Respuesta :
Mass of nitrogen = 14.0067
Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 / 14.0067 = 2.63
And oxygen = 63.14 / 15.9994 = 3.95
now we have: N----- 2.63 and O----3.95
by dividing both with the smallest number we get
N-------2.63/2.63 = 1
O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
N= 1 x 2 = 2 And O = 1.5 x 2= 3
So, the empirical formula is N₂O₃.
Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 / 14.0067 = 2.63
And oxygen = 63.14 / 15.9994 = 3.95
now we have: N----- 2.63 and O----3.95
by dividing both with the smallest number we get
N-------2.63/2.63 = 1
O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
N= 1 x 2 = 2 And O = 1.5 x 2= 3
So, the empirical formula is N₂O₃.
The empirical formula for a compound that is 36.86% N and 63.14% O by mass is N2O3.
HOW TO CALCULATE EMPIRICAL FORMULA?
The empirical formula of a compound can be calculated as follows:
- N = 36.86% = 36.86g
- O = 63.14% = 63.14g
First, we divide each value by its molar mass as follows:
- N = 36.86g ÷ 14g/mol = 2.63mol
- O = 63.14g ÷ 16g/mol = 3.95mol
Next, we divide by the smallest mole value;
- N = 2.63mol = 2.63 = 1
- O = 3.95mol ÷ 2.63 = 1.5
We multiply each value by 2 to get an approximate whole number ratio i.e. N = 2, O = 3
Therefore, the empirical formula of for a compound that is 36.86% N and 63.14% O by mass is N2O3.
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