How many atoms in methane? Please explain the rule.

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    That depends on how much methane you’re dealing with.
    The answer to your question depends on whether or not you meant in a molecule of methane or in a mole of methane (or in some specific mass of methane).
    A methane molecule ##CH_4## contains five atoms
    arranged in a tetrahedral .
    To get the number of atoms in a mole of methane you need to use the definition of a mole.
    In your case a mole of methane represents a collection of methane molecules. More specifically a mole of methane contains ##6.022 * 10^(23)## molecules of methane – this is known as Avogadro’s number.
    Basically to have one mole of methane you need to have a total of ##6.022 * 10^(23)## molecules of methane. Well if one molecule of methane contains five atoms you can say that one mole of methane will contain
    ##underbrace(6.022 * 10^(23))_(color(blue)(1 mole CH_4))color(red)(cancel(color(black)(molecules CH_4))) * 5 atoms/(1color(red)(cancel(color(black)(molecule CH_4)))) = 3.01 * 10^(24)atoms##
    What if you have a certain mass of methane? Let’s say that you need to find how many atoms you have in ##25.0 g## of methane.
    Since you already know how to use moles and molecules you need to find a way to convert the mass of methane into moles of methane.
    To do that use methane’s molar mass which tells you exactly what the mass of one mole of methane is.
    Methane’s molar mass is ##16.042 g/mol## which means that ##25.0 g## will be equivalent to a little over one mole
    ##25.0color(red)(cancel(color(black)(g CH_4))) * (1 mole CH_4)/(16.042color(red)(cancel(color(black)(g CH_4)))) = 1.56 moles CH_4##
    How many molecules of methane you have in this many moles? Use Avogadro’s number again
    ##1.56color(red)(cancel(color(black)(moles CH_4))) * (6.022 * 10^(23)molec. CH_4)/(1color(red)(cancel(color(black)(mole CH_4)))) = 9.39 * 10^(23)molec CH_4##
    Finally this number of molecules will contain
    ##underbrace(9.39 * 10^(23))_(color(blue)(1.56 moles CH_4))color(red)(cancel(color(black)(molec. CH_4))) * 5 atoms/(1color(red)(cancel(color(black)(molec. CH_4)))) = 4.70 * 10^(24)atoms##

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