Percentage composition and Molecular formula Questions in English

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

Element	Percentage	Moles (\% / Atomic Mass)	Simple Ratio
$C$	$60\%$	$60 / 12 = 5$	$5 / 1.66 \approx 3$
$H$	$13.3\%$	$13.3 / 1 = 13.3$	$13.3 / 1.66 \approx 8$
$O$	$26.7\%$	$26.7 / 16 = 1.66$	$1.66 / 1.66 = 1$

Thus,the empirical formula is $C_3H_8O$.

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

Element	Percentage	Moles $(Percentage/Atomic \ Mass)$	Simple Ratio
$C$	$85.72\%$	$85.72 / 12 = 7.14$	$7.14 / 7.14 = 1$
$H$	$14.28\%$	$14.28 / 1 = 14.28$	$14.28 / 7.14 = 2$

The empirical formula is $CH_2$.
Among the given options,$C_2H_4$ (ethene) has the empirical formula $CH_2$ and satisfies the composition.

(C) $1$. Calculate the mass of oxygen: $64 \, g - (24 \, g + 8 \, g) = 32 \, g$ of oxygen.
$2$. Determine the number of moles for each element:

Element	Moles	Simple Ratio
$C$	$24/12 = 2$	$2/2 = 1$
$H$	$8/1 = 8$	$8/2 = 4$
$O$	$32/16 = 2$	$2/2 = 1$

$3$. The empirical formula is $CH_4O$.

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

Element	Mass Ratio	Moles (Mass/Atomic Mass)	Simple Ratio
$C$	$6$	$6/12 = 0.5$	$0.5/0.5 = 1$
$H$	$1$	$1/1 = 1$	$1/0.5 = 2$
$O$	$8$	$8/16 = 0.5$	$0.5/0.5 = 1$

Thus,the empirical formula is $CH_2O$.
Step $2$: Calculate the empirical formula mass.
Empirical formula mass = $(1 \times 12) + (2 \times 1) + (1 \times 16) = 30 \ g/mol$.
Step $3$: Calculate the molecular mass.
Molecular mass = $2 \times$ Vapour Density = $2 \times 30 = 60 \ g/mol$.
Step $4$: Calculate the molecular formula.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2$.
Molecular formula = $n \times (CH_2O) = 2 \times (CH_2O) = C_2H_4O_2$.

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

Element	Moles (Percentage / Atomic Mass)	Simple Molar Ratio
$C$	$80 / 12 = 6.67$	$6.67 / 6.67 = 1$
$H$	$20 / 1 = 20$	$20 / 6.67 \approx 3$

Since the ratio of $C:H$ is $1:3$,the empirical formula is $CH_3$.

(B) The percentage of oxygen is calculated as: $O = 100 - (40 + 6.7) = 53.3\%$.
The calculation for the empirical formula is as follows:

Element	Percentage	Atomic Mass	Moles	Simple Ratio
$C$	$40$	$12$	$40/12 = 3.33$	$3.33/3.33 = 1$
$H$	$6.7$	$1$	$6.7/1 = 6.7$	$6.7/3.33 = 2$
$O$	$53.3$	$16$	$53.3/16 = 3.33$	$3.33/3.33 = 1$

Thus,the empirical formula is $CH_2O$.

(B) The relationship between molecular formula and empirical formula is given by: $\text{Molecular Formula} = n \times \text{Empirical Formula}$.
Therefore,the value of $n$ is calculated as: $n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}}$.

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

$C = 40\% \rightarrow 40/12 = 3.33$	$3.33/3.33 = 1$
$H = 13.33\% \rightarrow 13.33/1 = 13.33$	$13.33/3.33 = 4$
$N = 46.67\% \rightarrow 46.67/14 = 3.33$	$3.33/3.33 = 1$

The ratio of $C:H:N$ is $1:4:1$.
Therefore,the empirical formula is $CH_4N$.

