Lesson 1.1 · Unit 1 · CED topic 1.1
Moles and molar mass
A chemist cannot count atoms. A single drop of water holds more molecules than there are drops of water in all of Earth’s oceans, and no instrument can tally them one at a time. What a chemist can do is weigh a sample on a balance. The whole of quantitative chemistry rests on one idea that connects the two: the mole. Every calculation in this course, from reaction yields to gas volumes to equilibrium concentrations, starts by converting a measured mass into a number of moles. Master this lesson and the rest of the year has a firm floor under it.
Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.
- One cannot count particles directly while performing laboratory work, so there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes.
- Avogadro’s number (NA = 6.022 × 1023 mol−1) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance.
- The average mass in amu of one particle of a substance is numerically equal to the molar mass of that substance in grams. This gives a quantitative connection between the mass of a substance and the number of particles it contains.
The mole is a counting unit
A dozen means 12 of anything. A mole means 6.022 × 1023 of anything. That number is Avogadro’s number, NA, and it is a plain count, exactly like a dozen, only enormously larger. It was chosen so that the numbers work out neatly on a balance, as the next section shows. One mole of marbles would cover the whole Earth to a depth of many kilometers; one mole of atoms is a small, easily weighed pile.
The word “particle” covers whatever the substance is made of. In helium it is an atom; in water it is a molecule; in sodium chloride it is a formula unit, one Na+ and one Cl− taken together, because ionic solids are lattices and have no separate molecules. So “1 mol of NaCl” contains 6.022 × 1023 formula units, which is 6.022 × 1023 sodium ions and 6.022 × 1023 chloride ions.
Molar mass connects mass to moles
The atomic mass unit (amu) is a mass scale built for single atoms. Carbon-12 is defined as exactly 12 amu, and every other atom is measured against it. The periodic table lists the average atomic mass of each element in amu: 12.01 for carbon, 16.00 for oxygen, 1.008 for hydrogen. The remarkable fact, and the reason NA has the value it does, is that the same number is the mass in grams of one mole. One carbon atom averages 12.01 amu, and 1 mol of carbon atoms has a mass of 12.01 g. That mass of one mole is the molar mass, M, with units g/mol.
To find the molar mass of a compound, add the molar masses of every atom in the formula, counting each one as many times as the subscript says. For water, H2O: 2(1.008) + 15.999 = 18.015 g/mol. For glucose, C6H12O6: 6(12.011) + 12(1.008) + 6(15.999) = 180.156 g/mol. For a compound with parentheses, such as (NH4)2SO4, the subscript outside multiplies everything inside: 2 N, 8 H, 1 S and 4 O.
\[ n = \dfrac{m}{M} \] where \(n\) is the amount in moles, \(m\) is the mass in grams and \(M\) is the molar mass in g/mol. Rearranged, \(m = nM\). To convert between moles and particles, multiply or divide by \(N_A\): number of particles \(= n \times 6.022 \times 10^{23}\).
Dimensional analysis
Dimensional analysis means writing each conversion as a fraction equal to 1 and arranging the fractions so that the unwanted units cancel. The two relationships in this lesson give two such fractions: \(\dfrac{1\ \text{mol}}{M\ \text{g}}\) and \(\dfrac{6.022\times10^{23}\ \text{particles}}{1\ \text{mol}}\). The chain for “grams to particles” is always grams, then moles, then particles, never a shortcut. Writing the units and canceling them catches most errors before they happen: if the units on your answer are not the ones you wanted, the setup is wrong.
How many moles are in 25.0 g of ammonium sulfate, (NH4)2SO4?
Step 1, molar mass. 2(14.007) + 8(1.008) + 32.06 + 4(15.999) = 28.014 + 8.064 + 32.06 + 63.996 = 132.134 g/mol.
Step 2, convert. \(n = \dfrac{25.0\ \text{g}}{132.134\ \text{g/mol}} = 0.189\ \text{mol}\).
