I was reading the Wikipedia article about Sucrose, when I noticed that the skeletal formula was missing some $\ce{C}$ and $\ce{H}$ atoms. The chemical formula being $\ce{C12H22O11}$, I counted the $11$ Oxygen parts. But there seem to be only $3$ $\ce{C}$ atoms and $14\, \ce{H}$ atoms, if I counted correctly. My chemistry lessons date back about $15$ years, so I'd like to understand where the missing carbon and oxygen is hidden within the skeletal formula.

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    $\begingroup$ Every point in the bond is a carbon atom.. H is just added to satisfy valency after.. You could read up on the bond-line formula of a compound.. $\endgroup$ Aug 27, 2020 at 15:37
  • $\begingroup$ Ah, that makes sense. And the "bold" parts are CH ions then? That would account for the missing Hydrogen. $\endgroup$ Aug 27, 2020 at 15:39
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    $\begingroup$ This Libretexts page should help. $\endgroup$ Aug 27, 2020 at 15:40
  • $\begingroup$ Any formula is a condensed knowledge that must be unpacked by applying certain rules. For example, the mathematical formula for "eleven" is 11. One who does not know the rules might wonder why there are only two sticks and not eleven. Same thing here. $\endgroup$ Aug 27, 2020 at 16:28

1 Answer 1


Since your chemistry lessons date back about 15 years, I'd loke to refreash your organic chemistry knowledge. You may need to be aware that carbon can make maximum if 4 single bonds. Therefore, when you have a line structure of organic compound like sucrose (see below), keep in mind that an each corner represent a carbon atom and appropriate hydrogen atoms, which are not shown. For example, if a corner contains only three bonds (as shown in most of sucrose molecule), then the forth bond is $\ce{C-H}$ bond:

Sucrose Structure

I have put those $\ce{C-H}$ at each corner, which are representing them. However, the anomeric corner of fructose molecule already has 4 bonds. Therefore, that corner represents only $\ce{C}$ as indicated in turquoise color. Therefore, the molecular formula of sucrose can be counted as $\ce{C12H22O11}$.


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