Question

Answer the following questions using the chemical reaction shown below.   (A) Terephthalic acid+ (B) methanol -----> (C) C10O4H10 + (D) H2O           a. Identify the functional groups b. Balance the chemical equation   c. For the questions below you will need the molar mass of A, B, and C. Calculate the molar mass for A, B, and C.   d. The reaction was performed using 50.5mg of compound A. Calculate the amount of moles.   e. If 25.3 uL of compound B is needed to carry out the chemical reaction, how many moles of compound B they added in the reaction flask?    f. Determine the limiting reactant      g. What is the theoretical yield for compound C?   h. If 25.3 mg of compound C is isolated after purification, what is the % yield of this reaction?  Thank you for doing this for me I want to check my answers because it has been awhile since I've done chemistry.

Answer 1

Answer:

a) Answer:

Acid groups (COOH) is the functional group.

b) Balanced equation:

C₈H₆O₄ + 2CH₃OH -------------> C₁₀H₁₀O₄ +2H₂O

c) molar masses:

Note: Molar mass (red colored) are consider only for the one mole of the compound.

d) Answer:

A (Terephthalic acid) weight = 50.5 mg = 50.5 x 10^-3 g

moles = weight / molar mass

moles = 50.5 x 10^-3 / 166.13

moles = 0.3039 x 10^-3

e) Answer:

C₈H₆O₄ + 2CH₃OH -------------> C₁₀H₁₀O₄ +2H₂O

25.3 x 10^-6 L

In STP condition 22.4 L is consired as 1 mole

25.3 x 10^-6 L ......................................... ?

= (25.3 x 10^-6 x 1) / 22.4

= 1.129 x 10^-6 moles.

f) Limiting reactant:

C₈H₆O₄    + 2CH₃OH -------------> C₁₀H₁₀O₄ + 2H₂O

166 2 x 22.4 L 194 2 x 22.4 L

50.5 x 10^-3 g ......... ?

= (50.5 x 10^-3 x 2 x 22.4) / 166

= 13.628 x 10^-3 L

Then converting in to micro liter

= 13.628 x 10^-3 x 10^-6

= 13.628 x 10^-9 microliter

But in the question 25.3 micro liter given,

So, The MeOH is more than its required. Hence the Compound A is the limiting reagent of the reaction.

g) Theorical yield of C:

C₈H₆O₄    + 2CH₃OH -------------> C₁₀H₁₀O₄ + 2H₂O

166 194

50.5 mg ................................................................. ?

= (50.5 x 194) / 166

= 59.018 mg.

h) % yield of reaction:

we know the 100% yield 59.018 mg

59.018 mg ...................... 100

25.3 mg ........................... ?

= (25.3 x 100) / 59.018

= 2530 / 59.018

= 42.868

Hence the percentage of the yield = 42.868