Seven teaching hours at both levels, outcomes 1.4.1 to 1.4.6. There is no additional higher level content in Structure 1.4.
Guiding question: How do we quantify matter on the atomic scale?
Structure 1.4 · The mole
1The mole and the Avogadro constant 1.4.1 SL + HL Strong
A chemical equation is a statement about particles: 2H2 + O2 → 2H2O says that two molecules of hydrogen react with one of oxygen. But nobody can count molecules in a laboratory — a drop of water contains about 1021 of them. Chemists need a unit that converts a number of particles into something that can be weighed or measured. That unit is the mole.
The mole (mol) is the SI unit of amount of substance, n. One mole contains exactly 6.02214076 × 1023 elementary entities — the number given by the Avogadro constant, NA = 6.02 × 1023 mol−1.
An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or a specified group of particles. A question must name the entity, and the answer changes with it.
N = n × NA number of entities = amount in mol × Avogadro constant (mol−1)
The mole is chosen so that the mass of one mole of any substance, in grams, is numerically equal to its relative formula mass. One mole of carbon-12 atoms has a mass of exactly 12 g; one mole of water molecules, 18.02 g. The amount of substance is the hub from which every other quantity in this chapter — mass, number of particles, volume of gas and concentration of a solution — can be reached.
Counting different entities within the same substance is a matter of reading the formula. One mole of CO2 contains one mole of carbon atoms, two moles of oxygen atoms and three moles of atoms in total. One mole of (NH4)2SO4 contains two moles of NH4+ ions, one mole of SO42− ions, three moles of ions, and fifteen moles of atoms.
Question. What is the mass of one molecule of C60? Take NA = 6.0 × 1023 mol−1.
M = 60 × 12.01 = 720.6 g mol−1; mass of one molecule = 720.6 ÷ 6.0 × 1023 = 1.2 × 10−21 g
Reasonableness check. A single molecule must have a tiny mass; dividing the wrong way round gives 4 × 1026 g, heavier than a mountain.
Question. Calculate the number of atoms in 7.75 g of phosphorus. [1]
n = 7.75 ÷ 30.97 = 0.250 mol → N = 0.250 × 6.02 × 1023 = 1.51 × 1023
A published scheme prints the whole calculation as one expression, (7.75/30.97) × 6.02 × 1023. A report on a similar question records that about a quarter of the candidates calculated the number of atoms correctly while the majority made mistakes during the steps of the calculation.
Exam focus · what the published papers show
This is what the multiple-choice items actually test, and the distractors are built from the wrong count. In Ca3(PO4)2 there are two phosphate ions per formula unit, so 0.333 mol of the compound holds 0.67 mol of phosphate — and 0.33 is offered. In CuSO4·5H2O there are nine oxygen atoms per formula unit, four in the sulfate and five in the water of crystallization; counting only the sulfate gives 4.82 × 1022 instead of 1.08 × 1023 for the sample in that item. Read the noun.
2Relative atomic mass and relative formula mass 1.4.2 SL + HL Strong
Masses of atoms are compared on a scale relative to 12C and are expressed as relative atomic mass, Ar, and relative formula mass, Mr. Both are ratios, so both have no units.
On this scale an atom of carbon-12 has a mass of exactly 12, and the Ar of every other element is its weighted mean atomic mass compared with one-twelfth of the mass of a 12C atom (Structure 1.2). The relative formula mass is the sum of the relative atomic masses of all the atoms in the formula. For molecular substances it is often called the relative molecular mass; for ionic compounds the formula unit is used.
Mr(Ca(NO3)2) = 40.08 + 2(14.01) + 6(16.00) = 164.10
A bracket multiplies everything inside it, and water of crystallization is part of the formula: Mr(MgSO4·7H2O) = 24.31 + 32.07 + 4(16.00) + 7(18.02) = 246.52.
The syllabus states that the values of relative atomic masses given to two decimal places in the data booklet should be used in calculations. Rounding carbon to 12 or oxygen to 16 usually still lands on the right formula but can shift a borderline ratio, and it costs precision in a molar-mass answer.
3Molar mass: particles, amount and mass 1.4.3 SL + HL Repeated
The molar mass, M, is the mass of one mole of a substance. It is numerically equal to Ar or Mr but carries the units g mol−1: water has Mr = 18.02 and M = 18.02 g mol−1. One relationship, given in the data booklet, links amount and mass:
n = m ÷ M n in mol · m in g · M in g mol−1
Combined with N = n × NA, it solves every problem linking particles, amount and mass. Rearranged, it gives m = n × M and M = m ÷ n. The last is how molar masses are measured: find the mass and the amount of the same sample.
