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?

Question

Sarah Angotti · Accepted Answer

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

We evaluate the response. #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 the aforementioned substance. "napthol ethyl ether" is equal to (1.1*g)/(172.23*g*mol^-1) = 0.00638*mol. And thus #"yield"=(0.00638*mol)/(0.0146*mol)xx100%=44%#. Additionally, yield follows the quotient as usual. #"% 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.

Ella Allen · Answer

undefined To calculate the percentage yield:

1. Calculate the theoretical yield of the product using stoichiometry.
2. Divide the actual yield by the theoretical yield and multiply by 100 to get the percentage yield.

Given:
- 2 g of beta-naphthol reacted
- 1 g of naphthoethyl ether obtained

The molar mass of beta-naphthol is 144.18 g/mol.
The molar mass of ethyl bromide is 108.97 g/mol.
The molar mass of naphthoethyl ether is 172.18 g/mol.

Using the balanced chemical equation:
C10H7OH + C2H5Br -> C10H7OC2H5 + HBr

Theoretical yield:
1 mole of beta-naphthol reacts with 1 mole of ethyl bromide to produce 1 mole of naphthoethyl ether.
2 g of beta-naphthol is approximately 0.01386 moles.
So, the theoretical yield of naphthoethyl ether is 0.01386 moles * 172.18 g/mol = 2.39 g.

Percentage yield:
Actual yield / Theoretical yield * 100%
= (1 g / 2.39 g) * 100%
= 41.8% (rounded to one decimal place)