SJChO Notes · Chapter 4
4

Chemical Formulae,
Equations, the Mole and Gases

Singapore Junior Chemistry Olympiad

Atoms are far too small to count one by one, yet chemists need to know exactly how much of each substance reacts. This chapter writes reactions as balanced equations, counts particles in moles, and uses moles to work out masses, concentrations and gas volumes.

Learning objectives

By the end of this chapter, you can:

Foundation O Level
  • write balanced equations with state symbols, and ionic equations
  • calculate Ar, Mr, moles, masses and numbers of particles
  • find empirical and molecular formulae
  • use mole ratios for reacting masses, limiting reagents, percentage yield and purity
  • use concentrations and the molar gas volume
Core A Level
  • calculate Ar from isotopic abundances
  • find a formula from combustion data
  • use pV = nRT, including Mr and density of a gas
  • use Dalton's law of partial pressures
  • state the ideal gas assumptions and explain when real gases deviate
Prerequisites Chapter 1 (particles, diffusion, states of matter), Chapter 2 (isotopes and the mole as a counting unit) and Chapter 3 (ions, ionic formulae and intermolecular forces). Fill in the blanks as the lesson goes.

4.1Chemical equations

A chemical equation shows the reactants on the left and the products on the right. Atoms are not created or destroyed in a reaction, so a balanced equation has the same number of each kind of atom on both sides. Balance an equation by changing the numbers in front of the formulae, never the formulae themselves.

Worked exampleBalance the equation for burning methane: CH4 + O2 → CO2 + H2O.

1. C: 1 on each side already.
Start with an element that appears in one formula on each side.
2. H: 4 on the left, so 2H2O on the right.
CH4 + O2 → CO2 + 2H2O
3. O: 2 + 2 = 4 on the right, so 2O2: CH4 + 2O2 → CO2 + 2H2O
Balance the free element last.

State symbols show the state of each substance: (s) solid, (l) liquid, (g) gas and (aq) aqueous, dissolved in water.

4.1.1Ionic equations

An ionic compound dissolved in water is present as separate ions. An ionic equation shows only the particles that change. Ions that are in the same state on both sides take no part, and are called spectator ions.

  1. Write the balanced equation with state symbols.
  2. Split every aqueous ionic compound into its ions. Solids, liquids, gases and covalent molecules stay whole.
  3. Cross out the ions that appear unchanged on both sides.

Worked exampleWrite the ionic equation for AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq).

1. Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)
AgCl is a solid, so it stays whole.
2. Na+ and NO3− are spectators.
Unchanged on both sides.
Ag+(aq) + Cl−(aq) → AgCl(s)
Charges also balance: 0 on each side.

A metal placed in a solution of another metal's ions works the same way: Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) becomes Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s). An ionic equation must balance in charge as well as atoms.

Check your understanding
  1. Balance: Al + O2 → Al2O3.
  2. Write the ionic equation for BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq).

4.2Relative masses

Atoms are compared with the carbon-12 atom, which is given a mass of exactly 12. Relative masses have no units.

DefinitionThe relative atomic mass, Ar, of an element is the average mass of one atom of the element compared with 1/12 of the mass of one atom of carbon-12. The relative molecular mass, Mr, is the same comparison for a molecule. For an ionic compound it is called the relative formula mass, also written Mr.

Mr is the sum of the Ar of all the atoms in the formula. Values of Ar are given in the Periodic Table: for CO2, Mr = 12.0 + 2(16.0) = 44.0. For Ca(NO3)2, Mr = 40.1 + 2(14.0 + 3 × 16.0) = 164.1.

The percentage by mass of an element in a compound is the total Ar of that element in the formula divided by Mr, × 100%. In CO2, carbon is 12.0 ÷ 44.0 × 100% = 27.3% by mass.

4.2.1Ar from isotopes Core · A Level

Ar is rarely a whole number because it averages the isotopes, weighted by how common each is (Figure 4.1).

Ar = Σ (percentage abundance × isotopic mass)100
25 × 35Cl25 × 35Cl25 × 35Cl25 × 37Cl353637balance point at 35.5: it balances ✓
Figure 4.1 A weighted average: balanced at 36 the plank tips, as three in four chlorine atoms are chlorine-35; it balances at 35.5.

Worked exampleChlorine is 75.77% 35Cl and 24.23% 37Cl. Calculate Ar(Cl).

