What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?

Answer:

CrO₃.

Step-by-step explanation:

First, we will find the chromium percentage in the oxide by using the following process:

Oxygen (O) = 48%

Chromium (Cr) =?

The oxide consists solely of chromium and oxygen, and its chromium percentage is calculated as:

Cr = 100 – percentage of oxygen

Cr = 100 – 48

Cr = 52%

Next, we will determine the empirical formula for the oxide:

Chromium (Cr) = 52%

Oxygen (O) = 48%

Now, we divide by their molar mass:

Cr = 52/52 = 1

O = 48/16 = 3

Now, divide by the lowest value:

Cr = 1/1 = 1

O = 3/1 = 3

Thus, the empirical formula for the oxide is CrO₃.