Percentage composition and Molecular formula Questions in English

(C) According to the law of constant proportions,a chemical compound always contains the same elements combined together in the same proportion by mass,regardless of the source.
Given that $CaCO_3$ contains $40\%$ calcium by mass.
Therefore,in $4 \, g$ of $CaCO_3$,the mass of calcium is:
$\text{Mass of } Ca = \frac{40}{100} \times 4 \, g$
$\text{Mass of } Ca = 0.4 \times 4 \, g = 1.6 \, g$.

(C) Since the given sulphate is isomorphous with $ZnSO_4 \cdot 7H_2O$,its formula is $MSO_4 \cdot 7H_2O$.
Let $m$ be the atomic weight of $M$.
The molecular weight of $MSO_4 \cdot 7H_2O = m + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = m + 32 + 64 + 126 = m + 222$.
The percentage of $M$ is given by $\frac{m}{m + 222} \times 100 = 9.87$.
$100m = 9.87(m + 222)$.
$100m = 9.87m + 2191.14$.
$90.13m = 2191.14$.
$m = \frac{2191.14}{90.13} \approx 24.3$.

(A) $1$. Mass of $P = 1.24 \ g$. Mass of $S = 2.2 \ g - 1.24 \ g = 0.96 \ g$.
$2$. Moles of $P = \frac{1.24}{31} = 0.04 \ mol$.
$3$. Moles of $S = \frac{0.96}{32} = 0.03 \ mol$.
$4$. Ratio of $P:S = 0.04:0.03 = 4:3$.
$5$. The empirical formula is $P_4S_3$. Thus,the correct option is $A$.

(A) The molecular weight of caffeine is $194 \ g/mol$.
The mass of nitrogen in $1 \ mole$ of caffeine is calculated as: $\frac{28.9}{100} \times 194 = 56.06 \ g$.
The atomic mass of nitrogen is $14 \ g/mol$.
The number of nitrogen atoms per molecule is: $\frac{56.06 \ g/mol}{14 \ g/mol} \approx 4.004 \approx 4$.

(D) The molecular formula of ferrous fumarate is $(C_2H_2O_4)Fe$.
Atomic mass of $Fe = 56 \ g/mol$.
Molar mass of $(C_2H_2O_4)Fe = (2 \times 12) + (2 \times 1) + (4 \times 16) + 56 = 24 + 2 + 64 + 56 = 146 \ g/mol$.
Mass of $Fe$ in $100 \ mg$ of ferrous fumarate $= \frac{56}{146} \times 100 \ mg \approx 38.36 \ mg$.
Percentage of $Fe$ in the $400 \ mg$ capsule $= \frac{38.36 \ mg}{400 \ mg} \times 100 \approx 9.59 \%$.
Given the options provided,the closest value is $8 \%$.

(B) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as:
$M = (2 \times 23) + 32 + (4 \times 16) + 10 \times (2 \times 1 + 16) = 46 + 32 + 64 + 180 = 322\,g/mol$.
In one mole $(322\,g)$ of $Na_2SO_4 \cdot 10H_2O$,the total mass of oxygen is:
$O = (4 \times 16) + (10 \times 16) = 64 + 160 = 224\,g$.
Therefore,in $32.2\,g$ of $Na_2SO_4 \cdot 10H_2O$,the mass of oxygen is:
$\text{Mass of oxygen} = \frac{32.2\,g}{322\,g} \times 224\,g = 0.1 \times 224 = 22.4\,g$.

(A) The molar mass of $NaOH$ is calculated as: $23 + 16 + 1 = 40 \ g/mol$.
The mass of oxygen in one mole of $NaOH$ is $16 \ g$.
The percentage of oxygen is calculated as: $\frac{\text{mass of oxygen}}{\text{molar mass of } NaOH} \times 100$.
Percentage of oxygen $= \frac{16}{40} \times 100 = 40\%$.

(A) The chemical formula of urea is $NH_2CONH_2$.
The molar mass of urea is $(2 \times 14) + (4 \times 1) + 12 + 16 = 60 \ g/mol$.
There are $2$ nitrogen atoms in one molecule of urea,so the mass of nitrogen is $2 \times 14 = 28 \ g$.
The percentage of nitrogen is calculated as $\frac{\text{mass of nitrogen}}{\text{molar mass of urea}} \times 100 = \frac{28}{60} \times 100 = 46.66\%$.
Rounding to the nearest whole number,we get approximately $47\%$.