Question. A sample of the vapour of A at 200.0 °C and 1.00 × 105 Pa has a density of 2.544 × 103 g m−3. Determine the molar mass. [2]
Combining pV = nRT with n = m/M and density d = m/V gives M = dRT/p:
M = (2.544 × 103 × 8.31 × 473) ÷ 1.00 × 105 = 1.00 × 102 g mol−1
Or per cubic metre: n = pV/RT = 25.4 mol in 1.00 m3, which has a mass of 2544 g, so M = 2544 ÷ 25.4 = 100 g mol−1. The published scheme accepts either route. The temperature must be in kelvin and the density in the units the equation expects. A report calls this a discriminating question.
A report on a mass-spectrum question notes that it was not sufficient to state that 288 showed the molar mass. The fragment is the molecular ion or M+. The two are numerically equal for a singly charged ion, but a question asking what a peak is wants the species, not the quantity.
4Empirical and molecular formulas 1.4.4 SL + HL Strong
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.
The molecular formula gives the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula.
| Substance | Molecular formula | Empirical formula |
|---|---|---|
| Glucose | C6H12O6 | CH2O |
| Ethanoic acid | C2H4O2 | CH2O |
| Butane | C4H10 | C2H5 |
| Ethanol | C2H6O | C2H6O (already simplest) |
| Sodium chloride | — (ionic lattice) | NaCl |
Different compounds can share an empirical formula — glucose and ethanoic acid both have CH2O — so an empirical formula alone cannot identify a compound. Ionic and giant covalent substances have no molecules and are always represented by their empirical formula.
From composition to formula
The empirical formula is found from the percentage composition by mass or from the masses of the elements that combine. The routine converts masses into amounts, because formulas are ratios of atoms, not of masses. If percentages are given, assume a 100 g sample so that each percentage becomes a mass in grams.
The reverse calculation — percentage by mass from a formula — divides the mass of the element in one mole by the molar mass: in water, %O = 16.00 ÷ 18.02 × 100 = 88.79 %. To go from an empirical to a molecular formula, a molar mass is needed from another measurement (mass spectrometry, or the ideal gas equation in Structure 1.5): the multiplier is M ÷ M(empirical unit).
Experimental data: combustion and oxidation
Empirical formulas are often obtained from mass changes. Heating magnesium in a crucible until its mass is constant gives the mass of oxygen combined; burning an organic compound in excess oxygen converts all its carbon to CO2 and all its hydrogen to H2O, which are collected and weighed. One mole of CO2 contains one mole of C; one mole of H2O contains two moles of H. Oxygen in the original compound cannot be found this way, because the combustion supplies oxygen too; it is found by difference.
Empirical and molecular formula
Percentage to moles, moles to the simplest ratio, ratio to a formula. Set the composition and the working appears the way it has to appear on the page to earn the marks.
Amount of each element, in moles per 100 g. Dividing every one of these by the smallest is the only step in the calculation — everything before it is unit conversion and everything after it is rounding.
Exam focus · what the published papers show
Two sessions report this independently: the most common mistake was not identifying that oxygen was present in the compound; and, a year earlier, the empirical formula was well answered by many except for weaker candidates who forgot to include the % oxygen or gave no response.
The habit that prevents it: before any arithmetic, add up the percentages. If they do not total 100 %, the shortfall is an element. One paper asked, as its opening part, why a compound is not a hydrocarbon; the published answer, percentages do not add up to 100% OR contains oxygen, was a hint for the next part.
Question. An organic compound A is 71.93 % carbon and 12.10 % hydrogen by mass. Determine its empirical formula. [2]
| Missing element | 100 − 71.93 − 12.10 = 15.97 % O |
| Amounts in 100 g | C 71.93 ÷ 12.01 = 5.99; H 12.10 ÷ 1.01 = 11.98; O 15.97 ÷ 16.00 = 1.00 |
| Ratio | ÷ 1.00 → 5.99 : 11.98 : 1.00 ≈ 6 : 12 : 1 |
Empirical formula C6H12O. Omitting the oxygen gives CH2 — plausible-looking and wrong.
Question. 4.32 g of a compound containing only C, H and O was burned completely, producing 9.49 g of CO2 and 5.18 g of H2O. Determine the empirical formula. [3]
| Carbon | n(C) = n(CO2) = 9.49 ÷ 44.01 = 0.216 mol; 0.216 × 12.01 = 2.59 g |
| Hydrogen | n(H) = 2 × (5.18 ÷ 18.02) = 0.575 mol; 0.575 × 1.01 = 0.581 g |
| Oxygen | 4.32 − (2.59 + 0.581) = 1.15 g; 1.15 ÷ 16.00 = 0.0718 mol |
| Ratio | ÷ 0.0718 → 3.01 : 8.01 : 1.00 |
C3H8O. The scheme allocates M2 for finding mass of oxygen — a full mark for the subtraction alone. Forgetting the factor of two for hydrogen gives C3H4O.
