The equation for formation of 1-Bromobutane is
CH3(CH2)3OH + HBrCH3(CH2)Br + H2O
We are starting with 22 mL of 1-Butanol.
Density of 1-Butanol=0.810 g/mL
Mass of 1-Butanol used= volume of 1-Butanol used * Density
= 22 mL * 0.810 g/mL
=17.82 g
No. of moles of 1-Butanol used= Mass of 1-Butanol used / Molar mass of 1-Butanol
=17.82 g /74 (g/mol)
= 0.2408 moles
From the balanced equation for formation of 1-Bromobutane from 1-Butanol , we see that 1 mole of 1-Butanol produces 1 mole of 1-Bromobutane.
Therefore, 0.2408 moles of 1-Butanol will produce 0.2408 moles of 1-Bromobutane.
Mass of 0.2408 moles of 1-Bromobutane=0.2408 moles * molar mass of 1-Bromobutane.
=0.2408 moles * 137 (g/mol)
= 32.9896 g
Therefore, expected mass of 1-Bromobutane assuming 100% yeild = 32.9896 g
How to calculate the expected mass of 1-bromobutane assuming 100% yeild experiment 4 preparation...
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