(D) To find the empirical formula,we calculate the mole ratio of the constituent elements:
$1.$ Moles of $C = \frac{24 \ g}{12 \ g/mol} = 2 \ mol$
$2.$ Moles of $H = \frac{4 \ g}{1 \ g/mol} = 4 \ mol$
$3.$ Moles of $O = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$
The ratio of $C:H:O$ is $2:4:2$,which simplifies to $1:2:1$.
Therefore,the empirical formula is $CH_2O$.

(A) The empirical formula is $CH_{2}O$. Let the molecular formula be $(CH_{2}O)_{n}$.
This means $1 \ mol$ of the compound contains $2n \ mol$ of hydrogen atoms.
The molar mass of hydrogen is $1 \ g/mol$,so $1 \ mol$ of the compound contains $2n \ g$ of hydrogen.
Given that $0.0835 \ mol$ of the compound contains $1.0 \ g$ of hydrogen.
Therefore,$1 \ mol$ of the compound contains $\frac{1.0}{0.0835} \approx 11.97 \approx 12 \ g$ of hydrogen.
Equating the two: $2n = 12$,which gives $n = 6$.
Thus,the molecular formula is $(CH_{2}O)_{6} = C_{6}H_{12}O_{6}$.

(B) The molecular formula is given by the expression: $(Molecular \ formula) = n \times (Empirical \ formula)$,where $n$ is a positive integer $(n = 1, 2, 3, ...)$.
For $n = 1$,the molecular formula is $1 \times CH_2O_2 = CH_2O_2$.
For $n = 2$,the molecular formula is $2 \times CH_2O_2 = C_2H_4O_4$.
Comparing the given options,$CH_2O_2$ (formic acid) is a valid molecular formula corresponding to $n = 1$.

(B) The molecular formula of Glucose is $C_6H_{12}O_6$. The ratio of $C:H:O$ is $6:12:6$,which simplifies to $1:2:1$.
The molecular formula of Acetic acid is $CH_3COOH$,which is $C_2H_4O_2$. The ratio of $C:H:O$ is $2:4:2$,which simplifies to $1:2:1$.
Since both compounds have the same ratio of $C$,$H$,and $O$,the correct option is $B$.

(C) The molar mass of $(NH_4)_2HPO_4$ is $2 \times (14 + 4) + 1 + 31 + 4 \times 16 = 36 + 1 + 31 + 64 = 132 \ g/mol$.
Since $2$ moles of $(NH_4)_2HPO_4$ contain $1$ mole of $P_2O_5$,the mass of $2$ moles of $(NH_4)_2HPO_4$ is $2 \times 132 = 264 \ g$.
The molar mass of $P_2O_5$ is $2 \times 31 + 5 \times 16 = 62 + 80 = 142 \ g/mol$.
Percentage of $P_2O_5 = \frac{\text{Mass of } P_2O_5}{\text{Mass of } 2 \text{ moles of } (NH_4)_2HPO_4} \times 100$.
Percentage of $P_2O_5 = \frac{142}{264} \times 100 = 53.78\%$.

(A) The minimum molecular weight of an enzyme is calculated assuming that at least one atom of the element $(Se)$ is present in one molecule of the enzyme.
Given that $0.5 \ g$ of $Se$ is present in $100 \ g$ of the enzyme.
Therefore,$78.4 \ g$ (atomic weight of $Se$) of $Se$ will be present in $\frac{100 \times 78.4}{0.5} \ g$ of the enzyme.
Calculation: $\frac{7840}{0.5} = 15680 = 1.568 \times 10^4 \ g/mol$.
Thus,the minimum molecular weight is $1.568 \times 10^4$.

(C) The molar mass of $Fe(CNS)_3 \cdot 3H_2O$ is calculated as follows:
$Fe = 56$,$C = 12$,$N = 14$,$S = 32$,$H = 1$,$O = 16$.
$Molar \ mass = 56 + 3 \times (12 + 14 + 32) + 3 \times (2 \times 1 + 16) = 56 + 3 \times 58 + 3 \times 18 = 56 + 174 + 54 = 284 \ g/mol$.
The mass of $3H_2O$ is $3 \times 18 = 54 \ g/mol$.
The percentage of $H_2O$ is calculated as:
$\frac{Mass \ of \ 3H_2O}{Total \ molar \ mass} \times 100 = \frac{54}{284} \times 100 \approx 19.01\%$.
Thus,the correct option is $C$.