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

Element	Moles (Percentage / Atomic Mass)	Simple Ratio
$C$	$18.5 / 12 = 1.54$	$1.54 / 1.54 = 1$
$H$	$1.55 / 1 = 1.55$	$1.55 / 1.54 \approx 1$
$Cl$	$55.04 / 35.5 = 1.55$	$1.55 / 1.54 \approx 1$
$O$	$24.81 / 16 = 1.55$	$1.55 / 1.54 \approx 1$

The simple ratio of $C:H:Cl:O$ is $1:1:1:1$.
Therefore,the empirical formula is $CHClO$.

(A) The mass ratio of $C : H$ is $6 : 1$ and the mass ratio of $C : O$ is $3 : 4$. To compare them,we express $C : O$ in terms of the same $C$ mass $(6)$: $C : O = 6 : 8$.
Thus,the mass ratio $C : H : O = 6 : 1 : 8$.
The atomic masses are $C = 12 \ g/mol$,$H = 1 \ g/mol$,and $O = 16 \ g/mol$.
The number of moles for each element is:
Moles of $C = \frac{6}{12} = 0.5 \ mol$
Moles of $H = \frac{1}{1} = 1 \ mol$
Moles of $O = \frac{8}{16} = 0.5 \ mol$
The mole ratio $C : H : O = 0.5 : 1 : 0.5 = 1 : 2 : 1$.
The empirical formula is $CH_2O$,which corresponds to $HCHO$ (formaldehyde).

(B) Step $1$: Calculate the empirical formula.
$C: \frac{10.06}{12} = 0.838 \approx 1$
$H: \frac{0.84}{1} = 0.84 \approx 1$
$Cl: \frac{89.10}{35.5} = 2.51 \approx 3$
Empirical formula is $CHCl_3$.
Step $2$: Calculate the molecular weight.
$\text{Molecular weight} = 2 \times \text{Vapour density} = 2 \times 60.0 = 120$.
Step $3$: Calculate the empirical formula weight.
$\text{Empirical formula weight} = 12 + 1 + 3 \times 35.5 = 13 + 106.5 = 119.5 \approx 120$.
Since the molecular weight equals the empirical formula weight,the molecular formula is $CHCl_3$.

(C) The chemical formula of urea is $(NH_2CONH_2)$.
The molar mass of urea is $(2 \times 14) + (4 \times 1) + 12 + 16 = 60 \, g/mol$.
The mass of nitrogen in one mole of urea is $2 \times 14 = 28 \, g$.
The percentage of nitrogen is calculated as: $\frac{\text{Mass of Nitrogen}}{\text{Molar Mass of Urea}} \times 100 = \frac{28}{60} \times 100 = 46.66 \%$.

(B) The mass ratio of $C : H : N$ is $9 : 1 : 3.5$.
To find the molar ratio,divide the mass of each element by its atomic mass $(C = 12, H = 1, N = 14)$:
Molar ratio of $C : H : N = \frac{9}{12} : \frac{1}{1} : \frac{3.5}{14} = 0.75 : 1 : 0.25$.
To simplify the ratio,divide each value by the smallest value $(0.25)$:
$\frac{0.75}{0.25} : \frac{1}{0.25} : \frac{0.25}{0.25} = 3 : 4 : 1$.
Therefore,the empirical formula is $C_3H_4N$.

(A) The empirical formula is $CH_2O$.
The empirical formula mass is $(12 \times 1) + (1 \times 2) + (16 \times 1) = 12 + 2 + 16 = 30 \ g/mol$.
The ratio $n$ is calculated as $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{90}{30} = 3$.
The molecular formula is $n \times (\text{Empirical formula}) = 3 \times (CH_2O) = C_3H_6O_3$.

(A) $1$. Calculate the percentage of oxygen: $\%O = 100 - (52.20 + 13.04) = 34.76\%$.
$2$. Calculate the molar ratio of elements:
$C = 52.20 / 12 = 4.35$
$H = 13.04 / 1 = 13.04$
$O = 34.76 / 16 = 2.17$
$3$. Divide by the smallest value $(2.17)$:
$C = 4.35 / 2.17 \approx 2$
$H = 13.04 / 2.17 \approx 6$
$O = 2.17 / 2.17 = 1$
$4$. The empirical formula is $C_2H_6O$. The empirical formula mass is $(2 \times 12) + (6 \times 1) + 16 = 46$.
$5$. Molecular weight = $2 \times \text{Vapour Density} = 2 \times 23 = 46$.
$6$. Since molecular weight equals empirical formula mass,the molecular formula is $C_2H_6O$.

