O Level Chemistry: The Mole

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O Level Chemistry

Chap 9: The Mole


1)

 

 

 

Ar – Relative Atomic Mass (elements)

·       The relative atomic mass (Ar) of an element is the average mass of its one atom, compared to 1/12 the mass of one atom of carbon-12.

·       Ar of element= Nucleon number of element (e.g. Ar of  = 23)

·       Ar is a ratio and has no units.

·       Ar values may not always be whole numbers as some elements like chlorine occur as mixtures of isotopes.

2)

Mr – Relative Molecular/Formula Mass

·       The relative molecular mass (Mr) of an element or compound is the average mass of its one molecule, compared to 1/12 the mass of one atom of carbon-12.

·       Relative molecular mass is used for molecules (e.g. O2, H2O);

·       Relative formula mass is used for ionic compounds (e.g. CuSO4, CaCO3).

·       Mr = (Ar of 1st element x No. of atoms) + ……

·       E.g. Mr of CuSO4.5H2O= [(64 x 1) + (32 x 1) + (16 x 4)] + 5[(1 x 2) + (16 x 1)] = 250

 

 

2)

Calculating % composition of compounds

·         Given the compound and element to be calculated,

·         E.g. To find % of hydrogen in hydrogen peroxide.

Mr of H2O2 = (1 x 2) + (16 x 2)

        = 34

% of hydrogen in H2O2 =

 = 5.88% (3 sf)

 

 

3)

The Mole

·         The quantity (number) of atoms is measured by mass. The mole is the unit of measurement for atoms and molecules and is the SI unit for chemical quantity. Symbol for mole= mol

·         1 mole of particles (atoms, molecules, ions or electrons) = 6.02 x 1023 particles.

·         Equal amounts of substances (say 5 mol) will only contain the same number of particles.

(e.g., 5 mol of NaCl and 5 mol of MgO will contain the same number of particles)

 

 

4)

Molar Mass

·         Molar mass is the mass of 1 mole of a substance. Unit = grams.

·         Molar mass of element = Ar of element (e.g. molar mass of O = 16g).

·         Molar mass of molecule/compound = Mr of molecule/compound (e.g. O2 = 32g).

 

 

5)

Calculating the number of moles

·         Given the mass of the substance,

·         Given the number of particles in the substance,

 

 

6)

Empirical Formula

·         The empirical formula shows the types of elements, in the simplest ratio, present in the compound.

·         E.g. Calculate the empirical formula of a compound consisting of 3.5g of nitrogen combined with 8.0g of oxygen.

 

 

Nitrogen

Oxygen

Mass (from experiment)

3.5g

8.0g

Ar of element

14

16

Number of moles

Molar ratio

 The empirical formula of the compound is NO2.

 

 

7)

Molecular Formula

·         Given the empirical formula ad Mr of the compound,

·         E.g. Compound X contains 40.0% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. Calculate the molecular formula of X.

 

First, we find the empirical formula of X.

 

Carbon

Hydrogen

Oxygen

% (from experiment)

40.0

6.6

53.3

Ar of element

12

1

16

Number of moles

Molar ratio

 The empirical formula of X is CH2O

 

Molecular formula of X = C6H12O6

 

 

8)

Molar gas volume

 

·       Avogadro’s Law states that equal volumes of gas will contain the same no. of molecules when under the same temperature and pressure.

(e.g., 750cm3 of CO2 and CH4 contains the same number of molecules)

·       Molar volume is the volume occupied by 1 mole of gas. It is 24000cm3/24dm3 under rtp. In other words, under rtp, one mole of any gas will occupy a volume of 24dm3; 2 moles occupy 48dm3.

·        1 mol of CO2 and CH4 (any gas) each will have same no. of molecules, and a similar volume of 24dm3.

 

 

9)

Calculating the number of moles of a gas

 

·         Given the mass and the name/Mr of the gas,

 

·         Given the volume of the gas,

 

Notes:

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