% composition of bromine = (atomic mass of bromine / molar mass of NaBr) x 100
The atomic mass of bromine (Br) is 79.904 g/mol. The molar mass of NaBr is the sum of the atomic masses of sodium (Na) and bromine (Br). The atomic mass of sodium is 22.990 g/mol. Therefore, the molar mass of NaBr is 22.990 g/mol + 79.904 g/mol = 102.894 g/mol.
Substituting these values into the formula, we get:
% composition of bromine = (79.904 g/mol / 102.894 g/mol) x 100
% composition of bromine = 0.777 x 100
% composition of bromine = 77.7%
Therefore, the percent composition of bromine in the compound NaBr is 77.7%.
