Amount of Substance (Mole)

Divide the mass by the molar mass: $n =\frac{ m}{M}$. Insert the mass m in grams and the molar mass M in $\frac{g}{mol}$ from the periodic table, then the amount of substance comes out in moles. Example: 8.0 g of sodium hydroxide NaOH with $M =\frac{ 40.0 g}{mol}$ give $n =\frac{ 8.0}{40.0} = 0.20 mol$. For compounds you first calculate M as the sum of the atomic masses, for NaOH $22.99 + 16.00 + 1.008 \approx{}\frac{ 40.0 g}{mol}$. Make sure mass and molar mass refer to the same particle type: if you want the amount of oxygen gas, use $M(O_{2}) =\frac{ 32.00 g}{mol}$, not $M(O) =\frac{ 16.00 g}{mol}$.

A mole is a counting unit for particles, just as a dozen denotes twelve items. Since the 2019 SI reform, one mole contains exactly $6.02214076\times{}10^{23}$ particles, the Avogadro constant. The trick of this number: one mole of carbon-12 weighs almost exactly 12 g, one mole of water about 18 g. So the molar mass in $\frac{g}{mol}$ has the same numerical value as the atomic or molecular mass in u. This lets you count particles with a balance: weigh out 58.44 g of table salt and you have exactly one mole of NaCl formula units. Chemical reactions proceed in particle ratios, which is why all of stoichiometry calculates in moles.

Because reaction equations describe particle ratios, not mass ratios. In 2 H₂ + O₂ → 2 H₂O two hydrogen molecules react with one oxygen molecule, but by no means two grams with one gram: by mass it is 4 g of hydrogen per 32 g of oxygen. The amount of substance translates the unwieldy particle number into a weighable quantity and makes the coefficients of the equation directly usable: $\frac{n(H_{2})}{n}(O_{2}) =\frac{ 2}{1}$. The typical exam procedure is therefore: convert the given mass into an amount with $n =\frac{ m}{M}$, apply the ratio of coefficients, and finally translate back into the required mass with $m = n\cdot{}M$.

The amount of substance n is the central hub: $m = n\cdot{}M$ with the molar mass M leads to the mass, $N = n\cdot{}N_{A}$ with the Avogadro constant $N_{A} = 6.022\times{}10^{23} mol^{-1}$ leads to the particle number. Example water: 9.0 g correspond to $n =\frac{ 9.0}{18.0} = 0.50 mol$, which is $N = 0.5\cdot{}6.022\times{}10^{23} \approx{} 3.0\times{}10^{23}$ molecules. To go from particle number to mass, always route via the amount of substance: first $n =\frac{ N}{N_{A}}$, then $m = n\cdot{}M$. For gases there is a third bridge: at standard conditions one mole of gas occupies about 22.4 L, so amounts can also be obtained from volumes.

Three classics. First, mixing units: inserting the mass in kilograms while M is in $\frac{g}{mol}$ puts you off by a factor of 1000; always convert m to grams. Second, the wrong particle type: for chlorine gas $M(Cl_{2}) =\frac{ 70.9 g}{mol}$ applies, not $\frac{35.45 g}{mol}$; for salts the complete formula unit counts, including all indices, so $\frac{74.10 g}{mol}$ for Ca(OH)₂. Third, conceptually confusing amount and mass: 1 mol of lead and 1 mol of helium contain the same number of particles but weigh 207 g and 4 g respectively. A quick plausibility check helps: verify the result unit (mol) and estimate the order of magnitude in your head.