On a three-mark empirical-formula question the notes include Award [2] for the simplest ratio “1.5 C : 3 H : 1 O”. Write the ratio down before scaling it. A report on another session confirms the pattern: a strength being the calculation of the molar ratios of each element for the first two marks; some candidates then struggled with the deduction of the empirical formula when handling fractional ratios.
After dividing by the smallest amount you may get 1 : 1.5 or 1 : 1.33. Do not round: multiply everything by 2 for .5, by 3 for .33 or .67, by 4 for .25 or .75. Rounding 1.5 to 2 turns P2O3 into PO2, a different compound. Round only values within about 0.02 of a whole number, where the difference is experimental error.
Question. 7.75 g of phosphorus burned in a limited supply of oxygen gave 13.75 g of oxide. Determine the empirical formula, showing your working. [3] The Mr of the oxide is 219.88. Determine the molecular formula. [1]
Oxygen combined: 13.75 − 7.75 = 6.00 g. n(P) = 7.75 ÷ 30.97 = 0.250 mol; n(O) = 6.00 ÷ 16.00 = 0.375 mol. Ratio 1 : 1.5, scaled by 2: P2O3. Empirical unit mass 2(30.97) + 3(16.00) = 109.94; 219.88 ÷ 109.94 = 2: P4O6.
The scheme’s note: Award [1] for P2O3, if no working shown. Where a question says “show your working”, a bare correct answer collects one mark of three.
One report lists among the errors: candidates used masses totalling 100g and not 1g to calculate moles. Percentages are ratios, so any sample size gives the same formula — but use one size consistently.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
How many atoms of nitrogen are there in 0.50 mol of (NH4)2CO3?
What is the number of hydrogen atoms in 2.00 moles of Ca(HCO3)2? NA = 6.02 × 1023 mol−1
Which contains the most atoms of oxygen?
What is the molar mass, in g mol−1, of a compound if 0.200 mol of the compound has a mass of 13.2 g?
What are the units of molar mass?
Which of these molecular formulae are also empirical formulae?
I. C2H6O II. C2H4O2 III. C5H12
What is the molecular formula of a hydrocarbon containing 84.6 % carbon by mass with a molar mass of 142.3 g mol−1?
8.8 g of an oxide of nitrogen contains 3.2 g of oxygen. What is the empirical formula of the compound?
30 g of an organic compound produces 44 g CO2 and 18 g H2O as the only combustion products. Which of the following is the empirical formula for this compound? Mr CO2 = 44, Mr H2O = 18
A compound with Mr = 102 contains 58.8 % carbon, 9.80 % hydrogen and 31 % oxygen by mass. What is its molecular formula? Ar: C = 12.0; H = 1.0; O = 16.0
Urea, (H2N)2CO, is excreted by mammals and can be used as a fertilizer. Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet. [2]
There are many oxides of silver with the formula AgxOy. All of them decompose into their elements when heated strongly.
(i) After heating 3.760 g of a silver oxide 3.275 g of silver remained. Determine the empirical formula of AgxOy. [2]
(ii) Suggest why the final mass of solid obtained by heating 3.760 g of AgxOy may be greater than 3.275 g, giving one design improvement for your proposed suggestion. Ignore any possible errors in the weighing procedure. [2]
A 4.406 g sample of a compound containing only C, H and O was burnt in excess oxygen. 8.802 g of CO2 and 3.604 g of H2O were produced.
(a) Determine the empirical formula of the compound using section 6 of the data booklet. [3]
(b) Determine the molecular formula of this compound if its molar mass is 88.12 g mol−1. [1]
Menthol is an organic compound containing carbon, hydrogen and oxygen. Complete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working. [3]
An unknown organic compound, X, comprising only carbon, hydrogen and oxygen, was found to contain 48.6 % of carbon and 43.2 % of oxygen.
(a) Determine the empirical formula. [3]
(c) The molecular ion peak in the mass spectrum of X is at m/z 74. Determine the molecular formula of X. [1]
Solutions and mark-scheme guidance · Set 1G
1G.1 D
Two N per formula unit: n(N) = 2 × 0.50 = 1.0 mol = 6.02 × 1023 atoms. C counts formula units, not nitrogen atoms; A and B are amounts, not numbers of atoms.
1G.2 D
Two H per formula unit (the bracket doubles HCO3): 4.00 mol H × 6.02 × 1023 = 2.41 × 1024. B is the amount in mol, not a number of atoms.
