Chemical Formulae, Equations, Calculations (Edexcel IGCSE Science (Double Award): Chemistry): Flashcards

The relative formula mass (M_r) is the total mass of a molecule or formula unit, calculated by adding up the relative atomic masses (A_r) of all the atoms in the formula.

The M_r of Ca(OH)₂ is calculated as:

(1 × 40) + (2 × 16) + (2 × 1) = 40 + 32 + 2 = 74

To calculate the M_r of a substance, add up the relative atomic masses of all the atoms present in the formula.

True.

This follows from the Law of Conservation of Mass. No atoms are created or destroyed, so the sum of the relative formula masses of the reactants equals the sum for the products.

The relative atomic mass (A_r) is the weighted average mass of all the isotopes of an element, calculated from their mass numbers and relative abundances. It is the larger of the two numbers shown on the Periodic Table.

The M_r of (NH₄)₂SO₄ is calculated as:

(2 × 14) + (8 × 1) + (1 × 32) + (4 × 16) = 28 + 8 + 32 + 64 = 132

The M_r of potassium carbonate, K₂CO₃, is calculated as (2 × 39) + (1 × 12) + (3 × 16) = 138.

False.

The relative atomic mass is the larger of the two numbers on the Periodic Table. The smaller number is the atomic number (number of protons).

The mole (mol) is the unit for the amount of a substance. One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules or ions): the Avogadro constant.

The Avogadro constant is 6.02 × 10²³ per mole. It represents the number of atoms, molecules or ions in one mole of any substance.

The mass of one mole of a substance is called its molar mass and has units of g mol^-1.

True.

One mole of any substance always contains 6.02 × 10²³ particles. Whether those particles are atoms (as in carbon) or molecules (as in CO₂), the count is the same.

Molar mass is the mass of one mole of a substance in grams per mole (g mol^-1). For an element it equals the A_r; for a compound it equals the M_r.

The relationship is:

moles = mass ÷ molar mass

mass = moles × molar mass

molar mass = mass ÷ moles

The mass of 0.250 mol of zinc is:

1. Molar mass of Zn = 65.38 g mol^-1

2. Mass = moles × molar mass

3. Mass = 0.250 × 65.38 = 16.3 g

To find the number of moles from a mass, divide the mass by the molar mass.

True.

The M_r of H₂O = (2 × 1) + 16 = 18, so the molar mass is 18 g mol^-1.

The number of moles in 2.64 g of sucrose is:

1. Moles = mass ÷ molar mass

2. Moles = 2.64 ÷ 342.3

3. Moles = 7.71 × 10^-3 mol

The three steps are:

1. Calculate the moles of the known substance (mass ÷ M_r)

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

3. Calculate the mass of the unknown (moles × M_r)

In a reacting mass calculation, the molar ratio from the balanced equation is used to convert moles of one substance into moles of another.

The mass of MgO produced is:

1. Moles of Mg = 6.0 ÷ 24 = 0.25 mol

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

3. M_r of MgO = 24 + 16 = 40; mass = 0.25 × 40 = 10 g

False.

Reacting mass calculations work in any consistent mass unit (grams, kilograms, tonnes, etc.) because reacting masses are always in proportion to the balanced equation.

A molar ratio is the ratio of the number of moles of each substance in a reaction, shown by the coefficients in a balanced chemical equation.

To determine the balanced equation from masses:

1. Calculate the M_r of each substance

2. Convert each mass to moles (mass ÷ M_r)

3. Find the simplest whole-number ratio of moles

4. Use that ratio as the coefficients in the balanced equation

When calculating reacting masses, the chemical equation must be balanced before identifying molar ratios.

True.

The coefficients are 2 and 4, which simplify to a ratio of 1 : 2 — two moles of Al are produced for every one mole of Al₂O₃ decomposed.

Steps to find the balanced equation:

1. Write the unbalanced equation: C + ZnO → CO₂ + Zn

2. Calculate moles: C = 1.2/12 = 0.1; ZnO = 16.2/81 = 0.2; CO₂ = 4.4/44 = 0.1; Zn = 13/65 = 0.2

3. Simplest ratio: 1 : 2 : 1 : 2

4. Balanced equation: C + 2ZnO → CO₂ + 2Zn

To calculate the number of moles from a mass, use the formula: moles = mass ÷ M_r.

Percentage yield is the actual yield expressed as a percentage of the theoretical yield.\n\nIt measures how successful a chemical reaction or process has been.

Percentage yield = (actual yield / theoretical yield ) × 100

False.

In a real experiment, some product is always lost, so percentage yield is always less than 100%. A result above 100% indicates a calculation error — most commonly dividing theoretical yield by actual yield instead of the other way around.

100% yield is never achieved because product can be lost at several stages. Reactants may be left behind in equipment, the reaction may be reversible, and product can be lost during separation and purification steps such as filtration or distillation.

Theoretical yield is the maximum amount of product that could be obtained if the reaction went to completion with no losses.

It is calculated from the balanced equation and the reacting masses.

The actual yield is the recorded amount of product obtained in an experiment, while the theoretical yield assumes 100% conversion of reactants to products.

Percentage yield is calculated as follows:

1. Percentage yield = (actual yield / theoretical yield) × 100

2. Percentage yield = (1.6 / 2.0) × 100

3. Percentage yield = 80%

True.

