Percentage formula of oxygen in sodium trioxonitrate(v)
Please I need answer to the question
percent of what?
mass?
no. of atoms?
in any case, percent is (amount you are interested in) / total * 100%
Pleeeeeease
Na=23N=14, O=16
I am waiting someone please give me the answer
what, still no effort on your part? If you want the % of O atoms, then that is just
16/(23+14+16) = 16/53 = 0.30 = 30%
if you want % mass, then just multiply all the numbers by their respective atomic weights.
Na=23N=14, O=16
Thx
Note: %O = [(3*16)/(85)]*100
To calculate the percentage of oxygen in sodium trioxonitrate(V), we need to determine the molar mass of oxygen in the compound and divide it by the total molar mass of the compound, then multiply by 100 to get the percentage.
Here's how you can determine the percentage:
1. Determine the molar mass of oxygen (O):
- Look up the atomic mass of oxygen, which is approximately 16.00 grams per mole (g/mol).
2. Determine the molar mass of sodium trioxonitrate(V) (NaNO3):
- The molar mass of sodium (Na) is approximately 22.99 g/mol.
- The molar mass of nitrogen (N) is approximately 14.01 g/mol.
- The molar mass of three oxygen atoms (O) is calculated as 3 * 16.00 g/mol = 48.00 g/mol.
Therefore, the molar mass of sodium trioxonitrate(V) is 22.99 + 14.01 + 48.00 = 84.00 g/mol.
3. Calculate the percentage of oxygen:
- Divide the molar mass of oxygen (16.00 g/mol) by the molar mass of sodium trioxonitrate(V) (84.00 g/mol):
(16.00 g/mol / 84.00 g/mol) * 100 = 19.05%.
So, the percentage of oxygen in sodium trioxonitrate(V) is approximately 19.05%.