Answer:
The yield of the product in gram is [tex]\mathsf{{w_P}=0.26 \ gram}[/tex]
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 = [tex]2.7 \times 10^{-3} \ moles[/tex]
We know that:
Number of moles = mass/molar mass
Then:
[tex]2.7 \times 10^{-3} = \dfrac{ mass}{257.997}[/tex]
[tex]mass = 2.7 \times 10^{-3} \times 257.997[/tex]
mass = 0.697
Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100
i.e
Theoretical yield = [tex]\dfrac{n_P}{n_R}\times 100\%[/tex]
where;
[tex]n_P = \dfrac{w_P}{m_P}[/tex] and [tex]n_R = \dfrac{w_R}{m_R}[/tex]
Theoretical yield = [tex]\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%[/tex]
Given that the theoretical yield = 100%
Then:
[tex]100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%[/tex]
[tex]\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}[/tex]
[tex]{w_P}=\dfrac{w_R \times m_P}{m_R}[/tex]
where,
[tex]w_P[/tex] = derived weight of the product
[tex]m_P =[/tex]the molecular mass of the derived product
[tex]m_R =[/tex] the molecular mass of the reagent
[tex]w_R[/tex] = weight in a gram of the reagent
[tex]{w_P}=\dfrac{w_R \times m_P}{m_R}[/tex]
[tex]{w_P}=\dfrac{0.697 \times 96.2}{257.997}[/tex]
[tex]\mathsf{{w_P}=0.26 \ gram}[/tex]