(B) The atomic mass of sulphur is $32 \ g/mol$.
Given that the compound contains $8\%$ sulphur by mass,it means $8 \ g$ of sulphur is present in $100 \ g$ of the compound.
To find the least molecular mass,we assume there is at least one atom of sulphur in the molecule.
Therefore,$1 \ g$ of sulphur is present in $\frac{100}{8} \ g$ of the compound.
For $32 \ g$ of sulphur (one mole of atoms),the mass of the compound is $\frac{100}{8} \times 32 = 400 \ g/mol$.

(A) Mass of $C = \frac{12}{44} \times 2.63 = 0.717 \ g$.
Mass of $H = \frac{2}{18} \times 1.28 = 0.142 \ g$.
Total mass of $C + H = 0.717 + 0.142 = 0.859 \ g$,which is approximately equal to the mass of compound $X$ $(0.858 \ g)$. Thus,the compound contains only $C$ and $H$.
Moles of $C = \frac{0.717}{12} = 0.05975$.
Moles of $H = \frac{0.142}{1} = 0.142$.
Ratio $C:H = 0.05975 : 0.142 \approx 1 : 2.37$.
Converting to simple integer ratio: $1 : 2.37 \approx 3 : 7$.
The empirical formula is $C_3H_7$.
The empirical formula mass is $(3 \times 12) + (7 \times 1) = 43 \ g/mol$.
The lowest possible molecular mass is $43 \ g/mol$.

(B) The molecular formula of the chloride is $MCl_3$.
Since the valency of $Cl$ is $-1$,the valency of metal $M$ must be $+3$.
The carbonate ion is $CO_3^{2-}$,which has a valency of $-2$.
To form a neutral compound,the charges must be balanced: $2 \times (+3) + 3 \times (-2) = 0$.
Therefore,the formula of the carbonate is $M_2(CO_3)_3$.

(D) Step $1$: Calculate the number of moles of each element.
Number of moles of $I = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{254}{127} = 2 \ \text{mol}$.
Number of moles of $O = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{80}{16} = 5 \ \text{mol}$.
Step $2$: Determine the empirical formula.
The ratio of moles of $I : O$ is $2 : 5$.
Therefore,the formula of the compound is $I_2O_5$.

(C) The molar mass of a gas is given by the relation: $M = 2 \times \text{Vapour Density}$.
Given,$\text{Vapour Density} = 70$.
Therefore,$M = 2 \times 70 = 140 \ g/mol$.
The molar mass of the empirical unit $[CO]$ is $12 + 16 = 28 \ g/mol$.
Since the formula is $[CO]_x$,we have $x \times 28 = 140$.
Thus,$x = \frac{140}{28} = 5$.

(D) The chemical formula for calcium chlorite is derived from the calcium ion $(Ca^{2+})$ and the chlorite ion $(ClO_2^-)$.
To balance the charges,two chlorite ions are required for every one calcium ion,resulting in the formula $Ca(ClO_2)_2$.
Calcium chlorite is a white granular solid.

(A) Feldspar is a common mineral group. The molecular formula of potassium feldspar (orthoclase) is $K_2O \cdot Al_2O_3 \cdot 6SiO_2$.

(A) To find the percentage of nitrogen,we calculate the mass percentage for each compound:
$1$. Urea $(NH_2CONH_2)$: Molar mass = $60 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/60) \times 100 = 46.66 \%$.
$2$. Ammonium sulphate $(NH_4)_2SO_4$: Molar mass = $132 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/132) \times 100 = 21.21 \%$.
$3$. Ammonium nitrate $(NH_4NO_3)$: Molar mass = $80 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/80) \times 100 = 35 \%$.
$4$. Calcium nitrate $Ca(NO_3)_2$: Molar mass = $164 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/164) \times 100 = 17.07 \%$.
Thus,Urea has the highest percentage of nitrogen.

(D) The molar mass of urea $(NH_2CONH_2)$ is $60 \ g/mol$. It contains $2$ nitrogen atoms,so the percentage of nitrogen is $(28/60) \times 100 \approx 46.67\%$.
The molar mass of ammonium sulphate $(NH_4)_2SO_4$ is $132 \ g/mol$. It contains $2$ nitrogen atoms,so the percentage of nitrogen is $(28/132) \times 100 \approx 21.21\%$.
Comparing the two,$46.67\% / 21.21\% \approx 2.2$.
Therefore,urea contains more than twice the amount of nitrogen present in ammonium sulphate.

