Question

Calculate the theoretical yield and percent yield for experiments 7, 8, and 9. For experiment 9, determine the theoretical yield based on starting with 0.50g of 3-nitrobenzoic acid. Show work

Acid I (experiment 7): Weight of methyl 3-nitrobenzoate = 1.9 g
Acid II (experiment 8): Weight of 3-nitrobenzoic acid = 1.75 g
Acid III (experiment 9): Weight of 3-nitrobenzamide = 0.30 g

Experiment 7:

-2.5 mL concentrated H₂SO₄

-1.75 mL concentrated HNO₃ (16M)

-2.1 mL concentrated Methyl Benzoate

H2SO4 HNO3 H₂O 2= 0

Experiment 8:

-0.7g of sodium hydroxide pellets for each gram of ester

-2mL of HCL

Experiment 9:

Answer 1

kaperiment 7 O- onco-Chy Hysay, HNU Name: Methylbenzoate 3- utumethyltenzoate Molecular formula : CH 02 Cq Hy Nou Molar Mass : (12x8) + (1x8) + (16 x2) (12x8) + (1x7) +(14x1) __96 +8 + 32. = 136 9 mal. + (1684) - 96+ 7+ 14 + 64 = 181 groot Density of methylbenzoate = 1.0891 ml volume of methylbenzoate dred 2mk Mase of methylsinzoate lized Density X Volume = 1.08 x 2.1 2.2689 No.04 moles of methylbenzoate - 2268 mor 136 Now, I mal of methylbenzoate yields I mol M 3- Neteamethyl- benzoate 2:268 mol of methylbenzoatel 136 2:26.8 mil 17.3- Netuomethyl benzoats 136 . Theoretical yilld of 3- Nutuomethylbenzoate is calculated as. 2:268" X481' 3.0184 g 136 V Molar mass of 3- Niteromethylbenzoate Now, Eqvimentally obtained field 3- Nituomethylbenzoate is 1.9g so Lexantage yield - Kapimental Yield x.loo • 19 X 100 Theoretical yuld 3.0184 = 62.95%

Experiment -8. OH Co- CH₂ Naoh Name : 3- Nitromethylbenzoate Molecular formula: Cg Hy Nou Molar mass: 1819 mee 3- Nutusbenzoic acid Ca He Nou (1247) + (1x5) +(1431) + (1684) -84+5+14+64 : 167 g/mol Now, one mol of A fields one moe of B ii 1819 of o yields 167 g of i 1.9g of Guelds 167 x 19 of ③ + 1.7539 07 ® • Theoretical field Now, Emperimental Yild 1.759 .. luscentage yield 1:75 4400 9982.86% 1.753

- NH₂ Experiment -9. COOH y soch @ 4) aq. Nhis No _ (B) 3- Netwobenzamide Name: 3- Niteroseng Molecular formula: Cq Hg Noy CH NO Molau Mars: - 16791 mie (1281)+(1x6) + (2x14). + (163) = 84 +6 + 28 +48 = 166 g/mol 167 Now, 167 g of ④ yields 1669 of 6 . 0.5g of 4 juilds 166 x0.5 = 0.4979 of B - Theoretical field Espesiesental yild 0:309 i Percentage yild0.30 x Loo * 60-36.. 0.492 kapelissent ne Theoretical yield Percentage yield 62:95.1: 3.0184 g 8_ 1 1.7539 99.828.): 0.4979 60.3600