 Now we are going to put it into context by doing a question and calculate the percentage U of the following process, instead of question size, 21.2 grams of ammonia reacts with carbon dioxide to produce urea and water. Calculate the percentage U for this reaction if 22.8 grams of urea is collected in the end. The molar mass for ammonia is 17.031 grams per mole, and molar mass for urea has a chemical formula of Cn2OH4 and the molar mass is equal to 60.056 grams per mole. So firstly, we need to write out the chemical equation and balance it. It allows us to see the chemical composition of the compounds involved as well as the stoichiometry ratio between them. So you have ammonia NH3 plus carbon dioxide CO2 producing urea Cn2OH4 plus water H2O. So if you balance that, the chemical equation will become 2NH3 plus CO2, and on this side you have Cn2OH4 and H2O. So the stoichiometry in this equation is 2111. That means you need two ammonia and one carbon dioxide to make every urea and water. Now let's summarize the question. 21.2 grams of ammonia, so mass of ammonia is equal to 21.2 grams. Calculate the percentage U, so U is unknown and you calculate that by dividing the actual U by the theoretical yield. So what's missing? We don't know the theoretical yield yet, so we need to calculate that. So since you already got the mass of the actual U, so the easiest way to calculate this yield is to calculate the mass of the theoretical yield. Now the mass of theoretical yield is equal to the number of moles divided by molar mass. So we know the molar mass from the question which is 60.056 grams per mole for urea, but we don't have the theoretical number of moles of urea, so we need to calculate that. To do that though, we need to look at the stoichiometry of the chemical equation. We see that we need two moles of ammonia to make one mole of urea, so the theoretical number of moles of urea will be half of whatever the number of moles of ammonia that we put in. So the number of moles of 21.2 grams of ammonia is equal to 1.18 moles. So the theoretical number of moles of urea will be half of that. You can just chuck that into the calculation for the theoretical mass of the urea, or that 35.4 grams. So this 35.4 grams is what you should get if all of your ammonia reacted and went into the production of urea perfectly, and also assumed that you were able to purify and collect all the product without missing anything, which is really not realistic. So the percentage yield is equal to the mass of the measured collected urea, which is 22.8 grams, divided by the theoretical mass of urea, which is 35.4 grams. Multiply that ratio by 100% will give you 64%. And it is a rule that you always round your percentage yield to the whole number. So you got your percentage yield of the process, which is 64%.