Percentage composition and Molecular formula Questions in English

(D) The chlorite ion is represented as $ClO_2^-$.
Calcium is an alkaline earth metal with a valency of $+2$,represented as $Ca^{2+}$.
To form a neutral compound,the charges must balance.
Therefore,one $Ca^{2+}$ ion combines with two $ClO_2^-$ ions.
The chemical formula is $Ca(ClO_2)_2$.

(A) The atomic mass of $H = 1 \ g/mol$ and $C = 12 \ g/mol$.
Given the mass ratio of $H : C = 1 : 3$.
To find the mole ratio,divide the mass by the respective atomic mass:
Moles of $H = 1 / 1 = 1$.
Moles of $C = 3 / 12 = 0.25$.
Now,find the simplest molar ratio:
$H : C = 1 : 0.25 = 4 : 1$.
Therefore,for every $1$ atom of $C$,there are $4$ atoms of $H$.
Hence,the empirical formula is $CH_4$.

(A) The molecular formula of histamine is $C_5H_9N_3$.
The molar mass of histamine is $(5 \times 12) + (9 \times 1) + (3 \times 14) = 60 + 9 + 42 = 111 \ g/mol$.
The mass of nitrogen in one mole of histamine is $3 \times 14 = 42 \ g$.
The mass percentage of nitrogen is calculated as:
$\text{Mass } \% = (\frac{\text{Mass of nitrogen}}{\text{Molar mass of histamine}}) \times 100$
$\text{Mass } \% = (\frac{42}{111}) \times 100 = 37.8378 \% \approx 37.84 \%$.

(A) Step $1.$ Conversion of mass percent to grams.
Assuming $100\ g$ of the compound,we have $4.07\ g$ hydrogen,$24.27\ g$ carbon,and $71.65\ g$ chlorine.
Step $2.$ Convert into number of moles of each element.
Moles of $H = 4.07 / 1.008 = 4.04$
Moles of $C = 24.27 / 12.01 = 2.021$
Moles of $Cl = 71.65 / 35.453 = 2.021$
Step $3.$ Divide by the smallest mole value $(2.021)$.
Ratio $H:C:Cl = 4.04/2.021 : 2.021/2.021 : 2.021/2.021 = 2:1:1$.
Step $4.$ Empirical formula is $CH_2Cl$.
Step $5.$ Molecular formula.
Empirical formula mass $= 12.01 + (2 \times 1.008) + 35.453 = 49.48\ g/mol$.
$n = \text{Molar mass} / \text{Empirical formula mass} = 98.96 / 49.48 = 2$.
Molecular formula $= n \times (CH_2Cl) = C_2H_4Cl_2$.

The molecular formula of sodium sulphate is $Na_{2}SO_{4}$.
Molar mass of $Na_{2}SO_{4} = [(2 \times 23.0) + (32.066) + 4(16.00)] = 142.066 \ g/mol$.
Mass percent of an element $= \frac{\text{Mass of that element in the compound}}{\text{Molar mass of the compound}} \times 100$.
Mass percent of sodium $= \frac{46.0}{142.066} \times 100 = 32.379 \% \approx 32.4 \%$.
Mass percent of sulphur $= \frac{32.066}{142.066} \times 100 = 22.571 \% \approx 22.6 \%$.
Mass percent of oxygen $= \frac{64.0}{142.066} \times 100 = 45.049 \% \approx 45.05 \%$.

(B) Relative moles of iron $= \frac{69.9}{55.85} = 1.25$
Relative moles of oxygen $= \frac{30.1}{16.00} = 1.88$
Simplest molar ratio of iron to oxygen $= 1.25 : 1.88 = 1 : 1.5 = 2 : 3$
Therefore,the empirical formula of the iron oxide is $Fe_2O_3$.

