Question

Two binary oxides of the element manganese contain, respectively 30.40% and 36.81% oxygen by mass. Calculate empirical formulas of the two oxides.

Short answer

The empirical formulas of two compounds of manganese and oxygen are$M{n}_2{O}_3$and$Mn{O}_2$

Step-by-step solution

Number of moles of O2 and Mn calculation

The binary compound having$O_2$and Mn contain$\text{30.40\%}$and$\text{36.81\%}$oxygen respectively$O_2$and 69.60% of Mn.

According to this, 100g of first compound contain 30.40% of

$$\begin{gathered} O_2=30.40\% \\ Mn=100-30.40=69.60\% \end{gathered}$$

Number of moles of each element in the sample

Moles of$O_2=30.40g\times \frac{1mol.O_2}{16.00g}=1.900mol.$

Moles of Mn$=69.60g\times \frac{1molMn}{54.94g}=1.267mol$

First Compounds Empirical formula

Ratio of the Elements O: Mn(1.267:1.900)

Divide each of the amount by smallest amount by smallest amount 1.267

$$\begin{gathered} Mn:O=\frac{1.267}{1.267}:\frac{1.900}{1.267} \\ =1:1.5 \end{gathered}$$.

$$\begin{gathered} Multiplyeachtermby2i.eMn:O \\ 1\times 2:1.5\times 2 \\ 2:3 \end{gathered}$$

Hence the Empirical formula is${\mathbf{Mn}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$

Moles of O2 and Mn in Second Compound

Mass of elements present in 100.0gm of second compound having 36.81% is

0 – 36.81g, Mn = 100-36.81 = 63.19g

No. of moles of each element present is:

Moles of O =$36.81g\times \frac{1molO}{16g}=2.301mol$

Moles of Mn =$63.19g\times \frac{1molMn}{54.94g}=1.150mol$

Second Compounds empirical formula

Ratio of the element O: Mn (2.301:1.150)

$$\begin{gathered} Dividingeachwith1.150=O:Mn \\ \frac{2.301}{1.150}:\frac{1.150}{1.150} \\ 2:1 \end{gathered}$$

Hence, the empirical formula is${\mathbf{Mn}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$

Thus the empirical formula of both the compounds are ${\mathbf{Mn}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$and${\mathbf{MnO}}_{\mathbf{3}}$