In need of $\ce{K2SiO3}$ for my cucumber plants, I purchased a bag of potassium silicate. The molecular formula is $\ce{K2O-nSiO2-xH2O}$. Normally, my bags of fertilizer are easy to figure out. For example, $\ce{Ca(NO3)2}$, or calcium nitrate, has a molecular weight of 164.0878 g/mol. The purity is also 95%. Knowing the purity and the molecular weight, I normally can find out how many ppm of Ca or N I want to use in my hydroponic nutrient tank. With $\ce{K2O-nSiO2-xH2O}$, I don't understand $n$ and $x$. I've googled this only to be stumped. I just don't get it.

The bag of fertilizer ($\ce{K2O-nSiO2-xH2O}$) has the following information listed on it:

  • Wt% $\ce{K2O}$ 26.5~29.5,
  • Wt% $\ce{SiO2}$ 55~60,
  • Weight ratio ($\ce{SiO2/K2O}$) 2.0~2.2,
  • Mole ratio ($\ce{SiO2/K2O}$) 3.2~3.4,
  • Bulk Density g/L 400~600,
  • solid % 82~89,
  • particle size 270+/-50 and
  • purity 77%

I need to know the molecular weight, so I can figure out how many ppm of $\ce{Si}$ and $\ce{K}$ to add to the solution.


1 Answer 1


You're dealing with a mixture of similar hydrated salts instead of a single salt (as in the example of $\ce{Ca(NO3)2}$). The notation $\ce{K2O-nSiO2-xH2O}$ means that there are different crystals with varying proportions of silicon dioxide and water (hydration). i.e. $n$ could be $1$, or $2$, $3$, etc. and $x$ likewise, for any given molecular "speck". A hydrated salt can be de-hydrated with heat (thus changing relative weights of elements), so be careful where you keep the fertilizer, because it will change the math.

This means, to measure a reasonable concentration, you need to consider the reported bulk average quantity of each component. Based on the percent by weight you reported, you can approximate the material as being (by mass -- that is wt%)

  • $28\%~\ce{K2O}$
  • $57.5\%~\ce{SiO2}$
  • (Thus) $14.5\%~\text{water}$

Say you want to know the $\pu{ppm}$ of $\ce{K+}$ for a solution of $\pu{1 g}$ of this fertilizer in a gallon ($\pu{3.78 L}$) of $\ce{H2O}$. Using the good approximation $\pu{1 mg/L} = \pu{1 ppm}$, We have

$$\begin{multline} \pu{1 g} \times \frac{\pu{0.28 g}~\ce{K2O}}{\pu{1 g fertilizer}} \times \frac{\pu{1 mol}~\ce{K2O}}{\pu{94.2 g}~\ce{K2O}}\\ \times \frac{\pu{2 mol}~\ce{K+}}{\pu{1 mol}~\ce{K2O}} \times \frac{\pu{39.0983 g}~\ce{K+}}{\pu{1 mol}~\ce{K+}} \times \frac{\pu{1000 mg}}{\pu{1 g}} \times \frac{\pu{1 gal}}{\pu{3.78 L}}\\ = \pu{61.5 ppm}~\ce{K+} \end{multline}$$

Here I guessed a concentration that may be relevant for you, but you can see the way to move forward. I would note, though, that since purity is reported pretty low ($77\%$), you should multiply the above result by $0.77$. Hard to say what the impurities are though, if they are mostly other potassium salts, the concentration could be higher. At any rate it's safe to say the above example is $\approx \pu{45-65 ppm}~\ce{K+}$.

You can also think of the compound like this, based on component masses:

$$\text{Molar mass} = 0.28 \times M(\ce{K2O}) + 0.575 \times M(\ce{SiO2}) + 0.145 \times M(\ce{H2O}) = \pu{63.532 g/mol},$$ and the average molar formula is $\ce{K2O - 3.2SiO2 - 2.7H2O}$.

Finally, I would recall that dissolution of $\ce{K2O}$ in general produces a basic solution, so if cucumbers are fond of acid (I don't know), you may consider adding a little acid ($\ce{HNO3}$, perhaps) to the solution to avoid perturbing the greens. But, you're the farmer!


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