(C) The empirical formula mass of $CH_2O$ is $(12 \times 1) + (1 \times 2) + (16 \times 1) = 30 \ g/mol$.
The value of $n$ is calculated as: $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{120}{30} = 4$.
The molecular formula is given by $n \times (\text{Empirical formula}) = 4 \times (CH_2O) = C_4H_8O_4$.
Thus,the correct option is $C$.

(C) The $Structural \ formula$ of a molecule provides information about the connectivity and the relative spatial arrangement of atoms within the molecule.
While the $Molecular \ formula$ only gives the actual number of atoms of each element present,the $Structural \ formula$ explicitly shows how these atoms are bonded to one another.

(D) The empirical formula is $C_2H_4O$.
The empirical formula mass is $(2 \times 12) + (4 \times 1) + 16 = 44 \ g/mol$.
The ratio $n$ is calculated as $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{132.16}{44} \approx 3$.
The molecular formula is $n \times (\text{Empirical formula}) = 3 \times (C_2H_4O) = C_6H_{12}O_3$.

(D) The molecular weight is calculated as: $\text{Mol. wt.} = 2 \times \text{Vapour Density} = 2 \times 45 = 90$.
The empirical formula weight of $CH_2O$ is: $12 + (2 \times 1) + 16 = 30$.
The value of $n$ is calculated as: $n = \frac{\text{Mol. wt.}}{\text{Empirical formula wt.}} = \frac{90}{30} = 3$.
Therefore,the molecular formula is: $(CH_2O)_3 = C_3H_6O_3$.

(D) To calculate the percentage of carbon,use the formula: $\text{Percentage of C} = \frac{\text{Total mass of C}}{\text{Molar mass of compound}} \times 100$.
For $CH_3COOH$ $(C_2H_4O_2)$: Molar mass = $2(12) + 4(1) + 2(16) = 60 \ g/mol$. Percentage of $C$ = $\frac{24}{60} \times 100 = 40\%$.
For $C_6H_{12}O_6$: Molar mass = $6(12) + 12(1) + 6(16) = 180 \ g/mol$. Percentage of $C$ = $\frac{72}{180} \times 100 = 40\%$.
Since both compounds have $40\%$ carbon,the correct pair is $CH_3COOH$ and $C_6H_{12}O_6$.

(C) The percentage of oxygen is $100 - (20 + 6.67 + 46.67) = 26.66\%$.
The ratio of the number of gram atoms among $C$,$H$,$N$,and $O$ is:
$C : H : N : O = \frac{20}{12} : \frac{6.67}{1} : \frac{46.67}{14} : \frac{26.66}{16} = 1.66 : 6.67 : 3.33 : 1.66$.
Dividing by the smallest value $(1.66)$:
$C : H : N : O = 1 : 4 : 2 : 1$.
The empirical formula is $CH_4N_2O$.
The empirical formula weight is $12 + 4 + 28 + 16 = 60 \ g/mol$.
Since the molecular mass is $60$,the molecular formula is also $CH_4N_2O$,which is urea,$(NH_2)_2CO$.
Urea on heating gives biuret as a solid residue,which gives a violet color with alkaline copper sulphate (biuret test).

(B) The percentage of oxygen is $100 - (49.3 + 6.84) = 43.86\%$.

Element	Moles	Simple Ratio
$C$ $(49.3/12 = 4.1)$	$4.1/2.7 = 1.5$	$1.5 \times 2 = 3$
$H$ $(6.84/1 = 6.84)$	$6.84/2.7 = 2.5$	$2.5 \times 2 = 5$
$O$ $(43.86/16 = 2.7)$	$2.7/2.7 = 1.0$	$1.0 \times 2 = 2$

Empirical formula = $C_3H_5O_2$.
Empirical formula weight = $(12 \times 3) + (1 \times 5) + (16 \times 2) = 36 + 5 + 32 = 73$.
Molecular weight = $2 \times \text{Vapour Density} = 2 \times 73 = 146$.
$n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} = \frac{146}{73} = 2$.
Molecular formula = $(C_3H_5O_2)_2 = C_6H_{10}O_4$.

