Question

a chemist Burns 100.0g chromium (Cr) in Excess oxygen (O). There is only one product, and it has a mass of 192.31g. what is the empirical formula of the product? Burning is a reaction with oxygen. Since there is only one product, it must have both Cr and O. We started with 100.0g CR. Anything over 100.0g must be O that was added. What is the mass of O?

Answer 1

CrO3.

Step-by-step explanation:

The mass of O in the product can be calculated as follows:

Mass of product - mass of Cr = mass of O

192.31 g - 100.0 g = 92.31 g

Next, we'll calculate the number of moles of Cr and O in the product:

Moles of Cr = mass of Cr / molar mass of Cr = 100.0 g / 51.996 g/mol = 1.922 mol

Moles of O = mass of O / molar mass of O = 92.31 g / 15.999 g/mol = 5.77 mol

Finally, we'll divide each mole by the smallest number of moles, which is 1.922 mol:

Moles of Cr / 1.922 = 1.000 mol

Moles of O / 1.922 = 3.000 mol

So, the empirical formula of the product is CrO3.