(D) The most common and stable form of the element sulfur is the puckered ring structure,which consists of eight atoms of sulfur.
Therefore,its molecular formula is $S_8$.

(C) To find the percentage of nitrogen,we use the formula: $\text{Percentage of } N = \frac{\text{Total mass of } N}{\text{Molar mass of compound}} \times 100$.
$1$. Urea $(NH_2CONH_2)$: Molar mass $= 60 \ g/mol$. Mass of $N = 28 \ g$. $\% N = (28/60) \times 100 \approx 46.67\%$.
$2$. Ammonium nitrite $(NH_4NO_2)$: Molar mass $= 64 \ g/mol$. Mass of $N = 28 \ g$. $\% N = (28/64) \times 100 = 43.75\%$.
$3$. Ammonium nitrate $(NH_4NO_3)$: Molar mass $= 80 \ g/mol$. Mass of $N = 28 \ g$. $\% N = (28/80) \times 100 = 35.00\%$.
$4$. Ammonium chloride $(NH_4Cl)$: Molar mass $= 53.5 \ g/mol$. Mass of $N = 14 \ g$. $\% N = (14/53.5) \times 100 \approx 26.17\%$.
Thus,the correct decreasing order is $Urea > NH_4NO_2 > NH_4NO_3 > NH_4Cl$.

(C) To find the simplest formula,we calculate the mole ratio of the elements:
Moles of $X = \frac{50}{10} = 5 \text{ mol}$
Moles of $Y = \frac{50}{20} = 2.5 \text{ mol}$
Now,divide by the smallest number of moles to get the simplest ratio:
Ratio of $X : Y = \frac{5}{2.5} : \frac{2.5}{2.5} = 2 : 1$
Therefore,the simplest formula is $X_2Y$.

(B) The $Empirical \ formula$ is defined as the chemical formula that shows the simplest whole-number ratio of atoms of each element present in a compound.
Steps to determine it:
$1$. Obtain the mass of each element in grams.
$2$. Determine the number of moles of each type of atom.
$3$. Divide the number of moles of each element by the smallest number of moles obtained.
$4$. Convert these values to the simplest whole-number ratio.

(A) The chemical formula which represents the actual number of atoms of each element present in a molecule is known as molecular formula.

(D) To find the empirical formula,we calculate the molar ratio of the elements:

Element	Moles (Percentage / Atomic Mass)	Simplest Molar Ratio
$C$	$90 / 12 = 7.5$	$7.5 / 7.5 = 1$
$H$	$10 / 1 = 10$	$10 / 7.5 = 1.33$

To convert the ratio $1 : 1.33$ into whole numbers,multiply by $3$:
$C = 1 \times 3 = 3$
$H = 1.33 \times 3 = 4$
Therefore,the empirical formula is $C_3H_4$.

(A) $1$. Calculate the percentage of oxygen: $\% O = 100 - (36 + 6) = 58\%$.
$2$. Determine the number of moles for each element (assuming $100 \ g$ of compound):
$C = 36/12 = 3.0$
$H = 6/1 = 6.0$
$O = 58/16 = 3.625$
$3$. Determine the simplest molar ratio by dividing by the smallest value $(3.0)$:
$C = 3.0/3.0 = 1$
$H = 6.0/3.0 = 2$
$O = 3.625/3.0 \approx 1.2$
$4$. Since the ratio for oxygen is not an integer,we multiply by $5$ to get whole numbers:
$C = 1 \times 5 = 5$
$H = 2 \times 5 = 10$
$O = 1.2 \times 5 = 6$
Empirical formula = $C_5H_{10}O_6$.
Note: Given the options provided,there appears to be a discrepancy in the calculation or the provided options. Based on standard empirical formula calculation,the result is $C_5H_{10}O_6$. If we assume the oxygen percentage was intended to be $53.33\%$,the formula would be $CH_2O$.

(B) The empirical formula is $CH_2O$.
The empirical formula mass is calculated as: $12 + (2 \times 1) + 16 = 30 \ g/mol$.
The molecular mass is calculated using the relation: $\text{Molecular Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60 \ g/mol$.
Calculate the value of $n$: $n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{60}{30} = 2$.
The molecular formula is given by: $(Empirical \ Formula)_n = (CH_2O)_2 = C_2H_4O_2$.

(A) Step $1$: Determine the empirical formula.

