Quantitative chemistry Flashcards
(made w/ freesciencelessons) - left off at "Using concentration of solution 2 (Triple)"
What is the law of conservation of mass?
The idea that no atoms are lost or made during a chemical reaction, therefore, the mass of the products equals the mass of the reactants and the number of atoms in each element is the same.
What is a reactant?
The starting substances in a reaction.
What is a product?
The chemicals formed in a reactant.
Question:
92g of sodium reacts with 32g of oxygen. Calculate the mass of sodium oxide produced.
4Na + O₂ ⭢ 2Na₂O
92 + 32 = 124g
Total mass of product(s) = total mass of reactant(s)
Question:
A mass of calcium carbonate reacted to produce 112g of calcium oxide and 88g of carbon dioxide. Calculate the mass of calcium carbonate that reacted.
CaCO₃ ⭢ CaO + CO₂
112 + 88 = 200g
Total mass of reactant(s) = total mass of product(s)
Question:
A mass of magnesium oxide reacts with 73g of hydrogen chloride to produce 95g of magnesium chloride and 18g of water. Calculate the mass of magnesium oxide in the reaction.
MgO + 2HCl ⭢ MgCl₂ + H₂O
95 + 18 = 113g
113 - 73 = 40g
Total mass of product(s) = total mass of reactant(s)
How [when are] are ionic compounds formed?
When metals react with non-metals.
What are ions?
Atoms of an element with an overall charge that are formed when atoms gain or lose electrons.
What ions do metals / [most] non-metals form?
- Metals form positive ions.
- Most non-metals form negative ions.
What often correlates with the charge of an ion? What does not follow this pattern? Why?
The group number of the metals;
* transition metals as they can have [form] several different [ions] charges
* non-metal ionic compounds as they consist of several nonmetal atoms.
What are the charges of the non-metal ionic compounds?
Hydroxide: OH⁻ (1-)
Nitrate: NO₃⁻ (1-)
Sulfate: SO₄²⁻ (2-)
Carbonate: CO₃²⁻ (2-)
Ammonium: NH₄⁺ (1+)
Question:
Find the formula, using the charges, for sodium chloride.
Na⁺ + Cl⁻ ⭢ NaCl
Question:
Find the formula, using the charges, for sodium oxide.
Na⁺ + O²⁻ ⤚ (Na × 2) ⭢ Na₂O
Question:
Find the formula, using the charges, for magnesium iodide.
Mg²⁺ + I⁻ ⤚ (I × 2) ⭢ MgI₂
Question:
Find the formula, using the charges, for lithium carbonate.
Li⁺ + CO₃²⁻ ⤚ (Li × 2) ⭢ Li₂CO₃
Question:
Find the formula, using the charges, for calcium hydroxide.
Ca²⁺ + OH⁻ ⤚ (OH × 2) ⭢ Ca(OH)₂
Question:
Find the formula, using the charges, for magnesium nitrate.
Mg²⁺ + NO₃⁻ ⤚ (NO₃ × 2) ⭢ Mg(NO₃)₂
Question:
Balance the equation:
Calcium oxide + hydrochloric acid ⭢ Calcium chloride + water
CaO + HCl ⭢ CaCl₂ + H₂O
CaO + 2HCl ⭢ CaCl₂ + H₂O
Question:
Balance the equation:
Iron oxide + carbon monoxide ⭢ Iron + carbon dioxide
Fe₂O₃ + CO ⭢ Fe + CO₂
Fe₂O₃ + 3CO ⭢ 2Fe + 3CO₂
What is the Relative Atomic Mass (Ar)?
The average mass of the atoms of an element, taking into account naturally occurring isotopes, compared with carbon-12 (which is given a mass of exactly 12).
What is the Relative Formula Mass (Mr) of a compound?
The sum of the relative atomic masses (Ar) of the atoms in the numbers shown in the formula.
What are two things to watch out for when calculating the Relative Formula Mass (Mr)?
- They have no units
- Do not take into account coefficients (big numbers) when calculating the Mr.
Question:
Calculate the Mr of CaSO₄
(Ar of Ca = 40), (Ar of S = 32), (Ar of O = 16)
Mr = 40 + 32 + (16 × 4) = 136
Question:
Calculate the Mr of Mg(OH)₂
(Ar of Mg = 24), (Ar of O = 16), (Ar of H = 1)
Mr = 24 + (16 × 2) + (1 × 2) = 58
What is the equation for calculating the percentage composition (percentage by mass)?
