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(Note that this question has been fully edited for clarity, SE help advised not to delete the question and post another).

Background:

I have a computer model in which I feed it input data from various feedstocks usually containing sodium. My calculations input NaOH data (because it's an industry standard and easy to use for the user) but calculate Na2O for output quantities for the sake of consistency as everything else is in oxide form (e.g. SiO2, Al2O3, H2O etc.). Note that the full model incorporates the following metrics from each inputted feedstocks:

  • Feedstock Elemental Mol Quantities: (e.g. mols Si, mols Al and mols Na (etc.) from every feedstock, note that mols H instead of mols H2O is used for consistency in using elemental mols instead of changing to an oxide just for water).
  • Feedstock Oxide Weight Quantities: (e.g. wt% SiO2, wt% Al2O3, wt% Na2O (etc.) from every feedstock).
  • Feedstocks Solid wt%s: (i.e. $Solids_{i,wt\%s} = Feedstock_{i,Total,wt\%} - H_2O_{i,wt\%}$; the summed wt%s of all non-H20wt% quantities in a given feedstock)
  • Feedstock Liquids wt%: (i.e. $Liquids_{i,wt\%s} = Feedstock_{i,Total,wt\%} - Solids_{i,wt\%s} = H_2O_{i,wt\%}$

One such feedstock which is commonly used in the model are the standard reagent grade 97 wt% NaOH pellets. Knowing the reaction between NaOH and Na2O (i.e. 2NaOH -> Na2O + H2O), if I wanted to calculate the composition of a 1 g NaOH pellets I get the following (note that I did try to put everything in Chem SE table format but it was too difficult to do, format and the help page didn't provide too much clarity from my POV):

1. First, the pellet's composition with respect to NaOHwt%:

Calculations are as follows:

  • NaOHwt% = 97 wt% NaOH as per the specification sheet mentioned in the link above.

  • Note that it has been assumed that $H_2O_{wt} = 100_{wt\%} - Solids_{wt\%} = 100_{wt\%} - NaOH_{wt\%} = 3_{wt\%} H_2O$

  • $mols Na = \frac{97 wt\% NaOH}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g NaOH}{100 wt\% NaOH} \times \frac{1 mol NaOH}{40 g NaOH} \times \frac{1 mol Na}{1 mol NaOH} = 0.0243 \frac{mols Na}{1 g pellets}$

  • $mols H = \frac{3 wt\% H2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g H_2O}{100 wt\% H_2O} \times \frac{1 mol H_2O}{18.02 g H_2O} \times \frac{2 mols H}{1 mol H_2O} = 0.0033 \frac{mols H}{g pellets}$

  • $Solids_{i,wt\%s} = 100_{wt\%} - 3_{wt\%,H_2O,i} = 97 \%$

  • $Liquids_{i,wt\%s} = 3_{wt\%,H_2O,i} = 3 \%$

![enter image description here

2. Secondly, the pellet's composition with respect to Na2Owt%:

Calculations are as follows:

  • Note that Na2Owt = $ \frac{97 wt\% NaOH}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g NaOH}{100 wt\% NaOH} \times \frac{61.98 g Na2O / mol Na2O}{40 g NaOH /mol NaOH} \times \frac{1mol Na/ mols NaOH}{2mols Na/mols Na2O} = \frac{0.752gNa2O}{1 g Pellets} = 0.752 \frac{g Na_2O}{1 g pellets} \times 100\% = 75.2 wt\% Na_2O$

  • The resultant H2Owt% as per the above 3wt%, I presume instead needs to be of $H_2O_{wt} = 100_{wt\%} - Na_2O_{wt\%} = 24.8_{wt\%} H_2O$ which seems ludicrous...

  • $mols Na = \frac{75.2 wt\% Na_2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g Na_2O}{100 wt\% Na_2O} \times \frac{1 mol Na_2O}{62 g Na_2O} \times \frac{2 mols Na}{1 mol Na_2O} = 0.0243 \frac{mols Na}{1 g pellets}$

  • $mols H = \frac{24.8 wt\% H2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g H_2O}{100 wt\% H_2O} \times \frac{1 mol H_2O}{18.02 g H_2O} \times \frac{2 mols H}{1 mol H_2O} = 0.028 \frac{mols H}{g pellets}$

  • $Solids_{i,wt\%s} = 100_{wt\%} - 24.8_{wt\%,H_2O,i} = 75.2 \%$

  • $Liquids_{i,wt\%s} = 24.8_{wt\%,H_2O,i} = 24.8 \%$

![enter image description here

In can be seen that, although the mols Na are the same, neither the mols H nor the solids or liquids parameter match up. Which doesn't seem to make sense to me?

3. Finally, the pellet's nonsensical composition with respect to Na2O:

  • The only solution I've managed to balance with the initial NaOH composition with respect to the Na2Owt doesn't make sense as I've had to add a nonsensical "Other solids?" compound which accounts for the missing solids to ensure that the mols of Na, H and solids and liquids quantities remain consistent.

enter image description here

My questions from the above are as following:

  1. What is the missing component (maybe the "other solids" quantity) which resolves the inconsistency between the NaOH pellet's composition with respect to the NaOHwt and Na2Owt?

