Percentage composition and Molecular formula Questions in English

(B) $1$. Calculate the empirical formula:
$C = 20/12 = 1.67$,$H = 6.67/1 = 6.67$,$N = 46.67/14 = 3.33$,$O = (100 - (20 + 6.67 + 46.67))/16 = 26.66/16 = 1.67$.
$2$. Divide by the smallest value $(1.67)$: $C = 1, H = 4, N = 2, O = 1$.
$3$. The empirical formula is $CH_4N_2O$. The empirical formula weight is $12 + 4 + 28 + 16 = 60$.
$4$. Since the molecular weight is $60$,the molecular formula is $CH_4N_2O$,which corresponds to urea $(NH_2)_2CO$.
$5$. Urea,upon heating,forms biuret,which gives a violet color with alkaline copper sulfate (Biuret test).

(A) To find the percentage of hydrogen by mass,we use the formula: $\text{Percentage of H} = \frac{\text{Mass of H}}{\text{Molar mass of compound}} \times 100$.
$1$. For $CH_4$: Molar mass = $12 + 4 = 16 \ g/mol$. Mass of $H$ = $4 \ g$. Percentage = $(4/16) \times 100 = 25\%$.
$2$. For $C_2H_4$: Molar mass = $24 + 4 = 28 \ g/mol$. Mass of $H$ = $4 \ g$. Percentage = $(4/28) \times 100 \approx 14.28\%$.
$3$. For $C_6H_6$: Molar mass = $72 + 6 = 78 \ g/mol$. Mass of $H$ = $6 \ g$. Percentage = $(6/78) \times 100 \approx 7.69\%$.
$4$. For $C_2H_2$: Molar mass = $24 + 2 = 26 \ g/mol$. Mass of $H$ = $2 \ g$. Percentage = $(2/26) \times 100 \approx 7.69\%$.
Comparing these,$CH_4$ has the highest percentage of hydrogen.

(D) The combustion reaction is: $C_xH_y + (x + y/4) O_2 \to x CO_2 + (y/2) H_2O$
Moles of $H_2O = 0.72 / 18 = 0.04 \ mol$
Moles of $CO_2 = 3.08 / 44 = 0.07 \ mol$
From the stoichiometry:
Moles of $C = \text{moles of } CO_2 = 0.07 \ mol$
Moles of $H = 2 \times \text{moles of } H_2O = 2 \times 0.04 = 0.08 \ mol$
The ratio of $C:H = 0.07 : 0.08 = 7:8$
Therefore,the empirical formula is $C_7H_8$.

(B) $20$ carat gold means that out of $24$ parts,$20$ parts are gold and $4$ parts are copper or other metals.
$\% \, \text{gold} = \frac{20}{24} \times 100 = 83.33 \, \%$

(C) The empirical formula mass of $CH_2O$ is $(1 \times 12) + (2 \times 1) + (1 \times 16) = 30 \ g/mol$.
The value of $n$ is calculated as: $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{180}{30} = 6$.
The molecular formula is given by: $(\text{Empirical formula})_n = (CH_2O)_6 = C_6H_{12}O_6$.

(C) $1$. Calculate the relative number of moles of each element by dividing the weight ratio by their respective atomic weights:
$C = 9/12 = 0.75$,$H = 1/1 = 1$,$N = 3.5/14 = 0.25$.
$2$. Determine the simplest molar ratio by dividing by the smallest value $(0.25)$:
$C = 0.75/0.25 = 3$,$H = 1/0.25 = 4$,$N = 0.25/0.25 = 1$.
$3$. The empirical formula is $C_3H_4N$.
$4$. Calculate the empirical formula weight: $(3 \times 12) + (4 \times 1) + (1 \times 14) = 36 + 4 + 14 = 54$.
$5$. Determine the value of $n = \text{Molecular weight} / \text{Empirical formula weight} = 108 / 54 = 2$.
$6$. The molecular formula is $n \times (C_3H_4N) = 2 \times (C_3H_4N) = C_6H_8N_2$.

