Initial mass of methyl benzoate = 3.4 gms
Molar mass of methyl benzoate = 136.15 g/mol
No. of moles of methyl benzoate = Mass/molar mass = 3.4 g/136.15 g/mol = 0.02497 moles
We can observe in the above nitration reaction the molar ratio of methyl benzoate and meta nitro methyl benzoate is 1:1
Similarly the molar ratio of methyl benzoate and para nitro methyl benzoate is 1:1
Molar mass of meta nitro methyl benzoate / para nitro methyl benzoate is 181.15 g/mol
Hence moles of meta nitro methyl benzoate / para nitro methyl benzoate = 0.02497 moles
Theoretical Mass of meta nitro methyl benzoate / para nitro methyl benzoate = moles x molar mass = 0.02497 mol x 181.15 g/mol = 4.524 g
Percentage yield of product 1 (meta nitro methyl benzoate) = (actual yield/theoretical yield) x 100 = (3.9155g/4.524 g) x 100 = 86.5 %
Percentage yield of product 2 (para nitro methyl benzoate) = (2.1999 g/4.524 g/mol) x 100 = 48.6 %
the weight if the initial product was 3.4 grams of methyl benzoate. Who |o yald foR 3.91SSg st prodoc...
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-NITRATION OF Methyl benzoate -addditional information. .( 3.9 g of the product methyl 3-nitrobenzoate for the percentage yield Materials used: Amount Compound used/ Molecular weight Density (g/mL) Moles produced 3.05g used 136.15 g/mol N/A 0.0224 mol Methyl Benzoate HO-S-OH 6mL used 98.079 g/mol 1.83 g/cm3 0.112 Sulfuric Acid 2mL used 63.01 g/mol 1.51 g/cm3 0.047g o-N-OH Nitric Acid 3.034g of pure product 181.15 g/mol produced N/A 0.0167 mol NO2 Methyl-m- nitrobenzoate Calculating limiting reagent and theoretical yield: 0.0224 mol methylbenzoate...
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