Question

i need help.

what is the experimental yield, calculated of %yield, and %yield of the nitration of methyl benzoate?

this is my data:

Rart 1: Mass of Viaitcapt tontentg1 mass of vicel t(ae s 14.071 11.065 Ar&SSOt Acethy Benroate to 8.000 Part 2 ess of ViadCapt conternt as 12. 058 Finn(dy Consent a953 Mlting 1ange: 7e-77. Volumes of acids (hosH NdVIC Cid NHo no acid HNO Sulfent ocd

I also need to know how to get the moles in the gram row of H2O in this chart. my teacher said I need to work backwards, but i didnt understand.

this is how i got the calculations in the chart:

Answer 1

Moles of methyl benzoate taken = 3.006g/136.15g/mol = 0.0220786mol

Concentration of sulfuric acid and nitric acid is not provided, I am assuming that these are present in sufficient amount and moles of products would be same as moles of methyl benzoate .

Moles of product(theoretical) = 0.0220786mol

Mass of product(theoretical) = 0.0220786mol×182.15g/mol = 4.021615g

% experimental yield = (0.953g/4.021615g)×100% = 23.6969%

= 23.7 %. (Answer)

------------------------

Horizontal row Table data :

Since number of moles of m-nitro methyl benzoate (product) is equal to the number of moles of water . Hence moles of water produced = 0.00700mol

Mass of water = number of moles × molar mass of water = 0.00700mol× 18.01g/mol = 0.12607 g = 0.126 gram. (Answer)