Calculating theoretical yield and percent yield... help!?

Sharpen your pencil because this one takes some math. First you need to understand moles. One mole of an element is the amu(atomic m unit), the number below the element name. One mole of a compound is the combined amu's of all the individual elements. For example Ba has an amu of 137.327. So 1 mole of Ba weighs 137.327g. Cl has an amu of 35.453 and there are 2 of them in BaCl so we need to double that number and add it to Ba. That leaves us with 208.233g for 1 mole of BaCl2. Now do the same for BaSO4 and you should get 233.3896. Your numbers may vary slightly depending on your particular periodic table. Now you need to convert 12 ml to g. For that you need the density of BaCl which is 3.0979g/ml. So you multiply 3.0979(D)x12(V)=37.1748g(M). Now take that number divided by the grams per mole. 37.1748/208.233= the number of moles for that m of BaCl2, (.178525) If you look at the above balanced equation you can see that 1 mole of BaCl2 yields 1 mole of BaSO4. So that tells us that we will have .178525 moles of BaSO4. Take the moles times the number of grams per mole for BaSO4. .178525x233.3896= the number of grams of BaSO4 (41.6659g) This is the theoretical yield. Now take actual yield divided by theoretical yield times 100 to get %. (I am guessing this is 53.1g)/41.6659g x 100 = 127.44%. This number will only exceed 100% if you have contaminants in your BaSO4. The above procedure is correct, check the math and your values in your question. Hope I could help.