Atoms and MoleculesDate: 11/28/2001 at 19:54:25 From: Jenny Subject: Word problems A certain molecule contains twice as many atoms of hydrogen as oxygen and one more atom of carbon than hydrogen. If there are 21 atoms altogether in the molecule, how many atoms of carbon are there? Date: 11/29/2001 at 12:42:30 From: Doctor Ian Subject: Re: Word problems Hi Jenny, Suppose there is 1 hydrogen atom. Then there must be half as many ogygen atoms... oops! So there must be at least two hydrogen atoms. Okay, suppose there are 2 hydrogen atoms. Then there must be half as many oxygen atoms: 1. And there must be 1 more carbon atoms than hydrogen atoms: 3. That gives us hydrogen oxygen carbon total 2 1 3 6 So that's not the molecule we're looking for. But we can find the one we're looking for just by varying the hydrogen count. Note that we have to have an even number of hydrogen atoms, or we can't get an integer for the number of oxygen atoms: hydrogen oxygen carbon total 2 1 3 6 4 2 5 11 6 3 7 16 With a little imagination, we could write an expression to compute the total number of atoms in terms of the number of hydrogen atoms: hydrogen oxygen carbon total 2 1 3 6 4 2 5 11 6 3 7 16 h h/2 h+1 h + h/2 + (h+1) Once we've done that, we can specify the total that we want, h + h/2 + (h+1) = 21 and solve the equation to find h directly. (Then we have to add 1 to find the number of carbon atoms. Forgetting to do that would be an easy mistake to make, which is why the problem asks you for the number of _carbon_ atoms instead of the number of _hydrogen_ atoms. A well-constructed problem always gives you lots of different ways to make a mistake.) For a problem like this, where there aren't a lot of atoms involved, the extra work to set up and solve the equation may or may not be worth the trouble, compared to a little judicious guessing. But for much larger numbers (for example, suppose the total number of atoms is 251), using an equation is much, much quicker. Which is one of the things that makes equations so useful! Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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