Chemical Formulae,
Equations, the Mole
and Gases

Singapore Junior Chemistry Olympiad

By the end of this chapter, you can answer

1

Atoms are far too small to weigh one at a time. How can chemists still count them out with a balance?

2

A sugar burns to give only carbon dioxide and water. How do those two masses give its formula?

3

2.4 g of magnesium is heated with 0.80 g of oxygen. Which runs out first, and how much magnesium oxide can form?

4

Burn 10 cm3 of an unknown hydrocarbon gas and measure the gas volumes before and after. How does that give its formula?

5

How can you find the molar mass of a gas just by weighing a flask of it?

6

Why does ammonia stray further from ideal gas behaviour than helium does?

Recall: Chapters 1 to 3

1What are isotopes?

Atoms of the same element with the same number of protons but different numbers of neutrons.

2Write the formula of calcium chloride.

CaCl2: one Ca2+ balances two Cl−.

3Why can a gas be squashed into a smaller volume so easily?

Its particles are far apart, with large spaces between them.
1

Chemical Equations

Atoms are not created or destroyed in a reaction. A balanced equation has the same number of each kind of atom on both sides.

NoteBalance by changing the numbers in front of the formulae, never the formulae themselves.

State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous, dissolved in water.

Worked exampleBalance: CH4 + O2 → CO2 + H2O

C: 1 on each side already
start with an element in one formula on each side
H: 4 on the left, so 2H2O
CH4 + O2 → CO2 + 2H2O
O: 2 + 2 = 4 on the right, so 2O2
balance the free element last

CH4 + 2O2 → CO2 + 2H2O

1

Chemical Equations

1.1Ionic equations

Dissolved ionic compounds are present as separate ions. An ionic equation shows only the particles that change. Ions unchanged on both sides are spectator ions.

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

Worked exampleAgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Ag+ + NO3− + Na+ + Cl− → AgCl(s) + Na+ + NO3−

Na+ and NO3− are spectators

Ag+(aq) + Cl−(aq) → AgCl(s)

NoteAn ionic equation balances in charge as well as atoms: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s).
1

Chemical Equations

Your turnBalance: Al + O2 → Al2O3

O: 2 and 3 meet at 6, so 3O2 and 2Al2O3
Al: 4 on the right, so 4Al + 3O2 → 2Al2O3[1]

Your turnWrite the ionic equation for BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq).

Ba2+(aq) + SO42−(aq) → BaSO4(s)[1]
Na+ and Cl− are spectators.
1

Chemical Equations

Quick checkWhich equation is balanced?

ANa + Cl2 → NaCl
BH2 + O2 → H2O
C2Mg + O2 → 2MgO
DCaCO3 → CaO + 2CO2
NoteC: 2 Mg and 2 O on each side. A has 2 Cl on the left and 1 on the right, B has 2 O on the left and 1 on the right, and D has 3 O on the left and 5 on the right.
2

Relative Masses

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

Arthe average mass of one atom of an element compared with 1/12 of the mass of one atom of carbon-12
Mrthe same comparison for a molecule: add up the Ar of every atom in the formula
Mr (ionic)called the relative formula mass, worked out the same way from the formula unit

e.g. CO2: 12.0 + 2(16.0) = 44.0.  Ca(NO3)2: 40.1 + 2(14.0 + 3 × 16.0) = 164.1

2

Relative Masses

Percentage by mass

% of element = total Ar of that element in the formulaMr × 100%

In CO2: carbon = 12.044.0 × 100% = 27.3%

Your turnCalculate Mr of (NH4)2SO4. (Ar: H 1.0, N 14.0, O 16.0, S 32.1)

2(14.0 + 4 × 1.0) + 32.1 + 4 × 16.0 = 132.1[1]
2

Relative MassesCore · A Level

2.1Ar from isotopes

Ar averages the isotopes, weighted by how common each one is.