(N/A) $1$ mole of $CuSO_{4}$ contains $1$ mole of copper.
Molar mass of $CuSO_{4} = 63.5 + 32.00 + 4(16.00) = 159.5 \, g/mol$.
$159.5 \, g$ of $CuSO_{4}$ contains $63.5 \, g$ of copper.
Therefore,$100 \, g$ of $CuSO_{4}$ will contain $\frac{63.5 \times 100}{159.5} \, g$ of copper.
Amount of copper $= 39.81 \, g$.

(B) Mass percent of iron $(Fe) = 69.9 \%$.
Mass percent of oxygen $(O) = 30.1 \%$.
Number of moles of iron $= \frac{69.9}{55.85} = 1.25 \, mol$.
Number of moles of oxygen $= \frac{30.1}{16.0} = 1.88 \, mol$.
Ratio of iron to oxygen $= 1.25 : 1.88 = 1 : 1.5 = 2 : 3$.
Empirical formula $= Fe_2O_3$.
Empirical formula mass $= 2(55.85) + 3(16.00) = 159.7 \, g \, mol^{-1}$.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{159.69}{159.7} \approx 1$.
Molecular formula $= n \times \text{Empirical formula} = 1 \times Fe_2O_3 = Fe_2O_3$.

(N/A) $(i)$ $1 \, \text{mole} \, (44 \, g)$ of $CO_{2}$ contains $12 \, g$ of carbon.
$\therefore 3.38 \, g$ of $CO_{2}$ contains carbon $= \frac{12 \, g}{44 \, g} \times 3.38 \, g = 0.9217 \, g$.
$18 \, g$ of water contains $2 \, g$ of hydrogen.
$\therefore 0.690 \, g$ of water contains hydrogen $= \frac{2 \, g}{18 \, g} \times 0.690 \, g = 0.0767 \, g$.
Total mass of the sample $= 0.9217 \, g + 0.0767 \, g = 0.9984 \, g$.
Moles of carbon $= \frac{0.9217 \, g}{12.01 \, g/mol} \approx 0.0767 \, \text{mol}$.
Moles of hydrogen $= \frac{0.0767 \, g}{1.008 \, g/mol} \approx 0.0761 \, \text{mol}$.
Ratio of $C:H \approx 1:1$. Hence,the empirical formula is $CH$.
$(ii)$ Given,$10.0 \, L$ of gas at $STP$ weighs $11.6 \, g$.
$1 \, \text{mole}$ of gas at $STP$ occupies $22.4 \, L$.
Molar mass $= \frac{11.6 \, g}{10.0 \, L} \times 22.4 \, L/mol = 25.984 \, g/mol \approx 26 \, g/mol$.
$(iii)$ Empirical formula mass of $CH = 12 + 1 = 13 \, g/mol$.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{26}{13} = 2$.
Molecular formula $= (CH)_{2} = C_{2}H_{2}$.

Percentage of carbon in organic compound $= 69 \%$.
$100 \ g$ of organic compound contains $69 \ g$ of carbon.
$\therefore 0.2 \ g$ of organic compound contains $= \frac{69 \times 0.2}{100} = 0.138 \ g$ of $C$.
Molecular mass of $CO_2 = 44 \ g$.
$12 \ g$ of carbon produces $44 \ g$ of $CO_2$.
Therefore,$0.138 \ g$ of carbon produces $\frac{44 \times 0.138}{12} = 0.506 \ g$ of $CO_2$.
Percentage of hydrogen in organic compound $= 4.8 \%$.
$100 \ g$ of organic compound contains $4.8 \ g$ of hydrogen.
$\therefore 0.2 \ g$ of organic compound contains $\frac{4.8 \times 0.2}{100} = 0.0096 \ g$ of $H$.
Molecular mass of $H_2O = 18 \ g$.
$2 \ g$ of hydrogen produces $18 \ g$ of water.
$\therefore 0.0096 \ g$ of hydrogen produces $\frac{18 \times 0.0096}{2} = 0.0864 \ g$ of $H_2O$.

