Counting Particles by Mass: The Mole (DP IB Chemistry)

Relative molecular mass is used for molecules, while relative formula mass is used for compounds containing ions. They are calculated in the same way.

The equation for calculating the number of particles in terms of moles is:

Number of particles = number of moles × Avogadro constant

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 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

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

Moles = mass / molar mass

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

• There are 4 H atoms in 1 molecule of CH₃CHO

• So, there are 0.080 moles of H atoms in 0.020 moles of CH₃CHO

• The number of H atoms is the amount in moles x L

• This comes to 0.080 x (6.02 x 10²³) = 4.8 x 10²² atoms

• In 1.806 x 10²³ molecules of H₂O₂  there are 2 x (1.806 x 10²³) atoms of H

• So, there are 3.612 x 10²³ atoms of H

• The number of moles of H atoms is the number of particles ÷ L

• This comes to 3.612 x 10²³ ÷ (6.02 x 10²³) = 0.60 moles of H atoms

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.

The information from the Periodic Table that is required to calculate the empirical formula of a metal oxide is:

• The atomic mass of the metal

• The atomic mass of oxygen

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

True.

The percentage composition of elements in a compound always adds up to 100%.

The percentage composition by mass of hydrogen in water (H₂O) is:

(2 / 18) x 100 = 11.1 %.

False.

The percentage composition of oxygen in carbon dioxide, CO₂, is:

(32 / 44) x 100 = 72.7 %

This is because CO₂ contains 2 oxygen atoms.

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%.

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

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

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

27.50 cm³ is 0.0275 dm³.

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:

1. Write out the balanced equation for the reaction

2. Calculate the moles of the known solution given the volume and concentration

3. Use the equation to deduce the moles of the unknown solution

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.

True.

1 ppm is equivalent to 1 mg in 1 kg.

To convert concentration from g dm^-3 to mol dm^-3, you multiply by the molar mass:

concentration (g dm^-3) = concentration (mol dm^-3) × molar mass (g mol^-1)

True.

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

There are 2 moles of gas in 45.4 dm³.

moles = = 2

2.5 moles of gas will occupy 56.75dm³ at RTP.

• volume (dm³) = moles x 22.7

• volume (dm³) = 2.5 x 22.7 = 56.75dm³

The volume of ammonia is produced from 600 cm³ of hydrogen is 400 cm³.

N₂ (g) + 3H₂ (g)   2NH₃ (g)

• hydrogen : ammonia = 3:2 ratio

• volume of ammonia = 600 cm³ x = 400 cm³

If 250 cm³ of oxygen gas reacted reacted with propane, 150 cm³ of carbon dioxide is formed.

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• oxygen : carbon dioxide = 5:3 ratio

• volume of carbon dioxide = 250 cm³ x = 150 cm³

100 cm³ of propane was required to react with 500 cm³ oxygen.

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• propane : oxygen = 1:5 ratio

• volume of propane = = 100 cm³

The change in the number of moles of gas is 9 to 4 OR decreases by 5.

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (l)

At standard temperature and pressure (STP), one mole of any gas has a volume of 22.7 dm³.

Standard temperature and pressure, STP, refers to a temperature of 0°C (273 K) and a pressure of 100 kPa.

True.

Avogadro's Law applies only to ideal gases.

True.

The volume of a gas is directly proportional to the number of moles at constant temperature and pressure.