Calculating Mass of Substances (AQA GCSE Chemistry)

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

False.

The mass of 1 mole of a compound in grams is equal to its M_r.

The molar mass of carbon dioxide, CO₂, is 44 g/mol.

The number of moles in 2.64 g of sucrose is:

• Moles = mass / M_r

• Moles = 2.64 / 342 = 7.72 x 10^-3.

The number of atoms in 15.7 g of water is:

• Moles = mass / M_r

• Moles = 15.7 / 18 = 0.872

One mole of water contains:

• 3 x 6.02 x 10²³ = 18.06 x 10²³atoms

So, 0.872 moles of water contains:

• 0.872 x 18.06 x 10²³ atoms = 1.58 x 10²⁴ atoms.

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

6.0 g of magnesium = 6.0 / 24 = 0.25 moles

2Mg + O₂ 2MgO

The ratio of Mg : MgO is 1 : 1

So, 0.25 moles of magnesium oxide can be produced from 6.0 g of magnesium.

M_r (Al₂O₃) = 102.

Moles of aluminium oxide = 51 / 102 = 0.5.

The ratio of Al₂O₃ : Al is 1 : 2.

So, 0.5 moles of aluminium oxide will produce 1.0 moles of aluminium.

So, the maximum mass of aluminum that can be produced is 27 tonnes.

A chemical equation should be balanced before performing reacting mass calculations to ensure that the correct molar ratios are used in the calculations.

To balance a chemical equation given the masses of the chemicals involved:

1. Calculate the molar masses of the substances in the equation

2. Calculate the moles of each chemical by dividing the mass by the molar mass

3. Use the mole ratios to balance the equation.

A coefficient is a number placed in front of a chemical formula in an equation to balance it.

To convert uneven mole ratios to whole numbers, multiply all numbers by the same factor to find the smallest whole number for each coefficient.

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

To determine the empirical formula of a compound:

• Calculate the moles of each element

• Find the ratio of these moles

• Convert the ratio to the simplest whole number ratio

• Write the answer with each whole number written as a subscript to its element

The empirical formula shows the simplest whole number ratio of atoms, while the molecular formula shows the actual number of atoms in a molecule.

False.

The empirical formula can be the same as the molecular formula if the molecule is already in its simplest ratio.

In chemical equations, stoichiometry refers to the ratios of reactants and products.

It is shown by large numbers written in front of the reactants and products.

A balanced chemical equation is important because it:

• Shows the correct ratios of reactants and products

• Allows accurate calculations in chemical reactions.

A limiting reactant is the reactant that is completely consumed in a chemical reaction and limits the amount of product formed.

False.

The limiting reactant is the reactant that produces the least amount of product according to the balanced equation and the given quantities.

The first step in determining the limiting reactant is to convert the mass of each reactant into moles.

A reactant is in excess when there is still some of that reactant left after a chemical reaction is complete.

True.

In a reaction, the limiting reactant determines the maximum amount of product that can be formed.

False.

In a reaction with a limiting reactant, excess reactants are not completely consumed.

When the limiting reactant is completely consumed, the reaction stops and no more product is formed.

False.

There can only be one limiting reagent in a reaction.

An easy way to determine the limiting reactant is:

1. Find the moles of each substance.

2. Divide by the coefficient in the equation.

3. The lowest resulting number is the limiting reactant.

A solute is a solid substance that dissolves in a liquid / solvent.

To convert from dm³ to cm³, multiply by 1000.

A solvent is the liquid in which a solid / solute dissolves.

Concentration refers to the amount of solid / solute there is in a specific volume of the liquid / solvent.

A solution is the mixture formed when a solid / solute dissolves in a liquid / solvent.

The equation to calculate concentration is:

Concentration = Mass of solute / Volume of solution.

False.

1 dm³ is equal to 1000 cm³, not 100 cm³.

Concentration is typically measured in grams per cubic decimetre (g/dm³).

To convert from cm³ to dm³, divide by 1000.

False.

A higher concentration means more solute in a given volume, not less.

To calculate the mass of solute, multiply the concentration by the volume:

mass = concentration × volume.