Ar = Σ (% abundance × isotopic mass)100

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

Ar = 75.77 × 35 + 24.23 × 37100 = 35.5

35.5 lies nearer 35 because most chlorine atoms are chlorine-35.

Your turnCopper is 69% 63Cu and 31% 65Cu. Calculate Ar(Cu).

Ar = 69 × 63 + 31 × 65100
= 63.6[1]
3

The Mole

AnalogyA dozen is a number: 12 eggs, 12 pencils. A mole is also just a number, a very large one, because particles are tiny.

DefinitionMole: the amount of substance containing 6.02 × 10²³ particles, the Avogadro constant, L.

Molar mass, M: the mass of one mole, in g mol⁻¹, numerically equal to Ar or Mr. One mole of carbon-12 atoms is exactly 12 g.

NoteSay which particle you count: 1 mol of O2 molecules contains 2 mol of O atoms.
3

The Mole

Why the mole matters

  1. Atoms are far too small to count or weigh one at a time, but a balance measures grams. The mole is the bridge.
  2. A proton or a neutron has a mass of about 1 on the relative scale. How many of them make 1 gram? 6.02 × 1023. (Exactly: 6.02 × 1023 carbon-12 atoms weigh 12 g.)
  3. We know each atom's mass in these units, Ar. So one mole weighs Ar or Mr in grams: C 12.0 g, O 16.0 g, H2O 18.0 g.
  4. Each of those samples holds the same number of particles. Weighing 18.0 g of water counts out 6.02 × 1023 molecules.
3

The Mole

Everything goes through moles

  • Mass: n = mass ÷ M
  • Particles: N = n × L
  • Gas volume: V = n × 24 dm³ at r.t.p. (Section 7)
  • Solution: n = c × V, with V in dm³ (Section 6)
Exam tipKeep 4 significant figures in working and give the answer to 3, unless told otherwise.
3

The Mole

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

n = 4.40 ÷ 44.0 = 0.100 mol
mass to moles
molecules = 0.100 × 6.02 × 10²³ = 6.02 × 10²²
moles to particles
atoms = 3 × 6.02 × 10²² = 1.81 × 10²³
each CO2 has 3 atoms

Your turn(a) What is the mass of 0.250 mol of NaOH? (Ar: Na 23.0, O 16.0, H 1.0)  (b) How many chloride ions are in 0.10 mol of CaCl2?

(a)
M = 40.0, mass = 0.250 × 40.0 = 10.0 g[1]
(b)
0.10 × 2 = 0.20 mol Cl⁻ = 0.20 × 6.02 × 10²³ = 1.2 × 10²³[1]
3

The Mole

Quick checkWhich sample contains the greatest number of atoms? (Ar: H 1.0, He 4.0, N 14.0, O 16.0)

A2.0 g of H2
B4.0 g of He
C16.0 g of O2
D14.0 g of N2
NoteA: 1.0 mol of H2 molecules is 2.0 mol of atoms. He is 1.0 mol of atoms. O2 and N2 are 0.50 mol of molecules, which is 1.0 mol of atoms each.
4

Empirical and Molecular Formulae

  • Empirical formula: the simplest whole-number ratio of the atoms in a compound
  • Molecular formula: the actual number of each atom in one molecule, a whole-number multiple of the empirical formula

multiple = Mr ÷ empirical formula mass. e.g. butane: C2H5 and C4H10, glucose: CH2O and C6H12O6.

4

Empirical and Molecular Formulae

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

Step 1Take 100 g, so each % becomes a mass in grams.

Step 2Empirical C3H7N, mass 57.0

Step 3114 ÷ 57.0 = 2 ∴ C6H14N2

4

Empirical and Molecular FormulaeCore · A Level

4.1Formulae from combustion

PredictA compound of C, H and O burns in oxygen. Can we find its mass of O from the CO2 and H2O?

