Question

Consider the below-given reaction. The percentage yield of an amide product is (Round off to the Nearest Integer).
[Given: Atomic mass: C: 12.0 u, H : 1.0 u, N : 14.0, O : 16.0 u, Cl : 35.5 u]

Answer 1

$\text{Molar mass of amide(product)}=273g/mol$
$\text{Stoichiometric moles of amide}={10}^{–3}\text{mol}$
$\text{Theoretical mass of amide formed}={10}^{–3}\times 273=0.273g$
$\text{Observed mass of amide}=0.210g$

$\text{\% yield}=\frac{0.210}{0.273}\times 100\%$

$\text{\% yield}=76.9\%\approx 77\%$