today let's get some practice on determining molecular
formula so let's take a look at the question, nicotine is a compound found in
cigarettes if nicotine is seventy four point zero
two percent carbon eight point seven one zero percent hydrogen and 17 point two
seven percent nitrogen by mass and forty point five seven grams of nicotine is
equal to zero point two five zero zero moles what is the molecular formula of
nicotine? so if you're not sure how to get this going go ahead and check out my
tutorial on elemental analysis that is going to give you the strategy you need
to answer this question and when you're ready go ahead and give it a shot.
so we have our mass percentages up at the top here and we also have our information
that tells us that 40 point 5 7 grams of nicotine equals zero point two five zero
zero moles of nicotine so what are we going to do with this information well
the easiest way to do this is to go ahead and assume some mass of nicotine
to start doing some calculations since it's totally arbitrary what mass we
choose because we're dealing with percentages let's just use a hundred
grams and make the math nice and easy on ourselves so if we have a hundred grams
of nicotine that means we must have seventy four point zero two grams of
carbon eight point seven one zero grams of hydrogen and 17 point two seven grams
of nitrogen. we're just taking those percentages
which are out of a hundred and turning them into grams. now we need to convert
these into moles because masses don't tell us anything. whenever we are doing
calculations of this nature we always want moles so let's take our mass of
carbon and multiply by one mole over 12.01 grams, that is the atomic mass of
carbon. grams will cancel and we have 6 point 1 6 3 moles of carbon in our
hypothetical sample of nicotine now for hydrogen let's multiply by 1 mole over
1.008 grams that is the atomic mass of hydrogen, grams will cancel and we have
eight point six two four moles of hydrogen lastly for nitrogen we multiply
by one mole over fourteen point zero one grams that's the atomic mass of nitrogen
grams will cancel and we have one point two three three moles of nitrogen. so in
this sample we have this many moles of each of these elements so this tells us
something about the ratio of these elements in the compound so how can we
get this in a way that's a little more useful? what we can do is divide each of
these values by the smallest so that we get one for the smallest one and then
some ratio for the others and hopefully they'll turn out to be integers and that
will give us the empirical formula so if we divide each of these by the smallest
one of them which is one point two three three we end up with one mole of
nitrogen and we also end up with integer values for carbon and hydrogen, we have
five and seven so that means for every nitrogen atom in the sample we have five
carbon atoms and we have seven hydrogen atoms so that means the empirical
formula of nicotine must be C5H7N, that is the only way for these mass
percentages to work out the way that they are but once again we need the
molecular formula in order to get that we need the molecular mass or the molar
mass and so let's take this other bit of information we have if we want to know
the molar mass of nicotine in grams per mole well we have forty point five seven
grams and we can just divide that by zero point two five zero zero moles that
will give us a value in grams per mole and so we get 162.3 grams per mole so
how is that useful well we know that if we add up the masses of the 5 carbon
atoms and the 7 hydrogen atoms and the 1 nitrogen atom that adds up to 81 point 1
3 grams per mole so that is the molar mass of one empirical formula unit now
our molar mass is twice as much we have a two to one ratio so that means in
order to get the molecular formula which represents the actual number of atoms of
each element in the molecule all we need to do is double everything so let's take
our empirical formula and put it in parentheses and multiply it by two if we
double everything we'll get twice that mass and we will get our molar mass and
that means that the molecular formula must be C10H14N2