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Recently I've been reviewing concepts belonging to the history of chemistry. But I came stuck at trying to understand a passage which I read from wikipedia entry (and which it seems has been mentioned in different sources). This is related with the law of reciprocal proportions.

It states as this:

It took 615 parts by weight of magnesia ($\ce{MgO}$), for example, to neutralize 1000 parts by weight of sulfuric acid.

There's an existing answer which explains a similar concept (law of multiple proportions) but not specifically to Richter's.

From what I understood in the law of reciprocal proportions there is a set of three elements. Two of them react with a fixed amount of a third element. The ratio of these two elements is the same when they combine between.

For example:

When $63.5~\rm{g}$ copper combines with sulphur produces copper sulphide

$\begin{array}{lllll} \ce{Cu}&+&\ce{S}&\rightarrow&\ce{CuS}\\ 63.5~\rm{g}&&32~\rm{g}&&95.5~\rm{g}\\ \end{array}$

Conversely when those $63.5~\rm{g}$ of copper combines with oxygen produces cupric oxide.

$\begin{array}{lllll} 2\ce{Cu}&+&\ce{O_2}&\rightarrow&2\ce{CuO}\\ 63.5~\rm{g}&&16~\rm{g}&&79.5~\rm{g}\\ \end{array}$

Therefore the proportion between the mass of sulphur and oxygen masses is:

$\frac{m_{\ce{S}}}{m_{\ce{O_{2}}}}=\frac{32}{16}=2$

When sulphur and oxygen combine together they make sulphur dioxide.

$\begin{array}{lllll} \ce{S}&+&\ce{O_2}&\rightarrow&\ce{SO_{2}}\\ 32~\rm{g}&&32~\rm{g}&&64~\rm{g}\\ \end{array}$

Then the proportion between these two becomes:

$\frac{m_{\ce S}}{m_{\ce{O_{2}}}}=\frac{32}{32}=1$

The latter is in proportion which is half of the first ratio. Hence follows the Richter law.

But how can I use this information to understand what it was mentioned in the paragraph from above? How is exactly related with equivalent weights?

The second part of the question arises from the fact, that would it be okay to state this?

In an hypothetical reaction between $A$, $B$ producing $C$ and $D$.

$\begin{array}{ccccccc} aA&+&bB&\rightarrow&cC&+&dD\\ \textrm{1 eq gram of A}&&\textrm{1 eq gram of B}&&\textrm{1 eq gram of C}&&\textrm{1 eq gram of D}\\ \end{array}$

Becoming in

$\textrm{1 eq gram of A}=\textrm{1 eq gram of B}=\textrm{1 eq gram of C}=\textrm{1 eq gram of D}$

Would it be accurate to put it in that way?

or?

$\textrm{number of eq gram of A}=\textrm{number of eq gram of B}=\textrm{number of eq gram of C}=\textrm{number eq gram of D}$

where:

$\textrm{1 eq gram A} =\frac{\textrm{grams of A compound equal to the equivalent weight of A}}{\frac{\textrm{formula weight of A}}{\textrm{number of electrons transfered}}}$

which is different when considering moles:

$\begin{array}{ccccccc} aA&+&bB&\rightarrow&cC&+&dD\\ \textrm{a moles of A}&&\textrm{b moles of B}&&\textrm{c moles of C}&&\textrm{d moles of D}\\ \end{array}$

where:

$\textrm{a moles of A}=\textrm{b moles of B}=\textrm{c moles of C}=\textrm{d moles of D}$

So, is what I wrote correct? How can I relate Richter's law of reciprocal proportions to the equivalent weights? How is it related with the paragraph from above? Perhaps does it meant to say that there is a definite amount of reagent which can react with a certain amount of sulphuric acid? Then why isn't it related more with the concept of limiting reagent rather than the equivalent weight? Can someone guide me on that?

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I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time.

It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid

This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the concept of equivalents. All this observation shows is that it always took the same amount of X to react with Y. Hence the ratio of X/Y remained constant.

If we take 1000 parts of sulfuric acid as a standard, then 615 parts by weight of MgO will neutralize it. It will take 672 parts of ammonia to neutralize the same amount of acid.

Note carefully this concept does not even need the knowledge of atomic weights.

Addendum Also note that Richter's values are incorrect as shown in the comments i.e. if we use moles. Most likely the reason was purity of his magnesia and sulfuric acid. The inaccuracy of his values was realized in the late 1860s but we should respect his idea of constant ratio of reactants. I don't know enough German to check his entire three volumes of books nor have access to them. However a table from 1860s shows the comparison of corrected values (which are still slightly off):

An Introduction to Chemical Philosophy, according to the Modern Theories, by Adolphe Wurtz, 1865.

Table

"It will be noticed that for many bodies Fischer's numbers are widely different from the theoretical figures, and consequently the analyses of Richter from which they are calculated were not nearly so accurate as those of Wenzel. The inaccuracy of these analyses, and the obscurity of a perplexed explanation, have not been much noticed in giving Richter credit and some authority to his works."

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    $\begingroup$ I totally agree that moles are a more appropriate measure. I can't figure out the reaction however. I'm expecting $$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$ However let's take the parts as grams. 615 grams of $\ce{MgO}$ is 15.3 moles, but 1000 grams of sulfuric acid is 10.2 moles. $\endgroup$ – MaxW Mar 15 at 4:58
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    $\begingroup$ Thanks for the additional information to the answer. I'd say that definitely clears up the discrepancy. $\endgroup$ – MaxW Mar 15 at 6:30
  • $\begingroup$ @M.Farooq Interesting, but this doesn't clear up the last part of my question. Is it okay to state that the number of equivalents for each reagent and product to be the same? as indicated in my first example given with $A$, $B$, $C$ and $D$?. So overall the statement given by Richter was merely obtained from an experiment not calculated or anything related with that?. What was him referring with parts of?. Was it grams?, liters?, percentage by weight?. or was just any of those?. How did you obtained that it will take 672 parts of ammonia to neutralize the same amount of acid.? $\endgroup$ – Chris Steinbeck Bell Mar 15 at 21:21
  • $\begingroup$ @M.Farooq I was just checking with the table that you provided. This value was also obtained from an experiment right?. An all in all he compared it with reaction with sulphuric acid?. Did he used other acids to make a table?. Can you please attend the latter questions from my earlier comment?. $\endgroup$ – Chris Steinbeck Bell Mar 15 at 21:22
  • $\begingroup$ @ChrisSteinbeckBell, I used the value from the table. Yes those numbers must have come from the experiments of acid base neutralization. I am not a native English speaker either, however, " 1000 Parts of Sulfuric Acid by Weight" is an old English terminology. By current standards, it would simply mean 1000 [weight units] of sulfuric acid react with X [weight units] of a base B. "Parts of" is just a filler of old language style. $\endgroup$ – M. Farooq Mar 15 at 21:47

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