% composition: 100 (total Ar of element ÷ total Mr of compound)
Question:
Calculate the percentage by mass of calcium (Ca) in calcium chloride (CaCl₂).
Ar of Ca = 40
Mr of CaCl₂ = 111
Percentage by mass of Ca = (40 ÷ 111) × 100 = (0.3603) × 100 = 36%
% composition: 100 (total Ar of element ÷ total Mr of compound)
Question:
Calculate the percentage by mass of hydrogen (H) in calcium chloride (CH₄).
Ar of H = 1
Mr of CH₄ = 16
Percentage by mass of H = ((4 × 1) ÷ 16) × 100 = (0.25) × 100 = 25%
% composition: 100 (total Ar of element ÷ total Mr of compound)
Question:
Calculate the percentage by mass of sodium (Na) in sodium sulphate (Na₂SO₄).
Mr of Na₂SO₄ = (23 × 2) + 32 + (16 × 4) = 142
Ar of Na = 23
Percentage by mass of Na = ((23 × 2) ÷ 142) × 100 = (0.3239) × 100 = 32.4%
% composition: 100 (total Ar of element ÷ total Mr of compound)
What is the equation for calculating the number of moles of an element?
[Higher tier]
Number of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
You are given a sample of magnesium with a mass of 48g. How many moles of magnesium have you been given?
[Higher tier]
Number of moles = 48g ÷ 24 = 2 mol
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
You are given 120g of calcium. How many moles of calcium have you been given?
[Higher tier]
Number of moles = 120g ÷ 40 = 3 mol
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
A sample of rock contains 252g of iron. Calculate the number of moles of iron in the sample.
[Higher tier]
Number of moles = 252g ÷ 56 = 4.5 mol
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
You are given a sample of sulphur with a mass of 4064g. Calculate the number of moles of sulphur in the sample.
[Higher tier]
Number of moles = 4064g ÷ 32 = 127 mol
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
What is the equation for calculating the number of moles of a compound?
[Higher tier]
Number of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Question:
You are given a sample of calcium carbonate (CaCO₃) with a mass of 300g. Calculate the number of moles of calcium carbonate in the sample.
[Higher tier]
Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100
Number of moles = 300g ÷ 100 = 3 mol
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Question:
You are given 380g of magnesium chloride (MgCl₂). How many moles of magnesium chloride have you been given?
[Higher tier]
Mr of MgCl₂ = 24 + (35.5 × 2) = 95
Number of moles = 380g ÷ 95 = 4 mol
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Question:
You are given a sample of lithium sulfate (Li₂SO₄) with a mass of 990g. Calculate the number of moles of lithium sulfate in the sample.
[Higher tier]
Mr of Li₂SO₄ = (7 × 2) + 32 + (16 × 4) = 110
Number of moles = 990g ÷ 110 = 9 mol
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Question:
You are given 64.5g of beryllium hydroxide (Be(OH)₂). Calculate the number of moles of beryllium hydroxide that you have been given.
[Higher tier]
Mr of Be(OH)₂ = 9 + (16 × 2) + (1 × 2) = 43
Number of moles = 64.5g ÷ 43 = 1.5 mol
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Rearrange the equation for calculating the number of moles of a compound and an element to make mass the subject.
[Higher tier]
- Mass (g) = number of moles (mol) × Relative formula mass or Mr
- Mass (g) = number of moles (mol) × Relative atomic mass or Ar
Question:
Calcuate the mass of four moles of sodium chloride (NaCl).
[Higher tier]
Mr of NaCl = 23 + 35.5 = 58.5
Mass = 4 mol × 58.5 = 234g
Mass (g) = number of moles (mol) × Relative formula mass or Mr
Question:
Calculate the mass of three moles of potassium oxide (K₂O).
[Higher tier]
Mr of K₂O = (39 × 2) + 16 = 94
Mass = 3 mol × 94 = 282g
Mass (g) = number of moles (mol) × Relative formula mass or Mr
Question:
Calculate the mass of 0.1 moles of caesium nitrate (CsNO₃).
[Higher tier]
Mr of CsNO₃ = 133 + 14 + (16 × 3) = 195
Mass = 0.1 mol × 195 = 19.5g
Mass (g) = number of moles (mol) × Relative formula mass or Mr
Question:
Calculate the mass of 5 moles of coper sulfate (CuSO₄).
[Higher tier]
Mr of CuSO₄ = 63.5 + 32 + (16 × 4) = 159.5
Mass = 5 mol × 159.5 = 797.5g
Mass (g) = number of moles (mol) × Relative formula mass or Mr
Question:
2g of hydrogen reacts with 71g of chlorine to make 73g of hydrogen chloride. Using this information, balance the equation.
