Question

How many grams of potassium permanganate contain 2.40 x 10^24 oxygen atoms? Include units and name of atom/molecule.

Answer 1

157.64 grams of potassium permanganate contain 2.40x10^24 oxygen atoms.

1st) We need to know how many moles of oxygen are equivalent to 2.40x10^24 oxygen atoms. Here, we need to use the Avogadro's number (6.022x10^23) and with a mathematical Rule of Three we can calculate the moles of oxygen:

`\begin{gathered} 6.022\cdot10^(23)atoms-1mol \\ 2.40\cdot10^(24)atoms-x=(2.40\cdot10^(24)atoms\cdot1mol)/(6.022\cdot10^(23)atoms) \\ \\ x=3.99\text{ moles} \end{gathered}`

Now we now that there are 3.99 moles of oxygen.

2nd) The formula of potassium permanganate is KMnO4, so there are 4 moles of oxygen in the molecule, and the molar mass of potassium permanganate is 158.034 g/mol.

With a mathematical Rule of Three we can calculate how much potassium permanganate it is necessary for 3.99 moles of oxygen:

`\begin{gathered} 4\text{mol}-158.034g \\ 3.99\text{mol}-x=\frac{3.99\text{mol}\cdot158.034g}{4\text{mol}} \\ \\ x=157.64g \end{gathered}`

If 4 moles of oxygen are contained in 158.034g of potassium permanganate, then the 3.99 moles of oxygen will be contained in 157.64g of potassium permanganate.