Ar = (75.77 × 35 + 24.23 × 37) ÷ 100 = 35.5
Weight each isotope by its abundance.
Check your understanding
  1. Calculate Mr of (NH4)2SO4. (Ar: H 1.0, N 14.0, O 16.0, S 32.1)
  2. Copper is 69% 63Cu and 31% 65Cu. Calculate Ar(Cu).

4.3The mole

DefinitionA mole is the amount of substance containing 6.02 × 1023 particles. This number is the Avogadro constant, L. The molar mass, M, is the mass of one mole of a substance in g mol−1. It is numerically equal to Ar or Mr.
Why the mole mattersAtoms are far too small to count or weigh one at a time, but a balance measures grams. The mole is the bridge between the two. A proton and a neutron each have a mass of about 1 on the relative scale. Chemists asked how many of these units it takes to make 1 gram, a mass we can actually weigh. The answer is 6.02 × 1023: that many protons weigh about 1 g (the exact definition uses carbon-12, so 6.02 × 1023 carbon-12 atoms weigh exactly 12 g).

We already know how heavy every atom is in these units: its Ar. So a mole of any substance weighs its Ar or Mr in grams. One mole of carbon-12 atoms has a mass of exactly 12 g, one mole of oxygen atoms 16.0 g, and one mole of water molecules (Mr 18.0) 18.0 g. Each of these samples contains the same number of particles, 6.02 × 1023. Weighing out 18.0 g of water is therefore a way of counting out 6.02 × 1023 molecules. Always say which particle you are counting: 1 mol of O2 molecules contains 2 mol of O atoms.

amount (mol) n = mass (g)molar mass (g mol−1)     number of particles N = n × L
amountn / molmass / gm = n × M× M÷ Mno. of particlesN = n × L× L÷ Lgas volume / dm³V = n × 24 (r.t.p.)× 24÷ 24concentrationc = n ÷ V(dm³)÷ V× V
Figure 4.2 Every calculation goes through the amount in moles.

Worked exampleHow many molecules, and how many atoms, are in 4.40 g of carbon dioxide?

1. n = 4.40 ÷ 44.0 = 0.100 mol CO2
Mass to moles.
2. Molecules = 0.100 × 6.02 × 1023 = 6.02 × 1022
Moles to particles.
3. Atoms = 3 × 6.02 × 1022 = 1.81 × 1023
Each CO2 has 3 atoms.
Exam tipKeep 4 significant figures in working and give the answer to 3 significant figures unless told otherwise. A value read from the Periodic Table, such as Ar(O) = 16.0, has 3.
Check your understanding
  1. What is the mass of 0.250 mol of sodium hydroxide, NaOH? (Ar: Na 23.0, O 16.0, H 1.0)
  2. How many chloride ions are in 0.10 mol of CaCl2?

4.4Empirical and molecular formulae

DefinitionThe empirical formula shows the simplest whole-number ratio of the atoms in a compound. The molecular formula shows the actual number of each atom in one molecule.

Glucose has the molecular formula C6H12O6 and the empirical formula CH2O. The molecular formula is a whole-number multiple of the empirical formula: multiple = Mr ÷ empirical formula mass.

Worked exampleCompound X (Mr 114) is 63.2% C, 12.2% H and 24.6% N by mass. Find its molecular formula.

CHN
mass in 100 g / g63.212.224.6
amount / mol (÷ Ar)5.26712.201.757
ratio (÷ 1.757)371
Empirical C3H7N (mass 57.0). 114 ÷ 57.0 = 2, so C6H14N2.
Take 100 g so each percentage becomes a mass.

4.4.1Formulae from combustion Core · A Level

When a compound of carbon, hydrogen and oxygen burns completely in oxygen, all its carbon ends up in CO2 and all its hydrogen in H2O (Figure 4.3). The oxygen cannot be found this way, because some of the oxygen in the products came from the O2. Find it by difference.

sample+ O₂CO₂H₂Oall the Call the H
Figure 4.3 Where the carbon and hydrogen go on combustion.

Worked example2.36 g of a sugar (C, H and O only, Mr 180) burns completely to give 3.46 g of CO2 and 1.42 g of H2O. Find its molecular formula.

1. C: 12/44 × 3.46 = 0.9436 g. H: 2/18 × 1.42 = 0.1578 g.
12 g of C in every 44 g of CO2, 2 g of H in every 18 g of H2O.
2. O: 2.36 − 0.9436 − 0.1578 = 1.259 g
By difference.
3. Moles: C 0.07863, H 0.1578, O 0.07869, ratio 1 : 2 : 1
Empirical formula CH2O (mass 30).
4. 180 ÷ 30 = 6, so C6H12O6.
Multiply up.
Check your understanding
  1. A compound is 85.7% C and 14.3% H by mass, and Mr is 56. Find its molecular formula.

