For a Grignard reaction, if you start with 3.97 mL of pure bromobenzene and 0.733 g of 1-phenylethanone, what is the theoretical yield of 1,1-diphenylethanol?

Please help.

This is a typical limiting reagent problem. I solve these this way.

I assume the equation is 1:1 for all reactant and products but if it isn't you need to write the equation and balance it.
Take the first reagent, convert grams to mols (mols = grams/molar mass), then convert to mols of the product.
Take the other reagent, convert to moles of the reagent then to mols of the product.
You likely will obtain two different values for mols of the product and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value (and the reagent producing that value is the limiting reagent). Then convert the smaller value for mols into grams (g = mols x molar mass). That is the theoretical yield.

thanks!

To calculate the theoretical yield of 1,1-diphenylethanol in a Grignard reaction, we need to first determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.

Step 1: Convert the given volume of bromobenzene from mL to grams using its density.
The density of bromobenzene is typically around 1.50 g/mL. So, we can multiply the given volume (3.97 mL) by its density to get the mass in grams:
3.97 mL x 1.50 g/mL = 5.955 g

Step 2: Calculate the number of moles for each reactant.
The number of moles is calculated using the formula:
moles = mass (g) / molar mass (g/mol)

The molar mass of bromobenzene (C6H5Br) can be calculated by adding up the atomic masses of its constituents:
C (carbon): 12.01 g/mol
H (hydrogen): 1.01 g/mol (6 hydrogen atoms in bromobenzene)
Br (bromine): 79.90 g/mol

Molar mass of bromobenzene = (6 x 1.01 g/mol) + 12.01 g/mol + 79.90 g/mol = 157.97 g/mol

moles of bromobenzene = 5.955 g / 157.97 g/mol ≈ 0.038 moles (rounded to three decimal places)

For 1-phenylethanone, we are given the mass directly in grams.

moles of 1-phenylethanone = 0.733 g / molar mass of 1-phenylethanone

We now need to calculate the molar mass of 1-phenylethanone (C8H8O):
C (carbon): 12.01 g/mol (8 carbon atoms in 1-phenylethanone)
H (hydrogen): 1.01 g/mol (8 hydrogen atoms in 1-phenylethanone)
O (oxygen): 16.00 g/mol

Molar mass of 1-phenylethanone = (8 x 12.01 g/mol) + (8 x 1.01 g/mol) + 16.00 g/mol = 120.15 g/mol

moles of 1-phenylethanone = 0.733 g / 120.15 g/mol ≈ 0.006 moles (rounded to three decimal places)

Step 3: Determine the limiting reagent.
To determine the limiting reagent, compare the mole ratios of the reactants. The balanced reaction equation for the Grignard reaction is:

1-phenylethanone + bromobenzene + Mg → 1,1-diphenylethanol + MgBrOH

From the equation, we can see that the mole ratio of 1-phenylethanone to bromobenzene is 1:1. However, the actual mole ratio of the reactants is 0.006 moles (1-phenylethanone) to 0.038 moles (bromobenzene). The actual ratio of moles is 0.006/0.038, which is less than 1. Therefore, 1-phenylethanone is the limiting reagent.

Step 4: Calculate the theoretical yield.
The theoretical yield is the maximum amount of product that can be obtained from the limiting reagent. From the balanced equation, we can see that the mole ratio of 1-phenylethanone to 1,1-diphenylethanol is 1:1.

moles of 1,1-diphenylethanol = moles of 1-phenylethanone (limiting reagent)

moles of 1,1-diphenylethanol = 0.006 moles

Finally, we can calculate the mass of the theoretical yield of 1,1-diphenylethanol by multiplying the moles by its molar mass.

mass of 1,1-diphenylethanol = moles of 1,1-diphenylethanol x molar mass of 1,1-diphenylethanol

You will need to provide the molar mass of 1,1-diphenylethanol in order to complete the calculation.