# 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. Mn2O c. Mn2O2 d. MnO2 e. none of these?

Nov 17, 2015

The answer is (d) ${\text{MnO}}_{2}$

#### Explanation:

The key to this problem is the given percent composition of maganese-55 in the mineral.

What you have to do here is pick a sample of the mineral, then work out exactly how many grams of each isotope you get in that sample.

To make calculations easier, let's pick a $\text{100.0-g}$ sample of pyrolusite. You know that the percent composition of manganese-55 in pyrolusite is 63%.

This means that you get $\text{63 g}$ of manganese-55 for every $\text{100.0 g}$ of mineral. In other words, this $\text{100-g}$ sample will contain

• manganese-55 $\to \textcolor{w h i t e}{x} \text{63 g}$
• oxygen-16 $\to \textcolor{w h i t e}{x} \text{37 g}$

To get the empirical formula of the mineral, you need to determine how many moles of each isotope you have in this sample. Use the molar masses of manganese-55 and oxygen-16 to get

63color(red)(cancel(color(black)("g"))) * ("1 mole " ""^55"Mn")/(54.938color(red)(cancel(color(black)("g")))) = "1.1467 moles " ""^55"Mn"

and

37color(red)(cancel(color(black)("g"))) * ("1 mole " ""^16"O")/(15.995color(red)(cancel(color(black)("g")))) = "2.313 moles " ""^16"O"

Next, find the mole ratio that exists between the two isotopes by dividing these values by the smallest one

"For Mn: " (1.1467color(red)(cancel(color(black)("moles"))))/(1.1467color(red)(cancel(color(black)("moles")))) = 1

"For O: " (2.313color(red)(cancel(color(black)("moles"))))/(1.1467color(red)(cancel(color(black)("moles")))) = 2.017 ~~ 2

The empirical formula of pyrolusite will thus be

"Mn"_1"O"_2 implies color(green)("MnO"_2)