  2. Is the H2O wt% assumption a good one to make due to the pellet's hygroscopic property?

  3. Is it better to only assume the 2 NaOH <> Na2O + H2O reaction for solids and therefore only use it in aqueous solutions with NaOH or Na2O?

  4. Should NaOH pellets only be quantified by their NaOH composition and not Na2O for better representation and accuracy?

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  • $\begingroup$ Did you think about what is the last 3% of NaOH? I think it is water because NaOH is hydroscopic. $\endgroup$ Feb 6, 2023 at 16:45
  • $\begingroup$ I had thought of that but assumed instead it was just trace elements of something else that exist from manufacturing. $\endgroup$
    – Hendrix13
    Feb 7, 2023 at 0:48
  • $\begingroup$ There is near always some Na2CO3(s), as NaOH(s) reacts with aerial CO2. You obviously cannot consider Na2O taking 97% of NaOH. // As 2 NaOH <> Na2O + H2O is formal conversion, there is fixed mass ratio $\frac{2*M(\ce{NaOH})}{M(\ce{Na2O})}$ between them, and their molar ratio is 2 : 1. $\endgroup$
    – Poutnik
    Feb 7, 2023 at 9:17
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    $\begingroup$ The first table is wrong, regarding molar amounts of Na versus H. If it were pure NaOH, the molar ratio would be 1 : 1. // Remember the 2 NaOH <> Na2O + H2O is purely formal reaction, describing NaOH/Na2O equivalence. So about 75% Na2O(s) + 25 H2O(l) in solid NaOH is purely formal. // Similarly, P in fertilizers is expressed conventionally as P2O5 in spite of there is no P2O5 either in solid ones either in liquid ones. $\endgroup$
    – Poutnik
    Feb 8, 2023 at 12:38
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    $\begingroup$ FIRST THERE IS NO REACTION $\ce{2NaOH -> Na2O + H2O}$ NO ! This transformation from $\ce{NaOH}$ to $\ce{Na2O}$ never happens. As Poutnik says, it is purely formal. It dates from the $19$th century where chemists thought that all oxygenated mineral compounds were made of "jointed" oxides. For example $\ce{Na2CO3}$ was thought to be $\ce{Na2O + CO2}$. Then $\ce{H2SO4}$ was thought to be made of $\ce{H2O + SO3}$. And of course $\ce{NaOH}$ was thought to be made of $\ce{Na2O + H2O}$. But there are no oxides in $\ce{Na2CO3, Na2SO4, H2SO4}$ and of course in $\ce{NaOH}$. $\endgroup$
    – Maurice
    Feb 8, 2023 at 18:41

2 Answers 2

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There are two mistakes in this post.

  1. The formula used for calculate the amount of $\ce{Na2O}$ has a wrong unit. It does not give a weight as it should. It gives a percentage. This mistake is easy to prove by simply reasoning with units. In the first numerical example, a weight ($16.7$ g) is successively multiplied by a ratio of grams, and then by a ratio of moles. The final result must be expressed in grams, and not in percent. This observation is valid for both the first silicate solution and for the final pills.
  2. The second mistake will be shown with the second calculation. Here, one starts from $97$ g pure $\ce{NaOH}$, containing $97/40 = 2.425$ mol $\ce{NaOH}$. These $2.425$ mol NaOH are not formed, but may be considered as formed by hydratation of $1.2125$ mol $\ce{Na2O}$, which weighs $1.2125 · 62$ g = $75.175$ g The difference in weight $97 - 75.175$ g = $21.825$ g is water, but it is not the adsorbed water fixed at the outside of the $\ce{NaOH}$ pills. It is the amount of water that is theoretically necessary to add to $75.175$ g $\ce{Na2O}$ to get the final mass ($97$ g) $\ce{NaOH}$ after hydratation reaction. This theoretical water is not present in the pills. The amount of adsorbed water is only $100$ g - $97$ g = $3$ g.

The same mistakes have also been done in the first numerical example. But it is not so easily discovered, as water is used both as solvant and as theoretical water necessary to transform $\ce{Na2O}$ into $\ce{NaOH}$. These two sorts of water are not really separated in the calculations of the first numerical example.

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  • $\begingroup$ 1: I had simplified the formula but have added the exact steps in the newest edit just then, the resulting number is unchanged. 2: I'm still a bit confused here. From a mass balance POV, when I calculate the mols of Na for each scenario (i.e. using Na2O and NaOH) I get 0.0243 mols Na for both scenarios, however, when I do the same for water I get 0.0138 mols H2O and 0.0017 mols H2O respectively supposedly with the same system? How does that make sense with a different total number of mols? Should the reaction between Na2O and NaOH be ignored for the solid NaOH pellets? $\endgroup$
    – Hendrix13
    Feb 7, 2023 at 1:49
  • $\begingroup$ I have just made a complete edit to the question so it's (hopefully) more easier and direct :) $\endgroup$
    – Hendrix13
    Feb 8, 2023 at 11:44
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    $\begingroup$ NaOH does not react like you think and does not loose water and produce Na2O by heating. $\endgroup$
    – Maurice
    Feb 8, 2023 at 17:08
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Let me now comment your blue tables. They are particularly ridiculous.

