Aspirin or acetyl salicylic acid is synthesized by reacting...

Aspirin or acetyl salicylic acid is synthesized by reacting salicylic acid with acetic anhydride: $\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\quad\quad\quad+\quad\quad\quad\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\quad \quad \longrightarrow \quad\quad\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$ $\begin{array}{llll}\text { salicylic acid } & \text { acetic anhydride } & \quad\quad\quad\quad\quad\text { aspirin } & \text { acetic acid }\end{array}$ (a) How much salicylic acid is required to produce $0.400 \mathrm{g}$ of aspirin (about the content in a tablet), assuming acetic anhydride is present in excess? (b) Calculate the amount of salicylic acid needed if only 74.9 percent of salicylic acid is converted to aspirin. (c) In one experiment, 9.26 g of salicylic acid is reacted with $8.54 \mathrm{g}$ of acetic anhydride. Calculate the theoretical yield of aspirin and the percent yield if only $10.9 \mathrm{g}$ of aspirin is produced.

Aspirin or acetyl salicylic acid is synthesized by reacting salicylic acid with acetic anhydride:
$\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\quad\quad\quad+\quad\quad\quad\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\quad \quad \longrightarrow \quad\quad\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
$\begin{array}{llll}\text { salicylic acid } & \text { acetic anhydride } & \quad\quad\quad\quad\quad\text { aspirin } & \text { acetic acid }\end{array}$
(a) How much salicylic acid is required to produce $0.400 \mathrm{g}$ of aspirin (about the content in a tablet), assuming acetic anhydride is present in excess?
(b) Calculate the amount of salicylic acid needed if only 74.9 percent of salicylic acid is converted to aspirin. (c) In one experiment, 9.26 g of salicylic acid is reacted with $8.54 \mathrm{g}$ of acetic anhydride. Calculate the theoretical yield of aspirin and the percent yield if only $10.9 \mathrm{g}$ of aspirin is produced.

Key Concepts

Percent Yield

Percent yield quantifies the efficiency of a chemical reaction by comparing the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry. It is calculated by dividing the actual yield by the theoretical yield and multiplying by 100, thereby highlighting potential losses or side reactions in practical settings.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a reaction based on the stoichiometric calculations, assuming complete conversion of the limiting reactant. It serves as an ideal benchmark to compare with actual experimental yields.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed. Identifying the limiting reactant is fundamental in stoichiometry because it governs the theoretical yield of the reaction.

Molar Mass

Molar mass is the mass of one mole of a substance and is expressed in grams per mole. Knowledge of molar masses is essential in converting between the mass of a substance and the number of moles, which is a necessary step in applying stoichiometric calculations in any chemical reaction.

Stoichiometry

Stoichiometry involves using the mole ratios derived from a balanced chemical equation to relate quantities of reactants and products. It allows one to calculate the amounts, often in moles or grams, of substances consumed and produced during a chemical reaction, establishing the foundation for scaling reactions up or down.

Balanced Chemical Equation

A balanced chemical equation shows the precise stoichiometric relationships between reactants and products in a reaction. It ensures that the number of atoms for each element is conserved, which is crucial for determining the mole ratios needed for any quantitative analysis such as determining how much of one substance is needed to form a given amount of another.

Transcript

00:02 Given information is c7 -8603 plus c4 -8603 into c -9 -h8 -804 solution for part a, the mass of s8804 -2.

00:53 Sparing spleen c9 h804 is mass of sparing is is equal to 0 .4400 gram and we have to find the mass of sly cycolic acid c7 8 6 -03 needed to produce a given amount of spreen.

02:22 First, we will find the number of moles of spleen.

02:44 The molar mass of spleen is molar mass of spleen is equal to 9 into molar mass of carbon plus 8 into molar mass of hydrogen plus 4 into molar mass of oxygen is equal to 9 into molar mass of carbon is 12 .011 gram per mole plus 8 into molar mass of hydrogen is 1 .008 gram per mole plus 4 into molar mass of oxygen is 15 .99 gram per mole.

03:52 It is equal to the calculated melee of molar mass of spleen is 180 .159 gram per mole.

04:09 The number of moles of spreen is, spleen is number of moles of spreen.

04:32 Is equal to given mass of spleen divided by molar mass of a spring is equal to given mass of a spring is 0 .40 gram divided by molar mass is 180 .159 gram per mole is equal to 22 .22 .22 into to 10 raise to the power minus 3 mole.

05:31 And here, 1 mole of spleen is produced by 1 mole of sly cyclic acid.

05:41 Hence 2 .22 into 10 raise to the power minus 3 moles of spreen will be produced by 2 .22 into 10 raise to the power minus 3 mole of celly cyclical.

05:56 Acid.

05:57 Now we can find the mass of cellic cyclic acid.

06:24 So the molar mass of sally cyclic acid is molar mass of cellic cyclic acid is equal to seven into molar mass of carbon plus 6 into molar mass of hydrogen plus 3 into molar mass of oxygen is equal to 7 into molar mass of carbon is 12 .01 gram per mole plus 6 into molar mass ofulsion is 1 .008 gram per mole plus 6 into molar mass of hydrogen is 1 .008 gran per mole plus 3 into molar mass of oxygen is 15 .99 gram per mole.

07:36 So the calculated molar mass of cellicic acid is equal to 138 .122 gram per mole and the mass of cellicicolic acid acid mass of cellic cyclic acid acid is equal to number of moles of cellicylic acid into molar mass of salicyclic acid is equal to number of moles of cellicicolic acid is 2 into 2 .22 into 10 raised to the power minus 3 moles into 138 0 .122 gram per mole is the molar mass of celliclic acid.

09:03 The calculated mass of cellicic acid is 0 .307 gram and given information for part b, c7 -8603 3 plus c4 -8603 into c9 h804 plus c2 or 2 from part a we know that we need we need 0 .3 07 gram of cellicic acid.

10:25 Now we will find the amount of cellic acid needed if only 74 .9 % of cellic cyclic acid is converted to aspyrin so if 0 .307 grams is 74 .9 % the starting the starting 100 % would be would be mass of salarly cyclic acid is equal to 0 .307 gram divided by 74 .9 % into 100 % is equal to 0 .410 gram.

12:33 The calculated mass of cellic cyclic acid.

12:38 So the starting mass of the starting mass of of cellicic acid should be 0 .410 gram.

12:51 Given information for part c7, h6 -03 plus c4, h6 -03, c9, h8804, plus c2, h -4 -2, h -4 -o2.

13:26 The mass of cellicic acid is mass of cellic cyclic acid is equal to 9 .26 ground...

Please give Ace some feedback