  • All the C in the CO2 came from the sample.
  • All the H in the H2O came from the sample.
  • Some O in the products came from the O2, so find O by difference: mass of sample − mass of C − mass of H.
4

Empirical and Molecular FormulaeCore · A Level

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. 12 g of C in every 44 g of CO2: C = 1244 × 3.46 = 0.9436 g
  2. 2 g of H in every 18 g of H2O: H = 218 × 1.42 = 0.1578 g
  3. O = 2.36 − 0.9436 − 0.1578 = 1.259 g
CHO
n / mol0.078630.15780.07869
ratio121

Empirical CH2O (mass 30), 180 ÷ 30 = 6 ∴ C6H12O6

4

Empirical and Molecular Formulae

Your turnA compound is 85.7% C and 14.3% H by mass, and Mr is 56. Find its molecular formula.

C: 85.7 ÷ 12.0 = 7.14, H: 14.3 ÷ 1.0 = 14.3, ratio 1 : 2
Empirical formula CH2 (mass 14)[1]
56 ÷ 14 = 4, so the molecular formula is C4H8[1]
5

Reacting Masses

The numbers in a balanced equation give the mole ratio. 2H2 + O2 → 2H2O means 2 mol of H2 react with 1 mol of O2 to give 2 mol of H2O.

mass → moles → × ratio → moles → mass

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

nMg = 6.0 ÷ 24.0 = 0.25 mol
mass to moles
nMgO = 0.25 mol
ratio 2 : 2
mass = 0.25 × 40.0 = 10 g
moles to mass
5

Reacting Masses

5.1Limiting reagent and yield

AnalogyA sandwich needs 2 slices of bread and 1 slice of cheese. With 10 bread and 4 cheese, how many sandwiches?

The limiting reagent is used up first. It fixes the most product that can form, the theoretical yield.

The other reactant is in excess: some is left over.

5

Reacting Masses

percentage yield = actual yieldtheoretical yield × 100%     percentage purity = mass of pure substancemass of impure sample × 100%

Worked example2.4 g of Mg is heated with 0.80 g of O2. 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
  2. 0.10 mol Mg needs 0.050 mol O2, but only 0.025 mol is present → O2 is limiting
  1. nMgO = 2 × 0.025 = 0.050 mol, theoretical yield = 0.050 × 40.0 = 2.0 g
  2. % yield = 1.62.0 × 100% = 80%
5

Reacting Masses

Your turnCaCO3 → CaO + CO2. What mass of CaO forms from 25.0 g of CaCO3? (Ar: Ca 40.1, C 12.0, O 16.0)

nCaCO₃ = 25.0 ÷ 100.1 = 0.2498 mol = nCaO[1]
mass = 0.2498 × 56.1 = 14.0 g[1]

Your turnA 5.00 g sample of impure CaCO3 gives 1.76 g of CO2. What is its percentage purity?

nCO₂ = 1.76 ÷ 44.0 = 0.0400 mol = nCaCO₃[1]
mass of CaCO3 = 0.0400 × 100.1 = 4.004 g
purity = 4.0045.00 × 100% = 80.1%[1]
5

Reacting Masses

Quick check2.0 mol of N2 reacts with 3.0 mol of H2:  N2 + 3H2 → 2NH3. What is the greatest amount of NH3 that can form?

A4.0 mol
B2.0 mol
C6.0 mol
D1.0 mol
NoteB: 2.0 mol of N2 would need 6.0 mol of H2, so H2 is limiting. 3.0 mol of H2 gives ⅔ × 3.0 = 2.0 mol of NH3. Option A works from the N2, which is in excess.
6

Solutions

DefinitionConcentration: 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.

c = n (mol)V (dm³)    1 dm³ = 1000 cm³

  1. Weigh the solute and dissolve it in a little water.
  2. Transfer to a volumetric flask and add water up to the mark.

Dilution: adding water does not change the amount of solute, so c1V1 = c2V2.

Worked example4.00 g of NaOH (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?

n = 0.100 mol, c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³

0.400 × 50.0 = 0.100 × V2, so V2 = 200 cm³

water to add = 200 − 50.0 = 150 cm³

6

Solutions

Your turnHow many moles of solute are in 25.0 cm³ of 0.200 mol dm⁻³ solution?

