# Moles and Equations

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## Relative Atomic Mass (Ar)

Relative Atomic Mass: The weighted mean mass of an atom of an element compared with 1/12 of the mass of K,atom of C-12.

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## Relative Molecular Mass (Mr)

Relative Molecular Mass: The weighted mean mass of a molecule of an element compared with 1/12 of the mass of an atom of C-12.

To Calculate the relative molecular mass, Mr, of a molecule you add up all of the individual relative atomic masses for the elements contained within the molecule.

EG.

NaOH=

Na=23 + O=16 + H=1

= 40

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## Mole

Mole: the amount of substance containing as many particles as there are carbon atoms in exactly 12g of C-12.

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## Molar Mass

Molar Mass: The mass, in g, per mole of a substance. Units are gmol-1

Molar masses are calculated using the following equation:

Amount (n) = mass / M

EG

Calculate the amount, in moles, of carbon in 11g of carbon dioxide (CO2)

n=mass/m        n=11/m         C=12 and O=16       m= 12 + (16x2) = 44

n=11/44= 0.25mol

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EG 2

Calculate the mass, in g, in 6.0mol of SiO2

Amount (n) = mass/M

Si= 28.1 and O=16    M= 28.1 + (16x2) = 60.1

mass= nxM        mass= 6.0 x 60.1

mass= 360.6

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## The Avogardo Constant

The number of particles in one mole of any substance is called the Avogadro Constant. It is represented using the symbol NA and has the value 6.02x10^23mol-1.

Simply for every One Mole of any substance there are 6.02x10^23 atoms

The Formula is:    Atoms/Ions = amount of moles X Avogadro Constant

=             n              X            NA

EG Calculate the total number of atoms of chlorine in 35.5g of chlorine atoms

n = Mass/M   n= 35.5/35.5= 1mol

Atom = n x NA = 1 x 6.02 x 10^23

= 6.02 x 10^23

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## Empirical Formula

Empirical Formula: The simplest whole number ratio of atoms of each element present in a compound.

EG A chloride of Iron contains 65.5% chlorine. What is the empirical formula?

element present:                     Iron                       Chlorine

% by Mass (g):                       34.5                         65.5

Relative atomic mass (Ar):     55.8                         35.5

Moles (mass/M):         34.5/55.8= 0.618        65.5/35.5= 1.845

Ratio (moles/0.618)                  1               :               3

Empirical Formula = FeCl3

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## Water of Crystallisation

Hydrated: When water of crystallisation is present in a crystal compound.

Anhydrous: When all the waters of crystallisation have been removed from a compound.

Water of Crystallisation: The water present in a compound giving the compound a crystalline appearance.

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EG  A sample of 7.38g of MgSO4 crystals ost 3.78g of water on heating.Calculate x in the formula   MgSO4 . xH2O

MgSO4                                H2O

Mass                         3.60                                    3.78

Mr                24.3+32.1+(16x4)= 120.4        (1x2)+16= 18.0

Moles (mass/Mr)     0.0299                                  0.21

Ratio                            1                     :                  7

therefore x = 7

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## Molecular Formula

Molecular Formula: the actual number of atoms of each element in a molecule.

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