For example, consider the simple equation H2+O2{\displaystyle H_{2}+O_{2}} → H2O{\displaystyle H_{2}O}. There are two atoms of hydrogen on both the left and right. But there are two atoms of oxygen going in as a reactant and only one atom in the product on the right. To balance, double the product, to get H2+O2{\displaystyle H_{2}+O_{2}} → 2H2O{\displaystyle 2H_{2}O}. Check the balance. This change has corrected the oxygen, which now has two atoms on both sides. But you now have two atoms of hydrogen on the left with four atoms of hydrogen on the right. Double the hydrogen in the reactant. This will adjust the equation to 2H2+O2{\displaystyle 2H_{2}+O_{2}} → 2H2O{\displaystyle 2H_{2}O}. This change now has 4 atoms of hydrogen on both sides, and two atoms of oxygen. The equation is balanced. As a more complicated example, oxygen and glucose can react to form carbon dioxide and water: 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6CO2+6H2O{\displaystyle 6CO_{2}+6H_{2}O} In this equation, each side has exactly 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 18 oxygen (O) atoms. The equation is balanced.
For this example, one molecule of oxygen (O2{\displaystyle O_{2}}) contains two oxygen atoms. The molar mass of one atom of oxygen is about 16 g/mol. If necessary, you can find more precise values. ) 2 oxygen atoms x 16 g/mol per atom = 32 g/mol of O2{\displaystyle O_{2}}. The other reactant, glucose (C6H12O6{\displaystyle C_{6}H_{12}O_{6}}) has a molar mass of (6 atoms C x 12 g C/mol) + (12 atoms H x 1 g H/mol) + (6 atoms O x 16 g O/mol) = 180 g/mol.
For example, suppose you begin with 40 grams of oxygen and 25 grams of glucose. 40 g O2{\displaystyle O_{2}} / (32 g/mol) = 1. 25 moles of oxygen. 25g C6H12O6{\displaystyle C_{6}H_{12}O_{6}} / (180 g/mol) = about 0. 139 moles of glucose.
In this example, you are starting with 1. 25 moles of oxygen and 0. 139 moles of glucose. Thus, the ratio of oxygen to glucose molecules is 1. 25 / 0. 139 = 9. 0. This ratio means that you have 9 times as many molecules of oxygen as you have of glucose.
For this reaction, the reactants are given as 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}}. The coefficients indicate that you need 6 oxygen molecules for every 1 glucose molecule. The ideal ratio for this reaction is 6 oxygen / 1 glucose = 6. 0.
In this example, you are beginning with 9 times as much oxygen as glucose, when measured by number of moles. The formula tells you that your ideal ratio is 6 times as much oxygen as glucose. Therefore, you have more oxygen than required. Thus, the other reactant, glucose in this case, is the limiting reactant.
Continuing the example above, you are analyzing the reaction 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6CO2+6H2O{\displaystyle 6CO_{2}+6H_{2}O}. The two products shown on the right are carbon dioxide and water. You can begin with either product to calculate theoretical yield. In some cases, you may be concerned only with one product or the other. If so, that is the one you would start with.
In the example above, glucose is the limiting reactant. The molar mass calculations found that the initial 25g of glucose are equal to 0. 139 moles of glucose.
The balanced equation for this example is 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6CO2+6H2O{\displaystyle 6CO_{2}+6H_{2}O}. This equation tells you that you expect 6 molecules of the desired product, carbon dioxide (CO2{\displaystyle CO_{2}}), compared to 1 molecule of glucose (C6H12O6{\displaystyle C_{6}H_{12}O_{6}}). The ratio of carbon dioxide to glucose is 6/1 = 6. In other words, this reaction can produce 6 molecules of carbon dioxide from one molecule of glucose.
In this example, the 25g of glucose equate to 0. 139 moles of glucose. The ratio of carbon dioxide to glucose is 6:1. You expect to create six times as many moles of carbon dioxide as you have of glucose to begin with. The theoretical yield of carbon dioxide is (0. 139 moles glucose) x (6 moles carbon dioxide / mole glucose) = 0. 834 moles carbon dioxide.
In this example, the molar mass of CO2 is about 44 g/mol. (Carbon’s molar mass is ~12 g/mol and oxygen’s is ~16 g/mol, so the total is 12 + 16 + 16 = 44. ) Multiply 0. 834 moles CO2 x 44 g/mol CO2 = ~36. 7 grams. The theoretical yield of the experiment is 36. 7 grams of CO2.
In this example, the second product is water, H2O{\displaystyle H_{2}O}. According to the balanced equation, you expect 6 molecules of water to come from 1 molecule of glucose. This is a ratio of 6:1. Therefore, beginning with 0. 139 moles of glucose should result in 0. 834 moles of water. Multiply the number of moles of water by the molar mass of water. The molar mass is 2 + 16 = 18 g/mol. Multiplying by the product, this results in 0. 834 moles H2O x 18 g/mol H2O = ~15 grams. The theoretical yield of water for this experiment is 15 grams.
