Stoichiometry and the mole flashcards

Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. Central to this is the mole concept: one mole contains 6.02 x 10^23 particles (Avogadro's constant) and has a mass equal to the relative atomic or molecular mass in grams. Students must be able to convert between mass, moles, number of particles, and gas volumes at standard conditions. Molar ratios from balanced equations allow calculation of theoretical yields, limiting reagents, and percentage yields. Empirical and molecular formulae can be determined from percentage composition data. Solutions require understanding of concentration in mol per decimetre cubed. Mastery of these skills underpins almost every quantitative calculation in IB Chemistry.

1. Card 1

   What is one mole of a substance?

   One mole is the amount of substance containing exactly 6.02 x 10^23 particles, whether atoms, molecules, or ions.

2. Card 2

   What is Avogadro's constant and what is its value?

   Avogadro's constant is the number of particles in one mole of any substance; its value is 6.02 x 10^23 mol^-1.

3. Card 3

   How do you calculate the number of moles from a given mass?

   Use moles = mass divided by molar mass, where molar mass is the mass in grams of one mole of the substance.

4. Card 4

   What is the difference between empirical and molecular formulae?

   The empirical formula shows the simplest whole-number ratio of atoms; the molecular formula shows the actual number of atoms in one molecule.

5. Card 5

   How is the molecular formula determined from the empirical formula?

   Divide the known molar mass of the compound by the empirical formula mass, then multiply each subscript in the empirical formula by that whole number.

6. Card 6

   What is a limiting reagent in a chemical reaction?

   The limiting reagent is the reactant that is completely consumed first, determining the maximum amount of product that can form.

7. Card 7

   How do you calculate percentage yield?

   Percentage yield = (actual yield divided by theoretical yield) multiplied by 100; it shows how efficient the reaction was in practice.

8. Card 8

   What does concentration in mol dm^-3 mean?

   It is the number of moles of solute dissolved per decimetre cubed (litre) of solution, written as c = n divided by V.

9. Card 9

   What volume does one mole of any ideal gas occupy at STP (0 degrees C, 100 kPa)?

   At STP, one mole of an ideal gas occupies 22.7 dm^3, a value used to convert between moles and gas volumes.

10. Card 10

    How do molar ratios from a balanced equation guide stoichiometric calculations?

    The molar ratio from the balanced equation tells you the proportional amounts of each substance that react or are produced.

11. Card 11

    What information is needed to find the empirical formula from percentage composition?

    Divide each element's percentage by its relative atomic mass, then find the simplest whole-number ratio of the resulting values.

12. Card 12

    What is the difference between the relative atomic mass and the molar mass of an element?

    Relative atomic mass is a dimensionless ratio; molar mass has the same numerical value but carries units of g mol^-1.

• Students often confuse the empirical formula with the molecular formula, forgetting that the molecular formula can be a whole-number multiple of the empirical formula.
• A common error is using the mass of a substance directly in molar ratio calculations instead of first converting mass to moles.
• Many students assume that the reagent present in the smaller mass is always the limiting reagent, without accounting for differences in molar mass and stoichiometric coefficients.

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