H₂ + Cl₂ ⭢ HCl
[Higher tier]
Moles of hydrogen = 2g ÷ 2 = 1 mol
Moles of chlorine = 71g ÷ 71 = 1 mol
Moles of hydrogen chloride = 73g ÷ 36.5 = 2 mol
| 1 ÷ 1 = 1 | 1 ÷ 1 = 1 | 2 ÷ 1 = 2 |
H₂ + Cl₂ ⭢ 2HCl
Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient
Question:
54g of aluminium reacts with 216g of iron (II) oxide, forming 102g of aluminium oxide and 168g of iron. Using this information, balance the equation.
Al + FeO ⭢ Al₂O₃ + Fe
[Higher tier]
Moles of aluminium = 54g ÷ 27 = 2 mol
Moles of iron (II) oxide = 216g ÷ (56 + 16) = 3 mol
Moles of aluminium oxide = 102g ÷ ((27 × 2) + (16 × 3)) = 1 mol
Moles of iron = 168g ÷ 56 = 3 mol
| 2 ÷ 1 = 2 | 3 ÷ 1 = 3 | 1 ÷ 1 = 1 | 3 ÷ 1 = 3 |
2Al + 3FeO ⭢ Al₂O₃ + 3Fe
Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient
Question:
1248g of barium chloride reacts with 684g of aluminium sulfate, forming 1398g of barium sulfate and 534g of aluminium chloride. Using this information, balance the equation.
BaCl₂ + Al₂(SO₄)₃ ⭢ BaSO₄ + AlCl₃
[Higher tier]
Moles of barium chloride = 1248g ÷ (137 + (35.5 × 2)) = 6 mol
Moles of aluminium sulfate = 684g ÷ ((27 × 2) + (32 × 3) + 3(16 × 4)) = 2 mol
Moles of barium sulfate = 1398g ÷ (137 + 32 + (16 × 4)) = 6 mol
Moles of aluminium chloride = 534g ÷ (27 + (35.5 × 3)) = 4 mol
| 6 ÷ 2 = 3 | 2 ÷ 2 = 1 | 6 ÷ 2 = 3 | 4 ÷ 2 = 2 |
3BaCl₂ + Al₂(SO₄)₃ ⭢ 3BaSO₄ + 2AlCl₃
Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient
Question:
24g of magnesium reacts with 16g of oxygen, forming 40g of magnesium oxide. Using this information, balance the equation.
Mg + O₂ ⭢ MgO
[Higher tier]
Moles of magnesium = 24g ÷ 24 = 1 mol
Moles of oxygen = 16g ÷ (16 × 2) = 0.5 mol
Moles of magnesium oxide = 40g ÷ (24 + 16) = 1 mol
| 1 ÷ 0.5 = 2 | 0.5 ÷ 0.5 = 1 | 1 ÷ 0.5 = 2 |
2Mg + O₂ ⭢ 2MgO
Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient
What is Avogadro’s constant?
[Higher tier]
The number of atoms, molecules or ions in a mole of a given substance.
What is the value of Avogadro’s constant?
[Higher tier]
6.02 × 10²³
Question:
Calculate the number of moles of atoms in one mole of water molecules (H₂O).
1 molecule of H₂O = 3 atoms
∴ 1 mole of H₂O = 3 moles of atoms
Question:
Calculate the number of moles of atoms in one mole of methane (CH₄).
[Higher tier]
1 molecule of CH₄ = 5 atoms
∴ 1 mole of CH₄ = 5 moles of atoms
Question:
Calculate the number of moles of atoms in one mole of calcium hydroxide (Ca(OH)₂).
[Higher tier]
1 molecule of Ca(OH)₂ = 5 atoms
∴ 1 mole of Ca(OH)₂ = 5 moles of atoms
Question:
Calculate the number of atoms in one mole of hydrogen chloride (HCl).
[Higher tier]
1 molecule of HCl = 2 atoms
1 mole of HCl = 6.02 × 10²³
Number of atoms in one mole 2 × 6.02 × 10²³ = 1.204 × 10²⁴
Question:
Calculate the number of atoms in one mole of sodium oxide (Na₂O).
[Higher tier]
Number of atoms in 1 molecule of Na₂O = 3
1 mole = 6.02 × 10²³
Number of atoms in one mole = 3 × 6.02 × 10²³ = 1.806 × 10²⁴
Question:
Calculate the number of atoms in 48g of magnesium (Mg).
[Higher tier]
Moles (mol) = 48g ÷ 24 = 2 mol
1 atom in one element of magnesium.
