Question

A 0.568-g sample of fertilizer contained nitrogen as ammonium sulfate, . It was analyzed for nitrogen by heating with sodium hydroxide. The ammonia was collected in 45.2 mL of 0.192 M (hydrochloric acid), with which it reacted. This solution was titrated for excess hydrochloric acid with 42.4 mL of 0.133 M . What is the percentage of nitrogen in the fertilizer

Answer 1

7.49% of nitrogen

Step-by-step explanation:

The reaction of ammonium sulfate ((NH₄)₂SO₄) with NaOH is:

(NH₄)₂SO₄ + NaOH → Na₂SO₄ + 2 H₂O + 2 NH₃

Then, the ammonia is collected in HCl, thus:

NH₃ + HCl → NH₄Cl

And the excess of HCl is neutralized with NaOH, thus:

HCl + NaOH → H₂O + NaCl

That means moles of NH₃ are the difference between initial moles of HCl and the excess moles of HCl.

Initial moles of HCl are:

0.0452L × (0.192mol / L) = 8.68x10⁻³ moles of HCl.

Excess moles are:

0.0424L × (0.133mol / L) = 5.64x10⁻³ moles of HCl.

The difference is:

8.68x10⁻³ moles of HCl - 5.64x10⁻³ moles of HCl =

3.04x10⁻³ moles of HCl ≡ moles of NH₃ ≡ moles of nitrogen.

As nitrogen weights 14.0g/mol:

3.04x10⁻³ moles N × (14.0g / mol) = 0.04256g of nitrogen

Percentage of nitrogen is:

(0.04256g of nitrogen / 0.568g) × 100 = 7.49% of nitrogen

Answer 2

6.69%

Step-by-step explanation:

Given that:

Mass of the fertilizer = 0.568 g

The mass of HCl used in titration (45.2 mL of 0.192 M)

=
`0.192*(45.2)/(1000)* (36.5)/(1 \ mole)`

= 0.313 g HCl

The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)

=
`(44.3 \ mL * 1.0 \ L)/(1000 \ mL) *0.133 \ mole/L`

= 0.0058919 mole of NaOH

From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process

Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g

= 0.2151 g HCl

From above ; the total amount of HCl used = 0.313 g

The total amount that is used for complete neutralization = 0.2151 g

∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g

= 0.0979 g

We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g

Now; the amount of HCl neutralized by 0.0979 HCl =
`(17)/(36.5)*0.0979`

= 0.0456 g

Therefore, the mass of nitrogen present in the fertilizer is:

=
`0.0456 \ g \ NH_3 * (1 \ mol \ NH_3 )/(17.0 \ mol \ of \ NH_3) * (1 \mol \ (NH_4)_2SO_4)/(2 \ mol \ NH_3 ) * (2 \ mol \ N )/(1 \mol \ (NH_4)_2SO_4)* (14.0 g )/(1 \ mol \ N)`

= 0.038 g

∴ Mass percentage of Nitrogen in the fertilizer =
`(0.038 \ g)/(0.568 \ g) * 100`%

= 6.69%