(B) The general formula for an alkane is $C_nH_{2n+2}$.
The mass ratio of $C$ to $H$ is given by $\frac{12n}{1(2n+2)} = \frac{12n}{2n+2} = \frac{6n}{n+1}$.
Given the ratio is $5.1428$,we set up the equation:
$\frac{6n}{n+1} = 5.1428$
$6n = 5.1428(n + 1)$
$6n = 5.1428n + 5.1428$
$6n - 5.1428n = 5.1428$
$0.8572n = 5.1428$
$n = \frac{5.1428}{0.8572} = 6$
Substituting $n = 6$ into the general formula $C_nH_{2n+2}$,we get $C_6H_{2(6)+2} = C_6H_{14}$.

(C) $1$. Calculate the percentage of oxygen: $O = 100 - (38.71 + 9.67) = 51.62 \%$.
$2$. Calculate the mole ratio of each element:
$C: 38.71 / 12 = 3.226$
$H: 9.67 / 1 = 9.67$
$O: 51.62 / 16 = 3.226$
$3$. Divide by the smallest value $(3.226)$:
$C: 3.226 / 3.226 = 1$
$H: 9.67 / 3.226 = 3$
$O: 3.226 / 3.226 = 1$
$4$. The empirical formula is $CH_3O$.

(D) The molar mass of $H_{2}O$ is $18 \ g/mol$,which contains $2 \ g$ of $H$.
Therefore,$0.72 \ g$ of $H_{2}O$ contains $\frac{2}{18} \times 0.72 = 0.08 \ g$ of $H$.
The molar mass of $CO_{2}$ is $44 \ g/mol$,which contains $12 \ g$ of $C$.
Therefore,$3.08 \ g$ of $CO_{2}$ contains $\frac{12}{44} \times 3.08 = 0.84 \ g$ of $C$.
The molar ratio of $C:H$ is $\frac{0.84}{12} : \frac{0.08}{1} = 0.07 : 0.08 = 7 : 8$.
Thus,the empirical formula is $C_{7}H_{8}$.

(D) The general formula for an alkanoic acid (saturated carboxylic acid) is $C_nH_{2n}O_2$.
For $n=4$,the formula is $C_4H_8O_2$.
Therefore,$C_4H_6O_2$ cannot represent an alkanoic acid.
Examples of $C_4H_6O_2$ include:
$1$. Diketones: $CH_3-CO-CO-CH_3$ (butane$-2,3-$dione).
$2$. Dialdehydes: $OHC-CH_2-CH_2-CHO$ (butanedial).
$3$. Alkenoic acids: $CH_2=CH-CH_2-COOH$ (but$-3-$enoic acid).

(A) The purity of gold in carats is calculated based on the ratio of gold parts to the total $24$ parts.
Given that the gold content is $75\%$,we can set up the equation:
$\frac{x}{24} \times 100 = 75$
Solving for $x$:
$x = \frac{75 \times 24}{100}$
$x = 0.75 \times 24 = 18$
Therefore,the gold is $18$ carat.

(B) The term $925$ fine silver,commonly known as sterling silver,indicates that the alloy contains $92.5\%$ pure silver $(Ag)$ by weight.
The remaining $7.5\%$ of the alloy consists of other metals,typically copper $(Cu)$,added to increase the hardness and durability of the silver.
Therefore,the composition is $92.5\% \, Ag$ and $7.5\% \, Cu$.

(C) $1$. Calculate the molar mass: $\text{Molar mass} = 2 \times \text{Vapour density} = 2 \times 30 = 60 \ g/mol$.
$2$. Calculate the mass of carbon: $\text{Mass of } C = 80\% \text{ of } 60 = 0.80 \times 60 = 48 \ g$.
$3$. Calculate the number of carbon atoms: $\text{Number of } C \text{ atoms} = 48 / 12 = 4$.
$4$. Calculate the mass of hydrogen: $\text{Mass of } H = 60 - 48 = 12 \ g$.
$5$. Calculate the number of hydrogen atoms: $\text{Number of } H \text{ atoms} = 12 / 1 = 12$.
$6$. The molecular formula is $C_4H_{12}$.

