Element	%	No. of moles	Molar ratio	Simplest ratio
$$C$$	40	$$\frac{{40}}{{12}} = 3.33$$	$$\frac{{3.33}}{{3.33}} = 1$$	1
$$H$$	13.33	$$\frac{{13.33}}{1} = 13.33$$	$$\frac{{13.33}}{{3.33}} = 4$$	4
$$N$$	46.67	$$\frac{{46.67}}{{14}} = 3.33$$	$$\frac{{3.33}}{{3.33}} = 1$$	1

Thus, the empirical formula is $$C{H_4}N.$$

$$\eqalign{ & {\text{Seconds in 3 days can be calculated as}} \cr & {\text{3 days}} \times \frac{{24\,h}}{{1\,{\text{day}}}} \times \frac{{60\,\min }}{{1\,h}} \times \frac{{60\,s}}{{1\,\min }} \cr & = 3 \times 24 \times 60 \times 60\,s \cr & = 259200\,s \cr} $$

$$\eqalign{ & {\text{Mass percentage of}}\,\,C\,\,{\text{in}}\,\,C{O_2} \cr & = \frac{{{\text{Mass}}\,\,{\text{of}}\,\,C}}{{{\text{Mass}}\,\,{\text{of}}\,\,C{O_2}}} \times 100 \cr & = \frac{{12}}{{44}} \times 100 \cr & = 27.27\% \cr} $$

Molecular formula of ethanol $$ = {C_2}{H_5}OH$$
Molar mass of ethanol $$ = 2 \times 12 + 6 \times 1 + 16 = 46\,g$$
Mass per cent of oxygen $$ = \frac{{16}}{{46}} \times 100 = 34.78\% $$

$$\mathop {CaC{O_3}}\limits_{100\,g} \to CaO + \mathop {C{O_2}}\limits_{22.4\,L} $$
$$100\,g\,\,{\text{of}}\,\,CaC{O_3}\,\,{\text{at}}\,\,STP\,\,{\text{gives}}$$       $$22.4\,L\,\,{\text{of}}\,\,C{O_2}$$
$$50\,g\,\,{\text{of}}\,\,CaC{O_3}\,\,{\text{will produce}}\,\,\frac{{22.4}}{{100}} \times 50$$        $$ = 11.2\,L\,\,{\text{of}}\,\,C{O_2}$$

TIPS/Formulae:
$$\eqalign{ & {\text{Apply the formula }}d = M\left( {\frac{1}{m} + \frac{{{M_2}}}{{1000}}} \right) \cr & \therefore 1.02 = 2.05\left( {\frac{1}{m} + \frac{{60}}{{1000}}} \right) \cr & {\text{On solving we get, m = }}2.288{\text{ }}mol/kg \cr} $$

Relative no. of moles of $$'X' = \frac{{0.25}}{{{N_A}}} = 0.25$$     and $$'Y' = \frac{{0.5}}{{{N_A}}} = 0.5$$    $$\left[ {{\text{where}},{N_A}\,\,{\text{atom}} = 1\,mole} \right]$$
Mole ratio of $$'X' = \frac{{0.25}}{{0.25}} = 1$$
Mole ratio of $$'Y' = \frac{{0.5}}{{0.25}} = 2$$
Empirical formula $$ = X{Y_2}$$

Student A :
Average reading $$ = \frac{{3.01 + 2.99}}{2} = 3.0\,g$$
Student B : Average reading $$ = \frac{{3.05 + 2.95}}{2} = 3.0\,g$$
For both the students A and B, average reading is close to the correct reading ( i.e., $$3.0\,g$$  ). Hence, both recorded accurate readings. But the readings recorded by student A are more precise as they differ only by $$ \pm 0.01,$$  whereas readings recorded by the student B are differ by $$ \pm 0.05.$$  Thus, the results of student A are both precise and accurate.

TIPS/Formulae :
(i) In an ion sum of oxidation states of all atoms is equal to charge on ion and in a compound sum of oxidation states of all atoms is always zero.
Oxidation state of $$Mn$$ in $$MnO_4^ - = + 7$$
Oxidation state of $$Cr$$ in $$Cr\left( {CN} \right)_6^{3 - } = + 3$$
Oxidation state of $$Ni$$ in $$NiF_6^{2 - } = + 4$$
Oxidation state of $$Cr$$ in $$Cr{O_2}C{l_2} = + 6$$

$$\eqalign{ & 3F{e_2}{O_3}\left( s \right) \to 2F{e_3}{O_4} + \frac{1}{2}{O_2} \cr & 480\,g\,F{e_2}{O_3}\,{\text{provide}}\,16\,g\,{O_2}.\,{\text{For loss of}} \cr & 0.04\,g\,{O_2} \to 0.04 \times \frac{{480}}{{16}} = 1.2\,g\,F{e_2}{O_3} \cr & \% \,{\text{by mass of}}\,Si{O_2} = \frac{{0.8}}{{2.0}} \times 100 = 40\% \cr} $$