# What is meant by 'as Cl2'

I feel like I'm over thinking this one, but think I need a confirmation (and quantified example) to put my confusion to bed. I'm more a mechanical than chemical background but trying my best.

I really get stuck on papers measuring concentrations 'as XXX' when the 'XXX' is only a component of the compound.

Example 1: Hypochlorous Acid (HOCl) has a weight of 52.457 g/mol, while chlorine weighs 35.45 g/mol. Thus, the chlorine weight fraction would be ~67.5%. As this would be measured as Cl2 would we then halve the above weight fraction as we only have a single molecule of chlorine.

Example 2: If we add 1mol (143g) of Calcium Hypochlorite to make 1-litre of solution.

We would create 2mol (105g) of Hypochlorous Acid, which includes 2mol of Cl (aka 1mol of Cl2) weighing 70.9g. As we are interested in chlorine, would we express this as a measurement of HOCl 70.9g/l "as Cl2"? Or do we consider the entire mass of hypochlorous acid?

In general, sentences like "The concentration of X as Y is z mg/L" should be interpreted to mean "If you extract all the X from 1 L of solution, then convert the X completely to Y (throwing away any unused atoms), you will have z mg."

In example 2, you have a 2 M solution of hypochlorous acid. We want to know the concentration of hypochlorous acid as chlorine. This means we take all the hypochlorous acid (2 mol), then convert it to chlorine, throwing away the unneeded atoms. We end up with 70.9 g of chlorine so the concentration of hypochlorous acid as chlorine is 70.9 g/L. The fact that chlorine is $$\ce{Cl2}$$ is not relevant here; there's no need to divide by an additional 2.

It's probably worth mentioning why people use these units. There are several basically equivalent bleaches: $$\ce{NaOCl}, \ce{HOCl}, \ce{Ca(OCl)2}$$, etc., which can sometimes interconvert based on pH and other factors. A solution with 74.5 g of $$\ce{NaOCl}$$ has the same bleaching power as one with 52.5 g of $$\ce{HOCl}$$, but the distinct numbers hide the equivalence. Indeed, the chemical tests used to measure the bleaching power probably cannot even tell what form the chlorine is in. So people invented a measuring system where both systems have the same numeric value: 35.5 g of bleach "as chlorine." This emphasizes that the two systems behave similarly and lets a test report one unambiguous result.

They could also have said "0.5 molar equivalents of $$\ce{Cl2}$$", and I'm sure most chemists would find that much more intuitive these days. But the chemists don't always get their way.

EDIT: I hate to edit an accepted answer, but after consulting some references on chlorine specifically, I'm pretty sure that this answer, while correct for the vast majority of "X as Y," might actually be wrong for chlorine. The problem is that when you dissolve chlorine in water, only half the chlorine in $$\ce{Cl2}$$ is active as bleach, with the other half getting converted to inactive chloride. It appears that the convention does not "count" the inactive half of the chlorine and therefore 74.5 g of $$\ce{NaOCl}$$ is considered equivalent to 71 g of $$\ce{Cl2}$$, not 35.5 g. (Here's an example from the Ohio EPA.)

So yeah, you probably shouldn't have listened to me. Sorry.

• In clarification to your edit, would this depend on the form of chlorine being dissolved? The Ca(OCl)2 reaction posted in OP seems to retain all chlorine molecules as active HOCl. Commented Nov 30, 2023 at 0:07
• Link from Ohio EPA gives me 404 error. Commented Nov 30, 2023 at 0:08

Let imagine there are 3 bank accounts, one with 10000 Euros, one with 10000 US dollars, one with 10000 UK pounds. Let assume - just illustratively - there is exchange ratio 5 dollars per 4 Euros and 2 dollars per 1 UK pound. If there is requirement to express the money level "as dollars",
the first one is worthy "as 12500 dollars",
the third one "as 20000 dollars".

Back to chemistry, "as Cl2" is not limited to chlorine being a part of the considered compound. there is just expected some chemical equivalence, defined by chemical reactions.

$$\pu{1 mol}$$ $$\ce{Cl2}$$ is equivalent to $$\pu{1 mol}$$ of $$\ce{HClO}$$ or $$\ce{ClO-}$$ due the reactions

$$\ce{Cl2(aq) + H2O <=> H+(aq) + Cl-(aq) + HClO(aq)}$$ $$\ce{Cl2(aq) + 2 OH-(aq) <=>> ClO-(aq) + Cl-(aq) + H2O}$$

So if
$$M_\ce{Cl2} \approx \pu{71 g mol-1}$$
$$M_\ce{HClO} \approx \pu{52.5 g mol-1}$$

then $$\pu{52.5 g}$$ $$\ce{HClO}$$ means "as $$\ce{Cl2}$$" $$\pu{71 g} \ce{Cl2}$$.

