Chemical Quantities (AQA GCSE Combined Science: Synergy: Physical Sciences): Flashcards

When magnesium burns, it reacts with oxygen from the air to form magnesium oxide:

2Mg (s) + O₂ (g) ⟶ 2MgO (s)

The oxygen absorbed from the air adds to the mass of the solid product, so the total mass increases.

The percentage yield compares the actual yield to the theoretical yield: percentage yield = (actual yield ÷ theoretical yield) × 100. It is always 100% or less because some product is always lost during the reaction.

In a chemical reaction, atoms are neither created nor destroyed, only rearranged. This is why chemical equations must be balanced.

10 g is one quarter of 40 g, so:

1. mass of CaO = 56 ÷ 4 = 14 g

This follows directly from the Law of Conservation of Mass.

True.

This reaction takes place in solution — no gas is produced and no substance escapes. The system behaves as a closed system, so mass is conserved.

The relative atomic mass (A_r) of an element is an average value that compares the mass of atoms of that element with the mass of a carbon-12 atom, taking account of the isotopes of the element and their abundances.

The relative formula mass (M_r) of a compound is the sum of the relative atomic masses of all the atoms in the formula, in the numbers shown.

H₂O contains 2 hydrogen atoms and 1 oxygen atom:

1. (2 × 1) + (1 × 16) = 18

The relative formula mass (M_r) of water is 18.

False.

Relative atomic mass is a ratio comparing atomic masses to the carbon-12 isotope. Because it is a ratio, it has no units.

The percentage by mass of an element in a compound = × 100

There are 2 nitrogen atoms in NH₄NO₃:

1. % N = (2 × 14) ÷ 80 × 100 = 35%

Ca(OH)₂ contains 1 Ca, 2 O, and 2 H atoms:

1. (1 × 40) + (2 × 16) + (2 × 1) = 74

The relative formula mass (M_r) of Ca(OH)₂ is 74.

True.

Because a compound is made entirely of its constituent elements, the percentage by mass values of all elements must sum to 100%.

Stoichiometry refers to the ratio of moles of each substance in a balanced chemical equation. The coefficients in the equation show the proportions in which reactants and products react and form.

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. It is determined by dividing the moles of each element by the smallest value to give the simplest ratio.

The mole (mol) is the unit for amount of substance. One mole of any substance contains exactly 6.02 x 10²³ particles (atoms, molecules or ions). This number is called the Avogadro constant.

False.

One mole of carbon contains 6.02 × 10²³ atoms. One mole of CO₂ contains 6.02 × 10²³ molecules. Both contain the same number of their stated particles, but CO₂ consists of molecules, not atoms.

The Avogadro constant is 6.02 x 10²³ per mole. It represents the number of particles (atoms, molecules or ions) in exactly one mole of any substance.

The Avogadro constant has a value of 6.02 × 10²³ per mole. This is the number of particles in one mole of any substance.

Use the formula: moles = mass ÷ M_r

For example, the number of moles in 4.0 g of NaOH (M_r = 40): moles = 4.0 ÷ 40 = 0.10 mol.

False.

Each formula unit of MgCl₂ contains 2 chloride ions, so one mole of MgCl₂ contains 2 × 6.02 × 10²³ = 1.204 × 10²⁴ chloride ions.

Stoichiometry refers to the coefficients (numbers in front of reactants and products) in a balanced chemical equation. These numbers show the molar ratio in which substances react and are produced.

The empirical formula is the simplest whole-number ratio of atoms of each element in a substance. It can be deduced from the masses of each element using moles.

The law of conservation of mass states that no atoms are created or destroyed during a chemical reaction. The total mass of the reactants always equals the total mass of the products. This is why equations must be balanced.

The empirical formula shows the simplest ratio of atoms in a compound. To find it, divide moles of each element by the smallest value to give whole-number ratios.

False.

Coefficients must be whole numbers. If the ratios include a decimal such as 2.5, multiply all values by a suitable integer (here × 2) to give whole numbers: 2, 4, 5.

The relative formula mass (M_r) is the sum of the relative atomic masses (A_r) of all the atoms shown in the formula. For example, for H₂O: M_r = (2 x 1) + 16 = 18.

True.

Values very close to a whole number (e.g. 1.003 ≈ 1) can be rounded. Only values that are clearly not near a whole number (e.g. 0.5 or 0.33) require multiplying all values up to whole numbers.

