RIDDLES — THE CONCEPT OF THE MOLE
Relative Atomic Mass & Relative Molecular Mass
Riddle 1: I am a physical quantity used in chemistry. I compare the average mass of an atom of an element to one-twelfth the mass of a carbon-12 atom. I have no unit, since I am a ratio. I account for the natural abundance of an element's isotopes. Who am I?
ANSWER: Relative Atomic Mass (Ar)
Riddle 2: I am the molecular counterpart of relative atomic mass. I am the average mass of one molecule of a substance compared with one-twelfth the mass of a carbon-12 atom. I am found by summing the relative atomic masses of every atom in a molecule. For ionic compounds, I go by a different name — relative formula mass. Who am I?
ANSWER: Relative Molecular Mass (Mr)
Riddle 3: I am a unit used to measure mass on an atomic scale. I am defined as one-twelfth the mass of a single carbon-12 atom. I provide the reference scale on which relative atomic mass is built. I give atomic masses a common standard. Who am I?
ANSWER: Atomic Mass Unit (amu)
The Mole as a Unit of the Amount of Substance
Riddle 4: I am a unit of measurement in chemistry. I represent a fixed number of elementary entities — atoms, ions or molecules. I am defined by the number of atoms present in exactly 12 g of carbon-12. I am the base SI unit for the amount of substance. Who am I?
ANSWER: The Mole (mol)
Riddle 5: I am a general term used in chemistry. I refer to the basic units that make up a substance. I can be an atom, a molecule, an ion, an electron or a proton. Chemists count me in multiples of Avogadro's number. Who am I?
ANSWER: Elementary Entities
Riddle 6: I am a fundamental quantity in chemistry. My base SI unit is the mole. I am found by dividing the number of entities in a sample by Avogadro's constant. I am represented by the symbol n. Who am I?
ANSWER: Amount of Substance
Riddle 7: I am a constant used throughout chemistry. I represent the number of elementary entities contained in one mole of any substance. My value is 6.02 x 10²³ per mole. I am denoted by the symbol L or NA. Who am I?
ANSWER: Avogadro's Number (Avogadro's Constant)
Calculating the Number of Entities, Mass and Volume
Riddle 8: I am a count of particles. I am obtained by multiplying the amount of substance by Avogadro's constant. My formula is N = n × L. I tell you exactly how many atoms, molecules or ions are present in a sample. Who am I?
ANSWER: Number of Entities (N)
Riddle 9: I am a mass expressed per mole. I am numerically equal to the relative atomic or relative molecular mass of a substance. My unit is grams per mole (g mol⁻¹). I link the amount of substance in moles to its mass in grams through the formula m = n × M. Who am I?
ANSWER: Molar Mass
Riddle 10: I am a fixed volume in gas chemistry. I am the volume occupied by one mole of any gas at standard temperature and pressure. My value is 22.4 dm³ mol⁻¹. I apply strictly at s.t.p. Who am I?
ANSWER: Molar Volume (Vm)
Riddle 11: I am a set of reference conditions. I am fixed at a temperature of 273 K and a pressure of 101.3 kPa. Under my conditions, one mole of any gas occupies 22.4 dm³. Chemists use me as a benchmark for comparing gas volumes. Who am I?
ANSWER: Standard Temperature and Pressure (s.t.p.)
Concentration and Standard Solutions
Riddle 12: I am a type of concentration. I am defined as the number of moles of solute dissolved in one cubic decimetre of solution. I am calculated by dividing the amount of solute in moles by the volume of solution in dm³. I am represented by the symbol C. Who am I?
ANSWER: Molarity (Concentration in mol dm⁻³)
Riddle 13: I am also a type of concentration. I am defined as the mass of solute dissolved in one cubic decimetre of a solution. My unit is grams per cubic decimetre (g dm⁻³). I am represented by a symbol similar to the one used for density. Who am I?
ANSWER: Mass Concentration
Riddle 14: I am a formula. I connect molar concentration to mass concentration. I am written as C = ρ/M, where ρ is mass concentration and M is molar mass. I let chemists switch between grams-per-litre and moles-per-litre thinking. Who am I?
ANSWER: Relationship Between Molar and Mass Concentration (C = ρ/M)
Riddle 15: I am a solution used in analytical chemistry. My concentration is accurately known. I am prepared either from a primary standard or by diluting a concentrated stock solution. I am essential for accurate volumetric analysis. Who am I?
ANSWER: Standard Solution
Riddle 16: I am a substance used to prepare standard solutions. I am available in pure form or a state of known purity. I am stable, react speedily and completely, and have high solubility. Sodium carbonate and potassium iodate are examples of me. Who am I?
ANSWER: Primary Standard
Riddle 17: I am a set of requirements. A substance must meet me before it can serve as a primary standard. I demand purity, stability, a reasonably high relative formula mass, high solubility, and a fast, side-reaction-free response. Who am I?
ANSWER: Properties of a Primary Standard
Riddle 18: I am a laboratory procedure. I begin with weighing a solute accurately and dissolving it in a beaker. I end with topping up a volumetric flask to its calibration mark and inverting it to mix. I turn a weighed solid into a solution of known concentration. Who am I?
ANSWER: Preparation of a Standard Solution from a Solid Solute
Riddle 19: I am also a laboratory procedure. I start with a stock solution rather than a solid. I rely on the dilution formula to tell me how much stock to measure. I finish by topping up to the calibration mark, just like preparation from a solid. Who am I?
ANSWER: Preparation of a Standard Solution from a Concentrated Solution
Riddle 20: I am an equation. I relate the concentration and volume of a concentrated solution to those of a diluted one. I am written as C₁V₁ = C₂V₂. I am used to calculate how much stock solution is needed to prepare a solution of lower concentration. Who am I?
ANSWER: The Dilution Formula (C₁V₁ = C₂V₂)
Riddle 21: I am a calculation method. I use the density and percentage purity stated on a commercial stock solution's label. I find the mass of pure substance in one cubic decimetre before dividing by molar mass. I convert a bottle's assay information into a concentration in mol dm⁻³. Who am I?
ANSWER: Determination of the Concentration of a Stock Solution
Riddle 22: I am a percentage. I describe how much of a stock solution's mass is actually the substance of interest. I am multiplied by the total mass of solution to find the mass of pure solute. Commercial HCl is often only 37% me. Who am I?
ANSWER: Percentage Purity