A multi-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/mol Atomic mass of N: 14.007 g/mol Total 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/mol Atomic mass of N: 14.007 g / mol Total mass of N in the formula: 14.007 g/mol Amount 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/mol Atomic mass of N: 14.007 g / mol Total mass of N in the formula: 2*14.007 g/mol = 28.014 g/mol Amount 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 answer: 9.01 g of fertilizer should be applied to provide 2.90g of N to a plant.