0.200 × 0.0250 = 0.00500 mol[1]

Your turnExpress 0.500 mol dm⁻³ sodium chloride (Mr 58.5) in g dm⁻³.

0.500 × 58.5 = 29.3 g dm⁻³[1]
7

Gas Volumes

  • Avogadro's law: at the same temperature and pressure, equal volumes of all gases contain the same number of particles
  • So 1 mol of any gas occupies 24 dm³ at r.t.p. (22.7 dm³ at s.t.p., 0 °C and 1 bar)
  • For gases, volume ratio = mole ratio. Only gases count: water at r.t.p. is a liquid.
7

Gas Volumes

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

nNa = 0.230 ÷ 23.0 = 0.0100 mol
water is in excess, Na is limiting
nH₂ = 0.00500 mol
ratio 2 : 1
V = 0.00500 × 24 = 120 cm³
molar gas volume

Your turn(a) What volume does 8.0 g of O2 occupy at r.t.p.?  (b) What volume of O2 burns 20 cm³ of CH4 completely? CH4 + 2O2 → CO2 + 2H2O

(a)
n = 8.0 ÷ 32.0 = 0.25 mol, V = 0.25 × 24 = 6.0 dm³[1]
(b)
volume ratio 1 : 2, so 40 cm³[1]
7

Gas Volumes

7.1Burning gaseous hydrocarbons

CxHy + (x + y4)O2 → xCO2 + y2H2O

Worked example10 cm³ of a gaseous hydrocarbon is burnt in 110 cm³ of O2 (excess). After cooling to r.t.p. the volume is 85 cm³, falling to 35 cm³ after aqueous NaOH. Find the formula.

  1. NaOH absorbs CO2: V(CO2) = 85 − 35 = 50 cm³ = 10x → x = 5
  2. The 35 cm³ left is unused O2: O2 used = 110 − 35 = 75 cm³ = 10(x + y4)
  3. 7.5 = 5 + y4, so y = 10 ∴ C5H10
7

Gas Volumes

Quick checkWhat volume of carbon dioxide, measured at r.t.p., forms when 0.10 mol of propane, C3H8, burns completely?

A2.4 dm³
B4.8 dm³
C7.2 dm³
D9.6 dm³
NoteC: each C3H8 gives 3CO2, so 0.30 mol of CO2, and 0.30 × 24 = 7.2 dm³. D counts the water as a gas.
8

The Ideal Gas EquationCore · A Level

Pressure: the force of the particles' collisions with the walls, per unit area.
  • Volume ↓ → more frequent collisions → p ↑
  • Temperature ↑ → faster particles hit harder and more often → p ↑
8

The Ideal Gas EquationCore · A Level

pV = nRT
R = 8.31 J K⁻¹ mol⁻¹, every quantity in SI units
Notep, V, n and T are tied together: pV ÷ nT is always the same constant, R. Know any three and the equation gives the fourth.
quantitySI unitconversions
pressurePa1 atm = 101 325 Pa, 1 bar = 10⁵ Pa
volumem³1 dm³ = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³
temperatureKT / K = T / °C + 273

Your turnConvert into SI units: (a) 25 °C, (b) 250 cm³, (c) 2.00 atm.

(a)
298 K[1]
(b)
2.50 × 10⁻⁴ m³[1]
(c)
2.03 × 10⁵ Pa[1]
8

The Ideal Gas EquationCore · A Level

Finding Mr

Substitute n = mMr into pV = nRT:

Mr = mRTpV

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

V = 83.1 × 10⁻⁶ m³, T = 300 K

Mr = 0.10 × 8.31 × 3001.0 × 10⁵ × 83.1 × 10⁻⁶ = 30

Your turnWhat volume, in dm³, does 1.00 mol of an ideal gas occupy at 0 °C and 1.00 × 10⁵ Pa?

