Answer:- First choice is correct. Empirical formula is [tex]CH_4N[/tex] .
Solution:- Mass percentages are given for C, H and N for a compound and we are asked to find out it's empirical formula. First of all we calculate the moles of each element on dividing it's percentage by it's atomic mass;
C = [tex]39.97g(\frac{1mol}{12.01g})[/tex] = 3.33
H = [tex]13.41g(\frac{1mol}{1.01g})[/tex] = 13.3
N = [tex]46.62g(\frac{1mol}{14.01g})[/tex] = 3.33
Let's divide the moles of each by the least one of them which is 3.33.
So, C = [tex]\frac{3.33}{3.33}[/tex] = 1
H = [tex]\frac{13.3}{3.33}[/tex] = 4
N = [tex]\frac{3.33}{3.33}[/tex] = 1
Hence, the empirical formula of the compound is [tex]CH_4N[/tex] . First choice is correct.