4.5Reacting masses

The numbers in a balanced equation give the mole ratio of the substances. 2H2 + O2 → 2H2O means 2 mol of H2 react with 1 mol of O2 to give 2 mol of H2O. To find a reacting mass: mass to moles, moles across by the ratio, then moles back to mass.

Worked exampleWhat mass of magnesium oxide forms when 6.0 g of magnesium burns completely? 2Mg + O2 → 2MgO (Ar: Mg 24.0, O 16.0)

1. nMg = 6.0 ÷ 24.0 = 0.25 mol
Mass to moles.
2. nMgO = 0.25 mol
Ratio 2 : 2.
3. mass = 0.25 × 40.0 = 10 g
Moles to mass.

4.5.1Limiting reagent and yield

The limiting reagent is the reactant that is used up first. It fixes the greatest amount of product that can form, the theoretical yield. The other reactant is in excess (Figure 4.4).

recipe: 2 bread + 1 cheese → 1 sandwich10 breadExcess: 2 bread4 cheeseall used up: the limiting reagentsandwiches made4
Figure 4.4 The cheese limits the number of sandwiches. The bread is in excess.
percentage yield = actual yieldtheoretical yield × 100%     percentage purity = mass of pure substancemass of impure sample × 100%

Worked example2.4 g of magnesium is heated with 0.80 g of oxygen. 2Mg + O2 → 2MgO. 1.6 g of MgO is collected. Find the limiting reagent and the percentage yield.

1. nMg = 2.4 ÷ 24.0 = 0.10 mol. nO₂ = 0.80 ÷ 32.0 = 0.025 mol.
Convert both to moles.
2. 0.10 mol Mg needs 0.050 mol O2, but only 0.025 mol is present, so O2 is limiting.
Compare with the ratio 2 : 1.
3. nMgO = 2 × 0.025 = 0.050 mol, theoretical yield = 0.050 × 40.0 = 2.0 g
Work from the limiting reagent.
4. % yield = 1.6 ÷ 2.0 × 100% = 80%
Actual ÷ theoretical.
Check your understanding
  1. Calcium carbonate decomposes on heating: CaCO3 → CaO + CO2. What mass of CaO forms from 25.0 g of CaCO3? (Ar: Ca 40.1, C 12.0, O 16.0)
  2. A 5.00 g sample of impure CaCO3 gives 1.76 g of CO2. What is its percentage purity?

4.6Solutions

DefinitionThe concentration of a solution is the amount of solute in 1 dm3 of solution, in mol dm−3, or the mass of solute in 1 dm3, in g dm−3.
concentration (mol dm−3) = amount (mol)volume (dm3)     1 dm3 = 1000 cm3

To make a solution of known concentration, weigh the solute, dissolve it in a little water, transfer it to a volumetric flask and add water up to the mark (Figure 4.5). Adding water to a solution dilutes it: the amount of solute stays the same, so c1V1 = c2V2.

0.00 g1weigh2dissolvemark3make up to the mark, invert
Figure 4.5 Making a solution of known concentration.

Worked example4.00 g of sodium hydroxide (Mr 40.0) is dissolved to make 250 cm3 of solution. Find its concentration. What volume of water must be added to 50.0 cm3 of it to make a 0.100 mol dm−3 solution?

1. n = 4.00 ÷ 40.0 = 0.100 mol, c = 0.100 ÷ 0.250 = 0.400 mol dm−3
Convert 250 cm3 to 0.250 dm3.
2. 0.400 × 50.0 = 0.100 × V2, so V2 = 200 cm3
c1V1 = c2V2
3. Water to add = 200 − 50.0 = 150 cm3
V2 is the final total volume.
Check your understanding
  1. How many moles of solute are in 25.0 cm3 of 0.200 mol dm−3 solution?
  2. Express 0.500 mol dm−3 sodium chloride (Mr 58.5) in g dm−3.

4.7Gas volumes

DefinitionAvogadro's law: at the same temperature and pressure, equal volumes of all gases contain the same number of particles. So one mole of any gas occupies the same volume, the molar gas volume: 24 dm3 at room temperature and pressure (r.t.p.).
  • volume of gas (dm3) = amount (mol) × 24 at r.t.p. At standard conditions (s.t.p., 0 °C and 1 bar) one mole occupies 22.7 dm3. Use the value for the conditions the question gives.
  • For gases at the same temperature and pressure, the volume ratio equals the mole ratio. Only gases count: a liquid such as water at r.t.p. takes up almost no volume.

