An ammonium halide A contains 9.664% nitrogen. Identify A.
The formula has to be NH4F, NH4Cl, NH4I or NH4Br. NH4At is unstable. Those are the only halides of the ammonium ion. Which of them is 9.66% nitrogen?
Do the numbers. For NH4F, 14/37 = 37.8% is N. It is one of the others, with a much heavier halogen atom
To identify the ammonium halide A, we need to determine the percentage of nitrogen in the compound and compare it to the available options.
Given that ammonium halide A contains 9.664% nitrogen, we can calculate the molar mass of nitrogen and the molar mass of the other elements in the compound to find the correct ammonium halide.
The molar mass of nitrogen (N) is approximately 14.007 g/mol.
Now let's consider the various options:
1. Ammonium chloride (NH4Cl):
The molar mass of NH4Cl:
(N: 14.007 g/mol) + (H: 4 x 1.008 g/mol) + (Cl: 35.453 g/mol) = 53.491 g/mol
The percentage of nitrogen in NH4Cl:
(N: 14.007 g/mol / 53.491 g/mol) x 100% ≈ 26.172%
Since the experimentally observed percentage of nitrogen is 9.664%, we can conclude that the compound is not ammonium chloride.
2. Ammonium bromide (NH4Br):
The molar mass of NH4Br:
(N: 14.007 g/mol) + (H: 4 x 1.008 g/mol) + (Br: 79.904 g/mol) = 97.045 g/mol
The percentage of nitrogen in NH4Br:
(N: 14.007 g/mol / 97.045 g/mol) x 100% ≈ 14.408%
Again, the experimentally observed percentage of nitrogen (9.664%) does not match that of ammonium bromide.
3. Ammonium iodide (NH4I):
The molar mass of NH4I:
(N: 14.007 g/mol) + (H: 4 x 1.008 g/mol) + (I: 126.904 g/mol) = 144.994 g/mol
The percentage of nitrogen in NH4I:
(N: 14.007 g/mol / 144.994 g/mol) x 100% ≈ 9.660%
The experimentally observed percentage of nitrogen (9.664%) is closest to the percentage found in ammonium iodide (NH4I). Therefore, we can identify compound A as ammonium iodide (NH4I).
To identify the ammonium halide A, we need to determine its chemical formula based on the given information that it contains 9.664% nitrogen.
To do this, we can calculate the molar mass of the compound and then find its empirical formula. The empirical formula represents the simplest whole-number ratio of atoms in the compound.
Step 1: Calculate the molar mass of nitrogen (N):
Nitrogen (N) has an atomic mass of 14.01 g/mol.
Step 2: Find the number of moles of nitrogen in the compound:
To find the number of moles, we divide the given mass percentage by the molar mass of nitrogen.
Mass of nitrogen = 9.664% of A
Assuming the total molar mass of A is 100 g/mol:
Mass of nitrogen = 9.664/100 * 100 = 9.664 g
Number of moles of nitrogen = mass/molar mass
Number of moles of nitrogen = 9.664 g/14.01 g/mol = 0.690 mol
Step 3: Determine the empirical formula:
The empirical formula shows the simplest whole-number ratio of atoms in the compound. To determine it, we need to find the ratio of the moles of nitrogen to the respective element(s) in the ammonium halide.
Ammonium ion (NH4+) contains one nitrogen (N) atom.
Hydrogen halides (HCl, HBr, HI) contain one hydrogen (H) atom.
As the ratio of moles of nitrogen to moles of (NH4+) and (H) should be in whole numbers, let's consider a common factor to get whole numbers. In this case, multiplying all numbers by 10 will yield whole numbers.
Number of moles of nitrogen (N) = 0.690 * 10 = 6.90
Number of moles of (NH4+) = 0.690 * 10 = 6.90
Number of moles of (H) = 0.690 * 1 = 0.690
Now, we can write the empirical formula:
(NH4)7H
So, the compound A is ammonium heptahydride (NH4)7H.