chemistry - key concepts in chemistry: calculations involving masses (1.43 - 1.53)

notations

M_r = relative formula mass

A_r = relative atomic mass

1.43 calculate M_r mass given A_rs

M_r = sum of A_r of atoms in formula

e.g. M_r of CO₂:

1. A_r of C = 12, A_r of O = 16

2. 12 + (2 × 16) = 44

1.43 calculate % by mass of element in compound given A_rs

e.g. % by mass of oxygen in CO₂:

1. M_r of CO₂ = 12 + (2 × 16) = 44

2. O₂ = 2 × 16 = 32

3. oxygen/total = 32/44 = 0.727 × 100% = 72.7%

empirical formula

simplest whole number ratio of atoms of each element in substance

molecular formula

actual number of atoms of each element in one molecule of substance

1.44 calculate formulae of simple compounds (empirical formulae) from reacting masses

1.44 calculate formulae of simple compounds (empirical formulae) from % composition

1. assume 100g sample

2. convert % → grams

   • e.g. 75% = 75g

3. follow table

1.45 find empirical formula from molecular formula

e.g. empirical formula of C₆H₁₂O₆:

1. find HCF = 6

2. divide subscripts by 6: 6/6 = 1; 12/6 = 2

3. CH₂O

1.45 find molecular formula from empirical formula & M_r

e.g. molecular formula of glucose: empirical formula = CH₂O, M_r = 180

1. find empirical formula mass: 12 + (2 × 1) + 16 = 30

   • A_r of C = 12, A_r of H = 1, A_r of O = 16

2. M_r/empirical formula mass: 180/30 = 6

3. molecular formula = empirical formula x 6: C₆H₁₂O₆

1.46 how to find empirical formula of simple compound - e.g. magnesium oxide

1. heat magnesium ribbon in limited oxygen supply

2. weigh reactant & product

e.g. empirical formula of MgO: 0.576g Mg ribbon heated, produced 0.960g MgO

1. find mass O: 0.960 - 0.576 = 0.384

2. O = 0.384, Mg = 0.576

3. answer/smallest number: 0.384/0.384 = 1, 0.576/0.384 = 1.5

4. make whole number ratio: 1:1.5 = 2:3

5. empirical formula = Mg₂O₃

1.47 law of conservation of mass in closed system - precipitation reaction

lead nitrate solution + potassium iodide solution → lead iodide (yellow precipitate) + potassium nitrate (colourless solution)

Pb(NO₃)_{2 (aq)} + 2KI _(aq) → PbI_{2 (s)} + 2KNO_{3 (aq)}

number of atoms doesn’t change - mass can’t change

closed system - no new substances added/removed

1.47 law of conservation of mass in non-enclosed system - open flask

copper carbonate heated in air → copper oxide + carbon dioxide

CuCO_{3 (s)} → CuO _(s) + CO_{2 (g)}

mass of copper oxide left < mass copper carbonate at start

CO₂ gas escapes - mass decreases

non-enclosed system - gas can escape

1.48 calculate masses of reactants & products from balanced equations given mass of 1 substance

e.g. mass of chlorine needed to make 53.4g of aluminium chloride

1. write balanced equation: 2Al + 3Cl₂ → 2AlCl₃

2. fill in M_rs & known masses in table

   • M_r doesn’t take big numbers into account

3. find moles of aluminium chloride: n = m/Mr → 53.4/133.5 = 0.4

4. find moles of chlorine: 3:2 = x:0.4, 2/0.4 = 5, 3/5 = 0.6, x = 0.6 → 3:2 = 0.6:0.4

5. find mass of chorine: m = n x Mr → 0.6 × 71 = 42.6g

1.49 calculate concentrations of solutions in g dm^-3

1.50 Avogadro constant

one mole of particles of substance = 6.02 × 10²³ atoms/ molecules/ions

1.50 moles & mass

mass of one mole of substance = A_r or M_r in grams

e.g. mass of one mole of O₂ = 16 × 2 = 32g

1.51 calculate number of moles in given mass

mass/M_r = number of moles

1.51 calculate mass given number of moles

moles × M_r = mass

1.51 calculate number of particles given number of moles

number of moles × (6.02 × 10²³) = number of particles

1.51 calculate number of moles given number of particles

number of particles/(6.02 × 10²³) = number of moles

1.51 calculate number of particles given mass

mass/M_r = moles

number of moles × (6.02 × 10²³) = number of particles

1.51 calculative mass given number of particles

number of particles/(6.02 × 10²³) = number of moles

moles × M_r = mass

1.52 mass of product formed controlled

one of reactants often added in excess - not completely used up

mass of product formed controlled by mass of reactant not in excess (limiting reactant)

stoichiometry

ratio of moles of each substance

1.53 find stoichiometry of from masses of reactants & products (finding balanced equation)

e.g. 10.8g of aluminium reacted with 42.6g of chlorine to produce aluminium chloride (AlCl₃) - find balanced equation:

1. fill in M_rs & known masses in table

2. find moles aluminium & chlorine: n = m/M_r → 10.8/27 = 0.4, 42.6/71 = 0.6

3. divide both answers by smaller number: 0.4/0.4 = 1, 0.6/0.4 = 1.5

4. find simplest whole number ratio & write equation: 1:1.5 = 2:3 → 2Al + 3Cl₂ = xAlCl₃

5. balance normal way → 2Al + 3Cl₂ = 2AlCl₃