Question

QUESTIONS Copy of When a 0.2754 sample of manganese is heated in air/oxveen. 0.3957 g of manganese oxide is produced. What is the empirical formula of the product of this experiment? (the only product formed is the manganese oxide) Mno MnO2 Mngo Mn203 Mn304

Answer 1

Answer:

Mn2O3

Explanation:

Definition of emperical formula is the simplest whole number ratio of atoms in a compound.

Generally in a compound, the ratio of atoms is same as ratio of moles of atoms. So if we can calculate the molar ratio of Manganese (Mn) and Oxygen (O) we could find the emperical formula of compound.

We know that mass of manganese oxide should be equal to the mass of manganese plus mass of oxygen

Therefore, 0.3957 g = 0.2754 g + mass of Oxygen

Mass of oxygen= 0.3957 g - 0.2754 g = 0.1203 g

Now let us calculate the molar ratio of manganese and Oxygen.

Standard values required are Molar mass of Mn= 54.938 g/mol and Molar mass of O = 15.999 g/mol

Moles of Mn= mass/ molar mass

= 0.2754 g÷54.938 g/mol = 5.013E-3 mole Mn

Moles of O = 0.1203 g÷15.999 g/mol

=7.523E-3 mole O

To get the molar ratio in an interger values, we devide with the smaller value that is 5.013E-3

We get 1:1.5 molar ratio between Mn and O ( Mn:O)

This indicates that the emperical formula should be Mn2O3 because it has same 1:1.5 molar ratio of Mn and O . For a compound molar ratio also equal to ratio of atoms. 2 Manganese atoms and 3 Oxygen atoms (2×1.5= 3).

Therefore the emperical formula of manganese oxide is Mn2O3.