(D) Given that $0.0833 \, mol$ of the carbohydrate contains $1 \, g$ of hydrogen.
Therefore,$1 \, mol$ of the carbohydrate contains hydrogen $= \frac{1}{0.0833} \approx 12 \, g$.
The empirical formula is $CH_2O$,which contains $2 \, g$ of hydrogen per mole of the empirical unit.
To find the value of $n$ (where $\text{Molecular Formula} = n \times \text{Empirical Formula}$),we divide the total mass of hydrogen in the molecular formula by the mass of hydrogen in the empirical formula:
$n = \frac{12 \, g}{2 \, g} = 6$.
Thus,the molecular formula is $(CH_2O)_6 = C_6H_{12}O_6$.

(B) The mass of $H_2O$ produced is $0.9 \ g$.
The mass of hydrogen in $0.9 \ g$ of $H_2O$ is calculated as: $\text{Mass of } H = \frac{2}{18} \times 0.9 \ g = 0.1 \ g$.
The percentage of hydrogen in the hydrocarbon is: $\frac{0.1 \ g}{0.5 \ g} \times 100 = 20 \%$.
Since the compound is a hydrocarbon (containing only carbon and hydrogen),the percentage of carbon is $100 \% - 20 \% = 80 \%$.
Therefore,the correct option is $B$.

(D) Step $1$: Determine the empirical formula.
Moles of $C = \frac{10.5 \,g}{12 \,g/mol} = 0.875 \,mol$.
Moles of $H = \frac{1 \,g}{1 \,g/mol} = 1 \,mol$.
Ratio $C:H = 0.875:1 = 7:8$.
Thus,the empirical formula is $C_7H_8$.
Step $2$: Determine the molar mass using the ideal gas equation $PV = nRT = \frac{w}{M}RT$.
$1 \,atm \times 1 \,L = \frac{2.4 \,g}{M} \times 0.0821 \,L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times (127 + 273) \,K$.
$M = \frac{2.4 \times 0.0821 \times 400}{1} \approx 78.8 \,g/mol$.
Step $3$: Determine the molecular formula.
Empirical formula mass of $C_7H_8 = (7 \times 12) + (8 \times 1) = 92 \,g/mol$.
Since the calculated molar mass $(78.8)$ does not match the empirical formula mass $(92)$,and no option matches $C_7H_8$,the correct answer is $D$.

(A) The percentage of hydrogen is calculated as: $\text{Percentage of H} = \frac{\text{Mass of hydrogen}}{\text{Molar mass of compound}} \times 100$.
For $CH_4$: $\frac{4}{16} \times 100 = 25\%$.
For $C_2H_4$: $\frac{4}{28} \times 100 \approx 14.28\%$.
For $C_6H_6$: $\frac{6}{78} \times 100 \approx 7.69\%$.
For $C_2H_2$: $\frac{2}{26} \times 100 \approx 7.69\%$.
Thus,$CH_4$ has the highest percentage of hydrogen.

(C) At $STP$,$22.4 \, L$ of any gas corresponds to $1 \, mole$.
Given that $5.6 \, L$ of the alkane weighs $11 \, g$.
Therefore,the mass of $22.4 \, L$ of the alkane is calculated as:
$\text{Molar mass} = \frac{11 \, g}{5.6 \, L} \times 22.4 \, L/mol = 44 \, g/mol$.
The general formula for an alkane is $C_nH_{2n+2}$.
For $C_nH_{2n+2}$,the molar mass is $12n + 1(2n+2) = 14n + 2$.
Setting $14n + 2 = 44$,we get $14n = 42$,so $n = 3$.
Thus,the molecular formula is $C_3H_8$.

(D) In ferric sulphate,the iron ion is in the $+3$ oxidation state,represented as $Fe^{3+}$.
The sulphate ion has a charge of $-2$,represented as $SO_4^{2-}$.
To balance the charges,we need two $Fe^{3+}$ ions $(2 \times +3 = +6)$ and three $SO_4^{2-}$ ions $(3 \times -2 = -6)$.
Therefore,the chemical formula is $Fe_2(SO_4)_3$.

