Most transition metals can exist in more than one oxidation state, and as such, can form oxides with different formulas. If a basic solution containing 1.89 g of dissolved copper(II)chloride dihydrate (molar mass = 170.48 g/mol) is reacted with excess hydrazine, N2H4, a red form of copper oxide is created. If that solid is isolated, it has a mass of 0.79 g after drying.  What is the formula of this copper oxide?
Did the oxidation state of copper change during this reaction?

- The formula of the formed copper oxide is Cu₂O

- The oxidation state of copper does change from (II) to (I), because copper is reduced.

Explanation:

Hydrazine (N₂H₄) is a well known reducing agent, so it reduces copper(II)chloride (CuCl₂). The oxidation state of Cu in CuCl₂ is +2:

CuCl₂ → Cu²⁺ + 2 Cl⁻

Thus, when Cu²⁺ is reduced to an oxide, it is formed copper(I) oxide (Cu₂O), which is a red solid. According to this, we can conclude:

- The formula of the formed copper oxide is Cu₂O.

- The oxidation state of copper does change from (II) to (I), because copper is reduced.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally