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

Question

| General Chemistry 2

Accepted Answer

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

Answer

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

Mass-mole-number relationship - 2 | Chemical Calculations