The Mole & the Avogadro Constant (Cambridge (CIE) IGCSE Chemistry)

4 moles of hydrogen gas contains:

• Atoms = 4 x 2 x 6.02 x 10²³ = 48.16 x 10²³.

• Molecules = 4 x 6.02 x 10²³ = 24.08 x 10²³.

To convert from cm³ to dm³, you divide by 1000.

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

Molar gas volume = volume / moles

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

Moles = volume / 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.

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

One mole of any element is equal to the relative atomic mass of that element in grams.

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

The equation for calculating moles from mass and M_r is:

Moles = mass / Mr

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

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.

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.

True.

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

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.

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

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 two common units for concentration are:

• g / dm³

• mol / dm³

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

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

Moles = concentration x volume

To calculate the number of moles present in 2.5 dm³ of a solution which has a concentration of 0.30 mol dm^-3:

• Moles = concentration x volume

• Moles = 0.30 x 2.5 = 0.75 moles

The balanced equation is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

27.50 cm³ is 0.0275 dm³.

The ratio of acid to alkali is 1:3.

H₃PO₄ + 3NaOH → Na₂PO₄ + 3H₂O

The concentration in mol/dm³ of a solution of HCl with a volume of 23.55 cm³ and contains 0.00375 moles is:

Volume in dm³ = 0.02355 dm³

Concentration = = 0.16 mol/dm³

The four steps in a titration calculation are:

• Step 1: Write out the balanced equation for the reaction

• Step 2: Calculate the moles of the known solution given the volume and concentration

• Step 3: Use the equation to deduce the moles of the unknown solution

• Step 4: Use the moles and volume of the unknown solution to calculate the concentration

Using moles, volume in dm³ the equation is:

Concentration =

0.03905 dm³ is 39.05 cm³.

0.05 moles of HCl would be required as this reaction is a 1:1 ratio.

HCl + NaOH → NaCl + H₂O

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

The steps to calculate an empirical formula are:

1. Write the element

2. Write the value for each element

3. Write the relative atomic mass

4. Calculate moles

5. Calculate ratio

6. Write final formula.

The molecular formula gives the actual numbers of atoms of each element present in one molecule of a compound.

To calculate the molecular formula from the empirical formula, multiply the subscripts in the empirical formula by n, where n = (molecular mass / empirical formula mass).

True.

If the molecular formula of benzene is C₆H₆ the empirical formula of benzene is CH.

A hydrated salt is a crystallised salt that contains water molecules as part of its structure.

True.

The empirical formula is CH.

The mistake in the empirical formula calculation is that the mole calculations are incorrect / upside down.

The correct calculation is:

The empirical formula is correctly calculated to be MgCl₂.

The mistake in the empirical formula calculation is that the atomic number has been used, not the atomic mass.

The correct calculation is:

So, the empirical formula is CH₃.

The mass of the empirical formula, CH₃, is 12 + (1 x 3) = 15.

30 / 15 = 2, which means that two lots of the empirical formula are needed.

So, the molecular formula is C₂H₆.

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%

False.

A percentage yield of 100% is not common in chemical reactions.

Factors that can reduce percentage yield include:

• Incomplete reactions

• Side reactions

• Loss during processing

• Reversible reactions.

Percentage composition by mass is the mass of each element expressed as a percentage of the total mass of the compound.

The equation to calulcate percentage composition by mass is:

(total mass of the element in the compound / relative formula mass of the compound) × 100

True.

The sum of percentage compositions by mass for all elements in a compound should equal 100%.

False.

• A_r (Mg) = 24

• A_r (O) = 16

• M_r (MgO) = 40

So, the percentage by mass of Mg in MgO is 24 / 40 x 100 = 60%.

True.

The mass of nitrogen in ammonium nitrate, NH₄NO₃, is 28 because there are two nitrogen atoms in the compound.

The percentage of fluorine in tin(II) fluoride, SnF₂ is:

• Mass of fluorine = 2 x 19 = 38

• Mass of tin(II) fluoride = 119 + (2 x 19) = 157

• Percentage of fluorine = 38 / 157 x 100 = 24.2%.

Percentage purity is the mass of pure substance expressed as a percentage of the total mass of the impure substance.

The equation for percentage purity is:

(mass of pure substance / total mass of substance) × 100

True.

A sample with 100% purity contains no impurities.

The percentage purity of the calcium chloride is:

(9.8 / 12.0) x 100 = 81.7%