The nitrogen in sodium nitrate and in ammonium sulfate is available to plants as fertilizer. Which is the more economical source of nitrogen, a fertilizer containing 30.0% sodium nitrate by weight and costing $9.00 per 100 lb or one containing 20.0% ammonium sulfate by weight and costing $8.10 per 100 lb?

Video transcript

Welcome back everyone in this example. We're told that the nitrogen and urea and an ammonium nitrate Is available through plants as fertilizer and that one brand of fertilizer contains 25% of your area by mass and cost $3 per kilogram while another brand contains 30% of ammonium nitrate by mass and cost $2 per kilogram. So we need to determine which of these fertilizers is the more cost effective source of nitrogen. So we're going to label our first compound Urea as brand a. And then we'll label our second compound ammonium nitrate as Brand B. So we'll start our solution with Brand A. Which is our area. So according to the prompt we have 25% of area by mass. So we would understand that that is as a decimal 0.25 which we can express by the fraction 250 g of Iria. And this is going to be out of 1000 g of our fertilizer. And so because we want to end up with finding out how many grams of our nitrogen is from brand a. We're going to want to cancel out grams of Yuria in the numerator. So we're going to multiply by our conversion factor where we recall from the periodic table the molar mass of area equal to 60.06 g of area. And this is for one mole of Iria. So this allows us to cancel out grams of area. And now we're going to focus on getting into moles of nitrogen. And so we want to use a multiple conversion factor where we're going to see in one mole of Yuria that that for our atoms of nitrogen and urea we would see we have nitrogen here and then we have a subscript of two. So that would give us two moles of nitrogen in one mole of area. So we can use that in the numerator as a conversion factor B. C. We have two moles of nitrogen. So now we're able to cancel out moles of Iria. And now we want again grams of nitrogen to be our final unit. So we're going to recall the molar mass of nitrogen from the periodic table where we see that for one mole of nitrogen that corresponds to a mass of 14.01 g of nitrogen. So we can get rid of moles of nitrogen. And now we want to plug in everything into our calculators and we're going to get a value equal to 116.63 g of nitrogen. And we understand that this is out of 1000 g of our fertilizer. But we want to find out the dollar cost program of nitrogen from brand a so we want to go ahead and recognize that in the prompt we're told that we have a cost of $3 per kilogram of fertilizer for our Yuria. So we're going to use that information. So we're gonna take that $3. This is out of or this is per kilogram of our fertilizer. And so we would say that our prefix kilo tells us that we have 10 to the third power Of our g. So this is 1000 g of fertilizer. And we're going to multiply this by the value that we calculated above. But we want to go ahead and cancel out grams of fertilizers. So we're going to flip it and say we have 1000 g of fertilizer And this is for 116.63 g of nitrogen. And this allows us to cancel out grams of fertilizer leading us with dollars per gram of nitrogen to find our cost for brand a. And this is going to give us a cost equal to $0. per gram of nitrogen from fertilizer. A. So we want to see which is more cost effective. And we want to compare this value that we've just calculated to brand B which is ammonium nitrate or consists of ammonium nitrate. So we're going to go ahead and do our solution for brand being purple here below. So starting out with the same information according to the prompt, we have a mass of 30 for ammonium nitrate in fertilizer B. And so converting that to a decimal or as expressing this as a expressing this percentage as a fraction. Sorry we're going to write it out as 300 g of ammonium nitrate And this is going to be out of g of our fertilizer. And so next we want to go ahead and again cancel out our grams of ammonium nitrate so that we can end up with grams of nitrogen as our final unit. So we're going to recall from the periodic table that we have a molar mass of 80. g of ammonium nitrate In one mole of ammonium nitrate. So now we can get rid of grams of ammonium nitrate and focus on using that multiple conversion factor to analyze how many moles of nitrogen we have in our one mole of ammonium nitrate. So counting the atoms that we have in our molecular formula from ammonium nitrate, we see we have one atom of nitrogen here and another here. So that would be two moles total. Making up our ammonium nitrate of nitrogen. So we can say we have two moles of nitrogen in one mole of ammonium nitrate. This allows us to cancel out our moles of ammonium nitrate and now we want to end up with grams of nitrogen as our final unit. So we're going to multiply by our conversion factor, which we recall from the periodic table, our molar mass of nitrogen which is 14.01 g of nitrogen for one mole of nitrogen. So this allows us to cancel out moles of nitrogen. And we're left with grams of nitrogen per grams of our fertilizer here. And so this is going to give us a value that is equal to 105.06 g of nitrogen per 1000 g of fertilizer. And just so that's included. We'll scoot that over. So this is per 1000 g of fertilizer. And so now we want to find the cost per dollar or per gram of nitrogen By the dollar. And so we're going to use the information from the prompt which says that according to our ammonium nitrate, we have to pay $2 per kilogram in our fertilizer for Brand B. And so we're going to use that piece of information. So we have $2 and this is per kilogram of our fertilizer. So this is out of 1000 g of our fertilizer. And we're gonna multiply this so that we can cancel out grams of fertilizer by slipping our calculation from Brand B here so that we have g a fertilizer in the numerator. And this is for our 105.06 g of nitrogen. So now we can cancel out grams of fertilizer and we're left with dollars program of nitrogen for brand b. And this is going to give us a cost equal to zero point and sorry It's a dollar sign there. So zero 019037. And this is out of g of nitrogen. But we want around this to about zero .0200 dollars per gram of nitrogen. And so comparing these two values, we would see that compared to our cost for of nitrogen from brand A versus our cost of nitrogen for Brand B. We have a lower cost associated with Brand B. So we would say based on this value, therefore we have a lower cost to source nitrogen. And so this means that our ammonium nitrate. I'm sorry that's n. h. four. So our ammonium nitrate is going to be more cost effective since it was associated with the lower dollar cost of grams of nitrogen for the fertilizer in Brand B. And so that means that choice be is going to be the correct answer choice being that ammonium nitrate is, the more the more cost effective fertilizer. So I hope that everything that I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.