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Examples in Gravimetric Analysis

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Example: Calculate the amount of sulphate as barium sulphate from sodium sulphate.
Solution of sodium sulphate (Na2SO4) is treated with solution of barium chloride (BaCl2) to get precipitates of barium sulphate (BaSO4). The precipitates are then washed, dried and ignited to get free from impurities and then weighed.
Na2SO4 + BaCl2 → BaSO4 + 2 NaCl
Mol. Weight of BaSO4 = 233.42 gm
Mol. Weight of SO4- = 96.06 gm
Suppose obtained weight of BaSO4 precipitates = X · gm
233.42 gm of BaSO4 = 96.06 gm of SO4- ions
X · gm of BaSO4 = ?
BaSO4 = 96.06 · X / 233.32 = 0.411X gm of SO4- ions
Suppose 25 ml solution is consumed, then
25 ml solution contains = 0.411X gm of SO4- ions
1000ml solution conatins?
1000ml solution conatins = 0.411X · 1000 / 25  = 16.44X gm of SO4- ions
Example: Calculate the amount of zinc oxide from zinc sulphate.
A solution of zinc sulphate is boiled to convert it into zinc carbonate by adding solution of sodium carbonate. Sodium carbonate is added to precipitate zinc completely as zinc carbonate. Precipitates of zinc carbonate is boiled for few minutes to convert it into zinc oxide and collected in a tarred Gooch crucible. Precipitates are washed with hot water until it gets free from alkali and then dried, ignited and weighed to a constant weight.
ZnSO4 + Na2CO3 → ZnCO3 + Na2SO4
ZnCO3 → ZnO + CO2
ZnSO4 = ZnCO3 = ZnO
ZnSO4 = ZnO
Mol. Weight of ZnSO4 = 168 gm
Mol. Weight of ZnO = 81.38 gm
81.38 gm of ZnO = 168 gm of ZnSO4
1 gm of ZnO =?
= 168 · 1 / 81.38
= 1.984 gm
Example: Calculate the amount of Boric acid from Borax.
Borax is an alkaline substance, and reacts with conc. HCl to form Boric acid. Boric acid is freely soluble in boiling water and precipitated out in cold water. To get high grade of Boric acid, Borax is treated with conc. HCl as it is volatile in nature and won’t left any residual traces on crystal surface of Boric acid.
Weight and dissolve 5 gm of Borax in 15 ml of distilled water. Add 7 ml of conc. HCl, mix thoroughly with glass rod and mark the original volume with glass rod. Evaporate the solution till the volume reduces to half of the original volume. Allow to cool at room temperature. Keep it aside for few min and add ice water. Filter the residue under suction and dry it in air. Weight the compound preparation.
Na2B4O7.10H20 → 4H3BO3 + 5H20 + 2NaCl
Mol wt of Borax = 381.37gm
Mol wt of Boric acid = 61.83gm
Practical yield: X gm
381.37 gm of Borax = 4 × 61.83 gm of Boric acid
X gm of Borax = ?
= X · 4 · 61.83 / 381.37
= 0.674X gm of Boric acid © 2021 | Contact us | Terms of Use | Privacy Policy | Yellow Sparks Network
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