Mole Concept and Stoichiometry — Question 12

(a) Given, hydrocarbon contains 4.8 g of carbon per gram of hydrogen

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of carbon = $\dfrac{4.8}{12}$ = 0.4 and

g atom of hydrogen = $\dfrac{1}{1}$ = 1

(b)

Simplest ratio of whole numbers = H : C = $\dfrac{5 }{2}$ : 1 = 5 : 2

Hence, empirical formula is C₂H₅

(c) Empirical formula weight = (2 x 12) + (5 x 1) = 24 + 5 = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29 = 58

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{58}{29} = 2$

So, molecular formula = 2(C₂H₅) = C₄H₁₀