1G.3 D
A: 2.0 mol O2 = 4.0 mol O. B: 2.0 mol O2 = 4.0 mol O. C: 64 ÷ 89.08 = 0.72 mol × 3 = 2.2 mol O. D: 2.0 mol × 3 = 6.0 mol O.
1G.4 A
M = m ÷ n = 13.2 ÷ 0.200 = 66.0 g mol−1, to three significant figures to match the data. B is numerically right but reported to too few significant figures — the item tests precision as well as the relationship.
1G.5 D
Mass per amount: g mol−1. C is inverted; Ar and Mr have no units at all.
1G.6 B
C2H6O (2 : 6 : 1) and C5H12 share no common factor. C2H4O2 simplifies to CH2O.
1G.7 C
C 84.6 ÷ 12.01 = 7.04; H 15.4 ÷ 1.01 = 15.2; ratio 1 : 2.2 = 5 : 11, so the empirical formula is C5H11 (71.16). 142.3 ÷ 71.16 = 2: C10H22. D is the empirical formula; A has double the molar mass.
1G.8 B
Nitrogen: 8.8 − 3.2 = 5.6 g → 5.6 ÷ 14.01 = 0.40 mol. Oxygen: 3.2 ÷ 16.00 = 0.20 mol. Ratio 2 : 1 → N2O. Taking 8.8 g as the mass of nitrogen is the trap.
1G.9 D
1 mol CO2 → 1 mol C (12 g); 1 mol H2O → 2 mol H (2 g). 30 − 14 = 16 g O = 1 mol. C : H : O = 1 : 2 : 1. A ignores the oxygen found by difference.
1G.10 C
C 4.90 : H 9.80 : O 1.94 → 2.5 : 5 : 1 → C5H10O2 (M = 102), which is also the molecular formula. The fractional ratio must be doubled, not rounded.
1G.11 [2]
M = 4(1.01) + 2(14.01) + 12.01 + 16.00 = 60.07 g mol−1 ✔; %N = 2 × 14.01 ÷ 60.07 × 100 = 46.65 % ✔. Award [1 max] for final answer not to two decimal places.
1G.12 [4]
(i) n(Ag) = 3.275 ÷ 107.87 = 0.03036 mol AND n(O) = (3.760 − 3.275) ÷ 16.00 = 0.485 ÷ 16.00 = 0.03031 mol ✔; ratio ≈ 1 : 1, so AgO ✔. Award [1 max] for correct empirical formula if method not shown.
(ii) Temperature too low / heating time too short / oxide not decomposed completely ✔; heat the sample to constant mass ✔. The scheme also accepts moisture absorbed from the air (improvement: cool in a desiccator).
1G.13 [4]
(a) n(C) = 8.802 ÷ 44.01 = 0.2000 mol (2.402 g); n(H) = 2 × 3.604 ÷ 18.02 = 0.4000 mol (0.404 g) ✔; m(O) = 4.406 − 2.806 = 1.600 g ✔ = 0.1000 mol; ratio 2 : 4 : 1 → C2H4O ✔.
(b) Empirical unit 44.06; 88.12 ÷ 44.06 = 2 → C4H8O2 ✔.
1G.14 [3]
Carbon: 0.4490 ÷ 44.01 = 0.01020 mol (0.1225 g) OR hydrogen: 2 × 0.1840 ÷ 18.02 = 0.02042 mol (0.0206 g) ✔; oxygen: 0.1595 − (0.1225 + 0.0206) = 0.0164 g = 0.001025 mol ✔; ratio 10 : 20 : 1 → C10H20O ✔. Award [3] for correct final answer.
1G.15 [4]
(a) n(C) = 48.6 ÷ 12.01 = 4.05 AND n(O) = 43.2 ÷ 16.00 = 2.70 ✔; %H = 8.2 %, n(H) = 8.12 ✔; ratio 1.5 : 3 : 1 → C3H6O2 ✔. Award [2] for the simplest ratio “1.5 C : 3 H : 1 O”.
(c) C3H6O2 (M = 74.09) ✔. [Part (c) is adapted: the paper gives the mass spectrum; the molecular ion is at m/z 74.]
5Molar concentration and standard solutions 1.4.5 SL + HL Strong
The molar concentration of a solution is the amount of solute per unit volume of solution. Square brackets denote it: [NaOH] is the molar concentration of sodium hydroxide.
n = c × V n in mol · c in mol dm−3 · V in dm3
The volume is the volume of the solution, not of the solvent added. Concentration may also be expressed as a mass concentration in g dm−3; the two are linked by the molar mass: c(g dm−3) = c(mol dm−3) × M. A solution of NaOH containing 8.00 g dm−3 has [NaOH] = 8.00 ÷ 40.00 = 0.200 mol dm−3. Volumes measured in cm3 must be divided by 1000 before use: 25.00 cm3 = 0.02500 dm3.