Number of atoms = 2 × 6.02 × 10²³
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
Calculate the number of atoms in 28g of lithium (Li).
[Higher tier]
Moles (mol) = 28g ÷ 7 = 4 mol
1 atom in one element of lithium.
Number of atoms = 4 mol × 6.02 × 10²³ = 2.408 × 10²⁴
# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar
Question:
Calculate the number of atoms in 56g of calcium oxide (CaO).
[Higher tier]
Mole (mol) = 56g ÷ (40 + 16) = 1 mol
2 atoms in one molecule of calcium oxide.
Number of atoms = 2 × 6.02 × 10²³ = 1.204 × 10²⁴
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
Question:
Calculate the number of atoms in 54g of water (H₂O).
[Higher tier]
Moles = 54g ÷ ((1 × 2) + 16) = 3 mol
3 atoms in one molecule of water.
3 × 3 = 9
Number of atoms = 9 × 6.02 × 10²³ = 5.418 × 10²⁴
# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr
What are the steps to calculating the masses of products or reactants using reacting masses in a chemical equation?
[Higher tier]
- Identify the known and unknown substances from the question.
- Calculate Mr or Ar of the known substance using your periodic table.
- Calculate moles of known substance using Mr (or Ar) and the given mass.
- Workout the moles of the unknown substance using the ratio, which are the coefficients.
- Calculate the Mr or Ar of the unknown substance using your periodic table.
- Calculate the mass of the unknown substance using the moles and Mr (or Ar).
Question:
Calculate the mass of magnesium chloride that could be produced from 72g of magnesium. Assume that the chlorine is unlimited.
Mg + Cl₂ ⭢ MgCl₂
[Higher tier]
-
Identify known & unknown substance.
Known: Mg | Unknown: MgCl₂ -
Ar or Mr of known substance.
Ar of Mg = 24 -
Moles of known substance.
Moles (mol) = 72g ÷ 24 = 3 mol -
Moles of unknown substance from ratio.
Ratio = 1:1 | Moles (mol) = 3 moles -
Mr or Ar of unknown substance.
Mr of MgCl₂ = 95 -
Mass of unknown substance.
Mass (g) = 3 mol × 95 = 285g
Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Calculate the mass of calcium sulfate that could be produced from 80g of calcium. Assume that the sulfuric acid is unlimited.
Ca + H₂SO₄ ⭢ CaSO₄ + H₂
[Higher tier]
-
Identify known & unknown substance.
Known: Ca | Unknown: CaSO₄ -
Ar or Mr of known substance.
Ar of Ca = 40 -
Moles of known substance.
Moles (mol) = 80g ÷ 40 = 2 mol -
Moles of unknown substance from ratio.
Ratio = 1:1 | Moles (mol) = 2 moles -
Mr or Ar of unknown substance.
Mr of CaSO₄ = 136 -
Mass of unknown substance
Mass (g) = 2 mol × 136 = 272g
Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Calculate the mass of calcium carbonate that we would need to produce 224g of calcium oxide.
CaCO₃ ⭢ CaO + CO₂
[Higher tier]
-
Identify known & unknown substance.
Known: CaO | Unknown: CaCO₃ -
Mr or Ar of known substance.
Mr of CaO = 56 -
Moles of known substance
Moles (mol) = 224g ÷ 56 = 4 mol -
Moles of unknown substance from ratio.
Ratio: 1:1 | Moles (mol) = 4 moles -
Mr or Ar of unknown substance.
Mr of CaCO₃ = 100 -
Mass of unknown substance.
Mass (g) = 4 mol × 100 = 400g
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Calculate the mass of magnesium chloride that could be produced from 146g of hydrochloric acid. Assume that the magnesium hydroxide is unlimited.
Mg(OH)₂ + 2HCl ⭢ MgCl₂ + H₂O
[Higher tier]
-
Identify known and unknown substance.
Known: HCl | Unknown: MgCl₂ -
Mr or Ar of known substance.
Mr of HCl = 36.5 -
Moles of known substance
Moles (mol) = 146g ÷ 36.5 = 4 mol -
Moles of unknown substance from ratio.
Ratio = 2:1 | Moles (mol) = 4 ÷ (2 ÷ 1) = 2 moles -
Mr or Ar of unknown substance.
Mr of MgCl₂ = 95 -
Mass of unknown substance.
Mass (g) = 2 mol × 95 = 190g
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Calculate the mass of sodium sulfate that could be produced from 240g of sodium hydroxide. Assume that the sulfuric acid is unlimited.
