1.5 Chemical Formulae, Equations, Calculations (Edexcel IGCSE Chemistry)

The missing numbers to balance the equation are:

N₂ + 3H₂ 2NH₃

Spectator ions are ions which appear on both sides of an ionic equation and are eliminated from the final equation.

2KI + Br₂ 2KBr + I₂ is not an ionic equation because there are no ions.

M_r is the symbol for the relative formula mass and it refers to the total mass of the substance.

To calculate the relative formula mass of a substance, you have to add up the relative atomic masses of all the atoms present in the formula.

The equation for the percentage by mass of an element is:

False.

Relative molecular mass does not apply to ionic compounds.

Molar mass refers to the mass of one mole of a substance.

The relative formula mass of CuCO₃ is:

63.5 + 12 + (16 x 3) = 123.5

The relative molecular mass of C₅H₅N is:

(12 x 5) + (1 x 5) + 14 = 79

The difference between relative molecular mass and relative formula mass is that relative moleculas mass only applies to covalent compounds.

Relative formula mass applies to all compounds.

The relative molecular mass of NH₄NO₃ is:

14 + (1 x 4) + 14 + (16 x 3) = 80

True.

The relative formula mass of CaCO₃ is = 40+12 +(16 x 3) = 100

The total number of ions in one mole of NaCl is:

2 x 6.02 x 10²³ = 1.204 x 10²⁴

Using moles, mass in grams and the molar mass, the equation is:

Moles = mass / molar mass

True.

There are 6.02 x 10²³ carbon dioxide molecules in one mole of carbon dioxide.

The equation for molar mass using moles and mass is:

molar mass = mass / moles

To calculate the number of moles:

• Moles = mass / molar mass

• Moles = 45.0 / 18.0

• Moles = 2.5

To calculate the mass:

• Mass = moles x molar mass

• Mass = 0.25 x 123.5

• Mass = 30.9 (or 30.875) g

To calculate the molar mass:

Molar mass = mass / moles

Molar mass = 22.0 / 0.5

Molar mass = 44.0 g mol^-1

To calculate the molar mass of a compound for a moles, mass, molar mass calculation, you add the atomic masses of all the atoms in the compound.

The equation connecting moles, mass and molar mass is:

mass = moles x molar mass

False.

When completing reacting masses calculations, the mass unit (e.g. grams, tonnes) does not affect the calculation.

The mass unit only affects the final anwer as it must be given in the appropriate units.

The correct order of steps to complete a reacting mass calculation is:

1. Write a balanced symbol equation.

2. Calculate relevant molar masses.

3. Calculate the moles of the known chemical.

4. Check the molar ratio in the equation.

5. Determine the moles of the target chemical.

6. Calculate the mass of the target chemical.

CuCO₃ CuO + CO₂

The molar ratio of copper carbonate to copper oxide is 1 : 1.

2Na + Cl₂ 2NaCl

The molar ratio of chlorine to sodium chloride is 1 : 2.

False.

The masses of reactants and products in a reaction can be used to write a balanced equation for the reaction.

How the masses of the reactants and products be used to form a balanced chemical equation:

• Convert all masses to moles

• Find the molar ratio of all chemicals

• Simplify the molar ratio

• Place these values in front of each chemical in the equation

False.

3A + B 2C + 2D

The molar ratio of A : C is not 2 : 3.

There are 3 moles of A and 2 moles of C, which means the ratio is 3 : 2.

The moles, mass, molar mass equation that is needed for reacting mass questions is:

moles = mass / molar mass

Percentage yield compares the actual yield to the theoretical yield.

Actual yield is the recorded amount of product obtained from a chemical reaction.

The actual yield can be determined by experiment only.

Theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions.

The type of calculation lets you determine theoretical yield is a reacting mass calculation.

The equation for percentage yield is:

(actual yield theoretical yield) x 100

True.

For economic reasons, the objective is to have as high a percentage yield as possible to increase profits and reduce costs and waste.

If the actual yield is 1.50 g and theoretical yield is 2.00g, the percentage yield is:

(1.50 2.0) x 100 = 75%

A molecular formula shows the number and type of each atom in a molecule.

The molecular formula of ethanoic acid is C₂H₄O₂.

Empirical formula is the simplest whole number ratio of the atoms of each element present in one molecule or formula unit.

The empirical formula of propene is CH₂.

False.

Organic molecules often have different empirical and molecular formulae.

The formula of an ionic compound is always an empirical formula.

The empirical formula is calculated from the ratio of masses of each element in the compound.

The molecular formula is a multiple of the empirical formula.

It indicates the exact numbers of each atom in the compound.

False.

Empirical formula shows the simplest whole number ratio of atoms in a molecule.

To calculate the molecular formula from the empirical formula:

1. Calculate the relative formula mass of the empirical formula.

2. Divide the relative formula mass of the compound by the mass of the empirical formula.

3. Multiply each number of elements by this number to find the molecular formula.

Using moles, volume in dm³ and concentration, the equation is:

Moles = concentration x volume

To calculate the number of moles of solute present in 2.0 dm³ of a solution whose concentration is 0.15 mol dm^-3:

• Moles = concentration x volume

• Moles = 0.15 x 2.0 = 0.3 moles

The 2 common units for concentration are:

• g dm^-3

• mol dm^-3

To convert g dm^-3 to mol dm^-3, you divide the concentration in g dm^-3 by the molar mass.

The equation for concentration using moles and volume is:

concentration = moles / volume

The concentration of a solution where 1.0 mole of solute is dissolved in 2.0 dm³ of water is:

Concentration = moles / volume

Concentration = 1.0 / 2.0 = 0.5 mol dm^-3

False.

To convert dm³ into cm³, you multiply by 1000.

The equation to calculate the volume of a solution using moles and concentration is:

Volume = moles / concentration

True.

Avogadro’s Law states that equal amounts of gases occupy the same volume of space at the same temperature and pressure.

The equation for molar gas volume using moles and volume is:

Molar gas volume = volume / moles

The conditions for room temperature and pressure are:

• 20 ^oC

• 1 atmosphere

The equation to calculate moles using volume and molar gas volume is:

Moles = volume / molar gas volume

At rtp, the number of moles of gas present in 48 dm³ is:

• Moles = volume / molar gas volume

• Moles = 48 / 24 = 2

One mole of any gas at room temperature and pressure occupies:

• 24 dm³

• 24 000 cm³

The equation connecting moles, volume and molar gas volume is:

Volume = moles x molar gas volume

The amount of space, in dm³, that 1.5 moles of gas occupies is:

• Volume = moles x molar gas volume

• Volume = 1.5 x 24 = 36 dm³

True.

To convert volume from dm³ to cm³, you multiply by 1000.

A 300 dm³ cylinder contains 300 000 cm³ of gas.