Amulti-nutrient fertilizer contains several different nitrogen containing compounds. the fertilizer is 54.8% ch4n2o (urea), 26.3% kno3 , and 14.1% (nh4)2hpo4 by mass. the remainder of the fertilizer consists of substances that do not contain nitrogen. how much fertilizer should someone apply to provide 2.90 g n to a plant?

9.01 g of fertilizer

Explanation:

Assume initially that the amount in grams of the multinutrient fertilizer is 100 g, and calculate the amount of nitrogen supplied by every compound present in these 100 g of fertilizer.

1) Urea (CH₄N₂O):

Molar mass of urea: 60.06 g/molAtomic mass of N: 14.007 g/molTotal mass of N in the formula: 2 × 14.007 g/mol = 28.014 g/mol Amount of N in 100 g of compound: 100 g × 54.8% × 28.014 g / 60.06 g =  25.56 g

2) KNO₃

Molar mass of KNO₃: 101.1032 g/molAtomic mass of N: 14.007 g/ molTotal mass of N in the formula: 14.007 g/molAmount of N in 100 of the compound: 100 g × 26.3% × 14.007 / 101.1032 = 3.64 g

3) (NH₄)₂PO₄

Molar mass of (NH₄)₂PO₄: 132.06 g/molAtomic mass of N: 14.007 g/ molTotal mass of N in the formula: 2×14.007 g/mol = 28.014 g/molAmount of N in 100 of the compound: 100 g × 14.1% × 14.007 / 132.06 = 2.99 g

4) Total mass of N in 100 g of fertilizer:

Add all the amounts of N obtained above

25.56g + 3.69 g + 2.99 g = 32.19 g of N

5) Mass of fertilizer that should be applied to provide 2.90 g of N to a plant.

Set a proportion:

32.19 g of N / 100 g of fertilizer = 2.90 g of N / X

Solve for X:

X = 2.90 g of N × 100 g of fertilizer / 32.19 g of N = 9.01 g of fertilizer

That is the 9.01 g of fertilizer should be applied to provide 2.90g  of N to a plant.

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