Find the mass of oxygen in 4.00g of manganese dioxide, Mn02.?

There are 2 ways to do this problem.

The fastest way is to solve this is in terms of % composition. Here is the formula you use:

% O2 = mass of O2/Total mass x 100

First, look up the atomic weights for the elements on the periodic table:

O2 = 32 amu

Mn = 55 amu

O2 + Mn = 87 amu (molecular weight)

Then plug these values into the equation above:

% O2 = 32amu/87amu x 100 = 36.8%

Then just multiply this by 4g: .367x4g=1.47g of O2

(36.7% = .367)

Another way to do this is to convert the grams of MnO2 into moles of MnO2, which would give you:

4g / 87 g/mole = .046 moles of MnO2

Then you would convert this into moles of O2, using the molar ratio between O2 and Mn, which is 2 to 1, like this:

.046 moles x 2/1 = .092 moles of O2

Then convert this to grams of O2 by multiplying by the molecular weight of Oxygen:

.092 moles x 16 g/mole= 1.47grams of O2

And there are your answers done 2 ways!

1 mol MnO2 contains 1 mol Mn and 2 mol O.

1 mol Mn = 54.95 g

2 mol O = 2 x 16.0 = 32.00 g

1 mol MnO2 has a mass of 54.95 + 32.0 = 86.95 g

4.00 g MnO2 x (32.00 g O / 86.95 g MnO2) = 1.47 g O