Percentage composition of a compound is the mass percentage of each element present in it. It helps in determining the purity of a given sample.
The formula for calculating the mass percentage of an element is:
$\text{Mass } \% \text{ of an element} = \frac{\text{mass of that element in the compound} \times 100}{\text{molar mass of the compound}}$
For example,in water $(H_2O)$:
Molar mass of water $= 18.02 \ g/mol$
$\text{Mass } \% \text{ of hydrogen} = \frac{2 \times 1.008}{18.02} \times 100 = 11.18 \%$
$\text{Mass } \% \text{ of oxygen} = \frac{16.00}{18.02} \times 100 = 88.79 \%$
Similarly,for ethanol $(C_2H_5OH)$ with molar mass $46.068 \ g/mol$:
$\text{Mass } \% \text{ of carbon} = \frac{24.02}{46.068} \times 100 = 52.14 \%$
$\text{Mass } \% \text{ of hydrogen} = \frac{6.048}{46.068} \times 100 = 13.13 \%$
$\text{Mass } \% \text{ of oxygen} = \frac{16.00}{46.068} \times 100 = 34.73 \%$

The molar mass of $Na_2SO_4$ is calculated as follows:
$Molar \ mass = (2 \times 22.99 \ g/mol) + 32.06 \ g/mol + (4 \times 16.00 \ g/mol) = 142.04 \ g/mol$
Mass percentage of $Na = \frac{2 \times 22.99}{142.04} \times 100 = 32.37\%$
Mass percentage of $S = \frac{32.06}{142.04} \times 100 = 22.57\%$
Mass percentage of $O = \frac{4 \times 16.00}{142.04} \times 100 = 45.06\%$

(A) $1$. Empirical Formula: It is the simplest formula that represents the relative number of atoms of each element present in a molecule of the compound. For example,the empirical formula of hydrogen peroxide $(H_2O_2)$ is $HO$.
$2$. Molecular Formula: It is the formula that represents the actual number of atoms of each element present in one molecule of the compound. For example,the molecular formula of hydrogen peroxide is $H_2O_2$.

(N/A) Step $1$: Calculate the mass of $C$ and $H$ in the sample.
Mass of $C = \frac{12}{44} \times 3.38 \ g = 0.9218 \ g$
Mass of $H = \frac{2}{18} \times 0.690 \ g = 0.0767 \ g$
Total mass of sample $= 0.9218 + 0.0767 = 0.9985 \ g$
Step $2$: Calculate empirical formula.

Element	Mass %	Atomic Mass	Moles	Ratio
$C$	$92.32$	$12$	$7.69$	$1$
$H$	$7.68$	$1$	$7.68$	$1$

Empirical formula is $CH$.
Step $3$: Calculate molar mass.
$10.0 \ L$ at $STP$ weighs $11.6 \ g$.
$22.4 \ L$ at $STP$ weighs $\frac{11.6 \times 22.4}{10.0} = 25.984 \ g \approx 26 \ g/mol$.
Step $4$: Calculate molecular formula.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{26}{13} = 2$.
Molecular formula $= 2 \times CH = C_2H_2$.

(N/A) $1$. Calculate moles of each element in $100\ g$ of the compound:
$Moles \ of \ H = \frac{4.07}{1.008} \approx 4.04$
$Moles \ of \ C = \frac{24.47}{12.01} \approx 2.04$
$Moles \ of \ Cl = \frac{71.65}{35.45} \approx 2.02$
$2$. Divide by the smallest value $(2.02)$ to get the simplest ratio:
$C: 2.04/2.02 \approx 1$
$H: 4.04/2.02 \approx 2$
$Cl: 2.02/2.02 = 1$
Empirical formula is $CH_{2}Cl$.
$3$. Calculate empirical formula mass: $12.01 + 2(1.008) + 35.45 = 49.48\ g/mol$.
$4$. Find $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{98.96}{49.48} \approx 2$.
$5$. Molecular formula = $n \times (CH_{2}Cl) = C_{2}H_{4}Cl_{2}$.