2NaOH + H₂SO₄ ⭢ Na₂SO₄ + 2H₂O
[Higher tier]
-
Identify known and unknown substance.
Known: NaOH | Unknown: Na₂SO₄ -
Mr or Ar of known substance.
Mr of NaOH = 40 -
Moles of known substance
Moles (mol) = 240g ÷ 40 = 6 mol -
Moles of unknown substance from ratio.
Ratio = 2:1 | Moles (mol) = 6 ÷ (2 ÷ 1) = 3 moles -
Mr or Ar of unknown substance.
Mr of Na₂SO₄ = 142 -
Mass of unknown substance.
Mass (g) = 3 mol × 142 = 426g
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Calculate the mass of hydrogen peroxide that could produce 64g of oxygen.
2H₂O₂ ⭢ 2H₂O + O₂
[Higher tier]
-
Identify known and unknown substance.
Known: O₂ | Unknown: H₂O₂ -
Mr or Ar of known substance.
Ar of O₂ = 16 × 2 = 32 -
Moles of known substance
Moles (mol) = 64g ÷ 32 = 2 mol -
Moles of unknown substance from ratio.
Ratio = 1:2 | Moles (mol) = 2 × 2 = 4 moles -
Mr or Ar of unknown substance.
Mr of H₂O₂ = 34 -
Mass of unknown substance.
Mass (g) = 4 mol × 34 = 136g
Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr
Question:
Nitrogen and hydrogen form ammonia shown by the following equation:
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia.
[Higher tier]
-
Identify known and unknown substance.
Known: NH₃ | Unknown: N₂ -
Mr of known substance.
Mr of NH₃ = 17 -
Convert mass into grams
1 tonne = 1000kg = 1 × 10⁶g | Mass = 6.8 tonnes ⤚ (×1,000,000) ⭢ 6,800,000g -
Moles of known substance.
Moles (mol) = (6.8 × 10⁶)g ÷ 17 = 400,000 mol -
Moles of unknown from ratio.
Ratio = 2:1 | Moles = 400,000 ÷ (2 ÷ 1) = 200,000 moles -
Mr of unknown substance
Mr of N₂ = 28 -
Mass of unknown substance.
Mass (g) = 200,000 mol × 28 = 5,600,000g (5.6 tonnes)
Moles (mol) = mass (g) ÷ relative formula mass or Mr
What is a limited reactant (or reagent)?
[Higher tier]
The reactant (in a chemical equation) that is completely used up, limiting the amount of products produced.
What is the excess reactant (or reagent)?
[Higher tier]
The reactant or reagent that is left after a chemical reaction is complete.
What are the steps to calculating the masses of products or reactants using limited reactants in a chemical equation?
[Higher tier]
- Identify and ignore, for the rest of your calculations, the excess reactant.
- Identify the known and unknown substance from the limited reactant and what you’re trying to find.
- Find the moles of product from moles of limited reactant (they are the same).
- Find the mass of the product using moles and Mr or Ar.
Question:
How many moles of zinc iodide would be produced if we used 0.5 moles of zinc and 1 mole of iodine? Calculate the mass of product.
Zn + I₂ ⭢ ZnI₂
[Higher tier]
-
Identify and ignore the excess reactant.
Given moles: 1 > 0.5 | Excess reactant = I₂ -
Identify known and unknown substance.
Known: Zn | Unkown: ZnI₂ -
Find moles of product from limited reactant.
Moles of Zn = moles of ZnI₂ = 0.5 mol -
Mass of product.
Mass (g) = 0.5 mol × (65 + (127 × 2)) = 159.5g
Mass (g) = number of moles (mol) × relative formula mass or Mr
Question:
How many moles of sodium chloride will be produced if we used 1 mole of sodium hydroxide and 0.25 moles of hydrochloric acid? Calculate the mass of sodium chloride produced.
NaOH + HCl ⭢ NaCl + H₂O
[Higher tier]
-
Identify and ignore the excess reactant.
Given moles: 1 > 0.25 | Excess = NaOH -
Identify known and unknown substance.
Known: NaOH | Unknown: NaCl -
Moles of product from limited reactant.
Moles of NaOH = moles of NaCl | Moles = 0.25 mol -
Mass of product.
Mass (g) = 0.25 mol × (23 + 35.5) = 14.625g
Mass (g) = number of moles (mol) × relative formula mass or Mr
Question:
How many moles of copper will be produced if we used 0.5 moles of copper sulphate and 1 mole of magnesium? Calculate the mass of copper produced.
