Molar and Compositional Quantities

Understanding Amount of Substance

In stoichiometry, we often want to answer questions like: “How many particles are in this sample?” or “How much of each element is in this compound?” Directly counting atoms or molecules is impossible in practice, so chemistry uses special quantities to describe amounts in a practical way.

The central quantity is the amount of substance, measured in moles.

The Mole

The mole (symbol: mol) is the SI unit for the amount of substance. One mole of any substance contains the same number of specified elementary entities (atoms, molecules, ions, electrons, etc.).

By definition:

• 1 mole contains exactly $6.022\,140\,76 \times 10^{23}$ entities

This constant is called the Avogadro constant, usually written as $N_\text{A}$:
$$
N_\text{A} = 6.022 \times 10^{23}\ \text{mol}^{-1}
$$

So:

• 1 mol of H atoms contains $6.022 \times 10^{23}$ hydrogen atoms
• 1 mol of $\text{H}_2$ molecules contains $6.022 \times 10^{23}$ $\text{H}_2$ molecules
• 1 mol of $\text{Na}^+$ ions contains $6.022 \times 10^{23}$ sodium ions

The amount of substance $n$ (in moles) and the number of particles $N$ are related by:
$$
N = n \cdot N_\text{A}
$$
or
$$
n = \frac{N}{N_\text{A}}
$$

These relations connect the microscopic world (counting particles) to the macroscopic world (measuring in grams, liters, etc.).

Molar Mass

To connect moles with mass, we use the molar mass.

The molar mass $M$ of a substance is the mass of 1 mole of that substance:

• Unit: $\text{g mol}^{-1}$ (or $\text{kg mol}^{-1}$ in SI, but grams per mole is standard in chemistry).

The relationship between amount of substance $n$, mass $m$, and molar mass $M$ is:
$$
n = \frac{m}{M}
$$
and equivalently
$$
m = n \cdot M
$$

Determining Molar Mass from the Periodic Table

The molar mass of an element is numerically equal (for most purposes) to its relative atomic mass from the periodic table, but with the unit $\text{g mol}^{-1}$ instead of “u” (atomic mass units).

Examples:

• Hydrogen, H:
  Approximate atomic mass $\approx 1.0\ \text{u}$
  $\Rightarrow M(\text{H}) \approx 1.0\ \text{g mol}^{-1}$
• Carbon, C:
  Approximate atomic mass $\approx 12.0\ \text{u}$
  $\Rightarrow M(\text{C}) \approx 12.0\ \text{g mol}^{-1}$
• Oxygen, O:
  Approximate atomic mass $\approx 16.0\ \text{u}$
  $\Rightarrow M(\text{O}) \approx 16.0\ \text{g mol}^{-1}$

For compounds, the molar mass is the sum of the molar masses of all atoms in the formula.

Example: $\text{H}_2\text{O}$

• 2 H atoms: $2 \times 1.0 = 2.0\ \text{g mol}^{-1}$
• 1 O atom: $16.0\ \text{g mol}^{-1}$
  Total:
  $$
  M(\text{H}_2\text{O}) = 2.0 + 16.0 = 18.0\ \text{g mol}^{-1}
  $$

Example: $\text{NaCl}$

• Na: $\approx 23.0\ \text{g mol}^{-1}$
• Cl: $\approx 35.5\ \text{g mol}^{-1}$
  Total:
  $$
  M(\text{NaCl}) = 23.0 + 35.5 = 58.5\ \text{g mol}^{-1}
  $$

Amount of Substance and Mass in Practice

Working with moles usually involves converting between mass and amount of substance.

Mass → Moles

1. Determine the molar mass $M$ from the formula.
2. Measure or know the mass $m$ of the sample.
3. Calculate:
   $$
   n = \frac{m}{M}
   $$

Moles → Mass

1. Determine $M$.
2. Know $n$.
3. Calculate:
   $$
   m = n \cdot M
   $$

Because 1 mole always corresponds to $N_\text{A}$ particles, you can also include the number of particles:
$$
m \;\xleftrightarrow{\div M}\; n \;\xleftrightarrow{\times N_\text{A}}\; N
$$

Molar Volume of Gases (Overview for Stoichiometry)

For gases, besides mass and moles, volume is often used. At a given temperature and pressure, 1 mole of any ideal gas occupies the same volume – the molar gas volume.

