Question

What is the empirical formula of the compound having the following percentage composition.
N = 63.63% , O = 36.36%.

Answer 1

Answer:

N2O

Explanation:

first we convert percentage of elements in their grams respectively.

63.63%=63.63grams. 36.36%=36.36grams

then we calculate mole of each one.

$63.63 \times 1mol \: of \: nitrogen \div 14atmic \: mass = 4.54$

$36.36 \times 1mole \: of \: oxygen \div 16atomic \: mass = 2.27$

so, we get 4.54moles of nitrogen and 2.27moles of oxygen.

then we divided both values of moles by small ones.

$4.54 \div 2.27 = 2$

$2.27 \div 2.27 = 1$

now we get the exact numbers of atoms of a compound.

N2O1=N2O.