Question

Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol). Predict the theoretical 1 0096 yield of the product in grams. Round to the hundredths place.

Answer 1

The yield of the product in gram is
`\mathsf{{w_P}=0.26 \ gram}`

Step-by-step explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles =
`2.7 * 10^(-3) \ moles`

We know that:

Number of moles = mass/molar mass

Then:

`2.7 * 10^(-3) = ( mass)/(257.997)`

`mass = 2.7 * 10^(-3) * 257.997`

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield =
`(n_P)/(n_R)* 100\%`

where;

`n_P = (w_P)/(m_P)` and
`n_R = (w_R)/(m_R)`

Theoretical yield =
`(((w_P)/(m_P)))/(((w_R)/(m_R))) * 100\%`

Given that the theoretical yield = 100%

Then:

`100\% =(((w_P)/(m_P)))/(((w_R)/(m_R))) * 100\%`

`(w_P)/(m_P)=(w_R)/(m_R)`

`{w_P}=(w_R * m_P)/(m_R)`

where,

`w_P` = derived weight of the product

`m_P =`the molecular mass of the derived product

`m_R =` the molecular mass of the reagent

`w_R` = weight in a gram of the reagent

`{w_P}=(w_R * m_P)/(m_R)`

`{w_P}=(0.697 * 96.2)/(257.997)`

`\mathsf{{w_P}=0.26 \ gram}`