2SO_4$: $2(1.008) + 32.06 + 4(16) = 98.08$ g mol$^{-1}$
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Sucrose ($C_{12}H_{22}O_{11}$): $12(12) + 22(1) + 11(16) = 342$ g mol$^{-1}$
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$MgSO_4$: $24.3 + 32.06 + 4(16) = 120.4$ g mol$^{-1}$
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$Na_2CO_3$: $2(23) + 12 + 3(16) = 106$ g mol$^{-1}$
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$AgNO_3$: $107.87 + 14 + 3(16) = 169.87$ g mol$^{-1}$
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$H_3PO_4$: $3(1.008) + 30.97 + 4(16) = 98.0$ g mol$^{-1}$
The mole formula can be rearranged to find mass when the number of moles is known. This is essential when working with chemical reactions where mole ratios are given but actual masses are needed.
$$m = n \times M$$
Calculates the mass of a substance from its number of moles and molar mass
$m$=Mass of the substance(g)
$n$=Number of moles(mol)
$M$=Molar mass(g mol$^{-1}$)
Molar Volume and Gas Calculations
One mole of any ideal gas at STP (standard temperature = 0 °C or 273 K, standard pressure = 1 atm or 101.325 kPa) occupies a fixed volume of 22.414 dm³. This is called the molar volume. Different gases have vastly different molar masses, yet one mole of each occupies the same volume because in the gaseous state, the distance between molecules is roughly 300 times greater than their diameters — the masses and sizes of individual molecules do not significantly affect the total volume.
Molar Volume at STP: $V_m$ = 22.414 dm³ mol$^{-1}$ at STP. This is true only for ideal gases.
Equal Volume, Different Mass: 22.414 dm³ of $H_2$ (mass = 2.016 g) and 22.414 dm³ of $CH_4$ (mass = 16 g) both contain $6.02 \times 10^{23}$ molecules. The heavier gas has heavier molecules, but the same count in the same volume.
Unit Conversion: 1 dm³ = 1000 cm³ = 1 litre. So 22.414 dm³ = 22414 cm³ = 22.414 L.
Molar Volume Equivalents at STP
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$H_2$: 2.016 g = 1 mol = $6.02 \times 10^{23}$ molecules = 22.414 dm³
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$CH_4$: 16 g = 1 mol = $6.02 \times 10^{23}$ molecules = 22.414 dm³
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$O_2$: 32 g = 1 mol = $6.02 \times 10^{23}$ molecules = 22.414 dm³
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$CO_2$: 44 g = 1 mol = $6.02 \times 10^{23}$ molecules = 22.414 dm³
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$N_2$: 28 g = 1 mol = $6.02 \times 10^{23}$ molecules = 22.414 dm³
The number of moles of a gas at STP can be found directly from its volume using the molar volume relationship.
$$n = \frac{V}{22.414 \text{ dm}^3 \text{ mol}^{-1}}$$
Calculates the number of moles of an ideal gas at STP from its volume
$n$=Number of moles of the gas(mol)
$V$=Volume of the gas at STP(dm³)
$22.414$=Molar volume at STP(dm³ mol$^{-1}$)
Volume to Moles: Divide the gas volume by 22.414 dm³ mol$^{-1}$. A gas occupying 500 cm³ (= 0.5 dm³) at STP contains $0.5 / 22.414 = 0.0223$ moles.
Finding Molar Mass from Volume and Mass: Once moles are found from volume, the molar mass follows from $M = m/n$ where $m$ is the measured mass of the gas sample.
Calculations with Dissociated Substances
When an ionic or molecular compound dissolves and dissociates in water, the resulting ions can be counted using the stoichiometric ratio from the balanced dissociation equation. The number of each type of ion, their individual masses, and the total charges can all be calculated from the original amount of substance.
Ion Count from Moles: From the balanced dissociation equation, multiply the number of moles of the original compound by the coefficient of each ion. For $H_3PO_4 \rightarrow 3H^+ + PO_4^{3-}$, 1 mole of $H_3PO_4$ yields 3 moles of $H^+$ and 1 mole of $PO_4^{3-}$.
Charge Calculation: Multiply the number of each ion by its charge magnitude. 1 mole of $H_3PO_4$ gives $3 \times N_A$ positive charges (from 3 moles of $H^+$, each with charge +1) and $3 \times N_A$ negative charges (from 1 mole of $PO_4^{3-}$ with charge −3). Positive and negative charges are always equal in a dissociated solution.
Mass of Each Ion: The mass of individual ions after dissociation can be found using $m_{\text{ion}} = n_{\text{ion}} \times M_{\text{ion}}$ where $n_{\text{ion}}$ is the number of moles of that ion and $M_{\text{ion}}$ is its ionic mass.
Practical Example — $H_2SO_4$: Dissolving 9.8 g of $H_2SO_4$ (0.1 moles) gives 0.2 moles of $H^+$ and 0.1 moles of $SO_4^{2-}$. The mass of $H^+$ is $0.2 \times 1.008 = 0.202$ g, and the mass of $SO_4^{2-}$ is $0.1 \times 96 = 9.6$ g. Total positive charges = $0.2 \times N_A$, and total negative charges = $0.2 \times N_A$ (each $SO_4^{2-}$ carries 2 negative charges).