y the most complex substance. The most complex substance is the one with the largest number of different atoms, which is C7H16C7H16C_7H_{16}. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance. 2. Adjust the coefficients. a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side: C7H16(l)+O2(g)→7–CO2(g)+H2O(g)C7H16(l)+O2(g)→7_CO2(g)+H2O(g)\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber 7 carbon atoms on both reactant and product sides b. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: C7H16(l)+O2(g)→7CO2(g)+8–H2O(g)C7H16(l)+O2(g)→7CO2(g)+8_H2O(g)\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber 16 hydrogen atoms on both reactant and product sides 3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. 4. Balance the remaining atoms. The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side: C7H16(l)+11–––O2(g)→7CO2(g)+8H2O(g)C7H16(l)+11_O2(g)→7CO2(g)+8H2O(g)\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber 22 oxygen atoms on both reactant and product sides 5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.