Source: Chemistry by Overby, Chapters 3.1, 3.2, 3.5, 4.1, 4.5
Difficulty: Introductory to Intermediate | Prerequisites: Periodic table, basic algebra, understanding of isotopes and atomic structure.
This is where chemistry becomes quantitative. You already know the names and structures of compounds; now you need to count atoms and weigh them. The mole is the bridge between the atomic scale (individual atoms and molecules) and the laboratory scale (grams you can measure on a balance). Everything in stoichiometry, solution chemistry, and later topics depends on converting fluently between moles, mass, and number of particles. Chapter 4 adds concentration (molarity) to the toolkit, so you can describe how much solute is dissolved in a given volume of solution. If you are uncomfortable with dimensional analysis or unit conversion, brush up on that first.
Atomic masses on the periodic table are weighted averages based on isotopic abundance. One mole of anything contains 6.022 x 10^23 particles (Avogadro's number). Molar mass (g/mol) lets you convert between grams and moles. Percent composition tells you what fraction of a compound's mass comes from each element. Molarity (mol/L) measures solution concentration.
Atomic mass unit (amu)
The unit used to express the mass of individual atoms and isotopes. One amu is defined as 1/12 the mass of a carbon-12 atom. Think of it as the atomic-scale equivalent of grams.
Average atomic mass
The weighted average of the masses of all naturally occurring isotopes of an element, taking into account their relative abundances. This is the number on the periodic table. In simple terms, it is the mass you would expect if you picked one atom at random from a natural sample.
Mole (mol)
The SI unit for amount of substance. One mole contains exactly 6.022 x 10^23 particles (atoms, molecules, ions, or formula units). Think of it as the chemist's counting unit, the way a "dozen" means 12.
Avogadro's number (N_A)
6.022 x 10^23 per mole. The number of particles in one mole of any substance.
Molar mass
The mass of one mole of a substance, expressed in grams per mole (g/mol). For an element, it equals the average atomic mass in amu but with the unit g/mol. For a compound, add up the molar masses of all atoms in the formula.
Percent composition
The percentage by mass of each element in a compound. Calculated as (mass of element in one mole of compound / molar mass of compound) x 100%.
Solute
The substance that is dissolved in a solution. In simple terms, it is the minor component (e.g. salt in salt water).
Solvent
The substance that does the dissolving. In simple terms, it is the major component (e.g. water in salt water).
Molarity (M)
The concentration of a solution expressed as moles of solute per litre of solution. In simple terms, it tells you how many moles are packed into each litre.
Dilute vs. concentrated
A dilute solution has a relatively small amount of solute per unit volume. A concentrated solution has a relatively large amount. These are qualitative terms (unlike molarity, which is quantitative).
Formula unit
The simplest ratio of ions in an ionic compound. Used instead of "molecule" for ionic compounds because ionic compounds do not exist as discrete molecules.
Individual isotopes have exact masses measured in amu. Carbon-12 is defined as exactly 12 amu.
Most elements exist as a mixture of isotopes in nature. The periodic table mass is the weighted average of these isotope masses.
Multiply each isotope's mass by its fractional abundance (percent abundance / 100), then add the results.
Example: Chlorine has two isotopes.
Cl-35: mass = 34.969 amu, abundance = 75.77%
Cl-37: mass = 36.966 amu, abundance = 24.23%
Average = (34.969 x 0.7577) + (36.966 x 0.2423) = 35.45 amu
This matches the value on the periodic table.
One mole = 6.022 x 10^23 particles. This number connects the atomic scale to the gram scale.
The mole applies to anything: atoms, molecules, ions, formula units, even golf balls (though you would never have a mole of those in practice).
For an element: the molar mass in g/mol equals the average atomic mass in amu from the periodic table.
Carbon: 12.01 g/mol
Oxygen: 16.00 g/mol
For diatomic elements (H2, N2, O2, F2, Cl2, Br2, I2): remember to double the atomic mass. O2 has a molar mass of 32.00 g/mol, not 16.00.
For a compound: add the molar masses of all atoms in the formula.
H2O: 2(1.008) + 16.00 = 18.02 g/mol
Ca(NO3)2: 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol
Three quantities linked by two conversion factors:
Mass (g) <-> Moles: divide by molar mass to go from grams to moles; multiply to go the other way.
Moles <-> Number of particles: multiply by Avogadro's number to go from moles to particles; divide to go back.
Mass <-> Particles: go through moles as the intermediate step. There is no direct shortcut.
One mole of H2O contains 2 moles of H atoms and 1 mole of O atoms.
One formula unit of Ca(NO3)2 contains 1 Ca2+ ion, 2 NO3- ions, 2 N atoms, and 6 O atoms. Scale these by Avogadro's number for mole-level quantities.
From masses: % element = (mass of element / mass of compound sample) x 100%
From molar mass: % element = (total mass of that element in one mole of compound / molar mass of compound) x 100%
Example: What is the percent composition of oxygen in H2O?
Molar mass of H2O = 18.02 g/mol
Mass of O per mole = 16.00 g
% O = (16.00 / 18.02) x 100% = 88.79%
You can also work backwards: given the mass of a compound and its percent composition, find the mass of a specific element.
Example: How many grams of oxygen are in 50.0 g of H2O?