Worked exampleWhat volume of hydrogen at r.t.p. forms when 0.230 g of sodium reacts with excess water? 2Na + 2H2O → 2NaOH + H2

1. nNa = 0.230 ÷ 23.0 = 0.0100 mol
Water is in excess, so Na is limiting.
2. nH₂ = 0.00500 mol
Ratio 2 : 1.
3. V = 0.00500 × 24 = 0.120 dm3 = 120 cm3
Molar gas volume at r.t.p.

4.7.1Burning gaseous hydrocarbons

A hydrocarbon CxHy burns completely to carbon dioxide and water: CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O. After cooling to room temperature the water is liquid, so the gas volume shrinks. Aqueous sodium hydroxide absorbs carbon dioxide, so passing the gas through it removes the CO2 and shows its volume (Figure 4.6).

O₂ addedCₓHᵧO₂ leftCO₂O₂ leftbeforeafter combustion, cooledafter excess NaOHcontractionCO₂ absorbed
Figure 4.6 Following the gas volumes when a hydrocarbon burns. Bar lengths are drawn for illustration.

Worked example10 cm3 of a gaseous hydrocarbon is burnt in 110 cm3 of oxygen (excess). After cooling to r.t.p. the volume is 85 cm3, falling to 35 cm3 after passing through aqueous sodium hydroxide. Find the formula of the hydrocarbon.

1. V(CO2) = 85 − 35 = 50 cm3 = 10x, so x = 5
NaOH removes the CO2.
2. O2 used = 110 − 35 = 75 cm3 = 10(x + y/4)
The 35 cm3 left is unused O2.
3. 7.5 = 5 + y/4, so y = 10: C5H10
Volume ratio = mole ratio.
Check your understanding
  1. What volume does 8.0 g of oxygen gas, O2, occupy at r.t.p.?
  2. What volume of oxygen is needed to burn 20 cm3 of methane completely? CH4 + 2O2 → CO2 + 2H2O

4.8The ideal gas equation Core · A Level

A gas exerts pressure because its particles collide with the walls of the container: pressure is the force of these collisions per unit area (Figure 4.7). For a fixed amount of gas, pressure rises when the volume falls (the particles travel a shorter distance between collisions with the walls, so they hit the walls more often) and when the temperature rises (faster particles hit harder and more often).

force
Figure 4.7 Each collision pushes on the wall.
The ideal gas equation is pV = nRT, with every quantity in SI units: p in Pa, V in m3, n in mol, T in K, and the gas constant R = 8.31 J K−1 mol−1.
What the equation saysPressure, volume, amount and temperature are the four things you can measure about a gas. For every ideal gas they are tied together: pV ÷ nT always comes out to the same constant, R. So the four cannot change independently. If you know any three, the equation gives the fourth.
quantitySI unitconversions
pressurePa1 atm = 101 325 Pa, 1 bar = 105 Pa, 1 kPa = 103 Pa
volumem31 dm3 = 10−3 m3, 1 cm3 = 10−6 m3
temperatureKT / K = T / °C + 273
  • Substituting n = m ÷ Mr gives Mr = mRT ÷ pV, so a gas's Mr can be found from its mass, volume, temperature and pressure.
  • For a fixed amount of gas, p1V1/T1 = p2V2/T2. At constant temperature pV is constant (Boyle's law). At constant pressure V ∝ T (Charles's law).

Worked example0.10 g of a gas occupies 83.1 cm3 at 1.0 × 105 Pa and 27 °C. Find its Mr.

1. V = 83.1 × 10−6 m3, T = 300 K
Convert to SI units first.
2. Mr = (0.10 × 8.31 × 300) ÷ (1.0 × 105 × 83.1 × 10−6) = 30
Mr = mRT ÷ pV

4.8.1Dalton's law of partial pressures

In a mixture of gases that do not react, each gas behaves as if it were alone (Figure 4.8).

The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the container. Dalton's law: the total pressure is the sum of the partial pressures, ptotal = pA + pB + …

A gas's share of the pressure equals its share of the particles: pA = xA × ptotal, where the mole fraction xA = nA ÷ ntotal.

pA+pB=pA + pB
Figure 4.8 Dalton's law.

Worked exampleA 1 dm3 flask of helium at 2 kPa is joined to a 2 dm3 flask of neon at 1 kPa at constant temperature (Figure 4.9). Find the final total pressure.