(C) The chemical formula for calcium pyrophosphate is $Ca_2P_2O_7$.
This dissociates as $Ca_2P_2O_7 \rightarrow 2Ca^{2+} + (P_2O_7)^{4-}$.
Thus,the pyrophosphate anion is $(P_2O_7)^{4-}$.
Ferric ion is $Fe^{3+}$.
To form a neutral compound,the charges must balance: $4 \times (+3) + 3 \times (-4) = 0$.
Therefore,the formula for ferric pyrophosphate is $Fe_4(P_2O_7)_3$.

(B) Nickel steel is an alloy primarily composed of iron and nickel.
It typically contains $95-97 \%$ of $Fe$ and $3-5 \%$ of $Ni$.

(C) Rust is a hydrated iron$(III)$ oxide formed by the oxidation of iron in the presence of air and moisture.
Its chemical formula is represented as $Fe_2O_3 \cdot xH_2O$.

(B) The molar mass of $CH_3CH_2Br$ is calculated as: $(2 \times 12) + (5 \times 1) + 80 = 24 + 5 + 80 = 109 \ g/mol$.
The percentage of $Br$ is given by the formula: $\frac{\text{Mass of } Br}{\text{Molar mass of compound}} \times 100$.
Percentage of $Br = \frac{80}{109} \times 100 = 73.39 \%$.
Therefore,the correct option is $B$.

(B) The percentage of chlorine is calculated as: $\% \text{ of chlorine} = \frac{\text{Mass of chlorine}}{\text{Molar mass of compound}} \times 100$
$1$. Chloral $(CCl_3CHO)$: Molar mass $= 147.5 \ g/mol$. Mass of $Cl = 3 \times 35.5 = 106.5 \ g$. $\% \ Cl = (106.5 / 147.5) \times 100 \approx 72.20\%$
$2$. Pyrene $(CCl_4)$: Molar mass $= 154 \ g/mol$. Mass of $Cl = 4 \times 35.5 = 142 \ g$. $\% \ Cl = (142 / 154) \times 100 \approx 92.21\%$
$3$. $PVC$ $(C_2H_3Cl)_n$: The monomer is $CH_2=CHCl$. Molar mass $= 62.5 \ g/mol$. Mass of $Cl = 35.5 \ g$. $\% \ Cl = (35.5 / 62.5) \times 100 \approx 56.80\%$
$4$. Gammexene $(C_6H_6Cl_6)$: Molar mass $= 291 \ g/mol$. Mass of $Cl = 6 \times 35.5 = 213 \ g$. $\% \ Cl = (213 / 291) \times 100 \approx 73.19\%$
Comparing the values,Pyrene $(CCl_4)$ has the highest percentage of chlorine.

(C) The percentage of $C$ is calculated as: $\frac{\text{Mass of } C}{\text{Molar mass of substance}} \times 100$.
$1$. Pyrene $(CCl_4)$: $\frac{12}{154} \times 100 \approx 7.79\ \%$.
$2$. Gammexane $(C_6H_6Cl_6)$: $\frac{72}{291} \times 100 \approx 24.74\ \%$.
$3$. Ethylene glycol $(C_2H_6O_2)$: $\frac{24}{62} \times 100 \approx 38.71\ \%$.
$4$. $PVC$ $([C_2H_3Cl]_n)$: $\frac{24}{62.5} \times 100 \approx 38.40\ \%$.
Comparing the values,the carbon percentage is maximum in Ethylene glycol.

(D) Urea $(NH_2CONH_2)$ contains approximately $46.6\%$ nitrogen by mass,whereas ammonium sulphate $(NH_4)_2SO_4$ contains about $21.2\%$ nitrogen.
Thus,urea has a higher nitrogen content.
Additionally,ammonium sulphate is a salt of a strong acid $(H_2SO_4)$ and a weak base $(NH_4OH)$,which makes it acidic in nature and can increase soil acidity over time.
Urea,on the other hand,is neutral and does not increase soil acidity.
Therefore,both statements $(a)$ and $(b)$ are correct.