Mass of O = 50.0 g x 0.8879 = 44.4 g
An aqueous solution has water as the solvent.
The solute is the dissolved substance (usually in smaller amount). The solvent does the dissolving (usually in larger amount).
Molarity (M) = moles of solute / litres of solution
Units: mol/L (read as "molar")
"Dilute" means low molarity; "concentrated" means high molarity. These are relative, qualitative terms.
Example: What is the molarity of a solution made by dissolving 5.85 g of NaCl in enough water to make 500.0 mL of solution?
Molar mass of NaCl = 58.44 g/mol
Moles of NaCl = 5.85 / 58.44 = 0.1001 mol
Volume = 500.0 mL = 0.5000 L
M = 0.1001 / 0.5000 = 0.200 M
Average atomic mass: Avg mass = sum of (isotope mass x fractional abundance) for all isotopes
Moles from mass: n = m / M where n = moles, m = mass in grams, M = molar mass in g/mol
Number of particles from moles: N = n x N_A where N = number of particles, N_A = 6.022 x 10^23 mol^-1
Molar mass of a compound: M_compound = sum of (number of atoms of each element x molar mass of that element)
Percent composition: % element = (n x M_element / M_compound) x 100% where n = number of atoms of that element in the formula
Molarity: M = mol solute / L solution
Rearranged for moles: mol = M x L
Rearranged for volume: L = mol / M
Molar mass and percent composition are the basis of pharmaceutical dosing: knowing exactly how many milligrams of active ingredient are in a tablet requires the same mole-to-mass conversions you are learning here. Molarity is how every chemistry lab, hospital, and water-treatment plant describes solution strength. When a doctor orders a 0.9% saline IV drip, the pharmacist uses molarity-style calculations to prepare it correctly.
Using the atomic mass of a single isotope instead of the average. The periodic table gives the weighted average. Do not use the mass number (whole number) from a specific isotope unless the problem tells you to.
Forgetting diatomic elements when calculating molar mass. Oxygen gas is O2 (32.00 g/mol), not O (16.00 g/mol). If the problem says "oxygen gas," use O2.
Confusing mass and moles. 18 grams of water is 1 mole of water, but 18 moles of water is roughly 324 grams. Always check your units.
Using mass of the solute as the volume of the solution for molarity. Molarity uses litres of total solution, not the mass or volume of the solute alone.
⚠️ Mole conversions (mass to moles, moles to particles, and the reverse) are the most heavily tested skill on this exam. Practise until dimensional analysis is reflexive.
⚠️ Average atomic mass calculations with isotopic abundance data appear frequently. Set up the weighted average carefully.
⚠️ Molar mass of compounds: watch for polyatomic ions with subscripts outside parentheses. Students often forget to multiply all atoms inside the parentheses.
⚠️ Percent composition can be tested in both directions: given a formula, find percentages, or given a mass and a percentage, find the mass of one element.
⚠️ Molarity problems will likely require you to convert between mass, moles, volume, and concentration. Show units at every step.
Fill in the blank: One mole of any substance contains ______ particles. (6.022 x 10^23)
True or false: The molar mass of N2 is 14.01 g/mol. (False. N2 = 28.02 g/mol.)
Fill in the blank: Molarity is defined as moles of solute divided by ______ of solution. (litres)
True or false: The mass number listed on the periodic table for chlorine is the mass of the most common isotope. (False. It is the weighted average of all isotopes.)
Fill in the blank: In the formula Ca(OH)2, there are ______ oxygen atoms and ______ hydrogen atoms per formula unit. (2 oxygen, 2 hydrogen)
Q: An element has two isotopes. Isotope A has mass 10.013 amu and 19.9% abundance. Isotope B has mass 11.009 amu and 80.1% abundance. What is the average atomic mass?
A: (10.013 x 0.199) + (11.009 x 0.801) = 1.993 + 8.818 = 10.81 amu. (This is boron.)
Q: How many moles are in 36.04 g of water (H2O)?
A: Molar mass of H2O = 18.02 g/mol. Moles = 36.04 / 18.02 = 2.000 mol.
Q: How many molecules are in 2.00 moles of H2O?
A: 2.00 mol x 6.022 x 10^23 = 1.20 x 10^24 molecules.
Q: What is the percent composition of nitrogen in NH3?
A: Molar mass of NH3 = 14.01 + 3(1.008) = 17.03 g/mol. % N = (14.01 / 17.03) x 100% = 82.27%.
Q: What is the molarity of a solution containing 0.50 mol of NaOH in 250 mL of solution?
A: Convert volume: 250 mL = 0.250 L. M = 0.50 / 0.250 = 2.0 M.
Q: How many grams of glucose (C6H12O6, molar mass = 180.16 g/mol) are needed to prepare 500 mL of a 0.10 M solution?
A: Moles needed = 0.10 mol/L x 0.500 L = 0.050 mol. Mass = 0.050 x 180.16 = 9.0 g.
Mole conversions are the foundation for stoichiometry in later chapters: every balanced equation is interpreted in moles, so if you cannot convert between grams and moles, you cannot solve reaction problems. Molarity links to Chapter 4's broader treatment of reactions in aqueous solutions, including dilution, precipitation, and acid-base reactions. Percent composition connects back to Chapter 2 naming, because you need the correct formula to compute it, and forward to empirical and molecular formula determination.
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