(C) The percentage of phosphorus in the compound is given by: $\text{Percentage} = \frac{\text{Atomic mass of P}}{\text{Molecular mass of compound}} \times 100$
Given that the percentage is $10^{-2} \%$ and the atomic mass of phosphorus is $31$.
Let the molecular mass be $M$.
$10^{-2} = \frac{31}{M} \times 100$
$M = \frac{31 \times 100}{10^{-2}}$
$M = 31 \times 10^2 \times 10^2 = 31 \times 10^4$

(B) $1$. Calculate the moles of each element:
$n_{Na} = \frac{1.15 \ g}{23 \ g/mol} = 0.05 \ mol$
$n_C = \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.05 \ mol$
$n_O = 0.1 \ mol$
$2$. Determine the molar ratio:
$n_{Na} : n_C : n_O = 0.05 : 0.05 : 0.1$
$3$. Simplify the ratio by dividing by the smallest value $(0.05)$:
$1 : 1 : 2$
$4$. The empirical formula is $NaCO_2$.

(B) $1$. Calculate the moles of each element:
Moles of $N = \frac{2.8 \ g}{14 \ g/mol} = 0.2 \ mol$
Moles of $O = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$
$2$. Determine the molar ratio:
Ratio of $N : O = 0.2 : 0.2 = 1 : 1$
$3$. The simplest formula is $NO$.

(A) To find the minimum molecular mass of the enzyme,we assume that it contains at least one atom of $Se$.
Given that the percentage of $Se$ by mass is $0.5\%$ and the atomic mass of $Se$ is $78.4 \ amu$.
The formula for minimum molecular mass is:
$\text{Minimum molecular mass} = \frac{\text{Atomic mass of element} \times 100}{\text{Percentage of element}}$
Substituting the values:
$\text{Minimum molecular mass} = \frac{78.4 \times 100}{0.5}$
$= \frac{7840}{0.5}$
$= 15680 \ amu$
$= 1.568 \times 10^4 \ amu$

(B) To find the empirical formula,divide the mass ratio by the atomic mass of each element:
$H: 1 / 1 = 1$
$C: 3 / 12 = 0.25$
$O: 4 / 16 = 0.25$
$N: 7 / 14 = 0.5$
Now,divide each value by the smallest value $(0.25)$:
$H: 1 / 0.25 = 4$
$C: 0.25 / 0.25 = 1$
$O: 0.25 / 0.25 = 1$
$N: 0.5 / 0.25 = 2$
The ratio of atoms is $H:C:O:N = 4:1:1:2$.
Therefore,the empirical formula is $H_4CON_2$.

(C) To find the molecular formula,we first determine the empirical formula:

Element	$\%$	At. wt	Atomic ratio	Simplest ratio
$N$	$30.5$	$14$	$\frac{30.5}{14} = 2.18$	$\frac{2.18}{2.18} = 1$
$O$	$69.5$	$16$	$\frac{69.5}{16} = 4.34$	$\frac{4.34}{2.18} \approx 2$

The empirical formula is $NO_2$.
The empirical formula mass $= 14 + 2 \times 16 = 46$.
The value of $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{92}{46} = 2$.
Therefore,the molecular formula $= n \times (NO_2) = 2 \times NO_2 = N_2O_4$.

(C) Given that the compound contains $2.8\%$ $Fe$ by mass and one molecule contains one $Fe$ atom.
The atomic mass of $Fe$ is $56\, g/mol$.
Let the molar mass of the compound be $M\, g/mol$.
According to the problem,$2.8\%$ of $M = 56$.
$\frac{2.8}{100} \times M = 56$
$M = \frac{56 \times 100}{2.8}$
$M = \frac{5600}{2.8} = 2000\, g/mol$.

(D) Assume the mass of the compound is $100 \ g$.
Moles of $C = \frac{70.6 \ g}{12 \ g/mol} = 5.88 \ mol$.
Moles of $H = \frac{4.2 \ g}{1 \ g/mol} = 4.2 \ mol$.
Moles of $N = \frac{11.8 \ g}{14 \ g/mol} = 0.84 \ mol$.
Moles of $O = \frac{13.4 \ g}{16 \ g/mol} = 0.84 \ mol$.
Divide each by the smallest value $(0.84)$:
$C : H : N : O = \frac{5.88}{0.84} : \frac{4.2}{0.84} : \frac{0.84}{0.84} : \frac{0.84}{0.84} = 7 : 5 : 1 : 1$.
Therefore,the empirical formula of Kevlar is $C_7H_5NO$.