The first one is called "NaOH pellets (NaOH)". The $3$ first lines give the composition of the pellets in weight percent. OK. Then there is a subtitle called "Elements (moles). This information has no meaning. Or it would have a meaning if some more information was given, saying that the given values in the two next lines are related to $1$ g of substance. One gram ! Let me study these values to convince you about this information.

$1$ gram of your pellets contains $0.97$ g $\ce{NaOH}$ and also $0.97/40 = 0.024$ mole $\ce{NaOH}$. There is also $0.03$ g $\ce{H2O}$ and $0.03/18 = 0.00166$ mol $\ce{H2O}$, which correspond is to $0.00333$ mole H. These values are the same as in your table. But I don't understand what is the use of calculating the amount of $\ce{H}$ atoms in the $3$ % of adsorbed water. Why ?

The second table is called "NaOH Pellets (Na2O)". The same observation can be made about the percent and mass units. It should be specified that the given amounts in mole are only valid for a $1$ gram sample.

Now let's comment about the moles of $\ce{Na2O}$. I need now to recopy the comment I wrote half an hour ago about the correspondance $\ce{NaOH - Na2O}$. Here is my comment :

In the $19$th century chemists thought that all oxygenated mineral compounds were made of "jointed" oxides. The nature of this joint was a mystery. For example $\ce{Na2CO3}$ was thought to be made of $\ce{Na2O + CO2}$. Then $\ce{H2SO4}$ was thought to be made of $\ce{H2O + SO3}$. And of course $\ce{NaOH}$ was thought to be made of $\ce{Na2O + H2O}$. But there are no oxides in $\ce{Na2CO3, Na2SO4, H2SO4}$ and of course in $\ce{NaOH}$. This old theory is not stupid, but it is no more useful today in chemistry; and it is still in use in geology, because it respects the oxidation numbers and the valences.

Let's go back to the second table "NaOH Pellets (Na2O)". It mentions that the $97$ % $\ce{NaOH}$ "corresponds" to $75.2$% $\ce{Na2O}$. This is right, and you have done this calculation yourself. But this is of no interest any more, except for geologists. The rest ($24.8$ %) is the sum of the really adsorbed water at the surface of the pellet + the theoretical amount of water necessary to transform $\ce{Na2O}$ in $\ce{NaOH}$. Why doing this sum which is of no interest ?

Later on, in the middle of the blue table, under "Elements", there is $0.024$ mol $\ce{Na}$ and $0.028$ mol $\ce{H}$. It should be stated that these amounts are present in $1$ gram pellets. These numerical values are only true if the $0.028$ mole $\ce{H}$ atoms are coming from the theoretical water necessary to transform $\ce{Na2O}$ into $\ce{NaOH}$. Has it any meaning ?

The last lines of this second blue table are specially ridiculous. They state that the pellets are made of $24.8$% liquid, as if constitution water was liquid in $\ce{NaOH}$ as analyzed by a 19th century technique.

I begin to get tired tomato disagreeable comments about your tables.

The worst is in the last blue table, where the author believes that some unknown solid is necessary to explain the composition of the pellets. The author has simply given the weight percent ($75.2$%) of $\ce{Na2O}$ necessary to give the real $\ce{NaOH}$ believing that it was the weight percent of $\ce{NaOH}$. The "missing solid" is not a solid. It is simply the theoretical amount of water necessary to transform the non-existant $\ce{Na2O}$ into $\ce{NaOH}$. I will not go any further today. How silly !

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  • $\begingroup$ I had thought I was clear about the 'mols sections' due to their names but apparently not, my mistake, I have provided all the extended equations in the newest edit to be certain along with an explanation/context as to why mols H and not mols H2O was used in the background (it was because of other things in the model). I was also previously unaware of the whole NaOH and Na2O mix-up from 19th century geologists and am surprised that it's still a thing today as I regularly (and now confusingly) see some research literature with Na coming from Na2O instead of NaOH in some solutions (etc.). $\endgroup$
    – Hendrix13
    Feb 9, 2023 at 3:16
  • $\begingroup$ @Hendrix13 Consider all NaOH/Na2O thing as equivalent masses or molar amounts, without any reaction ever happening. Like, if I have 1 kg or 1 mol of NaOH, how much it would have been, if it had happened to be Na2O. // Similarly, table salt NaCl food income can be evaluated as Na, without thinking you swallow metallic explosive sodium. $\endgroup$
    – Poutnik
    Feb 9, 2023 at 6:55
  • $\begingroup$ Yes good explanation @Poutnik, and I see what you mean about the NaOH/Na2O. I just didn't understand fully how the relationship worked with the H2O. I knew something was wrong and didn't make sense which is why I specifically referred to the "nonsensical composition" in my final blue table. $\endgroup$
    – Hendrix13
    Feb 14, 2023 at 14:41

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