Q3.140CP - Chemistry: Molecular Nature Of Matter And Change

Question

Nitrogen (N), phosphorous(P), and potassium (K) are main nutrients in plant fertilizers. By industry convention, the number on the label refer to the mass percentage of N, P₂O₅ and K₂O, in that order. Calculate the N/P/K ratio of a 30/10/10 fertilizer in terms of moles of each element, and express it as x/y/1.0

Step-by-Step Solution

Answer

The molar ratio of N/P/Kin terms of each element is $10/0.66/1.0$

Step 1: Calculation of moles of nitrogen in fertilizer

Let us calculate the moles nitrogen present in the fertilizer

Given:

$\text{Molar mass of}$ ${\text{N}}_2$=$28.02\text{g}$

$28.02{\text{g of N}}_2\text{ contains = 1 mole}$

$\text{Mass percentage of}$ ${\text{N}}_2\text{=30\% }$

${\text{30\% of N}}_{\text{2}}{\text{means, 0.30 grams of N}}_{\text{2}}$

Therefore,

${\text{0.30g of N}}_{\text{2}}\text{ contains = }\frac{1\times 0.30}{{\displaystyle \text{28.02}}}$

$=0.0107\text{moles}$

Step 2: Calculation of moles of phosphorous in fertilizer

Let us calculate the moles phosphorous present in the fertilizer

Given:

${\text{Molar mass of P}}_2{\text{O}}_5=141.94\text{g/mol}$

${\text{Mass percentage of P}}_2{\text{O}}_5\text{ = 10\% }$

${\text{10\% of P}}_{\text{2}}{\text{O}}_{\text{5}}{\text{ means, 0.10 grams of P}}_{\text{2}}{\text{O}}_{\text{5}}$

$141.94{\text{g of P}}_2{\text{O}}_5\text{ contains = 1 mole }$

$$\begin{gathered} 0.10{\text{g of P}}_2{\text{O}}_5\text{ contains = }\frac{1\text{mole}\times 0.10\text{g}}{141.94\text{g}} \\ =0.0007\text{moles} \end{gathered}$$

Step 3: Calculation of moles of potassium in fertilizer

Let us calculate the moles potassium present in fertilizer

${\text{Molar mass of K}}_2\text{O}=94.2\text{g/mol}$

$94.2{\text{g of K}}_2\text{O contains = 1 mole }$$94.2{\text{g of K}}_2\text{O contains = 1 mole }$

$$\begin{gathered} \text{10}\%{\text{ of K}}_2{\text{O means, 0.10 grams of K}}_2\text{O} \\ 0.10{\text{g of K}}_2\text{O contains = }\frac{1\text{moles}\times 0.10\text{g}}{94.2\text{g}} \\ =0.0011\text{moles} \end{gathered}$$

Step4: Calculation of Molar ratio of N/P/K

Molar ratio of N/P/K = $\text{0.0107/0.0007/0.0011}$

Simple ratio: $$\begin{gathered} \frac{\text{0.0107}}{\text{0.0011}}:\frac{\text{0.0007}}{\text{0.0011}}:\frac{\text{0.0011}}{\text{0.0011}} \\ =10/0.66/1.0 \end{gathered}$$

Result:

The molar ratio of N/P/K in a fertilizer $=10/0.66/1.0$