(C) The chemical formula of $NaOH$ is $NaOH$. One mole of $NaOH$ contains $1 \ mole$ of $O$ atoms.
Given that the sample contains $3 \ moles$ of $O$ atoms,the amount of pure $NaOH$ is $3 \ moles$.
The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.
Mass of pure $NaOH = 3 \ mol \times 40 \ g/mol = 120 \ g$.
Percentage purity = $\frac{\text{Mass of pure } NaOH}{\text{Total mass of sample}} \times 100$.
Percentage purity = $\frac{120 \ g}{1000 \ g} \times 100 = 12 \%$.

(A) To determine the highest chlorine content,we calculate the percentage of chlorine in each compound:
$1$. Pyrene $(CCl_4)$: Molar mass = $12 + 4 \times 35.5 = 154 \ g/mol$. Chlorine mass = $4 \times 35.5 = 142 \ g$. Percentage = $(142/154) \times 100 \approx 92.2\%$.
$2$. $DDT$ $(C_{14}H_9Cl_5)$: Molar mass = $14 \times 12 + 9 \times 1 + 5 \times 35.5 = 168 + 9 + 177.5 = 354.5 \ g/mol$. Chlorine mass = $5 \times 35.5 = 177.5 \ g$. Percentage = $(177.5/354.5) \times 100 \approx 50.1\%$.
$3$. Chloral $(CCl_3CHO)$: Molar mass = $12 + 35.5 \times 3 + 12 + 1 + 16 + 1 = 147.5 \ g/mol$. Chlorine mass = $3 \times 35.5 = 106.5 \ g$. Percentage = $(106.5/147.5) \times 100 \approx 72.2\%$.
$4$. Gammaxene $(C_6H_6Cl_6)$: Molar mass = $6 \times 12 + 6 \times 1 + 6 \times 35.5 = 72 + 6 + 213 = 291 \ g/mol$. Chlorine mass = $6 \times 35.5 = 213 \ g$. Percentage = $(213/291) \times 100 \approx 73.2\%$.
Comparing the percentages,Pyrene $(CCl_4)$ has the highest chlorine content.

(C) The percentage of an element in a compound is given by the formula:
$\% \text{ of Sulphur} = \frac{\text{Atomic weight of } S \times \text{number of atoms}}{\text{Molecular weight of compound}} \times 100$
For the minimum molecular weight,we assume there is at least $1$ atom of sulphur per molecule.
$8 = \frac{32 \times 1}{\text{Molecular weight}} \times 100$
$\text{Molecular weight} = \frac{32 \times 100}{8} = 400\,g\,mol^{-1}$

(D) Mass of hydrated salt = $61 \ g$
Mass of anhydrous $BaCl_2$ = $52 \ g$
Mass of water lost = $61 \ g - 52 \ g = 9 \ g$
Molar mass of $BaCl_2 = 137 + 2 \times 35.5 = 208 \ g/mol$
Molar mass of $H_2O = 18 \ g/mol$
Moles of $BaCl_2 = \frac{52}{208} = 0.25 \ mol$
Moles of $H_2O = \frac{9}{18} = 0.5 \ mol$
Ratio of moles of $H_2O$ to $BaCl_2 = \frac{0.5}{0.25} = 2$
Therefore,the formula is $BaCl_2 \cdot 2H_2O$.

(D) Given,percentage of $H = 12.5\%$.
Therefore,percentage of $N = 100 - 12.5 = 87.5\%$.

Element	Percentage	Atomic ratio	Simple ratio
$H$	$12.5\%$	$\frac{12.5}{1} = 12.5$	$\frac{12.5}{6.25} = 2$
$N$	$87.5\%$	$\frac{87.5}{14} = 6.25$	$\frac{6.25}{6.25} = 1$

Empirical formula $= NH_2$.
Empirical formula mass $= 14 + (2 \times 1) = 16 \ g/mol$.
Molecular mass $= 2 \times \text{Vapour density} = 2 \times 16 = 32 \ g/mol$.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{32}{16} = 2$.
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (NH_2) = N_2H_4$.

(C) Given,vapour density $(V.D.)$ $= 94.8$.
Molecular weight of metal chloride $= 2 \times V.D. = 2 \times 94.8 = 189.6 \ g/mol$.
Mass of chlorine in $189.6 \ g$ of metal chloride $= 74.75\% \text{ of } 189.6 = 0.7475 \times 189.6 \approx 141.7 \ g$.
Number of chlorine atoms $= \frac{141.7}{35.5} \approx 4$.
Mass of metal $= 189.6 - 141.7 = 47.9 \ g$.
Since the metal chloride is $MCl_4$,the valency of the metal $M$ is $4$.
Therefore,the formula of the metal chloride is $MCl_4$.

