The empirical formula of an oxide of nitrogen containing 30.4 is unknown given that n=14,O=16 show full workings
If Nitrogen(N)=30.4%
Then Oxygen(O)=100 - 30.4%=69.6%
Divide both by their relative atomic mass
Nitrogen(N)= 30.4 ÷ 14 = 2.17
Oxygen(O)= 69.6 ÷ 16 = 4.35
Divide both by the smallest number= 2.17
Nitrogen(N)= 2.17 ÷ 2.17 = 1
Oxygen(O)= 4.35 ÷ 2.17 = 2.005 = 2
Empirical Formula = N1O2 = NO2
Assume this is 30.4% N which makes it 69.6% O. Take 100 g sample to give you
30.4 g N and 69.6 g O.
mols N = g/atomic mass N = 30.4/14 = 2.17
mols O = 69.6/16 = 4.35
Now find the ratio of N to O with the smallest number being so smaller than 1 and round to whole number.. Like this. The easy way to do this is to divide the smaller of the two numbers by itself. Then divide the other number by the same small number.
N = 2.17/2.17 = 1.00
O = 4.35/2.17 2.005 which rounds to 2.00 so you have N1O2 or NO2 as the empirical formula with the assumption made. You should make a habit of proofing your post.
Best
To determine the empirical formula of an oxide of nitrogen containing 30.4g, we would need to use the molar masses of nitrogen and oxygen. The molar mass of nitrogen (N) is approximately 14 g/mol, and the molar mass of oxygen (O) is approximately 16 g/mol.
To find the empirical formula, follow these steps:
Step 1: Calculate the number of moles of nitrogen and oxygen in the given mass.
To calculate the moles, we divide the given mass by the molar mass.
Moles of nitrogen (N):
Moles = Mass / Molar mass
Moles = 30.4 g / 14 g/mol
Moles of oxygen (O):
Moles = Mass / Molar mass
Moles = 30.4 g / 16 g/mol
Step 2: Find the ratio of moles.
To obtain the empirical formula, we need to find the simplest whole-number ratio between the moles of nitrogen and oxygen.
In this case, the moles of nitrogen and oxygen are not given explicitly. However, we can still find the ratio by dividing both by the smallest value.
Dividing the moles of nitrogen and oxygen by 2 (the lowest common denominator):
N moles = (30.4 g / 14 g/mol) / 2
O moles = (30.4 g / 16 g/mol) / 2
Simplifying this:
N moles = 1.0857 mol
O moles = 1.9 mol
Step 3: Determine the whole-number ratio.
Since the moles cannot have fractions in an empirical formula, we need to convert the decimals of each to the nearest whole number.
N ratio = 1 (rounded)
O ratio = 2 (rounded)
Thus, the empirical formula of the oxide of nitrogen is N1O2, which can be simplified as NO2.
Keep in mind that this is only the empirical formula, which provides the simplest ratio of atoms in a compound. The actual molecular formula may be a multiple of the empirical formula.
30.4 what?
N is indeed abut 14 g/mol and O is about 16 g/mol
What is a "working"?
You are in a science class. Speak science.