1. He spreads into 3 dm3: pHe = 2 × 1/3 = 0.667 kPa
Boyle's law for each gas alone.
2. Ne spreads into 3 dm3: pNe = 1 × 2/3 = 0.667 kPa
pV is constant.
3. ptotal = 0.667 + 0.667 = 1.33 kPa
Dalton's law.
He: 1 dm³, 2 kPaNe: 2 dm³, 1 kPaopen the tapeach gas fills all 3 dm³
Figure 4.9 Two flasks joined by a tap, before and after it opens.

4.8.2Ideal and real gases

An ideal gas obeys pV = nRT exactly. The kinetic theory explains its behaviour with these assumptions (Figure 4.10):

  1. The particles have negligible volume compared with the container.
  2. There are no intermolecular forces between the particles.
  3. The particles move randomly, and their collisions lose no kinetic energy.
tiny particles in random motionbeforeafterno kinetic energy lost
Figure 4.10 The ideal gas model.

Real gases come closest to ideal at high temperature and low pressure, where the particles are fast and far apart. They deviate most at low temperature and high pressure, for two reasons (Figure 4.11):

  • Intermolecular forces: a particle about to hit the wall is pulled back by its neighbours, so it hits more gently and pV is less than ideal. Stronger intermolecular forces give a larger deviation.
  • Particle volume: at high pressure the particles take up a real share of the container, so the gas occupies more volume than an ideal gas would and pV is more than ideal.
pulled back by neighboursparticles fill much of the volumeppV/nRTH₂CO₂H₂Oideal gas
Figure 4.11 Why real gases deviate, and pV/nRT against p for three gases (shapes for illustration). The more strongly the molecules attract, the larger the dip.

Worked examplePut CH4, HCl and HF in order, most ideal first, under the same conditions.

1. CH4 has only instantaneous dipole–induced dipole forces, HCl also has permanent dipole–permanent dipole forces, and HF has hydrogen bonds.
Chapter 3 intermolecular forces.
2. Stronger forces mean more deviation: CH4 > HCl > HF.
The weakest forces behave most ideally.
Check your understanding
  1. What volume, in dm3, does 1.00 mol of an ideal gas occupy at 0 °C and 1.00 × 105 Pa?
  2. A mixture contains 0.5 mol He and 1.5 mol Ar at a total pressure of 200 kPa. Find the partial pressure of He.
  3. Why does helium behave more ideally than ammonia?

Answers to Check your understanding

4.1

  1. 4Al + 3O2 → 2Al2O3
  2. Ba2+(aq) + SO42−(aq) → BaSO4(s). Na+ and Cl− are spectators.

4.2

  1. 2(14.0 + 4 × 1.0) + 32.1 + 4 × 16.0 = 132.1
  2. (69 × 63 + 31 × 65) ÷ 100 = 63.6

4.3

  1. Mr = 40.0, mass = 0.250 × 40.0 = 10.0 g
  2. 0.10 × 2 = 0.20 mol of Cl−, which is 0.20 × 6.02 × 1023 = 1.2 × 1023 ions

4.4

  1. C: 85.7 ÷ 12.0 = 7.14, H: 14.3 ÷ 1.0 = 14.3, ratio 1 : 2, empirical CH2 (14). 56 ÷ 14 = 4, so C4H8.

4.5

  1. nCaCO₃ = 25.0 ÷ 100.1 = 0.2498 mol, so nCaO = 0.2498 mol and mass = 0.2498 × 56.1 = 14.0 g
  2. nCO₂ = 1.76 ÷ 44.0 = 0.0400 mol = nCaCO₃. Mass of CaCO3 = 0.0400 × 100.1 = 4.004 g, purity = 4.004 ÷ 5.00 × 100% = 80.1%

4.6

  1. 0.200 × 0.0250 = 0.00500 mol
  2. 0.500 × 58.5 = 29.3 g dm−3

4.7

  1. n = 8.0 ÷ 32.0 = 0.25 mol, V = 0.25 × 24 = 6.0 dm3
  2. 40 cm3: the volume ratio CH4 : O2 is 1 : 2.

4.8

  1. V = nRT ÷ p = 1.00 × 8.31 × 273 ÷ 1.00 × 105 = 0.0227 m3 = 22.7 dm3
  2. x(He) = 0.5 ÷ 2.0 = 0.25, so p(He) = 0.25 × 200 = 50 kPa
  3. Helium atoms have only very weak instantaneous dipole–induced dipole forces and a small volume. Ammonia molecules are attracted by hydrogen bonds, so they deviate more.