Precision and Accuracy
When we investigate the quality of an experimental decision or calculation we take into account two major principles: The first principle, which deals with the reproducibility of our calculations, is called precision. The second principle called accuracy deals with how close our measured calculation is to the “actual” or “true” value. The idea of accuracy and precision can be illustrated with the images provided below, where the bullseye of the dart board representing the true value.
Throughout every measurement or calculation we do in chemistry there will be some level of uncertainty called experimental error. The accuracy of these measurements can be quantified by determining the percent error.
Within the percent error formula the experimental value is your calculated value, and the accepted value is your known or true value. In most cases, a percent error of less than 10% will be acceptable.
EXAMPLE: In lab you are asked to determine the density of a copper pellet. By weighing the mass of the copper pellet and the volume of water displaced you calculate the density as 9.12 g/cm3. When you look up the density of copper you find it to be 8.96 g/cm3. Calculate the percent error.
For non-geometric objects water displacement is used to determine the object’s volume.
STEP 1: Isolate the given variables from the question.
STEP 2: Utilize the Percent Error Formula for values provided in the question.
Stoichiometry and Percent Error
When we incorporate stoichiometry the percent error formula changes to where the experimental value becomes the actual yield and the accepted value becomes the theoretical yield.
The actual yield represents the amount of product that is made, isolated or produced when doing the reaction in real life. The theoretical yield represents the maximum amount of product that could be made, but in reality we never obtain that amount. Some product is always lost through isolation, purification, transfer or some other uncontrollable variable.
EXAMPLE: If 45.0 g benzene reacts with excess oxygen to produce 141 g carbon dioxide, what is the percent error of your measurements?
Within our calculations we will ignore both oxygen and water because they are either in excess or not addressed within the question.
STEP 1: Calculate the molar masses of C6H6 and CO2.
STEP 2: Convert the grams of C6H6 into moles of C6H6.
STEP 3: Convert the moles of C6H6 into moles of CO2 by doing a mole-to-mole ratio.
Bring down the coefficients from the balanced equation.
STEP 4: Convert the moles of CO2 into grams of CO2.
STEP 5: Calculate the percent error.
The actual yield would be the 141 grams of carbon dioxide given within the question, while the 152 grams of CO2, which we calculated, would be the theoretical yield.
The larger the percent error value then the less accurate our measurements. Since our percent error is less than 10% we can conclude that our experimental measurements were acceptable.