In this example problem, we balance a combustion reaction. Our problem reads, "Balance the following reaction: C15H28N403 liquid plus O2 gas produced CO2 gas plus H2O gas plus NO2 gas." Since we have molecular compounds, we can balance this reaction by atoms. We want to balance carbon, hydrogen, nitrogen, and oxygen atoms. We should balance any elements that show up in the equation as free elements last. So, we will balance oxygen last since it shows up as just O2 gas. We want to start balancing any element that shows up in only one place on each side of the reaction first, so we will balance carbon, hydrogen, and nitrogen first. We can start in any order since all three of these only show up in one place on each side. Let's start with carbon. Initially, we have 15 carbon on the left and one on the right. To get the same number of carbon on both sides, we need to add a coefficient of one in front of C15H28N4O3 and a coefficient of 15 in front of CO2. Now we have 15 carbon on each side. We have 28 hydrogen on the left right now. We can put a coefficient of 14 in front of H2O to get 28 hydrogen on the right. We have four nitrogen on the left right now. We can put a coefficient of 4 in front of NO2 to get 4 nitrogen on the right. Now we just need to solve for the coefficient of O2 gas. We can say the coefficient of O2 gas is "x" and then create an equation for the amount of oxygen on each side of our reaction. On the left, we have one times three oxygen in C15H28N403 plus x times two oxygen with O2 that needs to equal the amount of oxygen on the right, which is 15 CO2 times two oxygen in each CO2 plus 14 H2O times one oxygen in each H2O plus 4 NO2 times two oxygen in each NO2. On the left, we have 3 + 2x; on the right, combining the multiplication and addition, we get 52. Subtracting 3 from each side, we get 2x equals 49. So x must be 49/2. In our balanced chemical equation, we want the smallest whole number coefficients, so we'll need to multiply all of our coefficients by 2 to get rid of the halves. We end up with coefficients of 2, 49, 30, 28, and 8. To double-check that we balanced the reaction correctly, we should count up the number of atoms of each element on both sides to make sure it matches. With our new coefficients in red, we now have 30 carbon, 56 hydrogen, 8 nitrogen, and 104 oxygen on each side. This should be our correctly balanced chemical reaction.