(B) 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.
The mass of nitrogen in $1 \ mol$ of urea is $2 \times 14 = 28 \ g$.
The percentage of nitrogen is calculated as:
$\text{Percentage of N} = \frac{\text{Mass of N}}{\text{Molar mass of urea}} \times 100 = \frac{28}{60} \times 100 = 46.67\%$.

(A) The molecular formula of glucose is $C_6H_{12}O_6$.
To find the empirical formula,we determine the simplest whole-number ratio of the atoms present in the molecule.
The ratio of $C:H:O$ is $6:12:6$.
Dividing by the common factor $6$,we get the ratio $1:2:1$.
Thus,the empirical formula is $CH_2O$.

(D) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as:
$M = 2(23) + 32 + 4(16) + 10 \times (2 \times 1 + 16) = 46 + 32 + 64 + 180 = 322 \ g/mol$.
In one mole of $Na_2SO_4 \cdot 10H_2O$,there are $4$ oxygen atoms in $Na_2SO_4$ and $10$ oxygen atoms in $10H_2O$,totaling $14$ oxygen atoms.
Mass of oxygen in $1 \ mol$ $(322 \ g)$ of $Na_2SO_4 \cdot 10H_2O = 14 \times 16 \ g = 224 \ g$.
Therefore,the mass of oxygen in $32.2 \ g$ of $Na_2SO_4 \cdot 10H_2O = (224 / 322) \times 32.2 = 22.4 \ g$.

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

Element	Percentage	Atomic Mass	Relative Moles	Simplest Ratio
$C$	$38.71$	$12$	$38.71 / 12 = 3.22$	$3.22 / 3.22 = 1$
$H$	$9.67$	$1$	$9.67 / 1 = 9.67$	$9.67 / 3.22 = 3$
$O$	$51.62$	$16$	$51.62 / 16 = 3.22$	$3.22 / 3.22 = 1$

The simplest ratio of $C:H:O$ is $1:3:1$.
Therefore,the empirical formula is $CH_3O$.

(B) For the first oxide $(MO)$:
Metal = $50\%$,Oxygen = $50\%$.
Atomic mass of metal $(M)$ = $16 \ g/mol$ (since $1:1$ ratio).
For the second oxide:
Metal = $40\%$,Oxygen = $60\%$.
Ratio of moles of $M$ to $O$ = $\frac{40}{16} : \frac{60}{16} = 2.5 : 3.75$.
Dividing by the smaller value $(2.5)$:
$M : O = 1 : 1.5 = 2 : 3$.
Therefore,the formula is $M_2O_3$.

(B) The molecular formula of glucose is $C_6H_{12}O_6$. The ratio of atoms is $C:H:O = 6:12:6 = 1:2:1$. Thus,the empirical formula is $CH_2O$. For acetic acid $(CH_3COOH)$,the molecular formula is $C_2H_4O_2$. The ratio of atoms is $C:H:O = 2:4:2 = 1:2:1$. Therefore,the empirical formula of acetic acid is $CH_2O$,which is the same as glucose.

(D) Let the initial weight of the hydrated salt be $100 \ g$.
Weight of water lost $= 55.9 \ g$.
Weight of anhydrous $Na_2SO_4$ remaining $= 100 - 55.9 = 44.1 \ g$.
Molar mass of anhydrous $Na_2SO_4 = (2 \times 23) + 32 + (4 \times 16) = 142 \ g/mol$.
$44.1 \ g$ of $Na_2SO_4$ is associated with $55.9 \ g$ of $H_2O$.
Therefore,$142 \ g$ of $Na_2SO_4$ is associated with $\frac{142 \times 55.9}{44.1} \approx 180 \ g$ of $H_2O$.
Number of moles of water $(x)$ $= \frac{180}{18} = 10$.
Thus,the formula of the crystalline salt is $Na_2SO_4 \cdot 10H_2O$.

