One gram of Hydrogen reacts with exactly (almost) 8 grams of oxygen to produce $\ce{H2O}$. Another single gram of Hydrogen reacts with 16 grams of Oxygen to produce $\ce{H2O2}$.

Can the ratio $\frac{m_O}{m_H}$ be determined from this information? If not, what else do chemists need to know to find this ratio?

  • $\begingroup$ This appears to be a homework question, with no effort shown on how you would approach it. What is the concept that you are finding difficult? $\endgroup$ – Jon Custer Feb 13 '15 at 17:21
  • $\begingroup$ It is not actually a homework question. My question is how these mass ratios in general are deduced from reactions. I couldn't derive this ratio from these particular reactions. $\endgroup$ – richard Feb 13 '15 at 19:01
  • $\begingroup$ If you know that 1 gram of Hydrogen (well, H2) reacts with 8 grams of Oxygen (fortunately O2) to produce H2O, and you know it is H2O, that is kind of all the info you need to get the mass ratio of H:O. The tricky part is, without knowing that you start with H2 and O2, and end with H2O, how would you get there... $\endgroup$ – Jon Custer Feb 13 '15 at 20:23
  • $\begingroup$ yes exactly assumption is that you don't know. It seems to me the information is not enough and the first derivation is due to Avogadro. $\endgroup$ – richard Feb 13 '15 at 20:31
  • $\begingroup$ Well, it is kind of like the classic high school chemistry experiment mixing various salts to see what precipitates - looking at enough reactant and product combinations eventually gives you a pattern that gets you to integer ratios. But, yes, a key is being able to figure out moles of reactants and products. $\endgroup$ – Jon Custer Feb 13 '15 at 20:37

Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.

$$2n\space \rm{atoms \space H} = 1 \space g \space H$$ $$1n \space \rm{atoms \space O} = 8 \space g \space O$$

Divide the second equation by the first:

$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$

Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.

If you didn't know the formulae, then you could assume that the ratio was some integer multiple of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.

Incidentally, what you are describing is the law of multiple proportions, and it is one of the key findings that led to the development of atomic theory.

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    $\begingroup$ Incidentally, John Dalton (developer of the atomic theory) got all of his atomic masses and many of his formulas wrong because he assumed that water has hydrogen and oxygen in a 1:1 ratio of atoms (per his book A New System of Chemical Philosophy). You can't really fault him though, since the very existence of atoms was, at the time, both revolutionary and controversial. $\endgroup$ – j_foster Feb 14 '15 at 3:07

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