Question

In performing the Grignard reaction under the conditions of Experiment 16, if you were to use 121 mg of benzophenone (182.21 g/mol), 18 mg of magnesium (24.30 g/mol), and 76 microliters of bromobenzene (157.02 g/mol, 1.50 g/mL), calculate the theoretical yield of triphenylmethanol (260.33 g/mol), in mg. Benzophenone is the limiting reactant. Enter your answer as digits only, no units, using the proper number of significant figures.

Answer 1

Answer: The theoretical yield of triphenylmethanol is 0.173 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of benzophenone = 121 mg = 0.121 g (Conversion factor: 1 g = 1000 mg)

Molar mass of benzophenone = 182.21 g/mol

Putting values in equation 1, we get:

`\text{Moles of benzophenone}=(0.121g)/(182.21g/mol)=6.64* 10^(-4)mol`

The chemical equation for the formation of triphenylmethanol from benzophenone follows:

`\text{Benzophenone}+Mg+\text{Bromobenzene}\rightarrow \text{Triphenylmethanol}+MgBr`

As, bromobenzene is the limiting reagent. So, it will limit the formation of products

By Stoichiometry of the reaction:

1 mole of bromobenzene produces 1 mole of triphenylmethanol

So,
`6.64* 10^(-4)mol` of bromobenzene will produce =
`(1)/(1)* 6.64* 10^(-4)=6.64* 10^(-4)` moles of triphenylmethanol

Now, calculating the mass of triphenylmethanol from equation 1, we get:

Molar mass of triphenylmethanol = 260.33 g/mol

Moles of triphenylmethanol =
`6.64* 10^(-4)` moles

Putting values in equation 1, we get:

`6.64* 10^(-4)mol=\frac{\text{Mass of triphenylmethanol}}{260.33g/mol}\\\\\text{Mass of triphenylmethanol}=(6.64* 10^(-4)mol* 260.33g/mol)=0.173g`

Hence, the theoretical yield of triphenylmethanol is 0.173 grams