Question #4f14c

Oct 17, 2016

Molecular formula of Cadaverine is ${C}_{5} {H}_{14} {N}_{2}$

Explanation:

Given

Cadaverine contains 58.55% C, 13.81% H, and 27.40% N by mass, and has a molar mass of 102.2 g/mol.

Considering atomic masses as follows

$C \to 12 \text{ g/mol}$

$H \to 1 \text{ g/mol}$

$N \to 14 \text{ g/mol}$

Dividing the percentage of C,H and N by respective atomic mass and taking their ratio we get the ratio of number of atoms of C,H and N in the molecule of Cadaverine as follows

${n}_{C} : {n}_{H} : {n}_{N} = \frac{58.55}{12} : \frac{13.81}{1} : \frac{27.40}{14}$
$= 4.88 : 13.81 : 1.96$

$= \frac{4.88}{1.96} : \frac{13.81}{1.96} : \frac{1.96}{1.96}$

$\approx 2.5 : 7 : 1 = 5 : 14 : 2$

So empirical formula of Cadaverine is ${C}_{5} {H}_{14} {N}_{2}$

Let its molecular formula be ${\left({C}_{5} {H}_{14} {N}_{2}\right)}_{n}$

By this formula its molar mass is $\left(5 \cdot 12 + 14 \cdot 1 + 2 \cdot 14\right) n = 102 n \text{ g/mol}$

Equating this with the given molar mass we get

$102 n = 102.2 \implies n \approx 1$

So molecular formula of Cadaverine is ${C}_{5} {H}_{14} {N}_{2}$