# Whether copper can combine with oxygen in equal proportion?

Copper sulphide contains 66.6% copper and sulphur trioxide contains 40% sulphur. According to law of reciprocal proportions copper is 80% in copper oxide. The calculation: In the first, for 33.4% sulphur, there is 66.6% copper. In the second, there is 40% sulphur. Therefore, in the resultant there should be 80% copper. My question is why cannot we calculate the other way round:In the second, for 40% sulphur there is 60% oxygen; therefore, for 33.4% there should be 50% oxygen. Hence, the resultant combination should have equal proportion of copper and oxygen. Is there any such substance?

• Do you mean $\ce{CuO}$? This is just normal copper (II) oxide. – bon Jul 21 '16 at 9:20
• No. CuO is from the first calculation. I want to know the substance according to the second calculation and it ought to be CuO4. – Mathivanan Palraj Jul 21 '16 at 9:57

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.

• When did the mass of oxygen become 18.01?? – f'' Jul 21 '16 at 22:43
• @f'' Oops. You're right. I'll fix that. However, the fact it worked out in the end is just more evidence that the masses of the atoms doesn't really matter, as long as they're constant. – R.M. Jul 22 '16 at 14:56

You are not finishing the calculations properly and it is a coincidence that your first one got the right answer.

When you calculate the proportions in copper oxide, you have to find both the copper and the oxygen. You have correctly found that the masses of copper and sulfur are in a ratio of 80 to 40. From sulfur trioxide, you can find that the masses of sulfur and oxygen are in a ratio of 40 to 20 (one-third of 60 because there are three oxygens). Combining these, you know that the ratio of copper and oxygen's masses is 80 to 20. Therefore the percentage of copper in CuO is 80/(80+20), which is 80%. Your calculation only worked because the sum of the numbers (80+20) was 100.

Doing the calculation the other way, you need to remember to divide the oxygen by 3 because there are three of them in sulfur trioxide. You should find that the ratio of oxygen's mass to sulfur's mass is 16.7 to 33.4 and the ratio of sulfur's mass to iron's mass is 33.4 to 66.6. This makes the percentage by mass of oxygen in CuO equal to 16.7/(16.7+66.6), which is 20% as expected.

This is the law of proportion.

For chemicals it works much the same way as for other objects. consider this proportion: If 3 people have 6 eyes, then how many eyes 5 people have?

3 ->6
5 ->x

x=5*6/3 = 10 eyes. This is correct. You suggest to do something like

x=3*6/5 = 3.6 eyes. Which is incorrect.

Even your first calculation is not convincing. You don't account for valency difference in sulfur. You compare CuS (2 valent sulfur and 2 valent Cu) with SO3 (2 valent oxygen and 6 valent sulfur). Now you want to compare CuO (2 valent O and 2 valent Cu). Your step from CuS to SO3 is not justified because you don't account for the change of valency in sulfur. Please expand your calculation to make it easier to follow.

• You mentioned direct proportion and inverse proportion in mathematics. My logic is not about the above. In my first calculation I found volume of copper. In the second I tried to find the volume of oxygen. However, the two calculations produce different results. I don't think the law is violated. – Mathivanan Palraj Jul 21 '16 at 14:14
• @MathivananPalraj you can only use proportion if the oxidation of the atom is the same. You can go from MgS to MgCl2 to SCl2. Likewise you can go H2O, CO2, CH4. This is because in first example they all have valence 2 and in the second example H, O, and C have valences 1, 2, 4 respectively. What you are trying to do is to go from CO to CO2. In this case proportion law is not applicable. Does it make more sense now? – sixtytrees Jul 22 '16 at 22:32