Question

Calculate the theoretical yield of 1-bromobutane; base your calculations on using 1.0 g of 1-butanol (as the limiting reagent).

Answer 1

Answer:

Theoretical yield = 1.85 g

Explanation:

Moles of $1-butanol$:-

Mass = 1.0 g

Molar mass of $1-butanol$ = 74.12 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.0\ g}{74.12\ g/mol}$

$Moles_{1-butanol}= 0.0135\ mol$

According to the reaction:-

C₄H₉OH + HBr ⇒ C₄H₉Br + H₂O

1 mole of 1-butanol on reaction forms 1 mole of 1-bromobutane

0.0135 mole of 1-butanol on reaction forms 0.0135 mole of 1-bromobutane

Mole of 1-bromobutane = 0.0135 mole

Molar mass of 1-bromobutane = 137.02 g/mol

Mass = Moles * Molar mass = 0.0135 mole * 137.02 g/mol = 1.85 g

Theoretical yield = 1.85 g

Answer 2

C₄H₉OH + HBr = C₄H₉Br + H2O

Δmole of alcohol gives 1 mole of bromobutanol

HBr is in excess, so the yield of the product is limited by the alcohol

Wt. of 1 butanol = 18

Molar mass of the butanol = 74.12 g/mole

Moles of the alcohol = 1/74.12 = 0.01349 moles

So, moles of bromobutane = 0.01349 moles

Molar mass of C₄H₉Br = 137.018 g/moles

So, theoretical mass of bromobutane is = 0.01349 × 137.0.18

= 1.85 g