AP Chemistry

Counting particles by weighing

Atoms, molecules and ions are far too small to count one at a time, yet every laboratory measurement of a chemical change has to connect the mass of a sample to the number of particles that react. The link is the mole, a fixed count of particles. One mole of any substance contains the Avogadro constant of particles:

The particles may be atoms, molecules, ions or formula units, so you have to say which you mean. A mole of $\ce{H2O}$ holds $6.022 \times 10^{23}$ molecules, and a mole of $\ce{NaCl}$ holds $6.022 \times 10^{23}$ formula units, which is the same number of $\ce{Na+}$ ions and the same number of $\ce{Cl-}$ ions.

Atomic mass units and molar mass

The mass of a single atom or molecule is expressed in atomic mass units (amu). The useful fact about this unit is that the average mass of one particle in amu is numerically equal to the molar mass of the substance in grams per mole. A water molecule has an average mass of 18.02 amu, so one mole of water has a mass of 18.02 g. This one-to-one correspondence is what turns a balance reading into a count of particles.

The molar mass of a compound is the sum of the molar masses of its atoms, each multiplied by the number of times it appears in the formula. For calcium nitrate, $\ce{Ca(NO3)2}$, the subscript outside the bracket multiplies everything inside it:

Converting between mass, moles and particles

The amount in moles is the mass divided by the molar mass:

where $n$ is the amount in moles, $m$ the mass in grams and $M$ the molar mass in $\mathrm{g\,mol^{-1}}$. The number of particles is $n \times N_A$. A calculation that moves between these quantities is most reliable when it is set out as a chain of conversion factors, so that the units cancel and show whether the route is right.

For 45.0 g of water:

Each molecule contains two hydrogen atoms, which is why the last line doubles the number of molecules. Keeping a few extra digits through the working and rounding only at the end avoids a small error building up, and the answer takes the number of significant figures of the least precise measurement, here three.

The same relationship run backwards gives the mass of a sample from a number of moles: 0.250 mol of calcium nitrate has a mass of $0.250 \times 164.10 = 41.0$ g. The mass of one particle comes from dividing the molar mass by $N_A$, so one water molecule has a mass of $\dfrac{18.02\ \mathrm{g\,mol^{-1}}}{6.022 \times 10^{23}\ \mathrm{mol^{-1}}} = 2.99 \times 10^{-23}$ g.

Isotopes and the mass spectrum of an element

Atoms of one element all have the same number of protons, but they can differ in the number of neutrons. Atoms of the same element with different numbers of neutrons are isotopes, and each has its own mass. The atomic mass printed in the periodic table for an element is therefore not the mass of any single atom. It is the average atomic mass, weighted by how common each isotope is in nature.

Reading a mass spectrum

A mass spectrometer ionizes atoms of a sample, usually to ions with a single positive charge, separates the ions according to their mass and counts how many arrive at the detector. The result is a mass spectrum, a graph with the mass of the ion (in amu) along the horizontal axis and the relative abundance along the vertical axis. For a sample of one element:

• each peak is one isotope, and its horizontal position gives the mass of that isotope
• the height of each peak is proportional to the relative abundance of that isotope, so the tallest peak is the most common isotope
• the number of peaks is the number of isotopes present in the sample

If the heights are given as percentages they add up to 100. If they are given in arbitrary units, each fraction is found by dividing the height by the sum of all the heights.

Average atomic mass from abundances

The average atomic mass is the sum, over all isotopes, of the isotope's mass multiplied by its fractional abundance. Boron has two stable isotopes, of mass 10.013 amu (19.9 % of atoms) and 11.009 amu (80.1 %):

The result lies between the two isotope masses and is closer to the heavier one because that isotope is far more abundant. That check is worth making every time: an average that falls outside the range of the isotope masses, or that sits nearer the less common isotope, contains an error.

The same method works for more than two isotopes. Using whole-number mass numbers as an estimate, magnesium with isotopes of mass 24 (78.99 %), 25 (10.00 %) and 26 (11.01 %) gives $24(0.7899) + 25(0.1000) + 26(0.1101) = 24.32$, close to the accepted value of 24.31.

Working backwards

The calculation can also run the other way. If a two-isotope element has an average atomic mass of 79.90 amu and the isotopes have masses of 79 and 81, the average lies almost exactly halfway between them, so the two isotopes must be present in nearly equal amounts. Writing the abundance of the lighter isotope as $x$ and the heavier as $1 - x$ gives $79x + 81(1 - x) = 79.90$, so $81 - 2x = 79.90$ and $x = 0.55$, which means 55 % of the lighter isotope and 45 % of the heavier one.