What is the experimental and theoretical percent yield of synthesis of methyl orange? IntroductionMethyl orange is an azo dye whose thesis can be accomplished by a diazo coupling reactionbetween the diazonium salt of sulfanilic acid and N,N-dimethylaniline. Since the coupling reactionis conducted in an acidic solution, the first product to form is the acid form of methyl orange,called helianthin, which is bright red. The addition of a base leads to the conversion of helianthinto its sodium salt, which is methyl orange.HH;C.CH3OCH3H3COAcНОАсH,CČH3proton transferH3CH3CHelianthin (red)NaOHH3CN=N-SO,NaH3CMethyl oranIt is necessary to carry out the diazotization reaction in an acidic solution. However, sulfanilic acidis insoluble in acidic solutions. To overcome this problem, the sulfanilic acid is first dissolved ina basic solution: an aqueous solution of sodium carbonate- Sodium nitrite is then added to thisbasic solution. Finally, the solution is chilled and then acidified with hydrochloric acid. The instantthe sulfanilic acid precipitates from the solution, it becomes diazotized, yielding a finely dividedwhite precipitate of the diazonium salt of sulfanilic acid. This diazonium salt is allowed to reactimmediately with dimethylaniline in the solution in which it was precipitated. Running the reaction:0.065 g of anhydrous sodium carbonate were dissolved in 5 mL water in a 25 mLErlenmeyer flask0.202 g of sulfanilic acid were added and the mixture was heated which caused the solidsto dissolveAfter cooling the mixture to room temperature, 0.092 g of sodium nitrite were added whilestirringThe mixture was placed in an ice bath to cool it to 5 °C before 0.30 mL concentratedhydrochloric acid was added dropwiseSlowly, a fine, white precipitate was formedPart 2:Running the reaction:The suspension formed in part 1 was placed in an ice-bath0.17 mL of N,N-dimethylaniline and 0.12 mL of glacial acetic acid were placed in a 3-mLvialThis solution was added dropwise to the suspension obtained in part 1The mixture was stirred affording a red precipitateAfter 10 minutes, 1.5 mL of 10 % NaOH were added dropwiseThe mixture was stirred for 10 minutes and then the pH value was tested (pH=8-9)Product isolation:The suspension was heated to a gentle boil and 0.60 g of solid NaCl was addedThe solution could cool to room temperature and then be placed in an ice-bathThe solid was isolated by filtration on a Buchner funnelThe remaining solids were transferred using 5 mL of cold sodium chloride solutionThe solids and the filter paper were placed in 15 mL of boiling water to dissolve the productThe mixture could cool to room temperature before placing in an ice bath (~15 minutes)The solids are isolated by vacuum filtrationAfter allowing to dry, weigh the productProduct characterization:0.185 g of the product

In the given chemical equation, 1 mol sulfanilic acid reacts with 1 mol of N,N-dimethylaniline to form methyl orange dye. sulfanilic acid + N,N-dimethylaniline ------> Orange dye Since all the coefficients are 1, mole ratio is 1/1. Hence the reactant having the smallest mole is the limiting reactant. Given the mass of sulfanilic(C6H7NO3S) acid taken = 0.202 g Molar mass of sulfanilic acid (C6H7NO3S) = 173.2 g/mol Moles of sulfanilic(C6H7NO3S) acid taken = 0.202 g * 1 mol173.2 g= 0.001166282 molVolume of N,N-dimethylaniline taken, V = 0.17 mL Density of N,N-dimethylaniline, d = 0.956 g/mL => mass of N,N-dimethylaniline taken, m = V*d = 0.17 mL * 0.956 g/mL = 0.16252 g Molar mass of N,N-dimethylaniline(C8H11N) = 121.19 g/mol => Moles of N,N-dimethylaniline(C8H11N) taken = 0.16252 g * 1 mol121.19 g= 0.001341035 mol Since sulfanilic(C6H7NO3S) acid has the smaller mole, it is the limiting reactant. Hence we will use the moles of sulfanilic(C6H7NO3S) acid to find the theoretical mass of the product.Mole ratio between sulfanilic(C6H7NO3S) acid and product = 1/1 Hence moles of product, methyl orange(C14H14N3NaO3S) = moles of sulfanilic(C6H7NO3S) acid = 0.001166282 mol Molar mass of methyl orange(C14H14N3NaO3S) = 327.33 g/mol Theoretical yield of methyl orange = 0.001166282 mol * 327.33 g1 mol= 0.382 g (Answer) Theoretical yield of methyl orange = 0.382 g (Answer) Experimental yield of methyl orange = 0.185 g (Answer) Percent yield = Experimental yieldTheoretical yield*100= 0.185 g0.382 g*100= 48.4 % Percent yield = 48.4 % (Answer)