Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass. determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass. no

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 O3.95 
by dividing both with the smallest number we get
N2.63/2.63 = 1
O3.95/2.63 = 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% = 36.86gO = 63.14% = 63.14g

First, we divide each value by its molar mass as follows:

N = 36.86g ÷ 14g/mol = 2.63molO = 63.14g ÷ 16g/mol = 3.95mol

Next, we divide by the smallest mole value;

N = 2.63mol = 2.63 = 1O = 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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Crust,mantle,core,earth

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)