If a 2*g mass of beta-"napthol" is reacted with "ethyl bromide" to give a 1*g mass of "naptholethyl ether", what is the percentage yield?

Aug 5, 2017

Well, the molar equivalence is 1:1, i.e. one equiv ${C}_{10} {H}_{7} O H$...

Explanation:

We assess the reaction......

C_10H_7OH+H_3C-CH_2-Xstackrel("base")rarrC_10H_7OCH_2CH_3+"base"*HX

Well, the molar equivalence is 1:1, i.e. one equiv ${C}_{10} {H}_{7} O H$ to one equiv of ${C}_{10} {H}_{7} O C {H}_{2} C {H}_{3}$.

Moles of $\beta - \text{napthol} = \frac{2.1 \cdot g}{144.17 \cdot g \cdot m o {l}^{-} 1} = 0.0146 \cdot m o l$

Moles of ..............................

$\text{napthol ethyl ether} = \frac{1.1 \cdot g}{172.23 \cdot g \cdot m o {l}^{-} 1} = 0.00638 \cdot m o l$

And thus "yield"=(0.00638*mol)/(0.0146*mol)xx100%=44%.

And as is typical, yield follows the quotient,

"% Yield"="Moles of product"/"Moles of limiting reactant"xx100%

Of course, for different stoichiometry, we have to add the appropriate multiplier. Here it is $1 : 1$......Typically we would add the $\text{ethyl halide}$ in stoichiometric excess, and the base, $N E {t}_{3}$ or $\text{DBU}$ or $\text{DABCO}$ or something in stoichiometric quantity.