help with question 2? Look up (or determine using ChemDraw) the names and the mo

ID: 889690 • Letter: H

Question

help with question 2? Look up (or determine using ChemDraw) the names and the molecular masses of each of the five reagents and/or products and record them in the table below. If the chemical species is a liquid at room temperature, report its density; if it is a solid, report its melting point. (You will only report a density or a melting point for each species - leave the other space blank.) Make certain to include appropriate units for all numbers! Assume that the amounts of the reagents used in the reaction (from left to right on the reaction scheme shown above) are 1.26 grams, 1.14 grams, and 1.40 grams, respectively. Answer the following questions, and show your calculations on the back of this form or on an attached sheet. Which compound was your limiting reagent? grams of limiting reagent moles of limiting reagent theoretical yield of product (in grams) actual yield of product (in grams) 1.11 g percent yield

Explanation / Answer

Answer – 1)The given reaction with structures and we need to find out the name and their molar masses

The following table showing the name, molar mass and density or melting point –

Name

Molecular mass

Density

Melting point

4-methoxyphenol

124.1372 g/mol

52.5oC

acetyl chloride

78.49 g/mol

1.10 g/mL

triethylamine

101.19 g/mol

0.726 g/mL

4-methoxyphenyl acetate

166.17 g/mol

1.12 g/mL

triethylammonium chloride

137.65 g/mol

254-260oC

2) Given, mass of 4-methoxyphenol = 1.26 g ,

Mass of acetyl chloride = 1.14 g

Mass of trimethylamine = 1.40 g

We need to calculate moles of each reactant

Moles of 4-methoxyphenol = 1.26 g / 124.1372 g.mol-1

= 0.0102 moles

Moles of acetyl chloride = 1.14 g / 78.49 g.mol-1

= 0.0145 moles

Moles of trimethylamine = 1.40 g / 101.19 g.mol-1

= 0.0138 moles

We are given the mole ration 1:1 for all

So the moles of product lowest from the 4-methoxyphenol

So limiting reactant is 4-methoxyphenol

Grams of limiting reactant is 1.26 g and 0.0102 moles

Theoretical yield of product = 0.0102 moles *166.17 g/mol

= 1.69 g

Percent yield = 1.11 g /1.69 g*100%

= 65.8 %