What is the calculated empirical formula of the copper oxide if a 0.626 g sample of the copper oxide was reduced to 0.500 g of copper metal by heating in a stream of hydrogen gas?

CuO

Explanation:

Empirical formula is the simplest ratio of atoms presents in a compound.

The reduction of copper oxide to copper metal in hydrogen is:

CuₐOₓ + x H₂ → a Cu + x H₂O

Moles of copper metal after reaction (Cu = 63.546g/mol) are:

0.500g × (1mol / 63.546g) = 7.868x10⁻³mol Cu

Now, if mass of copper is 0.500g, mass of oxygen in the oxide is:

0.626g - 0.500g = 0.126g O. In moles:

0.126g × (1mol / 16g) = 7.875x10⁻³mol O

Thus, ratio between moles of Cu and moles of O is:

7.868x10⁻³mol Cu / 7.875x10⁻³mol O = 0.999 ≅ 1

That means empirical formula of copper oxide is Cu₁O₁ = CuO

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!