Phenol weighing 4.7 g was passed through a solution of bromine water.

Phenol weighing 4.7 g was passed through a solution of bromine water. How much was obtained practically by weight of 2,4,6 – tribromophenol, if the practical yield was 90%.

Given:
m (C6H5OH) = 4.7 g
η (C6H2Br3OH) = 90%

To find:
m pract. (C6H2Br3OH) -?

Decision:
1) C6H5OH + 3Br2 => C6H2Br3OH + 3HBr;
2) M (C6H5OH) = 94 g / mol;
M (C6H2Br3OH) = 331 g / mol;
3) n (C6H5OH) = m (C6H5OH) / M (C6H5OH) = 4.7 / 94 = 0.05 mol;
4) n (C6H2Br3OH) = n (C6H5OH) = 0.05 mol;
5) m theor. (C6H2Br3OH) = n (C6H2Br3OH) * M (C6H2Br3OH) = 0.05 * 331 = 16.55 g;
6) m practical. (C6H2Br3OH) = η (C6H2Br3OH) * m theor. (C6H2Br3OH) / 100% = 90% * 16.55 / 100% = 14.9 g.

Answer: The mass of C6H2Br3OH is 14.9 g.