At standard conditions (often taken in introductory courses as $0^\circ\text{C}$ and $1\ \text{bar}$; sometimes definitions vary slightly), the molar volume $V_\text{m}$ is approximately:
$$
V_\text{m} \approx 22.4\ \text{L mol}^{-1}
$$

Thus:

• 1 mol of $\text{O}_2$ gas $\approx 22.4\ \text{L}$ at standard conditions
• 0.5 mol of $\text{N}_2$ gas $\approx 11.2\ \text{L}$ at standard conditions

The relationship between gas volume $V$, amount of substance $n$, and molar volume $V_\text{m}$ (at fixed $T$ and $p$) is:
$$
n = \frac{V}{V_\text{m}}
$$
and
$$
V = n \cdot V_\text{m}
$$

More general gas laws and non‑standard conditions are treated elsewhere; here the key idea is that volume can act as a measure of the amount of gaseous substance via the molar volume.

Compositional Quantities

So far, we have described how much of a substance there is (in moles, mass, or gas volume).
Compositional quantities describe how that amount is distributed among the components of a mixture or among the elements in a compound.

Two common ways to express composition are:

• Absolute composition: gives actual amounts (e.g. grams of NaCl in a solution).
• Relative composition: gives ratios (e.g. mass percent, mole fraction).

In this chapter, we focus on quantitative expressions important for stoichiometric calculations.

Mass Fraction and Mass Percent

The mass fraction of a component is the ratio of its mass to the total mass of the mixture.

For a component A:
$$
w_\text{A} = \frac{m_\text{A}}{m_\text{total}}
$$

• $w_\text{A}$ is dimensionless (often written as a decimal between 0 and 1).
• To express it as a percentage, multiply by 100:
  $$
  w_\text{A}(\%) = \frac{m_\text{A}}{m_\text{total}} \times 100\%
  $$

If a solution contains 5.0 g of NaCl in 95.0 g of water:

• $m_\text{NaCl} = 5.0\ \text{g}$
• $m_\text{total} = 5.0\ \text{g} + 95.0\ \text{g} = 100.0\ \text{g}$
• Mass fraction:
  $$
  w_\text{NaCl} = \frac{5.0}{100.0} = 0.050
  $$
• Mass percent:
  $$
  w_\text{NaCl} = 5.0\%
  $$

For a mixture of several components, the sum of all mass fractions is:
$$
\sum_i w_i = 1
$$
or in percent:
$$
\sum_i w_i = 100\%
$$

Mass Fraction in Compounds

For pure compounds, we can calculate the mass fraction of each element from the molar mass and the chemical formula.

Consider water, $\text{H}_2\text{O}$:

• $M(\text{H}) \approx 1.0\ \text{g mol}^{-1}$
• $M(\text{O}) \approx 16.0\ \text{g mol}^{-1}$
• $M(\text{H}_2\text{O}) = 2 \times 1.0 + 16.0 = 18.0\ \text{g mol}^{-1}$

Mass fraction of hydrogen in water:
$$
w_\text{H} = \frac{2 \times M(\text{H})}{M(\text{H}_2\text{O})} = \frac{2.0}{18.0} \approx 0.111 \quad \Rightarrow \quad 11.1\%
$$
Mass fraction of oxygen:
$$
w_\text{O} = \frac{16.0}{18.0} \approx 0.889 \quad \Rightarrow \quad 88.9\%
$$

The sum is $11.1\% + 88.9\% = 100\%$.

Amount (Mole) Fraction

Instead of mass, we can describe composition using amount of substance (moles).

For component A:
$$
x_\text{A} = \frac{n_\text{A}}{n_\text{total}}
$$

• $x_\text{A}$ is the amount fraction (often called mole fraction).
• It is dimensionless and typically between 0 and 1.