Element	Mass $(g)$	Moles	Simple Ratio
$C$	$48$	$48/12 = 4$	$1$
$H$	$8$	$8/1 = 8$	$2$
$N$	$56$	$56/14 = 4$	$1$

Empirical formula = $CH_2N$.
Empirical formula mass = $12 + 2(1) + 14 = 28 \,g/mol$.
Step $2$: Determine the molar mass.
At $NTP$, $22400 \,mL$ of gas corresponds to $1 \,mole$.
Given $200 \,mL$ of the compound weighs $1 \,g$.
Therefore, $22400 \,mL$ weighs $\frac{1}{200} \times 22400 = 112 \,g$.
Molar mass = $112 \,g/mol$.
Step $3$: Calculate the molecular formula.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{112}{28} = 4$.
Molecular formula = $n \times (CH_2N) = 4 \times (CH_2N) = C_4H_8N_4$.

(D) The minimum molecular weight of a compound is calculated by assuming that it contains at least one atom of the element in question.
The atomic weight of sulphur $(S)$ is $32 \ u$.
Given that insulin contains $3.4\%$ sulphur by mass,this means $3.4 \ g$ of sulphur is present in $100 \ g$ of insulin.
To find the minimum molecular weight,we assume $32 \ g$ of sulphur is present in the molecule:
$\text{Minimum molecular weight} = \frac{100 \times 32}{3.4} \approx 941.17 \ g/mol$.
Rounding to the nearest provided option,the value is $940$.

(B) The combustion reaction of a hydrocarbon $C_xH_y$ is: $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$.
Mass of $H_2O$ produced = $1.8 \, g$.
Moles of $H_2O$ = $\frac{1.8 \, g}{18 \, g/mol} = 0.1 \, mol$.
Moles of $H$ atoms = $2 \times \text{moles of } H_2O = 2 \times 0.1 = 0.2 \, mol$.
Mass of $H$ = $0.2 \, mol \times 1 \, g/mol = 0.2 \, g$.
Mass of $C$ in hydrocarbon = $\text{Total mass} - \text{Mass of } H = 1.4 \, g - 0.2 \, g = 1.2 \, g$.
Moles of $C$ = $\frac{1.2 \, g}{12 \, g/mol} = 0.1 \, mol$.
Ratio of $C:H$ = $0.1 : 0.2 = 1 : 2$.
Therefore,the empirical formula is $CH_2$.

(A) Step $1$: Calculate the percentage of Carbon $(C)$:
$\% \text{ of } C = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of compound}} \times 100$
$\% \text{ of } C = \frac{12}{44} \times \frac{0.22}{0.24} \times 100 = 25 \%$
Step $2$: Given percentage of Hydrogen $(H)$ = $1.66 \%$.
Step $3$: Calculate the percentage of Oxygen $(O)$:
$\% \text{ of } O = 100 - (\% \text{ of } C + \% \text{ of } H)$
$\% \text{ of } O = 100 - (25 + 1.66) = 100 - 26.66 = 73.34 \%$.
Since $73.34 \%$ is not explicitly listed in the options,the closest logical calculation based on the provided data is $25 \%$ Carbon and $73.34 \%$ Oxygen. Given the options,there might be a typo in the question's provided options,but $25 \%$ is the correct value for Carbon.

(C) To find the empirical formula,we calculate the molar ratio of each element:

Element	Percentage	Atomic Mass	Moles (Percentage/Atomic Mass)	Simple Ratio
$C$	$74.0$	$12$	$74.0/12 = 6.16$	$6.16/1.23 = 5$
$H$	$8.65$	$1$	$8.65/1 = 8.65$	$8.65/1.23 = 7$
$N$	$17.3$	$14$	$17.3/14 = 1.23$	$1.23/1.23 = 1$

Based on the calculated ratios,the empirical formula is $C_5H_7N$.

(B) The empirical formula represents the simplest whole-number ratio of atoms in a compound.
If two compounds have the same empirical formula but different molecular formulae,their molecular weights must be different because the molecular formula is an integer multiple of the empirical formula $(Molecular \ Formula = n \times Empirical \ Formula)$.
Since the molecular weights are proportional to the molecular formulae,they will differ accordingly.

(C) The empirical formula mass of $C_2H_5O$ is calculated as: $(2 \times 12) + (5 \times 1) + 16 = 24 + 5 + 16 = 45 \ g/mol$.
The value of $n$ is calculated using the formula: $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{90}{45} = 2$.
The molecular formula is given by: $n \times (\text{Empirical formula}) = 2 \times (C_2H_5O) = C_4H_{10}O_2$.