(A) According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain an equal number of moles. Since the volumes and conditions are the same,the molar mass of the organic gas is equal to the molar mass of $N_2$.
$Molar \ mass \ of \ N_2 = 28 \ g/mol$.
Thus,the molar mass of the organic compound = $28 \ g/mol$.
The empirical formula is $CH_2$,so the empirical formula mass = $12 + (2 \times 1) = 14 \ g/mol$.
$n = \frac{Molar \ mass}{Empirical \ formula \ mass} = \frac{28}{14} = 2$.
Therefore,the molecular formula = $n \times (CH_2) = 2 \times (CH_2) = C_2H_4$.

(C) $1$. Calculate the molar mass of the hydrocarbon: $\text{Molar mass} = 2 \times \text{Vapor density} = 2 \times 15 = 30 \ \text{g/mol}$.
$2$. Determine the mass of carbon and hydrogen in one mole: $\text{Mass of C} = 80\% \text{ of } 30 = 24 \ \text{g}$. $\text{Mass of H} = 20\% \text{ of } 30 = 6 \ \text{g}$.
$3$. Calculate the number of moles of each atom: $\text{Moles of C} = \frac{24}{12} = 2$. $\text{Moles of H} = \frac{6}{1} = 6$.
$4$. The molecular formula is $C_2H_6$.

(A) Given: Carbon $(C)$ = $75\%$,Hydrogen $(H)$ = $100\% - 75\% = 25\%$.
Calculate the mole ratio:
$C = \frac{75}{12} = 6.25$
$H = \frac{25}{1} = 25$
Divide by the smallest value $(6.25)$:
$C = \frac{6.25}{6.25} = 1$
$H = \frac{25}{6.25} = 4$
Therefore,the empirical formula is $CH_4$.

(A) The minimum molecular weight is calculated assuming at least one atom of $Se$ is present in the enzyme molecule.
Given: Percentage of $Se = 0.5\%$,Atomic weight of $Se = 78.4$.
Formula: $\text{Minimum Molecular Weight} = \frac{\text{Atomic weight of element} \times 100}{\text{Percentage of element}}$.
Calculation: $\text{Molecular Weight} = \frac{78.4 \times 100}{0.5} = \frac{7840}{0.5} = 15680$.
In scientific notation,this is $1.568 \times 10^4$.

(B) The percentage of $Y$ is $100 - 75.8 = 24.2\%$.
The molar ratio of $X:Y$ is calculated as:
$X:Y = \frac{75.8}{75} : \frac{24.2}{16} = 1.01 : 1.51$.
Dividing by the smallest value $(1.01)$:
$X:Y = \frac{1.01}{1.01} : \frac{1.51}{1.01} = 1 : 1.5$.
Multiplying by $2$ to get whole numbers:
$X:Y = 2 : 3$.
Therefore,the empirical formula is $X_2Y_3$.

(A) Given that $0.1 \, \text{mol}$ of carbohydrate contains $1 \, g$ of hydrogen,then $1 \, \text{mol}$ of carbohydrate contains $10 \, g$ of hydrogen.
The empirical formula is $CH_2O$,so the molecular formula can be represented as $(CH_2O)_n$ or $C_nH_{2n}O_n$.
The number of hydrogen atoms in $1 \, \text{mol}$ of the compound is $2n \, \text{mol}$.
Since the molar mass of hydrogen is $1 \, \text{g/mol}$,the mass of hydrogen is $2n \times 1 \, \text{g} = 10 \, \text{g}$.
Solving for $n$: $2n = 10$,which gives $n = 5$.
Therefore,the molecular formula is $C_5H_{10}O_5$.

(C) Let the mass of each substance be $W$.
The number of moles of $X$ is $n_X = \frac{W}{30}$.
The number of moles of $Y$ is $n_Y = \frac{W}{20}$.
The ratio of moles $n_X : n_Y = \frac{W}{30} : \frac{W}{20} = \frac{1}{3} : \frac{1}{2} = 2 : 3$.
Thus,the empirical formula is $X_2Y_3$.
The empirical formula weight is $(2 \times 30) + (3 \times 20) = 60 + 60 = 120$.
Since the given molecular weight is $120$,the molecular formula is the same as the empirical formula,which is $X_2Y_3$.