Question

The mineral pyrolusite is a compound of manganese-55 and oxygen-16. If $63\%$ of the mass of pyrolusite is due to manganese, what is the empirical formula of pyrolusite?
a.$MnO$
b.$M{n_2}O$
c.$M{n_2}{O_2}$
d.$Mn{O_2}$
e.None of these

Answer 1

Pyrolusite is a mineral containing essentially manganese dioxide and is very important as an ore of manganese. Pyrolusite is of black colour, amorphous appearing mineral; often have granular, fibrous or columnar structure sometimes forming reniform crusts.

Complete answer:
To solve this problem we should know the term percent composition. The percent composition is calculated with the help of a molecular formula by dividing the mass of the single element in one mole of a compound by the mass of one mole of the entire compound.
Thus the main part of this question is the given percent composition of manganese-55 in the mineral. For this we will pick a sample of the mineral, then we compute exactly how many grams of each isotope we will get in this sample.
Let us take a $100.0g$ sample of pyrolusite. As it is given that the percent composition of manganese-55 in the pyrolusite is $63\%$. This means that we get $63g$ of manganese-55 for every $100.0g$ sample of pyrolusite. In other words, in this $100.0g$ of sample we have:
$manganese - 55 \to 63g$
$oxygen - 16 \to 37g$
First of all we need to determine how many moles of each isotope we have in this sample with the help of which we will get the empirical formula of the mineral. Thus by using the molar masses of manganese-55 and oxygen-16 we will get:

Element	Amount of element in gram	Relative no. of moles	Simple ratio moles	Simplest whole no. ratio
Mn	63	$63g \times \dfrac{{1\;mole{^{55}}Mn}}{{54.938g}} = 1.1467\;moles{\;^{55}}Mn$	$\dfrac{{1.1467}}{{1.1467}} = 1$	1
O	37	$37 \times \dfrac{{1\;mole{\;^{16}}O}}{{15.995g}} = 2.313\;moles{\;^{16}}O$	$\dfrac{{2.313}}{{1.1467}} = 2.017$	2

The empirical formula of pyrolusite will be $Mn{O_2}$. Thus our answer is option $d$.
Note:
The simplest positive integer ratio of atoms present in a compound is known as the empirical formula of a chemical compound. The ratio holds true on the molar level as well. For example, ${H_2}O$ is composed of two atoms of hydrogen and one atom of oxygen. Likewise, $1.0$ mole of ${H_2}O$ is composed of $2.0$ moles of hydrogen and $1.0$ mole of oxygen.