Atoms and Molecules

Molarity

A mole (mol) of a substance is 6.022 x 10²³ molecules of that substance; (6.022 x 10²³ is the Avogadro Number and equals the number of atoms of ¹²C in 12 g of ¹²C).

The molecular weight (MWt) of a substance is the weight in grams of 6.022 x 10²³ molecules of that substance; (i.e. it is the relative molecular mass in grams).

The relative molecular mass (RMM) is the mass of one molecule in atomic mass units (ie: relative to the mass of one twelfth of the mass of an atom of ¹²C). It may be calculated by adding the atomic masses (in amu) of all the constituent atoms, e.g.:

Thus: one mole of sulfuric acid is 98.087 g and 0.1 mol H₂SO₄ = 9.8087 g
One mole of hydrated calcium nitrate is 236.15 g and 0.2 mol Ca(NO₃)₂.4H₂O = 47.23 g

To calculate the percentage composition (w/w) of a compound, make use of the proportions present by weight, e.g.

1. there is one atom of sulfur in each molecule of sulfuric acid
   there is 32.066 g of sulfur in each 98.087 g of sulfuric acid
   the % (w/w) S in H₂SO₄ is 100x(32.066/98.087) = 32.69
2. there are 4 molecules of water in each molecule of hydrated calcium nitrate
   there are 4x18.015 g of water (=72.06 g) in each 236.15 g of Ca(NO₃)₂.4H₂O
   the % (w/w) H₂O in Ca(NO₃)₂.4H₂O is 100x(72.06/236.15) =30.51

To calculate the empirical formula from the percent w/w composition, determine the ratio of the number of molecules by dividing the % w/w of each element by its atomic mass.

E.g. what is the formula of the compound containing 74 % w/w sodium and 26 % w/w oxygen?
        for sodium (atomic mass Na = 22.990) 74/22.990 = 3.2
        for oxygen (atomic mass O = 15.999) 26/15.999 = 1.6
Therefore the ratio Na:O is 3.2:1.6 or 2:1 and the formula is Na₂O (sodium oxide).

If the ratio is not exactly a ratio of simple integers, choose the closest ratio of simple integers; remembering that most formulae do not involve ratios of large integers (as such they would be of little use), and small errors must be allowed for.

When substances are dissolved in water to make solutions the unit of their concentrations is molarity.

A one molar solution (1 M) has one mol of the substance dissolved in each litre (dm³) of solution; (equivalent to 1 mmol in a ml, and 1 mmol in a ml).

The amount of substance present is given by the concentration multiplied by the volume present in litres (dm³). A concentration of a material is usually indicated by the use of square brackets (i.e. [NaCl] means the concentration of sodium chloride).

number of moles = concentration (mol/L) x volume (L)

concentration (mol/L) = number of moles/volume (L)

1. to make a 1 M solution of sulfuric acid 98.087 g of sulfuric acid would be dissolved in each litre (dm³) of solution. 100 ml (cm³) of this solution would contain 98.087 (g/mol) x 1 (mol/L)  x 0.1 L = 9.8087  (g/mol) x (mol/L)  x  L = 9.8087g H₂SO₄.

2. to make 200 ml of a 0.5 M solution of calcium nitrate, first calculate how much Ca(NO₃)₂.4H₂O must be weighed out:
   
   The number of moles = concentration (0.5 mol/L) x volume (0.2 L) = 0.1 mol
   but                 0.1 mol Ca(NO₃)₂.4H₂O = 0.1 x 236.15 = 23.615 g
   therefore       23.615 g of Ca(NO₃)₂.4H₂O must be weighed out, dissolved in 150 ml (approximately) of water and the solution made up to 200 ml (exactly).
   

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This page was last updated by Martin Chaplin
on 10 February, 2005