Mass-mole-number relationship - 3 | Chemical Calculations

Chapter 2 - Chemical Calculations

Find the empirical formula of a compound containing 1.92 grams of Mn and 1.12 grams of O.

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Solution

n_Mn =

\frac{m_{Mn}}{M_{Mn}}

\frac{1.92}{54.94}

= 3.49 x 10^-2 g.mol^-1

n_O = $\frac{m_{O}}{M_{O}}$ = $\frac{1.12}{16.0}$ = 7.00 x 10^-2 g.mol^-1

n_Mn = 2 n_O, the empirical formula of this compound is MnO₂

General Chemistry 2 - Mass-mole-number relationship - 3