The molar ratio gives the ratio of the amounts in moles of each substance that reacts or is produced. It is also used as the coefficients in the balanced symbol equation.

The limiting reactant is the reactant that is completely used up in a chemical reaction. It limits the duration of the reaction and determines the maximum amount of product that can form.

The excess reactant is the reactant that is not completely used up — some remains once the reaction has stopped. The amount of product is determined by the limiting reactant, not the excess reactant.

1. Convert each mass to moles (mass ÷ molar mass)

2. Write the balanced equation and find the molar ratio

3. Determine which reactant would be used up first by comparing moles with the ratio

4. The reactant used up first is the limiting reactant

False.

The molar ratio from the balanced equation must also be considered. A reactant with more moles could still be limiting if the equation requires a much larger proportion of it.

1. Moles Na = 9.2 ÷ 23 = 0.40 mol; Moles S = 8.0 ÷ 32 = 0.25 mol

2. Molar ratio Na : S = 2 : 1, so 0.40 mol Na needs 0.20 mol S

3. 0.25 mol S is available — more than enough

4. Na is the limiting reactant; S is in excess

The amount of product obtainable from a reaction is directly proportional to the amount of the limiting reactant. If you double the amount of this reactant, you obtain double the product.

True.

Once the limiting reactant is completely consumed, there is no more of it available to react. The reaction stops regardless of how much excess reactant remains.

The molar ratio is the ratio of the moles of each substance in a balanced symbol equation. It is used to calculate the mass of reactants or products from a given mass of another substance.

1. Calculate the moles of the given substance: moles = mass ÷ M_r

2. Use the molar ratio from the balanced equation to find moles of the product.

3. Calculate the mass of the product: mass = moles × M_r

False.

The molar ratio depends on the balanced equation. For example, in 2Mg + O₂ → 2MgO, the ratio of Mg to MgO is 1:1, but in 2Al₂O₃ → 4Al + 3O₂, the ratio of Al₂O₃ to Al is 1:2.

To calculate moles from a mass, divide the mass by the relative formula mass (M_r). To calculate mass from moles, multiply the moles by the M_r.

1. Moles of Mg = 6 ÷ 24 = 0.25 mol

2. Ratio Mg : MgO = 1:1, so moles MgO = 0.25 mol

3. M_r of MgO = 24 + 16 = 40

4. Mass = 0.25 × 40 = 10 g

The relative formula mass (M_r) is the sum of the relative atomic masses of all atoms in a formula. It is used to convert between moles and mass in reacting mass calculations.

False.

As long as you are consistent, you can work in any mass unit (grams, tonnes, etc.). The reacting masses will always be in proportion to the balanced equation regardless of the unit used.

The molar ratio of Al₂O₃ to Al is 1:2. For every 2 moles of Al₂O₃ decomposed, 4 moles of Al are produced.

True.

A balanced symbol equation gives the molar ratio of each substance. Using the relative formula mass, you can convert moles to mass (or vice versa) for any substance in the equation.

Concentration is the amount of solute dissolved in a given volume of solution. It is measured in g/dm³ (grams per cubic decimetre).

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

True.

1 dm³ (cubic decimetre) = 1000 cm³. Since 1 dm³ = 1 litre, this is equivalent to saying 1 litre = 1000 cm³. Divide by 1000 to convert cm³ → dm³.

A student dissolves 10 g of sodium hydroxide in 2 dm³ of water. The concentration = 10 ÷ 2 = 5 g/dm³.

Divide by 1000. For example, 200 cm³ ÷ 1000 = 0.2 dm³. (To convert dm³ back to cm³, multiply by 1000.)

A solute is a substance that dissolves in a liquid (the solvent) to form a solution. The concentration of the solution depends on how much solute is dissolved in a given volume.

False.

The greater the amount of solute in a given volume, the greater the concentration. Concentration and the amount of dissolved solute are directly proportional.

Step 1: Convert volume — 200 cm³ ÷ 1000 = 0.2 dm³

Step 2: concentration = 4.0 ÷ 0.2 = 20 g/dm³

True.

g/dm³ means grams per cubic decimetre. This is the standard unit for concentration in mass per volume calculations and is worth stating explicitly in exam answers.