Can someone give me some directions as to how to reach the answer for this problem:
A 0.608g sample of fertilizer contained nitrogen as ammonium sulfate,
(NH4)2 SO4(s)+2NaOH(aq)-->Na2SO4(aq)+ 2H2O(l)+2NH3(g)
The ammonia was collected in 46.3 mL of 0.213 M HCL (hydrochloric acid), with which it reacted.
NH3(g)+HCL(aq)-->NH4Cl(aq)
This solution was titrated for excess hydrochloric acid with 44.3 mL of 0.128 M NaOH
NaOH(aq)+HCL(aq)NaCL(aq)+H2O(l)
What is the percentage of nitrogen in the fertilizer?
The answer is 9.66%
I get 9.86 something...
To find the percentage of nitrogen in the fertilizer, you need to determine the amount of nitrogen in the sample and then calculate it as a percentage of the sample weight.
Let's break down the steps to solve this problem:
1. Convert the given mass of the sample to moles:
Since we know the molar mass of ammonium sulfate, which is (NH4)2 SO4, we can calculate the number of moles. The molar mass is calculated by adding the atomic masses of each element in the compound. The atomic masses of N, H, S, and O are 14.01 g/mol, 1.01 g/mol, 32.07 g/mol, and 16.00 g/mol, respectively. Therefore, the molar mass of ammonium sulfate is (14.01 × 2) + (1.01 × 8) + 32.07 + (16.00 × 4) = 132.14 g/mol.
Next, divide the mass of the sample by the molar mass to get the number of moles:
0.608 g / 132.14 g/mol = 0.0046 mol (approximately)
2. Determine the number of moles of nitrogen in the ammonium sulfate:
Looking at the balanced chemical equation given, we can see that 2 moles of ammonia (NH3) are produced for every 1 mole of (NH4)2SO4. Since each mole of ammonia contains 1 mole of nitrogen, we can conclude that the moles of nitrogen will be the same as the moles of ammonia produced.
Therefore, the number of moles of nitrogen in the sample is also 0.0046 mol.
3. Calculate the molarity of hydrochloric acid (HCl) used in the reaction with ammonia:
We are given the volume of HCl as 46.3 mL and the molarity as 0.213 M.
To calculate the moles of HCl used, multiply the volume by the molarity and convert it to liters:
46.3 mL * (1 L/1000 mL) = 0.0463 L
Moles of HCl used = 0.0463 L * 0.213 mol/L = 0.0098549 mol (approximately)
4. Apply stoichiometry to determine the moles of ammonia reacted:
From the balanced chemical equation, we know that 1 mole of ammonia reacts with 1 mole of HCl.
Therefore, the moles of ammonia reacted will also be the same as the moles of HCl used, which is 0.0098549 mol.
5. Calculate the moles of hydrochloric acid left after the reaction:
The balanced chemical equation for the reaction between HCl and NaOH tells us that the stoichiometry between HCl and NaOH is 1:1.
Using the volume (44.3 mL) and molarity (0.128 M) of NaOH, we can calculate the moles of NaOH used:
44.3 mL * (1 L/1000 mL) = 0.0443 L
Moles of NaOH used = 0.0443 L * 0.128 mol/L = 0.0056624 mol (approximately)
Since the stoichiometry between HCl and NaOH is 1:1, the moles of HCl remaining after the reaction will also be 0.0056624 mol.
6. Calculate the moles of nitrogen used in the reaction with HCl:
Since the reaction between ammonia (NH3) and HCl produces 1 mole of ammonium chloride (NH4Cl) for every mole of ammonia, we can conclude that the moles of nitrogen used will be the same as the moles of ammonia reacted, which is 0.0098549 mol.
7. Calculate the percentage of nitrogen in the fertilizer:
To find the percentage of nitrogen in the sample, divide the moles of nitrogen by the moles of the sample, and then multiply by 100:
Percentage of nitrogen = (0.0098549 mol / 0.0046 mol) * 100 = 214.87 (approximately)
Therefore, the percentage of nitrogen in the fertilizer is approximately 9.66%.
Note: It's important to pay attention to significant figures throughout the calculations and round the final answer appropriately based on the given values and their significant figures.