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The problem:

Find out the amount of moles of a certain substance knowing that the amount of units of said substance is $\pu{1.8E23}$. I have to find how many moles of carbon, hydrogen, and oxygen that amounts to. The number $\pu{1.8E23}$ was calculated in a previous question where I was asked to find the amount of units in $n(\ce{C6H12O6}) = \pu{0.3 mol}$.

The answer key:

\begin{align} &\ce{C}: \pu{1.1E24} \\ &\ce{H}: \pu{2.2E24} \\ &\ce{O}: \pu{1.1E24} \end{align}

Try 1: \begin{align} m(\ce{C6H12O6}) &= M(\ce{C6H12O6}) \cdot n(\ce{C6H12O6}) \\ &= (\pu{180.156 g mol-1})(\pu{0.3 mol}) = \pu{54.0468 g} \end{align} so that \begin{equation} n(\ce{C}) = \frac{\pu{54.0468 g}}{(\pu{12.01 Da})(\pu{1.6603E-24 g/Da})} \approx \pu{2.71E24} \end{equation} Which is not the right answer. The right answer for carbon is $\ce{C}: \pu{1.1E24}$.

Try 2: \begin{equation} n(\ce{C}) = \left(\frac{\pu{1.8E23}}{\pu{1.6603E-24 g/Da}}\right) (\pu{12.01 Da}) \approx \pu{1.30E48} \end{equation} Which is not the right answer either.

How do I go about this problem to get to the right answer?

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  • $\begingroup$ Mere minutes after having spent a lot of time fiddling with MathJax to make this post I figure out how to get the right answer. I need to do this: multiply 1.8 times 10^23 with 6 (the amount of coal atoms in one glucose molecule). I'm unsure why that is, so an explanation would still be greatly appreciated... $\endgroup$
    – Nora
    Feb 4 at 9:42
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    $\begingroup$ Like many school chemistry problems, this one is so simple as to give the students a hard time for not believing the right answer could be as simple as it is. Don't be deceived: yes it is that simple, simpler than you think, and simpler than that, too. See: mass is not a part of the problem and not needed at all, as much so as the words like trousers, serendipity, or rhinoceros. Moles is all that matter. You have moles of glucose, you need moles of carbon. How many carbons are in one glucose? You'll never find out, unless you look at its formula. $\endgroup$ Feb 4 at 10:00

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Each molecule of glucose has 6 carbon atoms. Hence a given number of molecules of glucose would have 6 times the carbon atoms. Since you have $\pu{1.8E23}$ atoms of glucose, the number of atoms of carbon would be this number multiplied by 6 which gives us $\pu{1.08E24}$ atoms.

Similarly proceed for Hydrogen and Oxygen.

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