"Yield" from a chemical reaction means the amount of product you actually got compared to what is predicted from chemical theory. For example, we burned a piece of magnesium to produce magnesium oxide. The reaction is:
2Mg + O(sub2) --> 2MgO
If we started with 0.5 gram of magnesium, how much MgO should we get?
Use the mole equation to determine: moles = mass / atomic mass.
The mass is stated. Get the atomic mass from the periodic table: 24.
Then the number of moles is 0.5g / 24 = 0.0208
You can see from the equation that for every atom of magnesium reacted, one molecule of MgO is formed. Thus, 0.0208 moles of magnesium should produce 0.0208 moles of MgO. To find the number of grams of MgO, rearrange the mole equation to solve for mass:
mass = (moles)(atomic mass)
Find the atomic mass of MgO by adding the atomic weight of magnesium and of oxygen, or 24+ 16 = 40. Then, multiply 0.0208 x 40 = 0.833 gram.
But is that really how much MgO was recovered? No, because some was lost as a coating on the tongs I used to hold the magnesium strip and more was lost as a fume. (I showed you what happened if we inverted a beaker over the burning strip of magnesium: the fume coated the walls of the beaker so we collected more than if we just burned it in the open air. You saw that the burning it beneath the mouth of the beaker starved it for oxygen and the combustion stopped.)
I told you to weigh a beaker before the experiment. This gave you a "tare" weight. Then, when you burned the magnesium strip, you were to collect the resulting MgO in the beaker as it fell off the burning strip. Then you weighed the beaker with the MgO in it. This gave you a "gross" weight. You subtracted the tare from the gross to get the "net" weight of the MgO. This weight would be less than 0.833 gram because of the losses I discussed above. Let us say the gross weight was 108.52 grams and the tare weight was 108.1 grams. Then the amount of MeO you recovered would have been 0.42 grams. The yield, or "percent recovery" is the amount recovered divided by the expected amount times 100% or 0.42/0.833 x 100 = 50.4%.
Even if you had some difficulty with this, you could still use the amount of MgO actually recovered to determine how much MgCl(sub2) you would produce in the next step. The reaction is:
MgO + 2HCl --> MgCl(sub2) + H(sub2)O
You know how much MgO you recovered. From that determine the number of moles: it is 0.42/40 = 0.0105 in this example. Then, inspect the BALANCED equation. You can see that for every mole of MgO, you produce one mole of MgCl(sub2). Therefore, the number of grams of MgCl(sub2) is:
(0.0105) x [(24 + 2x35) = 0.987 grams.
After you evaporate the water from the beaker, let us say you reweigh it and find the gross weight is 108.6 grams. This means the amount of MgCl(sub2) in the beaker is 108.6 - 108.1 = 0.5 grams. What is the yield of MgCl(2)? It is 0.5gram/0.987gram x 100% = 50.66%.
You need to find how much magnesium you have at the end of the reaction (in the MgCl(sub2)) compared to what you started with. To do this, find the fractional amount of magnesium by dividing its atomic weight (24) by the total weight of the magnesium chloride (94). Then multiply that fraction by the amount of MgCl(sub2) recovered: 24/94x0.5gram = 0.13 gram. Compare that to the starting amount of magnesium, 0.5 gram. What is the yield? It is 0.13/0.5 x 100% = 26%
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