This concept has been illustrated for the reaction: The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction. The actual yield is stated in the problem, 6.
Thus, the percentage yield is 6. Due to chemical equilibrium or the mass action law , the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. A loss during the recovery process causes even lower actual yield.
Assume the percentage yield to be The reaction and relative masses of reagents and product are: Since the actual yield is only One can also calculate the theoretical yield of AgCl from the percentage yield of Na 11 Sodium Mg 12 Magnesium Al 13 Aluminum Si 14 Silicon P 15 Phosphorus S 16 Sulfur Cl 17 Chlorine Ar 18 Argon K 19 Potassium Ca 20 Calcium Sc 21 Scandium Ti 22 Titanium V 23 Vanadium Cr 24 Chromium Mn 25 Manganese Fe 26 Iron Co 27 Cobalt Ni 28 Nickel Cu 29 Copper Zn 30 Zinc Ga 31 Gallium Ge 32 Germanium As 33 Arsenic Se 34 Selenium Br 35 Bromine Kr 36 Krypton Rb 37 Rubidium Sr 38 Strontium Y 39 Yttrium Zr 40 Zirconium Nb 41 Niobium Mo 42 Molybdenum Tc 43 Technetium Ru 44 Ruthenium Rh 45 Rhodium Pd 46 Palladium Ag 47 Silver Cd 48 Cadmium In 49 Indium Sn 50 Tin Sb 51 Antimony Te 52 Tellurium I 53 Iodine Xe 54 Xenon Cs 55 Caesium Ba 56 Barium Hf 72 Hafnium Ta 73 Tantalum W 74 Tungsten Re 75 Rhenium Os 76 Osmium Ir 77 Iridium Pt 78 Platinum Au 79 Gold Hg 80 Mercury Ti 81 Thallium Pb 82 Lead Bi 83 Bismuth Po 84 Polonium At 85 Astatine Rn 86 Radon Fr 87 Francium Ra 88 Radium Rf Rutherfordium Db Dubnium Sg Seaborgium Bh Bohrium Hs Hassium Mt Meitnerium Ds Darmstadium Rg Roentgenium Cn Copernicium Uut Unutrium Uuq Flerovium UuP Ununpentium
Try an example: Calculate the percent yield of sodium sulfate when g of sulfuric acid reacts with excess sodium hydroxide to produce g of sodium sulfate. First, note that the question clearly states that sodium hydroxide is the excess reagent.
The discrepancy between the theoretical yield and the actual yield can be calculated using the percent yield, which uses this formula: To use this formula for percent yield, you need to make sure that your actual yield and theoretical yield are in the same units.
The equation for percent yield is percent yield = actual yield/theoretical yield x %. Let me show you how this works with an actual chemical reaction. Example One. If g of CO reacts with an excess of H 2 to form 75g CH 3 OH, what is the percent yield of the reaction? The balanced equation is CO + 2 H 2 = CH 3 OH. First, determine the theoretical yield. So, ideally, grams of CaO should have been produced in this reaction. This is the theoretical yield. However, the problem tells us that only 15 grams were produced. 15 grams is the actual yield. It is now a simple matter to find percent yield.
To determine percent yield, divide the actual yield by the theoretical yield and multiply by For this example, use the equation: 82 grams of silver / 85 grams of silver x = 96 percent. This percentage tells you the efficiency of the chemical reaction, or how good the reaction . You should be able to calculate the percentage yield of a product from the actual yield of a reaction or calculate the theoretical mass of a product from a given mass of reactant given the balanced equation for the reaction.