Question

a molecule contains 24.36 g n and 62.64 g ag.
what is its empirical formula?
agn₃
ag₃n
ag₆n₂
agn₂

Explanation:

Step1: Calculate moles of each element

Molar mass of \(N = 14.01\space g/mol\), moles of \(N=\frac{24.36\space g}{14.01\space g/mol}\approx1.74\space mol\)
Molar mass of \(Ag = 107.87\space g/mol\), moles of \(Ag=\frac{62.64\space g}{107.87\space g/mol}\approx0.58\space mol\)

Step2: Find mole ratio

Divide each mole value by the smaller mole value (\(0.58\))
For \(N\): \(\frac{1.74}{0.58}\approx3\)
For \(Ag\): \(\frac{0.58}{0.58} = 1\)

Answer:

\(AgN_{3}\)