Mass-wise, it does not matter if we consider $$\ce{Cl}$$ or $$\ce{Cl2}$$). Amount-wise, $$\ce{HClO}$$ is equivalent to twice as much amount of $$\ce{Cl}$$.

If we consider reactions

$$\ce{2 Fe^2+(aq) + H2O2(aq) + 2 H+(aq) -> 2 Fe^3+(aq) + 2 H2O}$$ $$\ce{2 Fe^2+(aq) + Cl2(aq) -> 2 Fe^3+(aq) + 2 Cl-(aq)}$$

then $$\pu{1 mol}$$ of $$\ce{Cl2}$$ is equivalent $$\ce{1 mol}$$ of $$\ce{H2O2}$$, so $$\pu{34 g}$$ of $$\ce{H2O2}$$, if expressed "as $$\ce{Cl2}$$", means (approximately) $$\pu{71 g}$$ $$\ce{Cl2}$$.

The notation mentioned in the other thread ($$\ce{NH3-N}$$) means something little different: "nitrogen in ammoniacal form, expressed as elemental nitrogen.". The direct equivalent, if we e.g. analyzed solution containing chloride $$\ce{Cl-}$$, hypochlorite $$\ce{ClO-}$$ and chlorate $$\ce{ClO3-}$$, would be the analogical syntax $$\ce{Cl-Cl}$$, $$\ce{ClO-Cl}$$, $$\ce{ClO3-Cl}$$, or verbose equivalents "chlorine in chloride/hypochlorite/chlorate form.".

It is not used to express equivalence of different chemicals, but rather to express content of chemical element (or compound), present in various forms, in a unified way. Typically to make their sum easily.

Like in elemental analysis, the mass of total nitrogen in a waste water sample is the sum of ammoniacal+nitrite+nitrate+organic nitrogen.

$$m_\text{Total-N}=m_\ce{NH3-N}+m_\ce{NO2-N}+m_\ce{NO3-N}+m_\ce{Org-N}$$

If we consider various chloramins:

\begin{align} \ce{1 NH2Cl} \equiv \ce{1 HOCl} \equiv \ce{1 Cl2}\\ \ce{1 NHCl2} \equiv \ce{2 HOCl} \equiv \ce{2 Cl2}\\ \ce{1 NCl3} \equiv \ce{3 HOCl} \equiv \ce{3 Cl2} \end{align}

As \begin{align} \ce{NH_{3-n}Cl_n + n H2O &<=> NH3 + n HOCl}\\ \ce{HOCl + Cl- + H+ &<=> H2O + Cl2} \end{align}

If we consider $$\ce{NHCl2-Cl}$$ as "chlorine in form of dichloramin, $$\pu{1 mg L-1} \ce{NHCl2} \equiv \frac{2 \cdot M_\ce{Cl}}{M_\ce{NHCl2}} \pu{mg L-1} \ce{NHCl2-Cl}$$

But, if we consider $$\ce{NHCl2}$$ "as $$\ce{Cl2}$$, or as active chlorine", $$\pu{1 mg L-1} \ce{NHCl2} \equiv \frac{4 \cdot M_\ce{Cl}}{M_\ce{NHCl2}} \pu{mg L-1} \ce{Cl}$$

• The former counts the equivalent mass concentration for the chlorine fraction contained in the substance.
• The latter counts the mass concentration for the equivalent free/active chlorine.
• Ok, that stirs the pot again. If it is possible to measure 'as X' without any content of 'X' makes it interesting indeed. I presume we need to know the original reaction we are equating to. I have only dealt with reactions from hypochlorous acid to different chloramines, how should I determine these (Monochloramine, Dichloramine and Nitrogen Trichloride) as Cl2 equivalents? Commented Dec 1, 2023 at 1:09
• If you were expressing the amount of money in your wallet "as US dollars", would it come weird if you had no US dollars there? Commented Dec 1, 2023 at 10:15