(C) The combustion of an organic compound containing $C, H,$ and $N$ follows the stoichiometry:
$C_xH_yN_z + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O + (z/2) N_2$
Given that $x = 6$ (moles of $CO_2$),$y/2 = 4$ (moles of $H_2O$),and $z/2 = 1$ (moles of $N_2$).
From $x = 6$,we get $6$ carbon atoms.
From $y/2 = 4$,we get $y = 8$ hydrogen atoms.
From $z/2 = 1$,we get $z = 2$ nitrogen atoms.
Thus,the empirical formula is $C_6H_8N_2$.

(B) In $CH_4$,the number of moles of carbon atoms is $n_C = 1$.
The number of moles of hydrogen atoms is $n_H = 4$.
The total number of moles of atoms in $1 \text{ mole}$ of $CH_4$ is $n_C + n_H = 1 + 4 = 5$.
The percentage composition of carbon by mole is calculated as:
$\text{Percentage of } C = \frac{n_C}{n_C + n_H} \times 100 = \frac{1}{5} \times 100 = 20 \% $.

(A) The combustion reaction of a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + y/4)O_2 \to xCO_2 + (y/2)H_2O$.
Mass of $C$ in $88 \ g$ of $CO_2 = (12/44) \times 88 = 24 \ g$.
Mass of $H$ in $9 \ g$ of $H_2O = (2/18) \times 9 = 1 \ g$.
Total mass of hydrocarbon = Mass of $C$ + Mass of $H = 24 \ g + 1 \ g = 25 \ g$.
Thus,the hydrocarbon contains $24 \ g$ of carbon and $1 \ g$ of hydrogen.

(A) Molecular formula $= X_4O_6$.
Mass of compound $= 10 \ g$.
Mass of $X = 5.72 \ g$.
Mass of $O = 10 - 5.72 = 4.28 \ g$.
Number of moles of $X$ $(n_X)$ $= \frac{5.72}{Mw_X}$ and number of moles of $O$ $(n_O)$ $= \frac{4.28}{16}$.
From the formula $X_4O_6$,the ratio of moles is $\frac{n_X}{n_O} = \frac{4}{6}$.
$\frac{5.72 / Mw_X}{4.28 / 16} = \frac{4}{6}$.
$Mw_X = \frac{5.72 \times 16 \times 6}{4.28 \times 4} = \frac{549.12}{17.12} \approx 32.07 \ g/mol$.
Thus,the atomic mass of $X$ is approximately $32 \ amu$.

(A) Pure gold is defined as $24$ carat.
To calculate the percentage of gold in $14$ carat gold,we use the formula:
$\text{Percentage of gold} = \left( \frac{\text{Carat value}}{24} \right) \times 100$
$\text{Percentage of gold} = \left( \frac{14}{24} \right) \times 100$
$\text{Percentage of gold} = 0.5833 \times 100 = 58.33\%$
Therefore,the correct option is $A$.

(B) Copper forms two common oxides: $CuO$ (copper$(II)$ oxide) and $Cu_2O$ (copper$(I)$ oxide).
$CuO$ is black in colour.
$Cu_2O$ is red in colour.
Therefore,the correct formula for the red-coloured oxide of copper is $Cu_2O$.

(D) The molar mass of $Ca(NO_3)_2$ is $40 + 2 \times (14 + 3 \times 16) = 40 + 2 \times 62 = 164 \ g/mol$.
In one mole of $Ca(NO_3)_2$,there are $2$ moles of nitrogen atoms $(2 \times 14 = 28 \ g)$ and $1$ mole of calcium $(40 \ g)$.
Using the unitary method:
$28 \ g$ of nitrogen corresponds to $40 \ g$ of calcium.
Therefore,$20 \ g$ of nitrogen corresponds to $\frac{20 \times 40}{28} \ g$ of calcium.
Mass of $Ca = \frac{800}{28} = \frac{200}{7} \approx 28.57 \ g$.

