Unit 2

Stoichiometry – Gravimetric analysis

Gravimetric AnalysisThe mass of an element or compound present in a substance is determined by changing that substance into another chemical substance of known chemical composition that can be easily isolated, purified and weighed.(e.g. precipitation followed by filtration)

The product of the reaction must be:

insoluble so that all the product is precipitated have a particle size large enough for it to be able

to be filtered stable at 100oC to allow drying before weighing

if required.

Analysis in this way can be difficult due toerrors in precipitation, filtering and weighing.

Another method of analysis is to use heat to change one substance into another.(e.g. dehydration of a hydrated salt to calculate the number of moles of water)

The product must be continually heated and weighed until a constant mass is achieved to ensure that the reaction is complete. Successive weighings, at room temperature, must be within 0.01g of each other.

A compound containing only carbon, hydrogen and oxygen was analysed by burning a sample and collecting the products. 1.365g of the compound burned to produce 1.98g of carbon dioxide and 0.945g of water.

Calculate:(a)The masses of carbon, hydrogen and oxygen

present in the compound.(b)The number of moles of carbon, hydrogen and oxygen present.(c)The empirical formula of the compound.

(a)The mass of carbon = RAM of C x mass of CO2

gfm of CO2

= 12 x 1.98 = 0.54g

44

The mass of hydrogen = gfm of H2 x mass of H2O

gfm of H2O

= 2 x 0.945 = 0.105g

18

The mass of oxygen = 1.365 – (0.54 + 0.105) = 0.72g

(b)Number of moles of C = m = 0.54 =

0.045molgfm 12

Number of moles of H = m = 0.105 = 0.105mol

gfm 1

Number of moles of O = m = 0.72 = 0.045mol

gfm 16

(c)Element C H O

Number of moles

0.045 0.105 0.045

Mole ratio 0.045 = 10.045

0.105 = 2.33 = 2 1/30.045

0.045 = 10.045

Whole number ratio

1 x 3 = 3 2 1/3 x 3 = 7 1 x 3 = 3

The empirical formula = C3H7O3

Example 2A 9.36g sample of hydrated copper(II)

sulphate, CuSO4.xH2O, was heated and weighed until no more loss of mass was detected. The remaining anhydrous copper(II) sulphate, CuSO4, weighed 5.985g

Calculate the value of x in the formula CuSO4.xH2O.

The mass of water in the hydrated copper(II) sulphate = 9.36 – 5.985 = 3.375g

So the value of x = 5 and the formula of the hydrated copper(II) sulphate is CuSO4.5H2O

Compound CuSO4 H2O

Mass 5.985g 3.375g

GFM 159.6g 18.0g

Number of moles 0.0375 0.1875

Mole ratio 0.0375 = 10.0375

0.1875 = 50.0375

Whole number ratio

1 5

Example 314.91g of an impure sample of aluminium

sulphate was dissolved in water and filtered to remove any insoluble material. Excess sodium hydroxide was then added to the filtrate forming a precipitate of aluminium hydroxide.

Al2(SO4)3(aq) + 6NaOH(aq) → 2Al(OH)3(s)+3Na2SO4(aq)

The precipitate was filtered, washed to remove any

soluble substances, dried and weighed. The mass of the precipitate was 5.69g.

Calculate:(a)the number of moles of aluminium hydroxide

in the precipitate.(b)the number of moles of aluminium sulphate in the original sample(c)the mass of aluminium sulphate in the original sample(d)the percentage purity by mass of aluminium sulphate in the original sample.

(a)n Al(OH)3 = m gfm

= 5.69 78

= 0.073mol

(b)From the balanced equation:

1 mol Al2(SO4) 3 →2 mol Al(OH)3

∴ 0.073mol Al(OH)3 will be formed from 0.0365 mol Al2(SO4) 3

(c)m Al(SO4)3 = n x gfm

= 0.0365 x 342.3

= 12.49g

(d)% purity = 12.49 x 100 = 83.77% 14.91

ExerciseAH Chemistry Calculations textbook P31 – 34