Stoichiometry will you help please

1)N & O form series of oxides with general formula NxOy.One of them contains 36.84% of N,what is empirical formula of this oxide?


2)A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine What is the
empirical formula of the indium compound?

Indium is a chemical element with chemical symbol In and atomic number 49.

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Well maybe not all that easy:
Do the calculations properly:

%N = 36.84
%O = 63.16

Divide by respective atomic mass:
N = 36.84*14 = 2.63
O = 63.16/16 = 3.95

Divide by smallest:
N = 1
O = 3.95/2.63 = 1.50
Multiply by 2 to remove fraction:
N = 2
O = 3
Formula = N2O3 dinitrogen trioxide.

Second question:
Mass In = 0.500-0.2404 = 0.2596
Mass of Cl = 0.2404

Divide by respective atomic mass
In = 0.2596/114.8 = 0.00226
Cl = 0.2404/35.5 = 0.006772
Divide by smallest:
In = 1
Cl = 0.006772/0.00226 = 3

Formula = InCl3

1)Its easy!!

If N is 36.84%,the O will be 100%-36.84%=63.26%
Divide N and O by their Ar. N=36.84/14=2.63 and O=63.26/16=3.95
Then divide the result with the simplest ratio. N=2.63/2.63=1 and O=3.95/2.63=1.50
Multiply by 2
N=2 and O=3
Empirical formula=N2O3

2)Its the same...
If Chlorine is 0.2404, Indium will be 0.5000-0.2404=0.2596
Divide Cl and In by their Ar. Cl=0.2404/35.5=0.0068 and In=0.2596/114.818=0.00226
Divide by smallest ratio..Cl=0.0068/0.00226=3 and In=0.00226/0.00226=1
Multiply the two number by 3 so as to remove the decimal.. Cl=3 and In=1
Empirical formula=InCl3