Question

If 0.890g of potassium and 0.590g of chromium react with 0.727g of oxygen, what is the empirical formula of the compound

Answer 1

K₂CrO₅

Step-by-step explanation

Given:

Mass of potassium = 0.890 g

Mass of chromium = 0.590 g

Mass of oxygen = 0.727 g

What to find:

The empirical formula of the compound.

Step-by-step solution:

Step 1: Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

The mole formula is given by:

`Moles=\frac{Mass}{Molar\text{ }mass}`

Molar mass of K = 39.098 g/mol

Molar mass of Cr = 51.996 g/mol

Molar mass of O = 15.999 g/mol

So,

`\begin{gathered} Moles\text{ }of\text{ }K=\frac{0.890\text{ }g}{39.098\text{ }g\text{/}mol}=0.02\text{ }mol\text{ }K \\ \\ Moles\text{ }of\text{ }Cr=\frac{0.590\text{ }g}{51.996\text{ }g\text{/}mol}=0.01\text{ }mol\text{ }Cr \\ \\ Moles\text{ }of\text{ }O=\frac{0.727\text{ }g}{15.999\text{ }g\text{/}mol}=0.05\text{ }mol\text{ }O \end{gathered}`

Step 2: Divide each number of moles by the lowest one, in this case

0.01.

`\begin{gathered} \frac{0.02\text{ }mol\text{ }K}{0.01}=2\text{ }mol\text{ }K \\ \\ \frac{0.01\text{ }mol\text{ }Cr}{0.01}=1\text{ }mol\text{ }Cr \\ \\ \frac{0.05\text{ }mol\text{ }O}{0.01}=5\text{ }mol\text{ }O \end{gathered}`

The final step is to use the mole ratio in step 2 to write the empirical formula of the compound.

The empirical formula of the compound is:

`K_2CrO_5`

.