Dilution
Adding water to a solution changes its volume and concentration but not the amount of solute. So the amount before equals the amount after:
c1V1 = c2V2 both volumes in the same unit; cm3 may be kept because only their ratio enters
Diluting 10.0 cm3 of 2.00 mol dm−3 acid to 250.0 cm3 gives 2.00 × 10.0 ÷ 250.0 = 0.0800 mol dm−3. A serial dilution repeats the process — for example tenfold at each step — to make a series of standards, as used for a calibration curve.
Standard solutions
A standard solution is one whose concentration is known accurately. It is prepared by dissolving an accurately known mass of a pure solid in distilled water and making the solution up to an accurately known volume in a volumetric flask. Every step is designed to keep the whole weighed mass in the flask and to fix the volume precisely.
The solid is weighed accurately, dissolved completely in a beaker of distilled or deionized water, and the solution transferred to the volumetric flask through a funnel. The beaker, stirring rod and funnel are rinsed into the flask with distilled water, so that no solute is left behind. Water is then added until the bottom of the meniscus sits on the graduation mark, read at eye level, and the stoppered flask is inverted several times to make the concentration uniform. A measuring cylinder is not accurate enough; tap water would introduce ions.
Exam focus · what the published papers show
n = cV takes dm3. A burette reading of 25.00 cm3 is 0.02500 dm3. This one conversion is behind more lost marks in this outcome than any chemistry; one report describes candidates who seemed not to know which values to substitute into the equation to find concentration. Any calculation that produces an amount in mol must use dm3.
Question. A student needs 500.0 cm3 of a 0.2500 mol dm−3 solution of NaOH from the solid. Calculate the mass needed. [1]
m = c × V × M = 0.2500 × 0.5000 × 40.00 = 5.000 g
Using 500 instead of 0.5000 would give five kilograms of sodium hydroxide, which should stop you before the answer is written. The data are given to four significant figures, so the answer is 5.000 g, not 5 g.
Any three, in any order: a known mass of solid, weighed accurately; «fully» dissolve in distilled/deionized/pure water; transfer to a volumetric flask with washings; fill up to the line/mark; «stopper and» turn over «several times»/shake/homogenize.
Three notes decide the marks: Penalize [1] mark if distilled/deionized/pure water not mentioned once in response; Do not accept stir for mix; and Do not award any marks for preparation of solution by dilution — describing dilution when asked for preparation from a solid scores zero.
One report advises: Students should master the correct procedure of making standard solutions and dilutions using volumetric glassware. Another describes what goes wrong: Students often used graduated cylinders instead of volumetric flasks and missed the idea of using deionized water.
Question. Explain why the concentration of a sodium hydroxide solution must be found by titration rather than from the mass weighed out. [2]
Solid NaOH is hygroscopic/will absorb water/moisture from air or reacts with/absorbs CO2/carbon dioxide from air; therefore the actual concentration cannot be determined by mass. The weighed mass is not all NaOH, so the calculated concentration is too high. The scheme refuses to ensure same concentration of NaOH throughout experiment — a claim about consistency, not accuracy.
6Avogadro’s law and reacting volumes of gases 1.4.6 SL + HL Strong
Equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules.
The law holds because, in a gas, the particles are so far apart that their own size is negligible: the volume depends on the number of particles, not on what they are. Its consequence is the whole of this section. For gases at the same temperature and pressure, a ratio of volumes is a ratio of amounts, and so a ratio of coefficients in the balanced equation:
N2(g) + 3H2(g) → 2NH3(g) 1 volume : 3 volumes : 2 volumes
So 50 cm3 of nitrogen reacts with 150 cm3 of hydrogen to give 100 cm3 of ammonia, all measured under the same conditions — without any moles, molar volume, temperature or pressure. A limiting reactant is identified in the same way as with amounts: divide each volume by its coefficient; the smaller quotient is limiting. Liquids and solids take no part in a volume ratio: in CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), the water occupies a negligible volume once condensed.
Because equal amounts of gas occupy equal volumes, one mole of any gas occupies the same volume at a given temperature and pressure: the molar volume, Vm. At STP (273 K, 100 kPa) Vm = 22.7 dm3 mol−1 (data booklet). The link to masses is n = V ÷ Vm; away from STP, the ideal gas equation is used (Structure 1.5).
Counting particles by mass, and the ideal gas
One quantity — amount, in moles — connects a mass you can weigh, a number you cannot count, and a volume you can measure. Change any input and the other three follow.