V = nRTp = 1.00 × 8.31 × 2731.00 × 10⁵[1]
= 0.0227 m³ = 22.7 dm³, the molar volume at s.t.p.[1]
8

The Ideal Gas EquationCore · A Level

Fixed amount of gas

p1V1T1 = p2V2T2

PredictThe gas is squeezed into half the volume at constant temperature. What happens to its pressure?

Boyle's law: at constant temperature, pV is constant. Halve V and p doubles.
8

The Ideal Gas EquationCore · A Level

PredictThe gas is heated from 300 K to 600 K at constant pressure (the piston is free to move). What happens to the volume?

Charles's law: at constant pressure, V ∝ T, with T in kelvin.
NotePlotted against T in °C, the line meets V = 0 at −273 °C (0 K). Always use kelvin in calculations.
8

The Ideal Gas EquationCore · A Level

8.1Dalton's law of partial pressures

Partial pressure: the pressure a gas would exert if it alone occupied the container.

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

ptotal = pA + pB + …

The total pressure is the sum of the partial pressures.

pA = xA × ptotal   mole fraction xA = nA ÷ ntotal

8

The Ideal Gas EquationCore · A Level

Worked exampleA 1 dm³ flask of He at 2 kPa is joined to a 2 dm³ flask of Ne at 1 kPa at constant temperature. Find the final total pressure.

He spreads into 3 dm³: pHe = 2 × 13 = 0.667 kPa

Ne spreads into 3 dm³: pNe = 1 × 23 = 0.667 kPa

ptotal = 1.33 kPa

8

The Ideal Gas EquationCore · A Level

Your turnA mixture contains 0.5 mol He and 1.5 mol Ar at a total pressure of 200 kPa. Find the partial pressure of He.

x(He) = 0.5 ÷ 2.0 = 0.25[1]
p(He) = 0.25 × 200 = 50 kPa[1]
8

The Ideal Gas EquationCore · A Level

8.2Ideal and real gases

An ideal gas obeys pV = nRT exactly. The kinetic theory assumes:

  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.
8

The Ideal Gas EquationCore · A Level

DescribeHow does the real gas differ from the ideal line?

For an ideal gas pV/nRT = 1 at every pressure.

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.

8

The Ideal Gas EquationCore · A Level

Intermolecular forces: pV dips

  1. A particle about to hit the wall is pulled back by its neighbours
  2. → it hits more gently → pV is less than ideal

Particle volume: pV rises at high p

  1. The particles take up a real share of the container
  2. → the gas occupies more volume than an ideal gas → pV is more than ideal
8

The Ideal Gas EquationCore · A Level

The more strongly the molecules attract, the larger the dip: H2O (hydrogen bonds) > CO2 > H2.

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.
  2. Stronger forces, more deviation ∴ CH4 > HCl > HF
8

The Ideal Gas EquationCore · A Level

Your turnWhy does helium behave more ideally than ammonia?

Helium atoms have only very weak instantaneous dipole–induced dipole forces and a small volume.[1]
Ammonia molecules are attracted by hydrogen bonds, so they deviate more.[1]

Quick checkUnder which conditions does a gas behave most like an ideal gas?

Ahigh pressure, low temperature
Blow pressure, high temperature
Chigh pressure, high temperature
Dlow pressure, low temperature
NoteB: at low pressure the particles are far apart, so their volume and attractions hardly matter, and at high temperature they move fast enough to overcome the attractions.

Exit ticket

1How many Cl⁻ ions are in 0.100 mol of CaCl2?

0.200 mol × 6.02 × 10²³ = 1.20 × 10²³[1]

2Why is the mass of O in a C, H, O compound found by difference after burning?

Some O in the products came from the O2 burnt.[1]

3What is the limiting reagent?

The reactant used up first. It fixes the most product that can form.[1]

4State the two main assumptions of an ideal gas.

Particles of negligible volume, and no intermolecular forces between them.[1]