We now hit the grammar (e.g., substance symbolism), syntax (e.g., "stoichiometry," balancing equations), and semantics (e.g., describing real reactions) of Chemistry. The changes in substances evidences by the rearrangement of their atoms from reactant kinds of molecules to product kinds, are symbolized by chemical reaction statements where the correct number of mols of reactant are shown combining "+" to yield "--->" the correct number of mols of the right products. In the statements below, we'll use "----->" for correct expressions and "--/-->" for incomplete (thus incorrect) ones. So the burning (oxidation) of glucose in oxygen, done in the very controlled aerobic respiratory reactions in your body, is codified in Chemistry as follows:
C6H12O6(s) + O2(g) --/--> CO2(g) + H2O(l)
where the suscripts code for numbers of atoms in each molecule and the parentheses enclose the physical state of each substance. Note that the water is shown as liquid by the L in those (l) parentheses; don't confuse the lowercase L with a one. While burning glucose in air would produce so much heat that the water would be best described as (g) for gas, in your body, no step in the respiratory chain (of which this is an overall summation) raises the temperature enough to vaporize the water product molecules. That "moleculeS" plural reminds us that this equation is broken (--/-->) by virtue of not conserving the number of atoms on each side. The chemical algebra is wrong. Us chemists just don't work at energies high enough to chunder atoms like the nuclear physicists do; so OUR atoms always add up!
But they don't add up as written; so we must "balance" this equation just as if it were an algebraic one. That task is called "stoichiometry" (in Greek it means "the measure of the elements" sort of).
The first stoichiometric step is taken when we notice that the carbons don't balance. That requires us to have 6 carbon dioxides on the right, as in:
C6H12O6(s) + O2(g) --/--> 6 CO2(g) + H2O(l)
But a similar hextupling of product water is necessary to balance the hydrogen atoms:
C6H12O6(s) + O2(g) --/--> 6 CO2(g) + 6 H2O(l)
We see now that there are 8 atoms of oxygen on the left but 18 on the right! So we need 10 more atoms in the form of 5 more oxygen (diatomic) molecules to give:
C6H12O6(s) + 6 O2(g) -----> 6 CO2(g) + 6 H2O(l)
where the yield arrow is no longer broken because this chemical equation is grammatically, syntactically, and semantically correct. Note that those "stoichiometric coefficients" (the numbers before the molecules) code for both the proper proportion of reactant and product molecules in this sequence but ALSO for the proper number of MOLES required and produced...a mere scaling difference of 6.022x10²³.
Return to the CHM 1341 Lecture Notes or Go To Previous or Next Lectures.
Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688 M/S BE2.6 (for snailmail)
Richardson, TX 75083-0688
Voice: (214) 883-2485
Fax: (214) 883-2925
BBS: (214) 883-2168 (HST) or -2932 (V.32bis)
Internet: parr@utdallas.edu (sends Chris e-mail.)
Modified 17 June 1996.