Amolecule contains 24.36g of n and 62.64 g of ag. what is its empirical formula

Answer is: empirical formula is AgN₃.
m(N) = 24,36 g.
n(N) = m(N) ÷ M(N).
n(N) = 24,36 g ÷ 14 g/mol.
n(N) = 1,74 mol.
m(Ag) = 62,64 g.
n(Ag) = m(Ag) ÷ M(Ag).
n(Ag) = 62,64 g ÷ 107,87 g/mol.
n(Ag) = 0,58 mol.
n(N) . n(Ag) = 1,74 mol : 0,58 mol /÷0,58.
n(N) . n(Ag) = 3 : 1.

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