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Axioma

Chemistry · Foundational · 30 min · Chapter 24 of 28

Amounts, stoichiometry, and gases

The mole, balancing reactions, concentration, and the gas laws chemists actually compute with.

Read first: Arithmetic and number sense, Algebra

The idea

Chemistry counts particles by the mole because the particles are too small and too numerous to count directly. One mole is Avogadro’s number of things, 6.022×10²³. The molar mass, in grams per mole, is numerically the same as the atomic or molecular mass in unified mass units. So 18.02 g of water is one mole of water, and it contains one mole of oxygen atoms and two moles of hydrogen atoms. That last sentence is stoichiometry already: the subscripts are mole ratios.

A balanced chemical equation equates atoms on both sides, and therefore equates moles of each substance in the ratios of the coefficients. If you are given grams, convert to moles, use the coefficient ratio, and convert back. The reactant that runs out first is the limiting reactant; the others are in excess. Percent yield compares what you isolated to what the limiting reactant could have produced. Theoretical yield is a mole calculation, not a hope.

Concentration in molarity is n/V, with V in liters of solution, not liters of solvent. Dilution at constant amount of solute is M₁V₁ = M₂V₂. Gases add the ideal gas law PV = nRT, with T in kelvin. At STP of 1 bar and 273.15 K, one mole of ideal gas occupies about 22.7 L; the older 22.4 L figure belongs to 1 atm. Dalton’s law says partial pressures add, and each gas behaves as if it occupied the volume alone.

Hold onto this

  1. The mole is a count: 6.022×10²³ particles. Molar mass converts between grams and moles.
  2. A balanced equation is a ratio of moles, not of grams.
  3. Molarity is moles of solute per liter of solution.
  4. The ideal gas law closes the set of gas computations once any three of P, V, n, T are known.

Formulas

Moles from mass

Divide the mass by the molar mass. Keep the units: grams over grams per mole.

m
Mass
M
Molar mass

Molarity

Moles of solute per liter of solution.

V
Volume of the solution, in liters

Dilution

Amount of solute is unchanged when you add solvent.

M
Molarity, not molar mass — a collision of symbols you must watch

Ideal gas law

The working equation for an ideal gas. T in kelvin.

R
0.082057 L·atm/(mol·K), or 8.314 J/(mol·K)

Dalton’s law

Total pressure is the sum of partial pressures. Each partial pressure is the mole fraction times the total.

xᵢ
Mole fraction of gas i

Worked examples

1. From grams to concentration

Dissolve 5.00 g of NaCl (molar mass 58.44 g/mol) in water and dilute to 250.0 mL. What is the molarity?

  1. n = 5.00 / 58.44 = 0.08556 mol.
  2. V = 0.2500 L.
  3. c = 0.08556 / 0.2500 = 0.342 M.

0.342 mol/L.

2. A limiting reactant, in outline

Hydrogen burns as 2 H₂ + O₂ → 2 H₂O. You have 4.0 mol of H₂ and 1.5 mol of O₂. Which runs out, and how much water forms?

  1. The equation wants 2 mol H₂ per 1 mol O₂.
  2. 1.5 mol O₂ would require 3.0 mol H₂. You have 4.0, so O₂ runs out.
  3. Water formed equals twice the O₂ consumed, from the coefficients: 3.0 mol.

Oxygen is limiting. You get 3.0 mol of water, and 1.0 mol of hydrogen is left over.

Try these

What volume does 0.25 mol of ideal gas occupy at 1.00 atm and 298 K? Use R = 0.0821 L·atm/(mol·K).

You dilute 10.0 mL of 2.00 M HCl to 250 mL. What is the new molarity?

Common mistakes

  • Coefficients are mole ratios, not gram ratios. Water is not “2 grams of hydrogen per 16 of oxygen” until you convert.
  • Molarity uses liters of solution. Adding 250 mL of water to a solid is not automatically a 250 mL solution; the solid has volume too. Volumetric flasks exist for this reason.
  • R has several values. If P is in atm and V in liters, use 0.0821. If you want joules, use 8.314 and measure pressure in pascals, volume in cubic meters.