For 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} Each side has exactly 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 18 oxygen (O) atoms. The equation is balanced. Read this guide if you are asked to balance an equation yourself.
For example, 1 molecule of oxygen (O2{\displaystyle O_{2}}) contains 2 oxygen atoms. Oxygen’s molar mass is about 16 g/mol. (You can find a more precise value on a periodic table. ) 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, say you started 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.
You started with 1. 25 moles of oxygen and 0. 139 moles of glucose. The ratio of oxygen to glucose molecules is 1. 25 / 0. 139 = 9. 0. This means you started with 9 molecules of oxygen for every 1 molecule of glucose.
The left side of the equation is 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}}. The coefficients tell you there are 6 oxygen molecules and 1 glucose molecule. The ideal ratio for this reaction is 6 oxygen / 1 glucose = 6. 0. Make sure you list the reactants in the same order you did for the other ratio. If you use oxygen/glucose for 1 and glucose/oxygen for the other, your next result will be wrong.
If the actual ratio is greater than the ideal ratio, then you have more of the top reactant than you need. The bottom reactant in the ratio is the limiting reactant. If the actual ratio is smaller than the ideal ratio, you don’t have enough of the top reactant, so it is the limiting reactant. In the example above, the actual ratio of oxygen/glucose (9. 0) is greater than the ideal ratio (6. 0). The bottom reactant, glucose, must be 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 right-hand side lists 2 products, carbon dioxide and water. Let’s calculate the yield of carbon dioxide, CO2{\displaystyle CO_{2}}.
In the example above, you discovered that glucose was the limiting reactant. You also calculated that you started with 0. 139 moles of glucose.
Your balanced equation is 6O2+C6H12O6{\displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6CO2+6H2O{\displaystyle 6CO_{2}+6H_{2}O}. There are 6 molecules of your desired product, carbon dioxide (CO2{\displaystyle CO_{2}}). There is 1 molecule of your limiting reactant, 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 1 molecule of glucose.
You started with 0. 139 moles of glucose and the ratio of carbon dioxide to glucose is 6. The theoretical yield of carbon dioxide is (0. 139 moles glucose) x (6 moles carbon dioxide / mole glucose) = 0. 834 moles carbon dioxide.
For 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.
The theoretical yield is the maximum amount of product the experiment could make. The actual yield is the actual amount you created, measured directly on a scale. The percent yield = ActualYieldTheoreticalYield∗100%{\displaystyle {\frac {ActualYield}{TheoreticalYield}}*100%}. A percent yield of 50%, for instance, means you ended up with 50% of the theoretical maximum.
Let’s say our actual reaction yields 29 grams of CO2.
The actual yield was 29 grams, while the theoretical yield was 36. 7 grams. 29g36. 7g=0. 79{\displaystyle {\frac {29g}{36. 7g}}=0. 79}.
0. 79 x 100 = 79, so the percent yield of the experiment is 79%. You created 79% of the maximum possible amount of CO2.
