Respuesta :
Answer:
27.06%
Explanation:
First let us calculate the molar mass of NaNO3. This is illustrated below:
MM of NaNO3 = 23 + 14 + (16x3) = 23 + 14 + 48 = 85g/mol
Now, we see that NaNO3 contains 1atom of Na.
Therefore the percentage composition of Na in NaNO3 is given by:
(23/85) x 100 = 27.06%
The percent composition in sodium nitrate has been 27.05% Na, 16.47% N, and 56.47% Oxygen.
Sodium nitrate has is the molecular compound that has been consisted of sodium, nitrogen, and oxygen.
1 mole of sodium nitrate has consisted of 1 sodium atom, 1 nitrogen, and 3 oxygen atoms.
The molecular mass of sodium nitrate has been:
[tex]\rm NaNO_3[/tex] = Mass of Na + Mass of N + Mass of O.
Molecular mass of [tex]\rm NaNO_3[/tex] = 23 + 14 + (16 × 3) g/mol
Molecular mass of [tex]\rm NaNO_3[/tex] = 85 g/mol.
The percent composition has been :
Percent composition = [tex]\rm \dfrac{Mass}{Total\;mass}\;\times\;100[/tex]
- Percent composition of sodium = [tex]\rm \dfrac{23}{85}\;\times\;100[/tex]
Percent composition of sodium = 27.05 %
- Percent composition of nitrogen = [tex]\rm \dfrac{14}{85}\;\times\;100[/tex]
Percent composition of nitrogen = 16.47 %.
- Percent composition of Oxygen = [tex]\rm \dfrac{48}{85}\;\times\;100[/tex]
Percent composition of Oxygen = 56.47%.
The percent composition in sodium nitrate has been 27.05% Na, 16.47% N, and 56.47% Oxygen.
For more information about percent composition, refer to the link:
https://brainly.com/question/12247957
