Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer:
C
Explanation:
I just aswered this question and i got it right
Answer:
% yield = 82.5%
Explanation:
HgO + 2Cl₂ → HgCl₂ + Cl₂O
Our reactants are:
Our products are:
We do not have information about moles of reactants, but we do know the theoretical yield and the grams of product, in this case Cl₂O, we have produced.
Percent yield = (Yield produced / Theoretical yield) . 100
Theoretical yield is the mass of product which is produced by sufficent reactant. We replace data:
% yield = (0.71 g/0.86g) . 100 = 82.5%
