Quantitative chemistry flashcards

Topic 3: Quantitative chemistry

Law of conservation of mass

No atoms are lost or made in a chemical reaction, so the mass of the products equals the mass of the reactants.

Calculate the Mr of Ca(OH)₂ (Ca = 40, O = 16, H = 1)

40 + 2 × (16 + 1) = 74

Percentage by mass of nitrogen in NH₄NO₃ (N = 14, H = 1, O = 16)

Mr = 14 + 4 + 14 + 48 = 80. % N = (2 × 14) ÷ 80 × 100 = 35%

Why does mass increase when magnesium is heated in an open crucible, but decrease when calcium carbonate is heated?

Magnesium: oxygen atoms from the air join it in the solid oxide. Calcium carbonate: carbon dioxide gas escapes into the air. Mass is still conserved in both.

How is the uncertainty of a mean worked out?

± half the range of the readings. Example: 31.2, 30.8, 31.6 cm³ → mean 31.2, range 0.8, result 31.2 ± 0.4 cm³.

Moles from mass, and the Avogadro constantHigher

moles = mass (g) ÷ Mr. One mole of any substance contains 6.02 × 10²³ of the stated particles.

Limiting reactantHigher

The reactant that is used up completely. It fixes the amount of product; the other reactant is in excess. Compare moles against the equation ratio, not masses.

15 g of solute is dissolved to make 250 cm³ of solution. Concentration in g/dm³?

15 ÷ 0.250 = 60 g/dm³ (divide cm³ by 1000 to get dm³).

Percentage yield: formula, and why it is less than 100%

mass of product actually made ÷ maximum theoretical mass × 100. Less because the reaction may be reversible, some product is lost when it is separated, or reactants react in unexpected ways.

Atom economy: formula, and the value for making CaO from CaCO₃ → CaO + CO₂

Mr of desired product ÷ sum of Mr of all reactants × 100. CaO: 56 ÷ 100 × 100 = 56%

Volume of one mole of gas at RTP. What volume does 3.2 g of O₂ occupy?Higher

24 dm³ at RTP (20 °C, 1 atmosphere). 3.2 ÷ 32 = 0.10 mol → 0.10 × 24 = 2.4 dm³

0.050 mol of solute is dissolved in 250 cm³. Concentration in mol/dm³?Higher

0.050 ÷ 0.250 = 0.20 mol/dm³. Multiply by Mr to convert to g/dm³.