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

Answer 1

The answer is (d) `"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 `"100.0-g"` sample of pyrolusite. You know that the percent composition of manganese-55 in pyrolusite is `63%`.

This means that you get `"63 g"` of manganese-55 for every `"100.0 g"` of mineral. In other words, this `"100-g"` sample will contain

• manganese-55 `-> color(white)(x)"63 g"`
• oxygen-16 `-> color(white)(x)"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)`