Quick review: Nutrients are present in a fertilizer; thus, the amount of nutrient in a fertilizer will depend on the fraction of the material that is the nutrient. Also, P and K are often expressed as P₂O₅ and K₂O, respectively. For example, P X 2.29 = P₂O₅; or P₂O₅/2.29 = P (see problem solutions below). In converting P₂O₅ to P, one may either divide by 2.29 or multiply by 0.44 and get the same answer.

• % nutrient = (pounds of nutrient/pounds of fertilizer) X 100
• pounds of nutrient = (% nutrient/100) X pounds of fertilizer
• pounds of fertilizer = pounds of nutrient/(% nutrient/100)

The molecular weight is:

N + H + O
2(14) + 4(1) + 3(16) = 80

Therefore, in 80 g of NH₄NO₃, one has 28 g of N

The percentage N is 28/80 (100) = 35%, or the analysis is (35-0-0) (most fertilizer grades are 33-34% N due to some impurity)

2. We must first know the percentage N in the fertilizer. If we are not given this, we must calculate it as above. For this problem, we will assume the fertilizer is 35-0-0.

120 lbs X 0.35 = 42 lbs N

3.  Unless otherwise given, we must assume the fertilizer is labeled on the oxide basis (standard). Therefore, this fertilizer is N - P₂O₅ - K₂O, and the numbers are percentages.

160 lbs fert. X 0.06 = 9.6 lbs N

160 lbs fert. X 0.10 = 16.0 lbs P₂O₅

160 lbs fert. X 0.08 = 12.8 lbs K₂O

4. This is a two-step problem. We must first determine (or know) the analysis of muriate of potash. From lecture, we learned it is a 0-0-60 fertilizer (oxide basis). Next we must calcuate the quantity of this fertilizer to give us 80 lbs of K₂O.

80 lbs of K₂O/0.60 = 133 lbs of fertilizer

5. This question is similar to problem #1. We must either know the conversion factor or (better yet) be able to calculate it. We are really asking the percentage of K in K₂O. at wts: K=39, O=16

The molecular weight is:

2(39) + 16 = 94

Therefore, in 94 g of K₂O, we have 78 g of K, or 0.83 (K₂0 X 0.83 = K)

40% K₂0 X 0.83 = 33.2% K

6. Here we are converting from the elemental to the oxide basis, but the principles are the same. at wts: P=32, 0=16

The molecular weight is:

2(31) + 5(16) = 142

Therefore, in 142 g of P₂O₅, we have 62 g of P. Or another way of stating: in 62 g of P (elemental), we have 142 g (oxide). This ratio of converting P to P₂O₅ is then 142/62 = 2.29

20% P X 2.29 = 45.8% P₂O₅

7. In a 50-lb bag of urea, we have 22.5 lbs of N (50 lbs fert. X 0.45 = 22.5)

Therefore, the cost per lb is:

$8.00/22.5 = $0.355/lb or 35.5 cents/lb

8. If P₂O₅ is worth 38¢ per lb and a ton of pure K₂HPO₄ sells for $484, how much are you paying per lb of K₂O? Hint: must determine the pounds of P₂O₅ and K₂O (convert both to oxide basis) per 2000 pounds of K₂HPO₄.

This problem requires several logical steps. First, we must determine the percentage of K and P in K₂HPO₄. Second, we must convert the K to K₂O and the P to P₂O₅. Third, we must determine how many pounds of K₂O and P₂O₅ are present in one ton of K₂HPO₄. Fourth, we must determine the value of the P₂O₅. Fifth, we must subtract the value of the P₂O₅ from $484. Lastly, we must divide the difference by the pounds of K₂O present to determine costs per lb of K₂O.

First: the molecular weight of K₂HPO₄ is 174. 78/174 = 0.45 K; 31/174 = 0.178 P

Second: 0.45 X 1.2 = 0.538 or 53.8% K₂O; 0.178 X 2.29 = 0.408 or 40.8% P₂O₅

Third: 2000 lb K₂HPO₄ X 0.538 = 1076 lb K₂O; 2000 lb K₂HPO₄ X 0.408 = 816 lb P₂O₅

Fourth: 816 lb P₂O₅ X 0.38 = $310.08

Fifth: $484.00 - $310.08 = $173.92 (this is the value of 1076 lb of K₂O)

Last: $173.92/1076 = 0.162 or 16.2 cents per lb K₂O

9. Assume that you wish to make a bulk blend and add 300 lbs of 0-46-0 and 200 lbs of 0-0-60. What is the analysis of this blend (assume everything is on the oxide basis)?

This is a relatively simple and very practical problem (most cooperatives do this many times in the fall and spring in loading trucks to send to the field). Remember that 0-46-0 is an analysis; i.e., the fertilizer contains 46% P₂O₅. If you dilute this with a material that contains no P₂O₅, the analysis will decrease. In 300 lbs of 0-46-0, we have 138 lbs P₂O₅. In 200 lbs of 0-0-60, we have 120 lbs K₂O. When we mix the two materials together to give us 500 lbs total weight, we have 138 lbs P₂O₅ and 120 lbs K₂O. The decimal equivalent of P₂O₅ is 138/500 = 0.276 or 27.6%.

Thus, the final analysis is 0-138/500(100)-120/500(100), or 0-27.6-24.

10. A soil test calls for 80 lbs P₂O₅ and 95 lbs K₂O per acre. You have 20 acres that need fertilized. If the source of P₂O₅ is 0-46-0 and the source of K₂O is 0-0-60, how many pounds of each fertilizer must you blend together to fill the truck going to the field? What will be the final analysis of the blended fertilizer?

P₂O₅: 80 lbs P₂O₅ per acre X 20 acres = 1600 lbs P₂O₅ needed. This requires 1600/0.46 = 3478 lb of a 0-46-0 fertilizer

K₂O: 95 lbs K₂O per acre X 20 acres = 1900 lbs K₂O needed. This requires 1900/0.60 = 3167 lb of a 0-0-60 fertilizer

The loaded truck will contian 6645 lbs fertilizer, which contains 1600 lbs P₂O₅ and 1900 lbs K₂O

The analysis is 0-1600/6645(100)-1900/6645(100), or 0-24.1-28.6