An unknown compound consists of Carbon, Hydrogen and Chlorine. 1.324g of the compound is divided into two equal samples. The first sample, when burnt in oxygen, produced 1.189g of Carbon Dioxide. The second sample, was oxidised with conc. nitric acid and treated with silver nitrate solution to yield 1.292g of silver chloride. The relative molecular mass of the unknown was found from freezing point depression measurements to be 147. Determine the:
a)the empirical formula
b). the molecular formula
HOW DO YOU DO THIS QUESTION - Urgent Empirical/Molecular Formula Question Chemistry?
2016-06-12 5:59 am
回答 (2)
2016-06-12 6:57 am
we have 1.19 g of CO2 so mass of C = 1.189 * 12/ 44 = 0.324 g
or 0.324 / 0.662 % C = 48.9% C
mass of Cl in 1.292 g of AgCl = 1.292g * 35.5g / 143.3212 g/mol = 0.320 g
% Cl = 0.320 g / 0.662 = 48.34%
%H = 100-97.20 = 2.80%
moles of C in 100 g of compound = 48.9 / 12 = 4.07 moles
moles of Cl in that mass = 1.36 moles
moles of H = 2.64 / 1 = 2.80 moles
molar ratio of C:H:Cl = 4.07:2.80:1.36 or 3.0:2.06:1 or 3:2:1
empirical formula is C3H2Cl
empirical formula mass = 73.5 g
which is 1/2 the molar mass
so the molecular formula is 2 * the empirical formula or C6H4Cl2
dichloro benzene
or 0.324 / 0.662 % C = 48.9% C
mass of Cl in 1.292 g of AgCl = 1.292g * 35.5g / 143.3212 g/mol = 0.320 g
% Cl = 0.320 g / 0.662 = 48.34%
%H = 100-97.20 = 2.80%
moles of C in 100 g of compound = 48.9 / 12 = 4.07 moles
moles of Cl in that mass = 1.36 moles
moles of H = 2.64 / 1 = 2.80 moles
molar ratio of C:H:Cl = 4.07:2.80:1.36 or 3.0:2.06:1 or 3:2:1
empirical formula is C3H2Cl
empirical formula mass = 73.5 g
which is 1/2 the molar mass
so the molecular formula is 2 * the empirical formula or C6H4Cl2
dichloro benzene
2016-06-12 6:54 am
(a)
Molar masses (in g/mol) :
C = 12.01, H = 1.008, Cl = 35.45, O = 16.00, Ag = 107.9
Mass fraction of C in CO₂ = 12.01/(12.01 + 16.00×2)
Mass fraction of Cl in AgCl = 35.45/(107.9 + 35.45)
Mass of each sample of the compound = 1.324 × (1/2) = 0.662
In the first sample :
Mass of C = Mass of C in CO₂ formed = 1.189 × [12.01/(12.01 + 16.00×2)] = 0.3245 g
Mass % of C = (0.3245/0.662) × 100% = 49.0%
In the second portion of the compound :
Mass of Cl = Mass of Cl in AgCl formed = 1.292 × [35.45/(107.9 + 35.45)] g = 0.3195 g
Mass % of Cl = (0.3195/0.662) × 100% = 48.3%
Mass % of H in the sample = (100 - 49.0 - 48.3)% = 2.7%
Molar ratio C : H : Cl
= (49.0/12.01) : (2.7/1.008) : (48.3/35.45)
= 4.08 : 2.7 : 1.36
= 3 : 2 : 1
Empirical formula = C₃H₂Cl
(b)
Let (C₃H₂Cl)ₓ be the molecular formula of the compound.
Relative molecular mass of the compound :
x(12.01×3 + 1.008×2 + 35.45) = 147
73.496x = 147
x = 2
Molecular formula = C₆H₄Cl₂
Molar masses (in g/mol) :
C = 12.01, H = 1.008, Cl = 35.45, O = 16.00, Ag = 107.9
Mass fraction of C in CO₂ = 12.01/(12.01 + 16.00×2)
Mass fraction of Cl in AgCl = 35.45/(107.9 + 35.45)
Mass of each sample of the compound = 1.324 × (1/2) = 0.662
In the first sample :
Mass of C = Mass of C in CO₂ formed = 1.189 × [12.01/(12.01 + 16.00×2)] = 0.3245 g
Mass % of C = (0.3245/0.662) × 100% = 49.0%
In the second portion of the compound :
Mass of Cl = Mass of Cl in AgCl formed = 1.292 × [35.45/(107.9 + 35.45)] g = 0.3195 g
Mass % of Cl = (0.3195/0.662) × 100% = 48.3%
Mass % of H in the sample = (100 - 49.0 - 48.3)% = 2.7%
Molar ratio C : H : Cl
= (49.0/12.01) : (2.7/1.008) : (48.3/35.45)
= 4.08 : 2.7 : 1.36
= 3 : 2 : 1
Empirical formula = C₃H₂Cl
(b)
Let (C₃H₂Cl)ₓ be the molecular formula of the compound.
Relative molecular mass of the compound :
x(12.01×3 + 1.008×2 + 35.45) = 147
73.496x = 147
x = 2
Molecular formula = C₆H₄Cl₂
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