CuSO₄ + Mg ⭢ MgSO₄ + Cu
[Higher tier]
-
Identify and ignore the excess reactant.
Given moles: 0.5 > 1 | Excess = Mg -
Identify known and unknown substance.
Known: CuSO₄ | Unknown: Cu -
Moles of product from limited reactant.
Moles of CuSO₄ = moles of Cu | Moles = 0.5 mol -
Mass of product.
Mass (g) = 0.5 mol × 63.5 = 31.75g
Mass (g) = number of moles (mol) × relative atomic mass or Ar
What is meant by concentration; what is the unit?
[Foundation & Higher tier]
Concentration tells us the mass of a solute in a given volume of solution; g/dm³.
What is a solute?
A chemical that is dissolved in a solvent.
What is the equation for calculating the concentration of a solution?
(not in specification but worth learning)
Concentration (g/dm³) = mass (g) ÷ volume (dm³)
Question:
200g of a chemical is dissolved in water to a final volume of 1 dm³. Calculate the concentration of the solution.
Concentration = 200g ÷ 1 dm³ = 200g/dm³
Concentration (g/dm³) = mass (g) ÷ volume (dm³)
Question:
150g of a chemical is dissolved in water to a final volume of 0.5 dm³. Calculate the concentration of the solution.
Concentration (g/dm³) = 150g ÷ 0.5 dm³ = 300 g/dm³
Concentration (g/dm³) = mass (g) ÷ volume (dm³)
Quetsion:
Calculate the mass of a chemical needed to dissolve in a final volume of 0.4 dm³ to give a final concentration of 600 g/dm³.
Mass (g) = 600 g/dm³ × 0.4 dm³ = 240g
Mass (g) =| Concentration (g/dm³) × volume (dm³)
Question:
Calculate the final volume of a solution containing 200g of a chemical with a concentration of 800 g/dm³.
Volume (dm³) = 200g ÷ 800g/dm³ = 0.25 dm³
Volume (dm³) = mass (g) ÷ concentration (g/dm³)
What 2 factors affect the concentration?
[Higher tier]
- Mass of the solute: increase in mass = increase in concentration (if volume of solution kept the same).
- Volume of the solution: increase in volume = decrease in concentration (if mass of solute kept the same).
What is yield?
Triple only
The mass of product that a chemical reaction produces.
What is the problem with calculating the yield, in practice?
Triple only
It is not always possible to achieve 100% yield in a chemical reaction.
What are the reasons why you cannot always achieve 100% yield in a chemical reaction? (3)
Triple only
- Some of the product may be lost when it is separated from the reaction mixture.
- Some of the reactants may react in different ways to the expected reaction so we do not get the product we expect.
- Reversible reactions may not go to completion.
What is the equation for calculating the percentage yield?
Triple only
Percentage yield = 100 (mass of product actually made ÷ maximum theoretical mass of product)
What are the steps to calculating the percentage yield?
Triple only
- Find out the theoretical mass by calculating the mass of the product provided in the question, using reacting masses.
- Calculate the percentage yield using your calculated maximum theoretical mass and the mass of the product given in the question.
Question:
A scientist reacted 48g of magnesium and produced 150g of magnesium sulphate. Calculate the percentage yield.
Mg + H₂SO₄ ⭢ MgSO₄ + H₂
Triple only
Step 1:
-
Identify known and unknown substances
Known: Mg | Unknown: MgSO₄ -
Mr of known and unknown substance.
Ar of Mg = 24 | Mr of MgSO₄ = 120 -
Moles of known substance.
Moles (mol) = 48g ÷ 24 = 2 mol -
Mole of unknown from ratio.
Ratio = 1:1 | Moles (mol) = 2 mol -
Mass of unknown.
[Theoretical] mass (g) = 2 mol × 120 = 240 g
Step 2:
Percentage yield = 150g ÷ 240g = 0.625 × 100 = 62.5%
- Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
- % yield = 100 (actual mass ÷ theoretical mass)
Question:
The theoretical yield of beryllium chloride was 10.7g. If the reaction actually yields 4.5g, what was the percent yield?
Be + 2HCl ⭢ BeCl₂ + H₂
Triple only
Percentage yield = (4.5g ÷ 10.7g) × 100 = 0.42 × 100 = 42%
% yield = 100 (actual mass ÷ theoretical mass)
Question:
2a) A scientist began this reaction with 20 grams of lithium hydroxide and an unlimited amount of KCl.
What is the theoretical yield of lithium chloride (grams)?
LiOH + KCl ⭢ LiCl + KOH
2b) The reaction actually produced 6 grams of lithium chloride. What is the percent yield?
