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?

1 Answer
Aug 5, 2017

Well, the molar equivalence is 1:1, i.e. one equiv C_10H_7OH...

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_10H_7OH to one equiv of C_10H_7OCH_2CH_3.

Moles of beta-"napthol"=(2.1*g)/(144.17*g*mol^-1)=0.0146*mol

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

"napthol ethyl ether"=(1.1*g)/(172.23*g*mol^-1)=0.00638*mol

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 "ethyl halide" in stoichiometric excess, and the base, NEt_3 or "DBU" or "DABCO" or something in stoichiometric quantity.