Question

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?

Answer 1

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.