(C) Mass of $Fe$ in $100 \, g$ of hemoglobin $= 0.334 \, g$.
Mass of $Fe$ in $67200 \, g$ of hemoglobin $= \frac{0.334}{100} \times 67200 = 224.448 \, g \approx 224 \, g$.
Number of $Fe$ atoms in one molecule $= \frac{\text{Mass of } Fe \text{ in one mole of hemoglobin}}{\text{Atomic mass of } Fe}$.
Number of $Fe$ atoms $= \frac{224}{56} = 4$.
Therefore,there are $4$ iron atoms in one molecule of hemoglobin.

(A) The enzyme contains at least one atom of $Se$.
The weight percentage of $Se$ in the enzyme is $0.5 \%$.
This means $0.5 \, g$ of $Se$ is present in $100 \, g$ of the enzyme.
Therefore,the minimum molecular weight of the enzyme containing $78.4 \, g$ of $Se$ (one mole of $Se$) is:
$\text{Minimum molecular weight} = \frac{78.4 \times 100}{0.5} \, g/mol$
$= 15680 \, g/mol$
$= 1.568 \times 10^{4} \, g/mol$.

(C) The mass of nitrogen in one molecule of caffeine is calculated as follows:
Mass of $N = \frac{28.9}{100} \times 194 \, a.m.u.$
Mass of $N = 56.066 \, a.m.u.$
Since the atomic mass of nitrogen is $14 \, a.m.u.$,the number of nitrogen atoms is:
Number of $N$ atoms $= \frac{56.066}{14} \approx 4$
Thus,there are $4$ nitrogen atoms in one molecule of caffeine.

(A) The minimum molecular mass is calculated by assuming that at least one atom of sulfur is present in one molecule of insulin.
Atomic mass of sulfur $(S)$ = $32 \, g/mol$.
Given that $3.4 \%$ of the total mass is sulfur.
Let the minimum molecular mass be $M$.
$0.034 \times M = 32$
$M = \frac{32}{0.034}$
$M \approx 941.176 \, g/mol$.

(C) Mass of metal oxide $= 7.2 \, g$.
Mass of metal $= 6.4 \, g$.
Mass of oxygen $= 7.2 - 6.4 = 0.8 \, g$.
Moles of metal $= \frac{6.4}{128} = 0.05 \, mol$.
Moles of oxygen $= \frac{0.8}{16} = 0.05 \, mol$.
Ratio of moles of metal to oxygen $= 0.05 : 0.05 = 1 : 1$.
Therefore,the formula of the oxide is $MO$.

(A) $1$. Calculate the molar mass of the gas: At $STP$,$22.4 \, L$ of any gas contains $1 \, mole$.
$2$. Given $2.8 \, L$ weighs $5.5 \, g$,so $22.4 \, L$ will weigh $(5.5 / 2.8) \times 22.4 = 44 \, g/mol$.
$3$. Check the molar masses of the options:
$N_2O = (2 \times 14) + 16 = 44 \, g/mol$.
$N_2O_5 = (2 \times 14) + (5 \times 16) = 108 \, g/mol$.
$NO_2 = 14 + (2 \times 16) = 46 \, g/mol$.
$N_2O_3 = (2 \times 14) + (3 \times 16) = 76 \, g/mol$.
$4$. Since the molar mass is $44 \, g/mol$,the correct formula is $N_2O$.

(C) The molecular mass of $(CO)_x$ is $2 \times \text{vapor density} = 2 \times 70 = 140$.
The molar mass of $CO$ is $12 + 16 = 28 \ g/mol$.
Therefore,$x \times 28 = 140$.
$x = \frac{140}{28} = 5$.

(B) The molar mass of oxalic acid $(H_2C_2O_4)$ is calculated as follows:
$M = (2 \times 1.008) + (2 \times 12.01) + (4 \times 16.00) \approx 2 + 24 + 64 = 90 \ g/mol$.
Percentage of $H = (2 / 90) \times 100 = 2.22 \%$.
Percentage of $C = (24 / 90) \times 100 = 26.67 \%$.
Percentage of $O = (64 / 90) \times 100 = 71.11 \%$.

(C) The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound.
If two compounds have the same empirical formula,they have the same percentage composition of elements.
However,the molecular formula is a multiple of the empirical formula $(Molecular \ Formula = n \times Empirical \ Formula)$.
Since the molecular formulas are different,the value of $n$ is different for each compound.
Consequently,the molecular weights $(M = n \times \text{Empirical Formula Weight})$ will be different.