The cylinder holds the amount of gas you set by mass. Its volume is V = nRT/p, drawn to scale against a 24.8 dm³ reference — the molar volume at 100 kPa and 298 K.
Boyle's law for this sample at the temperature you set: p against V, with your current state marked.
Exam focus · what the published papers show
A report on a gas-volume question: the item could be solved directly with volumes of reacting gases but was difficult for most candidates, some getting entangled with complex mole calculations. If every quantity is a volume of gas, work in volumes; the moment a mass appears, go through the mole.
Question. Which volume of butane will produce 40 cm3 of carbon dioxide when completely combusted? 2C4H10 + 13O2 → 8CO2 + 10H2O
Butane : carbon dioxide = 2 : 8 = 1 : 4, so 40 cm3 of CO2 needs 10 cm3 of butane. The 13 for oxygen is irrelevant here.
Question. What volume of ethane is produced when 0.25 dm3 of ethyne reacts with 0.40 dm3 of hydrogen, all volumes under the same conditions? C2H2 + 2H2 → C2H6
Divide by coefficients: ethyne 0.25 ÷ 1 = 0.25; hydrogen 0.40 ÷ 2 = 0.20. Hydrogen is limiting, so 0.20 dm3 of ethane forms and 0.05 dm3 of ethyne is left. 0.25 is the offered trap. A report notes candidates struggled to identify the limiting reagent and hence the gasses left after the reaction had finished.
Question. Equal volumes of X2 and Y2 react to form XY3: X2 + 3Y2 → 2XY3. After completion 40 cm3 of XY3 has formed. What was the starting volume of X2?
40 cm3 of XY3 needs 20 cm3 of X2 and 60 cm3 of Y2. Y2 is consumed three times as fast, so it is limiting and was all used: the container started with 60 cm3 of Y2 — and therefore 60 cm3 of X2, of which 40 cm3 remain. 20 is the answer you get by forgetting that the question asked what was there at the start.
Use Vm = 22.7 dm3 mol−1 at STP only to cross between an amount and a gas volume. A published item asks for the volume of NO2 when 1.28 g of copper reacts with excess nitric acid, Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O: 1.28 ÷ 63.55 = 0.0201 mol Cu → 0.0403 mol NO2 → × 22.7 = 0.914 dm3. Forgetting the 1 : 2 ratio gives 457 cm3, an offered option.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
Which amount, in mol, of sodium chloride is needed to make 250 cm3 of 0.10 mol dm−3 solution?
5.0 cm3 of 2.00 mol dm−3 sodium carbonate solution, Na2CO3(aq), was added to a volumetric flask and the volume was made up to 500 cm3 with water. What is the concentration, in mol dm−3, of the solution?
What is the volume, in cm3, of the final solution if 100 cm3 of a solution containing 1.42 g of sodium sulfate, Na2SO4, is diluted to the concentration of 0.020 mol dm−3? Mr(Na2SO4) = 142
What is the concentration of chloride ions, in mol dm−3, in a solution formed by mixing 200 cm3 of 1 mol dm−3 HCl with 200 cm3 of 5 mol dm−3 NaCl?
The complete combustion of 15.0 cm3 of a gaseous hydrocarbon X produces 60.0 cm3 of carbon dioxide gas and 75.0 cm3 of water vapour. What is the molecular formula of X? (All volumes are measured at the same temperature and pressure.)
Which volume of ethane gas, in cm3, will produce 40 cm3 of carbon dioxide gas when mixed with 140 cm3 of oxygen gas, assuming the reaction goes to completion?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
20 cm3 of gas A reacts with 20 cm3 of gas B to produce 10 cm3 of gas AxBy and 10 cm3 of excess gas A. What are the correct values for subscripts x and y in the empirical formula of the product AxBy(g)?
| x | y | |
|---|---|---|
| A. | 2 | 1 |
| B. | 2 | 2 |
| C. | 1 | 1 |
| D. | 1 | 2 |
What volume of carbon dioxide, CO2(g), can be obtained by reacting 1 dm3 of methane, CH4(g), with 1 dm3 of oxygen, O2(g)?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Which sample contains the fewest moles of HCl? NA = 6.02 × 1023 mol−1. Molar volume of an ideal gas at STP = 22.7 dm3 mol−1.
Calcium carbonate is heated to produce calcium oxide, CaO: CaCO3(s) → CaO(s) + CO2(g). Calculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet. [2]
Lithium reacts with water: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g). A 0.200 g piece of lithium was placed in 500.0 cm3 of water. Calculate the molar concentration of the resulting solution of lithium hydroxide. [2]
Solutions and mark-scheme guidance · Set 1H
1H.1 B
n = cV = 0.10 × 0.250 = 0.025 mol. D uses the volume in cm3; A and C divide instead of multiplying.
