what is the empirical formula of a compound that has the percentage composition 63.1% oxygen and 36.8% nitrogen??
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Answered by
5
Find out the mass percent of the oxygen.
100-36.8 = 63.2
2. Because percentages are just ratios, you can pretend that the given percentages have g as their units.
N = 36.8 g O= 63.2 g
3. Convert this to moles
N: (36.8 g) (mol/ 14.01g) =2.63 mol
O: (63.8 g) (mol/ 16.00 g) =3.95 mol
4. Divide both amounts by the smallest number (2.63) in order to find the ratio.
N: 2.63/2.63= 1
O: 3.95/2.63 = about 1.5
5. So the ratio in the smallest whole numbers of nitrogen to oxygen is 2:3. So your formula is N2O3
100-36.8 = 63.2
2. Because percentages are just ratios, you can pretend that the given percentages have g as their units.
N = 36.8 g O= 63.2 g
3. Convert this to moles
N: (36.8 g) (mol/ 14.01g) =2.63 mol
O: (63.8 g) (mol/ 16.00 g) =3.95 mol
4. Divide both amounts by the smallest number (2.63) in order to find the ratio.
N: 2.63/2.63= 1
O: 3.95/2.63 = about 1.5
5. So the ratio in the smallest whole numbers of nitrogen to oxygen is 2:3. So your formula is N2O3
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how
Answered by
1
Answer:
n2o3
Explanation:
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