(D) To find the empirical formula,we calculate the mole ratio of the elements:
$1$. For element $A$: Mass = $50$,Atomic mass = $10$. Moles = $\frac{50}{10} = 5$.
$2$. For element $B$: Mass = $50$,Atomic mass = $40$. Moles = $\frac{50}{40} = 1.25$.
$3$. Divide by the smallest number of moles $(1.25)$:
For $A$: $\frac{5}{1.25} = 4$.
For $B$: $\frac{1.25}{1.25} = 1$.
Therefore,the empirical formula is $A_4B$.

(A) Given that $0.1 \ mol$ of the carbohydrate contains $1 \ g$ of hydrogen.
Therefore,$1 \ mol$ of the carbohydrate contains $10 \ g$ of hydrogen.
The empirical formula is $CH_2O$.
Let the molecular formula be $(CH_2O)_n$,which contains $2n$ moles of hydrogen atoms.
The molar mass of hydrogen in $1 \ mol$ of the compound is $2n \times 1 \ g/mol = 2n \ g$.
Equating this to the given mass: $2n = 10$,so $n = 5$.
Thus,the molecular formula is $(CH_2O)_5 = C_5H_{10}O_5$.

(C) Let the total mass of the compound be $100 \ g$. Since masses are equal,mass of $X = 50 \ g$ and mass of $Y = 50 \ g$.
Number of moles of $X = \frac{50}{30} = \frac{5}{3} \approx 1.67$.
Number of moles of $Y = \frac{50}{20} = \frac{5}{2} = 2.5$.
Ratio of moles $X:Y = \frac{5}{3} : \frac{5}{2} = 2 : 3$.
Empirical formula is $X_2Y_3$.
Empirical formula mass $= (2 \times 30) + (3 \times 20) = 60 + 60 = 120$.
Since the given molecular mass is $120$,the molecular formula is the same as the empirical formula,which is $X_2Y_3$.

(D) The empirical formula mass of $CH$ is $12 + 1 = 13 \ g/mol$.
The value of $n$ is calculated as: $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{78}{13} = 6$.
The molecular formula is given by: $(\text{Empirical formula})_n = (CH)_6 = C_6H_6$.

(B) To find the percentage of chlorine by weight,we use the formula: $\text{Percentage of Cl} = (\text{Number of Cl atoms} \times 35.5 / \text{Molar mass of compound}) \times 100$.
$I.$ $DDT$ $(C_{14}H_9Cl_5)$: Molar mass $\approx 354.5 \ g/mol$. $\%Cl = (5 \times 35.5 / 354.5) \times 100 \approx 50.06\%$.
$II.$ Gammaxene $(C_6H_6Cl_6)$: Molar mass $\approx 291 \ g/mol$. $\%Cl = (6 \times 35.5 / 291) \times 100 \approx 73.19\%$.
$III.$ Carbon tetrachloride $(CCl_4)$: Molar mass $\approx 154 \ g/mol$. $\%Cl = (4 \times 35.5 / 154) \times 100 \approx 92.20\%$.
$IV.$ Chlorobenzene $(C_6H_5Cl)$: Molar mass $\approx 112.5 \ g/mol$. $\%Cl = (1 \times 35.5 / 112.5) \times 100 \approx 31.55\%$.
Comparing the percentages: $31.55\% (IV) < 50.06\% (I) < 73.19\% (II) < 92.20\% (III)$.
Thus,the increasing order is $IV < I < II < III$.

(D) To find the molecular formula,we first determine the empirical formula:

Element	Percentage	Atomic Mass	Moles	Simplest Ratio
$C$	$60\%$	$12$	$60 / 12 = 5$	$5 / 1.667 \approx 3$
$H$	$13.33\%$	$1$	$13.33 / 1 = 13.33$	$13.33 / 1.667 \approx 8$
$O$	$26.67\%$	$16$	$26.67 / 16 = 1.667$	$1.667 / 1.667 = 1$

Empirical formula = $C_3H_8O$
Empirical formula mass = $(3 \times 12) + (8 \times 1) + (1 \times 16) = 36 + 8 + 16 = 60$
Given Molecular weight = $60$
$n = \text{Molecular weight} / \text{Empirical formula mass} = 60 / 60 = 1$
Molecular formula = $n \times (\text{Empirical formula}) = 1 \times (C_3H_8O) = C_3H_8O$.