Chapter : 3. Atoms & Molecules
Determination molecular formula
Determination molecular formula :
Molecular formula = Empirical formula × n
Example. What is the simplest formula of the compound which has the following percentage composition : Carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula.
Solution. Calculation of empirical formula :
∴ Empirical formula is CH3.
Calculation of molecular formula :
Empirical formula mass = 12 × 1 + 1 × 3 = 15
= 30/15 = 2
Molecular formula
= Empirical formula × 2 = CH3 × 2 = C2H6.
Example. On heating a sample of CaCO3, volume of CO2 evolved at NTP is 112 cc. Calculate
(i) Weight of CO2 produced
(ii) Weight of CaCO3 taken
(iii) Weight of CaO remaining
Solution. (i) Mole of CO2 produced
= 
mole
Mass of CO2 = × 44 = 0.22 gm
(ii) CaCO3 → CaO + CO2
(1/200 mole)
Mole of CaCO3 = mole
∴ mass of CaCO3 = × 100 = 0.5 gm
(iii) mole of CaO produced = mole
mass of CaO = × 56 = 0.28 gm
* Interesting by we can apply Conversation of mass or wt. of CaO
= wt. of CaCO3 taken – wt. of CO2 produced
= 0.5 – 0.22 = 0.28 gm
Molecular formula = Empirical formula × n
Example. What is the simplest formula of the compound which has the following percentage composition : Carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula.
Solution. Calculation of empirical formula :
∴ Empirical formula is CH3.
Calculation of molecular formula :
Empirical formula mass = 12 × 1 + 1 × 3 = 15
= 30/15 = 2 Molecular formula
= Empirical formula × 2 = CH3 × 2 = C2H6.
Example. On heating a sample of CaCO3, volume of CO2 evolved at NTP is 112 cc. Calculate
(i) Weight of CO2 produced
(ii) Weight of CaCO3 taken
(iii) Weight of CaO remaining
Solution. (i) Mole of CO2 produced
= mole Mass of CO2 =
× 44 = 0.22 gm (ii) CaCO3 → CaO + CO2
(1/200 mole)
Mole of CaCO3 =
mole ∴ mass of CaCO3 =
× 100 = 0.5 gm (iii) mole of CaO produced =
mole mass of CaO =
× 56 = 0.28 gm * Interesting by we can apply Conversation of mass or wt. of CaO
= wt. of CaCO3 taken – wt. of CO2 produced
= 0.5 – 0.22 = 0.28 gm
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