Flashcards · 9202 · Topic 6

Quantitative chemistry flashcards

21 cards on Topic 6, subtopics 3.6.1 to 3.6.4. Pick a subtopic or study them all.

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Why is mass conserved in a chemical reaction?: Atoms are rearranged, not created or destroyed, so the total mass of the products equals the total mass of the reactants.
A metal carbonate is heated in an open crucible and its mass falls. Explain.: Carbon dioxide gas is produced and escapes. In a closed system the total mass would be unchanged.
A metal is heated in air and its mass increases. Explain.: A gas from the air (oxygen) is a reactant, so its mass is added to the product.
What does 2Mg + O₂ → 2MgO tell you quantitatively?: 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide. The coefficients, not the subscripts, give the reacting ratio.
Relative formula mass, Mr: The sum of the relative atomic masses of all the atoms in the formula.
Why are relative atomic masses not whole numbers?: They are weighted averages of the masses of an element’s isotopes, taking their abundances into account.

Equation linking moles, mass and Mr: moles = mass (g) ÷ Mr, so mass = moles × Mr.
Three steps of a reacting-mass calculation: 1. Convert the known mass to moles. 2. Use the equation’s mole ratio to find moles of the wanted substance. 3. Convert back to mass. Never use the mass ratio directly.
Mass of MgO from burning 4.8 g of magnesium (2Mg + O₂ → 2MgO; Ar: Mg = 24, O = 16): 4.8 ÷ 24 = 0.2 mol Mg → 0.2 mol MgO. 0.2 × 40 = 8.0 g
Finding an empirical formula from masses or percentages: Divide each by its Ar, divide all results by the smallest, then scale to whole numbers. Molecular formula = empirical formula × (Mr ÷ empirical formula mass).
A compound is 40.0% C, 6.7% H and 53.3% O, with Mr 180. Empirical and molecular formulae?: C 3.33, H 6.7, O 3.33 → 1 : 2 : 1, CH₂O (mass 30). 180 ÷ 30 = 6, so C₆H₁₂O₆.
Why might you not obtain the calculated mass of product?: The reaction may not go to completion (it is reversible); some product may be lost when it is separated; some reactants may react in unexpected ways. No atoms are lost.

The mole and the Avogadro constant: One mole contains 6.02 × 10²³ particles (the Avogadro constant). Its mass in grams equals the relative formula mass.
Number of particles from moles: number of particles = moles × 6.02 × 10²³
Volume of one mole of any gas at room temperature and pressure: 24 dm³ (24 000 cm³). volume (dm³) = moles × 24.
Why do gas volume ratios equal the mole ratios in an equation?: Equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules.
Volume of CO₂ at rtp from 2.50 g of CaCO₃ with excess HCl (CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂; Mr CaCO₃ = 100): 2.50 ÷ 100 = 0.0250 mol CaCO₃ → 0.0250 mol CO₂. 0.0250 × 24 = 0.600 dm³

Concentration in g/dm³ and in mol/dm³: g/dm³ = mass ÷ volume. mol/dm³ = moles ÷ volume. g/dm³ = mol/dm³ × Mr. Volume must be in dm³ (cm³ ÷ 1000).
Moles of HCl in 25.0 cm³ of 2.00 mol/dm³ hydrochloric acid: 2.00 × (25.0 ÷ 1000) = 0.0500 mol
25.0 cm³ of 0.0500 mol/dm³ H₂SO₄ neutralises 20.0 cm³ of NaOH (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O). Concentration of NaOH?: H₂SO₄: 1.25 × 10⁻³ mol → NaOH: 2.50 × 10⁻³ mol (1 : 2). 2.50 × 10⁻³ ÷ 0.0200 = 0.125 mol/dm³
Which titres are used to calculate the mean?: Only concordant titres (within 0.10 cm³ of each other). Leave out the rough titre.