Molar Mass

If every carbon atom has a mass of about 12 u, why is the table value $12.01$ and not exactly 12? Because most elements exist as a mixture of isotopes, atoms with the same number of protons but different numbers of neutrons. The periodic-table value is a weighted average of the isotopes' masses, weighted by how common each isotope is in nature. Chlorine is the classic example: about three-quarters of chlorine atoms are chlorine-35 and one-quarter are chlorine-37, so the average lands at $35.45$, closer to 35. No single chlorine atom has a mass of 35.45 u, but a mole of natural chlorine atoms really does have a mass of $35.45\text{ g}$, and a mole is what you weigh in the lab. Because the natural mix of isotopes is almost the same everywhere on Earth, one average value works for every sample you'll meet in this course.

For a compound, add up the molar mass of every atom in the formula, multiplying each element's atomic mass by its subscript. A subscript placed outside brackets multiplies every atom inside those brackets — for ${\text{Mg(OH)}}_2$, the subscript 2 applies to both the O and the H inside the brackets, giving 1 Mg, 2 O, and 2 H atoms total. This works because a mole of a compound contains a mole of formula units, and each formula unit carries exactly the atoms its formula shows: one mole of ${\text{H}}_2\text{O}$ contains two moles of H atoms and one mole of O atoms, so its mass is $2(1.01)+16.00$ grams.

1. Write each element in the formula once, in a column.
2. Count its atoms, distributing any bracket subscript: in ${\text{Al}}_2{({\text{SO}}_4)}_3$ the O count is $4\times 3=12$.
3. Multiply each count by the atomic mass from the periodic table (2 decimal places: C 12.01, H 1.01, O 16.00).
4. Add the products, round the total to 2 decimal places, and attach the unit $\text{g/mol}$.
5. Sanity check: the heaviest element or the one with the most atoms should dominate the total.

Here is the tally for calcium nitrate, \text{Ca{(NO}_3)}_2, which trips up many students. The bracket holds one nitrate ion, ${\text{NO}}_3^{-}$, and the 2 outside says there are two of them, so the formula unit contains 1 Ca, $1\times 2=2$ N and $3\times 2=6$ O. Then $M=40.08+2(14.01)+6(16.00)=40.08+28.02+96.00=164.10\text{ g/mol}\text{.}$ A student who forgets to distribute the 2 gets $40.08+14.01+48.00=102.09$, an error of almost 40%, which would throw off every calculation built on it. On a test, writing the atom count for each element in a small table before multiplying is the single most effective way to avoid this. Keep atomic masses at 2 decimal places (C 12.01, H 1.01, O 16.00, N 14.01) so your answers match the answer key.

Hydrates follow the same rule, with the centred dot meaning "add": $M({\text{CuSO}}_4\cdot 5{\text{H}}_2\text{O})=M({\text{CuSO}}_4)+5M({\text{H}}_2\text{O})=159.62+90.10=249.72\text{ g/mol}\text{.}$ Molar mass is the gateway to everything that follows in this unit. A pharmacist converting a dose in milligrams to an amount of drug molecules, a farmer comparing how much nitrogen two fertilizers deliver, and a chemist predicting how much product a reaction will make all start the same way: find $M$, then convert between grams and moles.