1H.2 C
c2 = c1V1 ÷ V2 = 2.00 × 5.0 ÷ 500 = 0.020 mol dm−3 — a hundredfold dilution.
1H.3 C
n = 1.42 ÷ 142 = 0.0100 mol, which is unchanged by dilution. V = n ÷ c = 0.0100 ÷ 0.020 = 0.500 dm3 = 500 cm3. B is the volume of water added, not the final volume.
1H.4 C
n(Cl−) = 0.200 × 1 + 0.200 × 5 = 1.2 mol in 0.400 dm3: 3 mol dm−3. D adds the concentrations without allowing for the doubled volume.
1H.5 C
Volumes ratio 15 : 60 : 75 = 1 : 4 : 5. One molecule of X gives 4 CO2 (4 C) and 5 H2O (10 H): C4H10. Here water is a vapour, so it counts in the volume ratio.
1H.6 B
C2H6 : CO2 = 2 : 4, so 40 cm3 CO2 needs 20 cm3 ethane, which uses 70 cm3 of the 140 cm3 O2 — oxygen is in excess, so the ethane is fully converted.
1H.7 D
10 cm3 A reacted with 20 cm3 B to give 10 cm3 product: 1 volume of A + 2 volumes of B → 1 volume of AxBy. With A and B of the same atomicity (both monatomic or both diatomic), the product contains A and B atoms in the ratio 1 : 2, so the empirical formula is AB2. B (A2B2) is not an empirical formula at all.
1H.8 A
Oxygen is limiting: 1 ÷ 2 = 0.5 against methane 1 ÷ 1 = 1. 1 dm3 O2 gives 0.5 dm3 CO2. B assumes methane is limiting.
1H.9 A
A: 0.0100 × 0.1 = 0.001 mol. B: 10 mol. C: 0.365 ÷ 36.46 = 0.0100 mol. D: 2.27 ÷ 22.7 = 0.100 mol. Four routes to an amount, one on each spoke of the mole map.
1H.10 [2]
n(CaCO3) = 555 ÷ 100.09 = 5.55 mol ✔; V = 5.55 × 22.7 = 126 dm3 ✔. Accept a method using pV = nRT with p = 100 kPa (126 dm3) or 101.3 kPa (125 dm3).
1H.11 [2]
n(Li) = 0.200 ÷ 6.94 = 0.0288 mol = n(LiOH) ✔; [LiOH] = 0.0288 ÷ 0.5000 = 0.0576 mol dm−3 ✔. The volume is converted to dm3, and the change in volume of the liquid is neglected.
Review · Structure 1.4
7Misconceptions, the examiner’s view, and the question types
- “A mole of a compound contains NA atoms.” Why it is wrong: it contains NA formula units. Correct model: multiply by the number of the specified entity per formula. Consequence: every counting distractor.
- “Ar and Mr are in g mol−1.” They have no units; molar mass carries g mol−1.
- “The empirical formula is the ratio of masses.” It is the ratio of amounts of atoms.
- “1 : 1.5 rounds to 1 : 2.” Scale to whole numbers; never round a fractional ratio.
- “The volume in n = cV is the volume of water added.” It is the volume of solution, in dm3.
- “Equal volumes of gases have equal masses.” Equal numbers of molecules; their masses differ.
- “The reactant present in the smaller amount is limiting.” Divide by the coefficients first.
Evidence base: the IB Diploma chemistry subject reports quoted in the exam-focus sections of this page.
Answered well: calculating the molar ratios of elements for the first marks of an empirical-formula question; straightforward mass-to-amount conversions.
Found difficult: (1) recognising that oxygen, not listed, must be found by difference — named as the most common mistake in two sessions; (2) turning a fractional ratio into a formula; (3) multi-step atom counts, where most candidates made mistakes in the steps; (4) the standard-solution procedure: graduated cylinders instead of volumetric flasks, and water not described as distilled or deionized; (5) substituting correctly into concentration relationships; (6) identifying the limiting reactant, and solving gas-volume problems directly instead of through moles; (7) distinguishing molar mass from the molecular ion.
What successful answers did: checked that percentages totalled 100 before starting; wrote the unscaled ratio as a line of working; converted every volume to dm3 before forming an amount; named each piece of glassware and the purpose of each rinse; and divided by coefficients before comparing amounts.