Side reactions consume reactants to produce unwanted products, meaning less reactant is available to form the desired product and overall yield is reduced.

Manufacturers aim to maximise percentage yield to increase profits and reduce costs. A higher yield means more product is obtained from the same amount of reactants, which also reduces waste.

Water of crystallisation is water that is chemically bonded within the crystal structure of a hydrated salt.

It is represented in the formula as .xH₂O, for example CuSO₄.5H₂O.

The formula can be determined by measuring masses before and after a reaction, then converting those masses into moles. The molar ratio of the reactants and products gives the empirical formula of the compound.

To find the formula of hydrated copper sulfate, the salt is heated until it turns from blue to white, indicating that all the water of crystallisation has been removed.

True.

Overheating can cause the anhydrous salt itself to decompose, so gases other than water vapour are also lost. This makes the final mass lower than expected, giving a larger calculated mass of water than was actually present.

The steps are:

1. Calculate the mass of the anhydrous salt and the mass of water lost

2. Divide each mass by the M_r of the salt and water respectively to find moles

3. Divide both values by the smaller number of moles to get the simplest ratio

4. Write the formula in the form salt.xH₂O

The mass of water lost from a hydrated salt is calculated by subtracting the mass of the anhydrous salt from the mass of the original hydrated salt.

False.

In practice, experimental measurements produce ratios that are close to but not exactly whole numbers. You must round to the nearest whole number to write the formula.

When a gas is lost or gained, the change in mass is easy to measure accurately by weighing before and after the reaction. This allows the mass of the gas involved to be calculated without needing to collect it directly.

Anhydrous describes a substance that contains no water of crystallisation.\n\nFor example, anhydrous copper sulfate is the white powder that remains after hydrated copper sulfate is heated.

The empirical formula is the simplest whole number ratio of atoms of each element present in a compound.

For example, the empirical formula of magnesium oxide is MgO.

The steps are:

1. Find the mass of the metal and the mass of oxygen that combined

2. Divide each mass by the A_r of the element to find the moles

3. Divide both values by the smaller number of moles to find the simplest ratio

4. Write the empirical formula in the form M_xO_y

In the magnesium oxide experiment, the crucible lid is lifted frequently to allow air in while preventing magnesium oxide smoke from escaping.

True.

When the mass remains constant, all the magnesium has reacted with oxygen. Heating should stop once constant mass is reached — prolonged heating beyond this point risks side reactions such as magnesium nitride formation.

The mass of oxygen is determined by subtracting the mass of magnesium used from the mass of the magnesium oxide produced.

Mass of magnesium oxide is calculated by subtracting the mass of the empty crucible from the mass of the crucible and its contents after heating.

In the copper(II) oxide reduction experiment, a steady stream of methane is passed over the heated metal oxide, and the excess gas is burned off.

The mass of oxygen is determined by subtracting the mass of the remaining metal powder from the original mass of the copper(II) oxide.

This gives the mass of oxygen that was removed during the reduction.

True.

When the black copper(II) oxide has fully changed to pink/brown copper metal, all the oxygen has been removed and the reaction is complete.

The boiling tube is held horizontal so that the gas flows evenly over the surface of the copper(II) oxide, ensuring thorough contact between the methane and the metal oxide throughout the experiment.

The molecular formula shows the actual number and type of each atom in one molecule of a compound.

For example, the molecular formula of ethanoic acid is C₂H₄O₂.

The empirical formula is the simplest whole number ratio of the atoms of each element present in a compound.

For example, the empirical formula of ethanoic acid is CH₂O.

True.

Ionic compounds exist as giant lattices rather than discrete molecules, so their formula always represents the simplest whole number ratio of ions — which is the empirical formula.

The steps are:

1. Write the element symbols and their given values (mass or percentage)

2. Write the A_r of each element

3. Calculate moles for each element: moles = value / A_r

4. Divide all mole values by the smallest to get the ratio

5. If any ratio is not a whole number, multiply all ratios to make them whole

6. Write the empirical formula using these whole number ratios

A sample contains 10 g of hydrogen and 80 g of oxygen. Moles of H = 10/1 = 10; moles of O = 80/16 = 5. The ratio H:O is 2 : 1, so the empirical formula is H₂O.

The steps are:

1. Calculate the M_r of the empirical formula by adding the A_r values of all atoms in it

2. Divide the M_r of the molecular formula by the M_r of the empirical formula to get a whole number multiplier

3. Multiply the number of each element in the empirical formula by that multiplier

A compound has the empirical formula C₄H₁₀S and an M_r of 180. Determine the multiplier and hence the molecular formula.

The multiplier is 2 and the molecular formula is C₈H₂₀S₂.

False.

Organic molecules often have different empirical and molecular formulae. For example, ethene has the molecular formula C₂H₄ but the empirical formula CH₂.

The steps are:

1. Find the mass of the anhydrous salt and the mass of water lost

2. Divide each mass by its M_r (salt and water) to find moles

3. Divide both values by the moles of the salt to get the ratio

4. Round to the nearest whole number — this is x in the formula

If the molar ratio of moles when dividing all values by the smallest gives a non-whole number such as 1.5, you must multiply all ratio values by 2 to get whole numbers.