Particles and amount of substance

Two samples each contain one mole, yet one is noticeably heavier. What do they have in common? We will learn to count particles by measuring mass, without counting each molecule separately.

Ten steel balls have more mass than ten plastic balls, although their counts are equal. For atoms and molecules, it is also useful to define a large counting unit. The mole is a unit of amount of substance containing a fixed number of specified particles. First state what is being counted: atoms, molecules, ions or other entities. One mole of oxygen molecules contains two moles of oxygen atoms because each molecule contains two atoms.

We denote amount of substance by \nu, measured in moles. By definition, one mole contains exactly 6{,}02214076\cdot10^{23} specified particles. The Avogadro constant N_A equals this number per mole; school calculations usually use 6{,}02\cdot10^{23}\,\text{моль}^{-1}. This defines the size of a counting unit, not a particle's mass. Half a mole contains half as many particles, whether they are light or heavy.

Weigh one mole of specified particles: its mass is the molar mass M, expressed in kg/mol. For identical particles, dividing this mass by the particle count gives the mass of one particle, m_0=M/N_A, in kg. Heavier particles have a larger molar mass. For a natural mixture of isotopes, tabulated M represents the average composition, and the calculated m_0 is the mean mass of the specified particle in the sample.

Molecular masses in kilograms are inconveniently small. We compare them with the atomic mass unit: one twelfth of the mass of a carbon-12 atom. The ratio of a molecule's mass to this unit is its relative molecular mass M_r, a dimensionless quantity. At school-level accuracy, add the relative masses of its atoms. For water, two hydrogens and one oxygen give approximately 2\times1+16=18. Here, 18 means neither 18 kg nor 18 mol.

How does this relative scale connect to a portion we can weigh? At school-level accuracy, the numerical value of molar mass in g/mol equals M_r: M\approx M_r\,\text{г/моль}. For water, M_r\approx18, so M\approx18 g/mol = 0.018 kg/mol. This relates different quantities with an explicit unit and approximation; it does not identify them. Using 18 instead of 0.018 when the sample mass is in kilograms would cause a thousandfold error.

A sample has mass m and contains N particles. How many moles does it contain? Using mass, divide the sample mass by molar mass: \nu=m/M. Using particle count, divide by the number per mole: \nu=N/N_A. Together, \nu=m/M=N/N_A. The mass units in m and M must match. Doubling the mass of the same substance doubles its amount in moles.

We can now count invisible particles using a balance: N=\nu N_A=(m/M)N_A. The count N has no physical unit. For the same substance, doubling the mass doubles N. At fixed mass, a greater molar mass means fewer particles. As with balls, a fixed mass contains fewer heavy balls. A sample of zero mass gives zero particles, as expected.

Particle count alone does not tell us how closely particles are packed. Divide the count by the occupied volume: number density n=N/V, in m⁻³ when V is in m³. Compressing a sealed sample to half its volume without chemical change leaves N unchanged and doubles n. Doubling both particle count and volume leaves number density unchanged. Distinguish N, the count; \nu, the amount in moles; and n, the count per unit volume.

What can we simplify in a sufficiently dilute gas? Particles are far apart compared with their own size and spend most of their time moving between collisions. The ideal-gas model neglects their own volume and interactions outside elastic collisions. Strong compression and approaching condensation make these assumptions less accurate. The model retains particle count and motion, which we will use to explain pressure and temperature.

A sample contains 0.036 kg of water. Its molar mass is 0.018 kg/mol. How many molecules does it contain? Take NA≈6.02·10²³ mol⁻¹.

First find the amount of substance from its mass: ν=m/M.

ν=0,036/0,018=2 моль.

The molecule count is N=νNA=2×6.02·10²³=1.204·10²⁴.

Check: two moles must contain twice as many molecules as one mole.

The sample contains approximately 1.20·10²⁴ water molecules. A mass of 36 g contains two portions of 18 g, or two moles. Its atom count is three times larger, since each water molecule has three atoms.

The mole counts specified particles; molar mass connects that count to a balance reading. Equal amounts in moles do not imply equal masses.

The chain m → ν → N distinguishes sample mass, amount of substance and particle count. For m/M, the mass units must match.

Compression without leakage increases number density n = N/V without creating particles. The ideal-gas model requires a sufficiently dilute gas.

Next, particle motion will explain gas pressure and temperature.