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Cerin, C30H50O2

Hello Folks

I am not now nor have I ever been any kind of chemist.

My question is, how is the formula for Cerin, C30H50O2, consistent with this diagram taken from Chemspider?

I can count the 30 Carbon atoms, assuming that each vertex of a hexagon is a single C. I can obviously see the two Oxygen atoms. But when I count the Hydrogens, I can only get to 28. Where are the other 22?

Thank you for your patience.

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    $\begingroup$ You also need to assume that each implicit carbon (like the corners of each ring) is saturated, so the topmost ring has 6 implicit hydrogens. $\endgroup$
    – pentavalentcarbon
    Sep 11, 2017 at 14:46
  • $\begingroup$ @Raditz_35 so that's an additional 2 H''s for every completely 'unclaimed' corner, ie 2 x 10 = 20, plus another one each for the methyl and hydroxyl groups on the bottom left ring, making up the 22? Is that right? And have I got those group names right? Thanks also to pentavalentcarbon, I'm only allowed to tag one user per comment, it seems $\endgroup$
    – Mark
    Sep 11, 2017 at 15:00
  • $\begingroup$ @Mark I haven't noticed anything wrong, so I'd say you are right! $\endgroup$
    – Raditz_35
    Sep 11, 2017 at 15:01
  • $\begingroup$ @Raditz_35 OK, thank you very much! If you turn this into an answer I'd be happy to make it the accepted. $\endgroup$
    – Mark
    Sep 11, 2017 at 15:03

1 Answer 1

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Each carbon atom (you seem to have identified them) needs four of those stick-like thingies (the lines, sometimes called "bonds"). These include the dashed lines and thick triangles as well as the "double bond" you see at the lower oxygen which looks like this ($=$). That one counts twice. Every time you only see 3 or even 2 for one carbon atom, that means 1 (or 2) hydrogen atoms are implied.

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    $\begingroup$ As an explanatory addendum for original poster; for large molecules it is extremely normal to not write out the hydrogens, unless their position or orientation is vital. It would be incredibly tedious and not add any value, nor clarity to the picture (the opposite, actually) to write H's all over that structure. $\endgroup$
    – Stian
    Sep 11, 2017 at 21:49

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