1. Find the molar mass of MnCl₂:
* Mn (Manganese): 54.94 g/mol
* Cl (Chlorine): 35.45 g/mol (x2 since there are two chlorine atoms)
Molar mass of MnCl₂ = 54.94 g/mol + (35.45 g/mol * 2) = 125.84 g/mol
2. Calculate the mass percentage of each element:
* Manganese (Mn):
(54.94 g/mol / 125.84 g/mol) * 100% = 43.67%
* Chlorine (Cl):
(70.90 g/mol / 125.84 g/mol) * 100% = 56.33%
Therefore, the percent composition of MnCl₂ is:
* Manganese (Mn): 43.67%
* Chlorine (Cl): 56.33%
