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.
The empirical formula of an oxide of nitrogen containing 30.4 is unknown given that n=14,O=16 show full workings
4 answers
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.
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.
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
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
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