The Preparation of Calcium Carbonate
Purpose: To create chalk (calcium carbonate) and to find the percentage yield in order to see the amounts of anhydrous sodium carbonate and calcium chloride were used up. Also to see if there’s any alterations like mass differentials.
1. To introduce the concept of “limiting factor” in a chemical reaction 2. To practice a. Writing a balanced equation b. Determining the number of moles of each reactant and product c. Deciding which chemical is the limiting factor d. Predict theoretical yield e. Determine actual yield f. Use error discussion
* 2 beakers * 2 watch glasses * Stirring rods * Filter …show more content…
Percentage yield= Experimental yield x 100 Theoretical yield = 3.01g x 100 3.64g = 82.7 %
4. The class average mean was 89.41 % which is fairly close to best possible yield of 98%. Many groups were close to this result with yields such as 93.87%, 92.50% and 90.00% and other groups were farther like 82.70 and 83.72.
5. Percent Error = 98% - 82.70% x 100 98 = 15.91%
My group received a 82.7% yield percentage comparing to the 98% at best there. There was a 15.91% error percentage, and so when comparing yields to yield it was an ok amount of a differential. This error may be because of earlier mention experimental errors. 6. The class data was 89.41% which is close to the best yield of 98%, the class yield was only 8.59% off of being accurate.
7. Yes, there is a expediential l difference between the answers to question 5 and 6. The percentage yield I got from my experiment is 6.71% less than the class average percentage yield, and the yield I got is 15.3% off the yield (98%) and the classes average is only 8.59% off. It seems that both averages are reasonable considering they