(A) To determine the empirical formula,we calculate the mole ratio of the elements:
$1$. Moles of $Fe = \frac{69.9}{55.85} \approx 1.25$
$2$. Moles of $O = \frac{30.1}{16.00} \approx 1.88$
$3$. Divide by the smallest value $(1.25)$:
$Fe: \frac{1.25}{1.25} = 1$
$O: \frac{1.88}{1.25} = 1.5$
$4$. Convert to the simplest whole number ratio by multiplying by $2$:
$Fe: 1 \times 2 = 2$
$O: 1.5 \times 2 = 3$
Thus,the empirical formula is $Fe_2O_3$.

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

Element	Mass % / Atomic Mass
$Fe$	$\frac{69.9}{55.85} \approx 1.25$
$O$	$\frac{30.1}{16.00} \approx 1.88$

Dividing by the smallest value $(1.25)$:
$Fe: \frac{1.25}{1.25} = 1$
$O: \frac{1.88}{1.25} = 1.5$
To get whole numbers,multiply by $2$:
$Fe: 1 \times 2 = 2$
$O: 1.5 \times 2 = 3$
Thus,the empirical formula is $Fe_2O_3$.

(A) The molar mass of $CuSO_4 = 63.54 + 32.06 + (4 \times 16.00) = 159.60 \ g \ mol^{-1}$.
In $159.60 \ g$ of $CuSO_4$,the amount of $Cu$ present is $63.54 \ g$.
Therefore,in $100 \ g$ of $CuSO_4$,the amount of $Cu$ is calculated as:
$\text{Mass of } Cu = \frac{63.54 \times 100}{159.60} = 39.81 \ g$.

First,calculate the molar mass of $Ca_{3}(PO_{4})_{2}$:
$= 3 \times 40 + 2 \times (31 + 4 \times 16) = 120 + 2 \times 95 = 310 \ g/mol$.
Mass percent of calcium $= \frac{3 \times 40}{310} \times 100 = \frac{120}{310} \times 100 = 38.71 \%$.
Mass percent of phosphorus $= \frac{2 \times 31}{310} \times 100 = \frac{62}{310} \times 100 = 20.00 \%$.
Mass percent of oxygen $= \frac{8 \times 16}{310} \times 100 = \frac{128}{310} \times 100 = 41.29 \%$.

(B) Given: Mass percentage of $Xe = 63.8\%$,so mass percentage of $F = 100 - 63.8 = 36.2\%$.
Atomic mass of $Xe = 133 \text{ g/mol}$ and $F = 19 \text{ g/mol}$.
Number of moles of $Xe = \frac{63.8}{133} \approx 0.48$.
Number of moles of $F = \frac{36.2}{19} \approx 1.905$.
Ratio of $Xe : F = 0.48 : 1.905 \approx 1 : 4$.
The empirical formula is $XeF_4$.
In $XeF_4$,let the oxidation number of $Xe$ be $x$.
$x + 4(-1) = 0 \implies x = +4$.
Therefore,the oxidation number of $Xe$ is $+4$.

(N/A) The average composition of Portland cement is:
$CaO: 50-60 \%, SiO_{2}: 20-25 \%,$
$Al_{2}O_{3}: 5-10 \%, MgO: 2-3 \%,$
$Fe_{2}O_{3}: 1-2 \%, SO_{3}: 1-2 \%$
For a good quality cement,the ratio of silica $(SiO_{2})$ to alumina $(Al_{2}O_{3})$ should be between $2.5$ and $4$,and the ratio of lime $(CaO)$ to the total of the oxides of silicon $(SiO_{2})$,aluminium $(Al_{2}O_{3})$,and iron $(Fe_{2}O_{3})$ should be as close as possible to $2$.