Check. 25 g is a little less than 132 g divided by 5, so the answer should be a little under 0.2 mol. It is.
How many oxygen atoms are in 10.0 g of carbon dioxide?
Step 1. M(CO2) = 12.011 + 2(15.999) = 44.009 g/mol.
Step 2, molecules. \(\dfrac{10.0\ \text{g}}{44.009\ \text{g/mol}} \times 6.022\times10^{23}\ \text{mol}^{-1} = 1.37\times10^{23}\) molecules.
Step 3, atoms. Each molecule has 2 O atoms, so \(2 \times 1.37\times10^{23} = 2.74\times10^{23}\) oxygen atoms.
A sample of sodium chloride contains \(3.50\times10^{22}\) formula units. What is its mass?
\(n = \dfrac{3.50\times10^{22}}{6.022\times10^{23}\ \text{mol}^{-1}} = 0.0581\ \text{mol}\). With M(NaCl) = 22.990 + 35.45 = 58.440 g/mol, \(m = nM = 0.0581 \times 58.440 = 3.40\ \text{g}\).
Notice the size of a single particle. One water molecule has a mass of \(\dfrac{18.015\ \text{g/mol}}{6.022\times10^{23}\ \text{mol}^{-1}} = 2.99\times10^{-23}\) g. A balance that reads to 0.001 g cannot see anything smaller than about \(10^{19}\) molecules, which is exactly why chemists count by the mole.
- Count the atoms in the formula. Many missed points come from using the molar mass of the compound when the question asks about one element inside it (or the reverse).
- Grams, then moles, then particles. Mass never converts straight to a particle count; it always passes through moles.
- Significant figures. Use the molar mass to at least one more digit than the data, and report the answer with the same number of significant figures as the least precise measurement.
- Show the units. On free-response questions the setup with units is often what earns the point, even if the arithmetic slips.
- Formula units are particles too. Say “formula units” for an ionic compound, not “molecules.”
- Calculate the molar mass of C12H22O11 (sucrose) and the number of moles in 5.00 g.
Show answer
M = 12(12.011) + 22(1.008) + 11(15.999) = 144.132 + 22.176 + 175.989 = 342.297 g/mol. \(n = 5.00 / 342.297 = 0.0146\) mol.
- How many copper atoms are in a 1.00 g copper wire? (M = 63.546 g/mol)
Show answer
\(n = 1.00 / 63.546 = 0.01574\) mol. Atoms \(= 0.01574 \times 6.022\times10^{23} = 9.48\times10^{21}\) atoms.
- What is the mass, in grams, of one atom of copper?
Show answer
One mole has a mass of 63.546 g and holds \(6.022\times10^{23}\) atoms, so one atom has a mass of \(63.546 / 6.022\times10^{23} = 1.06\times10^{-22}\) g.
- Explain why the molar mass of carbon in g/mol has the same numerical value as the average mass of one carbon atom in amu.
Show answer
The mole is defined so that a mole of carbon-12 atoms has a mass of exactly 12 g, and one carbon-12 atom has a mass of exactly 12 amu. Avogadro’s number is the conversion factor between grams and amu, so for every substance the average mass of one particle in amu equals the mass of a mole in grams.
- A student says that 18 g of water and 18 g of glucose contain the same number of molecules because they have the same mass. Evaluate the claim.
Show answer
The claim is incorrect. Water has M = 18.015 g/mol, so 18 g is about 1.0 mol. Glucose has M = 180.156 g/mol, so 18 g is about 0.10 mol. The water sample has about ten times as many molecules, because each glucose molecule is about ten times heavier. Equal numbers of particles require equal numbers of moles, not equal masses.
- A sample of an ionic compound contains 0.250 mol of calcium chloride, CaCl2. How many chloride ions does it contain?
Show answer
Each formula unit has 2 Cl−, so there are 0.500 mol of chloride ions. \(0.500 \times 6.022\times10^{23} = 3.01\times10^{23}\) chloride ions.