A. The student used 1.2ml of component X in an organic chemistry lab experiement

Question

A. The student used 1.2ml of component X in an organic chemistry lab experiement

and 1ml of component Y

in 3ml solvent component S

for a reaction

X - MW is 181.31g/mol; density is 1.14g/cm3

Y - MW is 120.15g/mol; densiuty is 1.028 g/cm3

S - MW is 74.12g/mol; density is 713kg/m3

Using some or all of the information, the student came up with a theoretical yield of 1.69g. How did the student come up with 1.69g as the theoretical yield? Show all work.

B. Pretend the reaction in A was 1.2ml phenylmagnesium bromide reacting with 1ml acetophenone in a 3ml solvent of diethyl ether to produce the 1.69g theoretical yield. This reaction created 0.65g of 1,1-diphenylethanol (MW 198.26g/mol). Determine the Percent Yield of this experiment reaction. Show all work.

********Please be legible!!!!!!!!!**************

Explanation / Answer

A. The equation can be written as follows:

X + Y => Product

Given information

X = 1.2 mL

Y = 1 mL

X - MW is 181.31g/mol; density is 1.14g/cm3

Y - MW is 120.15g/mol; densiuty is 1.028 g/cm3

S - MW is 74.12g/mol; density is 713kg/m3

Using d = mass/volume we will first calculate mass followed by the calculation of number of moles

First let us evaluate moles of X

density of X = 1.14g/cm3 = 1.14 g/mL = mass / volume = mass / 1.2

So mass of X = 1.14 x 1.2 = 1.368 g

Now moles of X = 1.368/181.31 = 0.00754 moles

Now let us evaluate moles of Y

density of Y = 1.028g/cm3 = 1.028 g/mL = mass / volume = mass / 1

So mass of X = 1.028 x 1 = 1.028 g

Now moles of X = 1.028/120.15 = 0.00855 moles

So now it means 0.00754 moles of X reacts with 0.00855 moles of Y in order to produce 0.00754 moles of product as the one which is having lesser amount in terms of moles will act as limiting reagent and will limit the formation of product.

Once we will know the molecular weight of product we can calculate the theoritical yield

and according to the theoritical yield mentioned in the question the molecular weight of the prodcut should be 224.13 g/mol and when this quantity is multiplied by 0.00754 moles it will give rise to theoritical yield as 1.69 g

B. Percentage yield = (Practical yield (Experimental yield) / Theoritical yield) x 100

= (0.65 / 1.69) x 100

= 38.46 %

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