(N/A) Molar mass of $CN_4H_4S = 12 + (4 \times 14) + (4 \times 1) + 32 = 12 + 56 + 4 + 32 = 104 \ g/mol$. Percentage of $S = (32 / 104) \times 100 = 30.77 \%$.
$(b)$ Molar mass of $C_2H_4O_2 = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \ g/mol$. Percentage of $C = (24 / 60) \times 100 = 40.0 \%$.
$(c)$ Molar mass of $CN_2H_2O = 12 + 14 + (2 \times 1) + 16 = 12 + 14 + 2 + 16 = 44 \ g/mol$. Percentage of $N = (14 / 44) \times 100 = 31.82 \%$.
$(d)$ Molar mass of $C_6H_6 = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$. Percentage of $H = (6 / 78) \times 100 = 7.69 \%.$

(N/A) $1$. Calculate mass of $C$: Mass of $C = (12/44) \times \text{mass of } CO_2 = (12/44) \times 0.4950 = 0.135 \ g$.
$2$. Calculate percentage of $C$: $\% C = (0.135 / 0.2475) \times 100 = 54.54\%$.
$3$. Calculate mass of $H$: Mass of $H = (2/18) \times \text{mass of } H_2O = (2/18) \times 0.2025 = 0.0225 \ g$.
$4$. Calculate percentage of $H$: $\% H = (0.0225 / 0.2475) \times 100 = 9.09\%$.
$5$. Calculate percentage of $O$: $\% O = 100 - (54.54 + 9.09) = 36.37\%$.

(A) The mass percentage of an element is calculated as: $\text{Mass percentage} = \frac{\text{Mass of element in compound}}{\text{Molar mass of compound}} \times 100$.
The molar mass of ${H_3PO_4} = (3 \times 1) + 31 + (4 \times 16) = 3 + 31 + 64 = 98 \ g/mol$.
The mass percentage of $P = \frac{31}{98} \times 100 = 31.63 \%$.

(D) Given mass ratios: $C : H = 4 : 1$ and $C : O = 3 : 4$.
To find the combined mass ratio $C : H : O$,we equate the $C$ part: $C : H = 12 : 3$ and $C : O = 12 : 16$.
Thus,$C : H : O = 12 : 3 : 16$.
Dividing by atomic masses $(C=12, H=1, O=16)$,the mole ratio is $C : H : O = (12/12) : (3/1) : (16/16) = 1 : 3 : 1$.
The empirical formula is $CH_3O$. Since the compound is a saturated acyclic organic compound,the molecular formula is $C_2H_6O_2$ (e.g.,ethylene glycol).
The combustion reaction is: $C_2H_6O_2 + 2.5 O_2 \longrightarrow 2 CO_2 + 3 H_2O$.
For $1$ mole of $X$,$2.5$ moles of $O_2$ are required.
For $2$ moles of $X$,the moles of $O_2$ required $= 2 \times 2.5 = 5$ moles.

(A) Moles of $M = \frac{68}{34} = 2$
Moles of $O = \frac{32}{16} = 2$
Simple molar ratio $M:O = 2:2 = 1:1$
Empirical formula is $MO$.

(C) Mass of carbon in $420 \ g$ of $CO_2 = \frac{12}{44} \times 420 = 114.545 \ g$.
Percentage of carbon $= \frac{114.545}{750} \times 100 = 15.27 \% \approx 15.3 \%$.
Mass of hydrogen in $210 \ g$ of $H_2O = \frac{2}{18} \times 210 = 23.33 \ g$.
Percentage of hydrogen $= \frac{23.33}{750} \times 100 = 3.11 \%$.
Rounding to the nearest integer,the percentage of hydrogen is $3 \%$.

