I placed 1 gram of NaBH4 in a balloon and placed it over a glass round bottle filled with water and acetic acid. While secured, I emptied the balloon into the bottle and made sure to wash the inside of the balloon. The reaction taking place is $\ce{NaBH4 + 2(H2O) -> NaBO2 + 4(H2)}$.

The yield in this case should be 213 grams of $\ce{H2}$. According to Wikipedia, "hydrogen" has a density of 0.08988 g/L. That should leave me with 2.369 L of $\ce{H2}$. But my little experiment showed a balloon with the about 13 cm in diameter, which a volume of sphere that wide would be 1.15 liters of H2. This would mean a yield of .103 grams or 48.5%. But that just doesn't make sense. My NaBH4 has not degraded by half.

If you were to double the given density of $\ce{H2}$ to 0.17976 g/L, the results are much more in line with what I'd expect. The yield of $\ce{H2}$ would be .206 grams or 97%, which would make sense since not everything is ideal and my $\ce{NaBH4}$ is a bit old. Obviously there will be minor errors in measurement, and its not a perfect sphere and I didn't control for temperature, but that cannot account for a 50% reduction in yield. Maybe the balloon is compressing the gas? But I don't think so.

So the only other thing I can think of is that the presented density of 0.08988 g/L for hydrogen is technically only for monoatomic Hydrogen, or $\ce{H}$, which doesn't make sense and no one ever uses. The density of $\ce{H2}$ at STP should be 0.17976 g/L then to match my observations. There must be something I'm missing here entirely that I'm not understanding.

  • 1
    $\begingroup$ It is always good to summarize and/or complement the verbal description by algebra. Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. You may find useful formatting mathematical/chemical expressions/formulas. $\endgroup$
    – Poutnik
    Dec 30, 2022 at 7:07
  • 3
    $\begingroup$ Balloons leak hydrogen easily. The pressure inside an inflated balloon is greater than ambient pressure. $\endgroup$
    – Jon Custer
    Dec 30, 2022 at 15:05
  • $\begingroup$ What is missing is that you have lept to assume the issue is with the density despite having chosen a way of measuring density that is a crude approximation and a way of collecting the gas that is unreliable for this purpose. Good technique would collect hydrogen by gas displacement (eg pushing water from an upturned measuring cylinder) which would give a reliable idea of the volume of hydrogen, a far better way of measuring the amount than estimating the approximate density. $\endgroup$
    – matt_black
    Jan 1, 2023 at 11:54

2 Answers 2


The pressure inside your balloon is not approximately 1 atmosphere, even at ambient temperature. It is substantially greater, as it has to counteract atmospheric pressure plus the elastic force of the balloon latex trying to pull itself together. Therefore, your sample of hydrogen gas is not close to STP conditions, and its molar volume and density should not match. Also annoyingly, the pressure inside a balloon is sensitively dependent on how inflated the balloon is, and it varies non-monotonically - it starts at 1 bar, increases up to a maximum, then decreases again. This is not a very good way to measure gas volumes.

If you want a easy and fairly reliable way of measuring gas volumes on the order of a liter, a typical procedure is to fill a container (usually a graduated cylinder for extra precision) with water, invert it, and measure the water displaced by the evolved gas. For volumes on the order of milliliters, you can displace the plunger of a gas-tight syringe.

Edit: Actually when I wrote this answer, I trusted the calculations were overall correct, but Karsten's comment tipped me off to some errors, so I decided to double-check. So let's calculate the theoretical volume of hydrogen gas produced:

$$\ce{NaBH4 (s) + 4 H2O (l) -> NaB(OH)4 (aq) + 4 H2 (g)}$$

(Arguably your proposed sodium metaborate product is actually best described as sodium tetrahydroxyborate in aqueous solution, though that doesn't affect the proportionality between moles of borohydride used and moles of hydrogen gas evolved.)

$$\mathrm{1.00\ g\ NaBH_4 \times \frac{1\ mol\ NaBH_4}{37.83\ g\ NaBH_4 }\times \frac{4\ mol\ H_2}{1\ mol\ NaBH_4}\times \frac{22.4\ L\ H_2(@0\ °C, 1\ atm)}{1\ mol\ H_2}=2.37\ L\ H_2(@0\ °C, 1\ atm)}$$

(This volume can change by ~10% depending on the actual temperature of the gas inside your balloon. At 25 °C, you would get 2.60 L instead.)

So even though your calculation is a bit roundabout and contains some mistakes, it does seem the major error is experimental.


First, a minor consideration about gases: In an experimental setup like yours, I would not really try to deal with a gas mass and density. The molar volume of any ordinary gas is ~22.4 liter. Easy to remember and one size fits all.

Second, the experiment results: Armed with the above consideration about the molar volume, I would ask myself first not "what's wrong with my hydrogen?", but instead "why I am getting too little gas?" .

This alternative question has two important advantages:

  • It does not challenge a well-established fact
  • It is way easier to propose answers.

The possible answers may be:

  1. The purity of the substances used (you state that "not everything is ideal with my..." . How can the possible impurities react in similar conditions?
  2. Parasitic reaction paths that can yield less hydrogen (well, I cannot think of one in your particular case)
  3. Did you collect all of the gas? Can some of it be dissolved in the solution? (there is solubility data available online - at least about hydrogen in water, you can get this as an estimate for your final solution)
  4. What are my margins of error in both measurement and calculation? You start with 1.0g +/- 1% or 1.0g +/- 20% ?

For the point 1 question, you can stage a much easier experiment that actually involves measuring the mass of the hydrogen: weight your reacting substances, mix them and wait until the reaction finishes, weight again. Subtract the final mass from the initial mass.

You may, or may not get something like 213mg .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.