Six question types cover most of this topic.
| If the question asks… | …then |
|---|---|
| How many atoms / ions / molecules? | n × (number per formula unit) × NA. Read the noun. |
| Mr or molar mass | Sum Ar (two decimal places), including brackets and water of crystallization; M = m ÷ n. |
| Empirical formula | Total 100 %? → ÷ Ar → ÷ smallest → scale; for combustion, C from CO2, 2H from H2O, O by difference. |
| Molecular formula | M ÷ M(empirical unit), a whole number. |
| Concentration or dilution | n = cV with V in dm3; c1V1 = c2V2; g dm−3 = mol dm−3 × M. |
| Volumes of reacting gases | Work in volumes; divide by coefficients to find the limiting reactant; Vm only to cross to mass. |
8Quick check
- Calculate the relative formula mass of Ca(NO3)2.
- How many molecules, and how many atoms in total, are there in 0.500 mol of CO2?
- A compound is 40.0 % C, 6.7 % H and 53.3 % O by mass. Determine its empirical formula, and its molecular formula if Mr = 180.
- Calculate the mass of sodium chloride needed to make 250 cm3 of a 0.100 mol dm−3 solution.
- Convert 0.150 mol dm−3 sulfuric acid into g dm−3.
- 50 cm3 of nitrogen reacts with 100 cm3 of hydrogen: N2 + 3H2 → 2NH3. Identify the limiting reactant and the volume of ammonia formed.
- A student writes: “weigh the solid, dissolve it in water, pour into a volumetric flask and fill to the line”. State two things that must be added.
- Explain why the percentages in an empirical-formula question sometimes total less than 100 %.
| 1 | 40.08 + 2(14.01) + 6(16.00) = 164.10. |
| 2 | 3.01 × 1023 molecules; 9.03 × 1023 atoms. |
| 3 | 3.33 : 6.63 : 3.33 → 1 : 2 : 1: CH2O (30.03); 180 ÷ 30.03 = 6 → C6H12O6. |
| 4 | 0.100 × 0.250 × 58.44 = 1.46 g. |
| 5 | 0.150 × 98.08 = 14.7 g dm−3. |
| 6 | Hydrogen (100 ÷ 3 = 33.3 < 50 ÷ 1); 100 × 2/3 = 66.7 cm3 NH3. |
| 7 | “Distilled/deionized” water; transfer with washings; then stopper and invert to mix (not “stir”). |
| 8 | An element present was not listed — usually oxygen, found by difference. |
9Summary and knowledge organiser
Essential knowledge
- The mole is the SI unit of amount of substance; N = n × NA, NA = 6.02 × 1023 mol−1; specify the entity.
- Ar and Mr are relative to 12C and have no units; molar mass M has units g mol−1; n = m ÷ M.
- The empirical formula is the simplest whole-number ratio of atoms; the molecular formula the actual numbers; molecular = empirical × (M ÷ Mempirical).
- Empirical formulas come from percentage composition or mass data (including combustion: C from CO2, H from H2O, O by difference).
- n = cV (V in dm3); [X] denotes molar concentration; g dm−3 = mol dm−3 × M; c1V1 = c2V2 for dilution.
- A standard solution is made from an accurately weighed solid, dissolved in distilled water and made up to the mark in a volumetric flask with all washings.
- Avogadro’s law: equal volumes of gases at the same T and p contain equal numbers of molecules, so volume ratios = mole ratios; Vm = 22.7 dm3 mol−1 at STP.
Examination checklist
- Read which entity is being counted.
- Use Ar to two decimal places; give answers to the precision of the data.
- Check that the percentages total 100 %; never round a fractional ratio.
- Convert cm3 to dm3 before n = cV.
- Standard solution: “accurately”, “distilled”, “washings”, “to the mark”, “invert”.
- Gas volumes: work in volumes; divide by coefficients to find the limiting reactant.
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Mole 1.4.1 | N = n × NA; NA = 6.02 × 1023 mol−1. | Atoms ≠ molecules ≠ ions ≠ formula units. |
| Relative masses 1.4.2 | Mr = ΣAr; no units. | Brackets and water of crystallization count. |
| Molar mass 1.4.3 | n = m ÷ M; M in g mol−1. | Molar mass (quantity) ≠ molecular ion (species). |
| Formulas 1.4.4 | % → ÷Ar → ÷smallest → scale; × (M ÷ MEF). | Oxygen by difference; 2 H per H2O; scale, don’t round. |
| Concentration 1.4.5 | n = cV; [ ]; g dm−3 ↔ mol dm−3; c1V1 = c2V2. | V in dm3; volumetric flask; distilled water. |
| Avogadro’s law 1.4.6 | Volume ratio = mole ratio (same T, p); Vm = 22.7 dm3 mol−1 at STP. | Liquids excluded from volume ratios; smallest amount ≠ limiting. |