Whether copper can combine with oxygen in equal proportion?
The Law of Reciprocal Proportions states: "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."

Wikipedia has a different phrasing: "If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A."

Note this is a statement about the weights of substances, not about the molar fractions or volumes.

For example, 

Copper sulfide ($\ce{Cu2S}$) contains 66% copper by moles. It does not, however, contain 66% copper by weight. Instead it contains $(2\cdot63.55)/( 2\cdot63.55+32.06) = 79.9\%$ copper by weight, or 20.1% sulfur by weight.

Likewise, while sulfur trioxide ($\ce{SO3}$) contains 25% sulfur on a molar basis, it contains $(32.06)/( 32.06+3\cdot16.00) = 40.0\%$ sulfur on a mass basis, or 60.0% oxygen by weight.

So when we apply the law of reciprocal proportions to copper oxide, we need to make use of the 20.1% and 40.0% numbers for sulfur. For example, if we had two masses of $\ce{Cu2S}$ and $\ce{SO3}$ each containing 1 g of sulfur, then we would have $1\rm{g}\cdot(79.9/20.1) = 3.98$ grams of copper and $1\rm{g}\cdot(60.0/40.0) = 1.50$ grams of oxygen. The ratio here is $3.98/1.50 = 2.65$.

Now, when we examine copper (II) oxide ($\ce{CuO2}$), for example we see that the ratio of masses for copper and oxygen are $63.55/(2\cdot16.00) = 1.99$ ... and 2.65 is very close to a simple multiple of 1.99 (specifically, 4/3 - the difference is due to rounding.)



The fact that this is always true is trivial, once you know atomic theory. So while the Law of Reciprocal Proportions is important in a historic context - as it helped lead to the development of modern atomic theory - basically no one uses it as such these days.

Say we have the compounds $A_hB_j$, $A_kC_l$ and $B_mC_n$. 

In $A_hB_j$ the masses occur in the ratio $(j\cdot M_b) / (h\cdot M_a)$. In $A_kC_l$ the masses are in the ratio $(l\cdot M_c)/(k\cdot M_a)$, so for a fixed mass (say $M_a$g) of A, the ratio of the masses is $(k\cdot j\cdot M_b)/(h\cdot l\cdot M_c)$

Now for the compound $B_mC_n$, the masses occur in the ratio $(m\cdot M_b) / (n\cdot M_c)$. Taking the proportion of this with the previous proportion, we see that the masses cancel, leading to $(k\cdot j\cdot n)/(h\cdot l\cdot m)$. Now because molecules form with small, whole number coefficients in their molecular formulas, this final quantity will typically be a simple, small fraction. (a.k.a. a "simple multiple" of each other.)

So all the Law of Reciprocal Proportions is doing is saying that 1) all atoms of a given element have about the same average weight 2) molecules form from (typically) small, integer numbers of atoms. -- That's it.

This is why you don't see many people discussing it. It's not all that interesting to people who are accustom to thinking atomically.