Triple only
Part A:
-
Identify known and unknown substances.
1Known: LiOH | Unknown: LiCl -
Mr of known and unknown substance.
Mr of LiOH = 24 | Mr of LiCl = 42.5 -
Moles of known substance.
Moles (mol) = 20g ÷ 24 = 0.8 mol -
Mole of unknown from ratio.
Ratio = 1:1 | Moles (mol) = 0.8 mol -
Mass of unknown.
[Theoretical] mass (g) = 0.8 mol × 42.5 = 34 g
Part B:
Percentage yield = 6g ÷ 34g = 0.176 × 100 = 17.6%
- Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
- % yield = 100 (actual mass ÷ theoretical mass)
Question:
Nitrogen and hydrogen at high temperatures are converted to ammonia using the following reaction known as the Haber Process.
3H₂ + N₂ ⭢ 2NH₃
When 400 grams of H₂ are added to an excess amount of N₂, 104 grams of NH₃ are formed. Calculate the percent yield.
Triple only
Step 1:
-
Identify known and unknown substances.
Known: H₂ | Unknown: NH₃ -
Mr of known and unknown substance.
Mr of H₂ = 2 | Mr of NH₃ = 17 -
Moles of known substance.
Moles (mol) = 400g ÷ 2 = 200 mol -
Mole of unknown from ratio.
Ratio = 3:2 | Moles = 200 ÷ (3 ÷ 2) = 133.3 mol -
Mass of unknown.
[Theoretical] mass (g) = 133.3 mol × 17 = 2266g
Step 2:
Percentage yield = 104g ÷ 2266g = 0.046 × 100 = 4.6%
- Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
- % yield = 100 (actual mass ÷ theoretical mass)
Question:
If the typical yield is 86.78%, how much SO₂ should be expected if 4897 grams of ZnS are used? (2803 of SO₂)
ZnS + O₂ ⭢ ZnO + SO₂
Triple only
Step 1:
-
Identify known and unknown substances.
Known: ZnS | Unknown: SO₂ -
Mr of known and unknown substance.
Mr of ZnS = 97 | Mr of SO₂ = 64 -
Moles of known substance.
Moles (mol) = 4897g ÷ 97 = 50.5 mol -
Mole of unknown from ratio.
Ratio = 1:1 | Moles = 50.5 mol -
Mass of unknown.
[Theoretical] mass (g) = 50.5 mol × 64 = 3232g
Step 2:
Actual mass = 86.78% × 3232g = 2804.7g
- Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
- Actual mass = % yield × theoretical mass
Question:
A scientist reacted 130g of zinc with unlimited copper sulfate. They produced 25.4g of copper. Calculate the percentage yield.
Zn + CuSO₄ ⭢ Cu + ZnSO₄
Triple only
Step 1:
-
Identify the known and unknown.
Known: Zn | Unknown: Cu -
Ar of known and unknown.
Zn = 65 | Cu = 63.5 -
Moles of known.
Moles = 130g ÷ 65 = 2 mol -
Moles of unknown from ratio.
Ratio = 1:1 | Moles = 2 mol -
Mass of unknown.
[Theoretical] mass = 2 mol × 63.5 = 127g
Step 2:
Percentage yield = 25.4g ÷ 127g = 0.2 × 100 = 20%
- Number of moles (mol) = mass (g) ÷ relative atomic mass or Ar
- % yield = actual mass ÷ theoretical mass
Question:
A scientist reacted 27g of beryllium with unlimited hydrochloric acid. They produced 144g of beryllium chloride. Calculate the percentage yield.
Be + 2HCl ⭢ BeCl₂ + H₂
Triple only
Step 1:
-
Identify known and unknown.
Known: Be | Unknown: BeCl₂ -
Mr and Ar of known and unknown.
Be = 9 | BeCl₂ = 80 -
Moles of known.
Moles = 27g ÷ 9 = 3 mol -
Find moles of unknown from ratio.
Ratio = 1:1 | Moles = 3 mol -
Mass of unknown.
[Theoretical] mass = 3 mol × 80 = 240g
Step 2:
Percentage yield = 144g ÷ 240g = 0.6 × 100 = 60%
- Number of moles (mol) = mass (g) ÷ relative atomic mass or Ar
- % yield = actual mass ÷ theoretical mass
What is atom economy
Triple only
A measure of the amount of starting materials that end up as useful products.
Why is a high atom economy important in industry?
Triple only
- By minimising the production of unwanted products, money is saved.
- By minimising the production of unwanted products, sustainability increases as resources are not wasted.
What is the equation for calculating atom economy?
Triple only
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
What is the greatest atom economy we can have? Why?
Triple only
100% because anything greater would mean we created atoms, which is not possible.
Question:
Calculate the percentage atom economy for the production of silver iodide in this reaction.
The equation for the reaction is:
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq)
Give your answer to 3 significant figures.
Relative formula masses:
AgNO₃ = 170 | NaI = 150 | AgI = 235 | NaNO₃ = 85
Triple only
Atom economy = 235 ÷ 320 = (0.7340) × 100 = 73.4%
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
Question:
This question is about the extraction of metals.
Tungsten (W) is a metal. Tungsten is produced from tungsten oxide by reaction with hydrogen.
The equation for the reaction is:
WO₃ + 3H₂ → W + 3H₂O
Calculate the percentage atom economy when tungsten is produced in this reaction.
Relative formula masses (Mr):
WO₃ = 232 | H₂ = 2
Triple only
Ar of W = 184
Atom economy = 184 ÷ (232 + (3 × 2)) = 0.773 × 100 = 77.3%
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
How do chemists try to reduce waste when unwanted side products have been produced?
They use the side products in other reactions.
Question:
This question is about reversible reactions and equilibrium.
Hydrogen is used to produce ammonia in the Haber process.
The hydrogen is made in two stages.
Stage 1 is the reaction of methane and steam to produce carbon monoxide and hydrogen.
The equation for the reaction is:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
Calculate the atom economy for the formation of hydrogen in stage 1.
Relative atomic masses (Ar): H = 1 | C = 12 | O = 16
Triple only
Atom economy = 3(1 × 2) ÷ 34 = 0.17647 × 100 = 17.7%
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
Question:
An equation for the reaction is:
NiO + C ⟶ Ni + CO
Calculate the percentage atom economy for the reaction to produce nickel.
Relative atomic masses (Ar): C = 12 | Ni = 59
Relative formula mass (Mr): NiO = 75
Give your answer to 3 significant figures.
Triple only
Atom economy = 59 ÷ (75 + 12) = 0.6781 × 100 = 67.8%
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
Question:
(d) Look at the equations for the two reactions:
Reaction 1: CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + CO₂(g)
Reaction 2: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l)
Calculate the percentage atom economy for Reaction 2 [Copper chloride].
Reactive formula masses: CuO = 79.5 | HCl = 36.5 | CuCl₂ = 134.5 | H₂O = 18
(e) The atom economy for Reaction 1 is 68.45%.
Compare the atom economies of the two reactions for making copper chloride.
Give a reason for the difference.
Triple only
Part D:
Atom economy = 134.5 ÷ (79.5 + (2 × 36.5)) = 0.882 × 100 = 88.2%
Part E:
Reaction 1 has a lower atom economy than reaction 2 because more side products were produced.
Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100
What is meant by concentration; what is the unit?
Triple only
Concentration tells us the number of moles of a solute in a given volume of solution; mol/dm³.
What is the equation for calculating the concentration of a solution?
Triple only
Concentration (mol/dm³) = number of moles ÷ volume (dm³)
Question:
A solution has a concentration of 0.5 mol/dm³. Calculate the number of moles in 0.2 dm³.
Triple only
Number of moles (mol) = 0.5 mol/dm³ × 0.2 dm³ = 0.01 moles
# of moles = concentration (mol/dm³) × volume (dm³)
Question:
A solution has a concentration of 0.2 mol/dm³. Calculate the number of moles in 1.5 dm³.
Triple only
Number of moles = 0.2 mol/dm³ × 1.5 dm³ = 0.3 moles
# of moles = concentration (mol/dm³) × volume (dm³)
Question:
A solution of calcium chloride has a concentration of 0.4 mol/dm³. Calculate the mass of calcium chloride in 2 dm³.
Mr: CaCl₂ = 111
Triple only
Step 1: Calculate the number of moles.
Number of moles = 0.4 mol/dm³ × 2 dm³ = 0.8 moles
Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.8 moles × 111 = 88.8g (1.d.p)
# of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass or Mr
Question:
A solution of sodium nitrate has a concentration of 0.8 mol/dm³. Calculate the mass of sodium nitrate in 0.5 dm³.
Mr: NaNO₃ = 85
Triple only
Step 1: Calculate the number of moles.
Number of moles = 0.8 mol/dm³ × 0.5 dm³ = 0.4 moles
Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.4 moles × 85 = 34g
# of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass or Mr
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“Using concentration of solution 2 (Triple)”