(D) To find the empirical formula,we calculate the mole ratio of each element:
$1$. Moles of $C = \frac{24 \ g}{12 \ g/mol} = 2 \ mol$
$2$. Moles of $H = \frac{4 \ g}{1 \ g/mol} = 4 \ mol$
$3$. Moles of $O = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$
Divide each by the smallest number of moles $(2)$:
$C = \frac{2}{2} = 1$
$H = \frac{4}{2} = 2$
$O = \frac{2}{2} = 1$
Thus,the empirical formula is $CH_2O$.

(A) To find the empirical formula,we calculate the molar ratio of each element:
$1$. For $C$: $\frac{38.8}{12} = 3.23$
$2$. For $H$: $\frac{16}{1} = 16$
$3$. For $N$: $\frac{45.2}{14} = 3.23$
Dividing by the smallest value $(3.23)$:
$C: \frac{3.23}{3.23} = 1$
$H: \frac{16}{3.23} \approx 5$
$N: \frac{3.23}{3.23} = 1$
The empirical formula is $CH_5N$,which can be written as $CH_3NH_2$.

(A) To find the empirical formula,we calculate the mole ratio of each element:

Element	Percentage / Atomic Mass	Moles	Simple Ratio
$C$	$54.5 / 12$	$4.54$	$4.54 / 2.27 = 2$
$H$	$9.1 / 1$	$9.1$	$9.1 / 2.27 = 4$
$O$	$36.4 / 16$	$2.27$	$2.27 / 2.27 = 1$

The ratio of $C:H:O$ is $2:4:1$. Therefore,the empirical formula is $C_2H_4O$.

(A)

Element	Calculation
$C$	$92.31 / 12 = 7.69$ moles; Ratio $= 7.69 / 7.69 = 1$
$H$	$7.69 / 1 = 7.69$ moles; Ratio $= 7.69 / 7.69 = 1$

The empirical formula is $CH$.
Empirical formula mass $= 12 + 1 = 13$.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{78}{13} = 6$.
Molecular formula $= (CH)_6 = C_6H_6$.

(C) To find the empirical formula,we calculate the number of moles of each element by dividing the percentage by its atomic mass:
$C: 53.3 / 12 = 4.44$
$H: 15.6 / 1 = 15.6$
$N: 31.1 / 14 = 2.22$
Dividing by the smallest value $(2.22)$:
$C: 4.44 / 2.22 = 2$
$H: 15.6 / 2.22 \approx 7$
$N: 2.22 / 2.22 = 1$
The empirical formula is $C_2H_7N$.
The empirical formula mass is $(2 \times 12) + (7 \times 1) + 14 = 24 + 7 + 14 = 45$.
Since the empirical formula mass equals the given molecular weight $(45)$,the molecular formula is the same as the empirical formula,which is $C_2H_7N$.

(C) To determine the empirical formula,we calculate the mole ratio of the elements present in the compound:

$Element$	$Moles \ (\% / \text{Atomic Mass})$
$C$	$80 / 12 = 6.67$
$H$	$20 / 1 = 20$

Dividing by the smallest value $(6.67)$:
$C = 6.67 / 6.67 = 1$
$H = 20 / 6.67 \approx 3$
The empirical formula is $CH_3$. The molecular formula must be a multiple of $CH_3$,such as $C_2H_6$ (Ethane). Therefore,the correct option is $C$.

(C) The empirical formula is determined by the ratio of moles of each element:

Element	Mole Ratio
$C = 50/12 = 4.16$	$4.16/3.125 \approx 1.33$
$O = 50/16 = 3.125$	$3.125/3.125 = 1$

Multiplying by $3$ to get whole numbers: $C = 4, O = 3$.
Empirical formula = $C_4O_3$.
Empirical formula mass = $(4 \times 12) + (3 \times 16) = 48 + 48 = 96$.
$n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{290}{96} \approx 3$.
Molecular formula = $(C_4O_3)_3 = C_{12}O_9$.

(C) To find the empirical formula,we calculate the mole ratio of the elements present:

Element	Moles $(\text{mass}/\text{at. wt.})$	Simple Ratio
$C = 83.70\%$	$83.70 / 12 = 6.975$	$6.975 / 6.975 = 1$
$H = 16.30\%$	$16.30 / 1 = 16.30$	$16.30 / 6.975 \approx 2.33$

To convert the ratio to whole numbers,multiply by $3$:
$C: 1 \times 3 = 3$
$H: 2.33 \times 3 \approx 7$
Thus,the empirical formula is $C_3H_7$.