With several components:
$$
\sum_i x_i = 1
$$

Mole fractions are particularly important for:

• Gas mixtures
• Theoretical calculations in thermodynamics and equilibria

Relationship Between Mass Fraction and Mole Fraction

To move between mass and mole-based descriptions of composition, use molar masses.

For a two-component mixture A and B:

• Given $m_\text{A}$ and $m_\text{B}$:
• $n_\text{A} = \dfrac{m_\text{A}}{M_\text{A}}$,
• $n_\text{B} = \dfrac{m_\text{B}}{M_\text{B}}$,
• $n_\text{total} = n_\text{A} + n_\text{B}$
• Mole fraction of A:
  $$
  x_\text{A} = \frac{n_\text{A}}{n_\text{A} + n_\text{B}}
  $$

In the opposite direction, if you know $x_\text{A}$ and $x_\text{B}$ and choose a convenient total amount (for example $n_\text{total} = 1\ \text{mol}$), then:

• $n_\text{A} = x_\text{A}\, n_\text{total}$, etc.
• $m_\text{A} = n_\text{A} M_\text{A}$

This allows conversion between composition by mass and by amount.

Mass Concentration and Amount Concentration (Overview)

In solutions and other mixtures where volume matters, composition is often expressed per unit volume.

Two basic quantities:

• Mass concentration of A:
  $$
  \rho_\text{A} = \frac{m_\text{A}}{V}
  $$
  Units: e.g. $\text{g L}^{-1}$
• Amount concentration (molar concentration) of A, written $c_\text{A}$:
  $$
  c_\text{A} = \frac{n_\text{A}}{V}
  $$
  Units: $\text{mol L}^{-1}$, also written as mol/dm^3 or M (“molar”).

These are especially useful when we later relate composition to chemical reaction rates and equilibria. The detailed use of concentrations in reactions is treated in other chapters; here you need only the idea that:

• Mass concentration links composition to mass.
• Amount concentration links composition to moles.

Relative Atomic and Molecular Quantities

Stoichiometry makes frequent use of relative masses and compositions:

• Relative atomic mass $A_\text{r}$ of an element:
• Dimensionless.
• Compared to $1/12$ of the mass of a carbon-12 atom.
• Relative molecular mass $M_\text{r}$ of a molecule:
• Sum of the relative atomic masses of the atoms in the molecular formula.

These quantities are numerically equal to the corresponding molar masses in $\text{g mol}^{-1}$, but they carry no units.
For example, for water:

• $M_\text{r}(\text{H}_2\text{O}) = 18.0$ (dimensionless)
• $M(\text{H}_2\text{O}) = 18.0\ \text{g mol}^{-1}$

They are mainly used to:

• Quickly estimate mass relationships.
• Derive formula composition (for example, in empirical formula determination from experimental data).

Summary of Key Relationships

For a substance with molar mass $M$:

• Between particles and moles:
  $$
  N = n \cdot N_\text{A}; \quad n = \frac{N}{N_\text{A}}
  $$
• Between mass and moles:
  $$
  n = \frac{m}{M}; \quad m = n \cdot M
  $$
• For gases at fixed $T$ and $p$ (e.g. standard conditions):
  $$
  n = \frac{V}{V_\text{m}}; \quad V = n \cdot V_\text{m}
  $$
• Composition by mass (mass fraction $w_\text{A}$):
  $$
  w_\text{A} = \frac{m_\text{A}}{m_\text{total}}
  $$
• Composition by amount (mole fraction $x_\text{A}$):
  $$
  x_\text{A} = \frac{n_\text{A}}{n_\text{total}}
  $$
• Concentrations in a volume $V$:
  $$
  \rho_\text{A} = \frac{m_\text{A}}{V}; \quad c_\text{A} = \frac{n_\text{A}}{V}
  $$

These molar and compositional quantities form the quantitative language of chemistry. They are the foundation on which stoichiometric calculations with chemical equations are built.