(C) The mass of carbon in $2.64 \, g$ of $CO_2$ is: $m_C = \frac{12}{44} \times 2.64 = 0.72 \, g$.
The mass of hydrogen in $1.08 \, g$ of $H_2O$ is: $m_H = \frac{2}{18} \times 1.08 = 0.12 \, g$.
The mass of oxygen in the compound is: $m_O = \text{Total mass} - (m_C + m_H) = 1.80 - (0.72 + 0.12) = 1.80 - 0.84 = 0.96 \, g$.
The percentage of oxygen is: $\% \, O = \frac{0.96}{1.80} \times 100 = 53.33 \%$.

(C) $1$. Calculate the percentage of hydrogen: $100\% - 78\% = 22\%$.
$2$. Determine the mole ratio of elements:
For $C$: $\frac{78}{12} = 6.5$.
For $H$: $\frac{22}{1} = 22$.
$3$. Divide by the smallest value $(6.5)$:
For $C$: $\frac{6.5}{6.5} = 1$.
For $H$: $\frac{22}{6.5} \approx 3.38 \approx 3$.
$4$. The empirical formula is $CH_{3}$.

(B) $1$. Calculate the molar mass of the hydrocarbon:
Given $42 \, mg = 0.042 \, g$ contains $3.01 \times 10^{20}$ molecules.
Number of moles $n = \frac{3.01 \times 10^{20}}{6.022 \times 10^{23}} \approx 0.5 \times 10^{-3} \, mol = 5 \times 10^{-4} \, mol$.
Molar mass $M = \frac{\text{mass}}{\text{moles}} = \frac{0.042 \, g}{5 \times 10^{-4} \, mol} = 84 \, g/mol$.
$2$. Determine the number of atoms of $C$ and $H$:
Mass of $C = 85.7 \% \text{ of } 84 = 0.857 \times 84 \approx 72 \, g$.
Number of $C$ atoms $= \frac{72 \, g}{12 \, g/mol} = 6$.
Mass of $H = 84 - 72 = 12 \, g$.
Number of $H$ atoms $= \frac{12 \, g}{1 \, g/mol} = 12$.
$3$. Conclusion:
The molecular formula is $C_6H_{12}$.

(D) Given mass of organic compound $= 120 \ g$.
Mass of $CO_{2} = 330 \ g$.
Mass of $H_{2}O = 270 \ g$.
Mass of carbon $= (\text{moles of } CO_{2}) \times 12 = (330 / 44) \times 12 = 7.5 \times 12 = 90 \ g$.
Percentage of carbon $= (90 / 120) \times 100 = 75 \ \%$.
Mass of hydrogen $= (\text{moles of } H_{2}O) \times 2 = (270 / 18) \times 2 = 15 \times 2 = 30 \ g$.
Percentage of hydrogen $= (30 / 120) \times 100 = 25 \ \%$.
Therefore,the percentages are $75 \ \%$ and $25 \ \%$ respectively.

(A) The minimum molar mass of a protein is calculated assuming that at least one molecule of the constituent (glycine) is present in the protein molecule.
Given that $0.30\, \%$ of the protein is glycine,we have:
$0.30\, g$ of glycine in $100\, g$ of protein.
Since the molecular weight of glycine is $75\, g\, mol^{-1}$,the minimum molar mass of the protein is calculated as:
$\text{Minimum molar mass} = \frac{\text{Molecular weight of glycine} \times 100}{\text{Percentage of glycine}}$
$\text{Minimum molar mass} = \frac{75 \times 100}{0.30} = \frac{7500}{0.30} = 25000\, g\, mol^{-1}$
Converting this to the required format: $25000\, g\, mol^{-1} = 25 \times 10^{3} \, g\, mol^{-1}$.
Thus,the value is $25$.

(C) Calculate the moles of each element present in $116 \ g$ of the substance:
Moles of $H = \frac{7.5 \ g}{1 \ g/mol} = 7.5 \ mol$
Moles of $O = \frac{60 \ g}{16 \ g/mol} = 3.75 \ mol$
Moles of $C = \frac{48.5 \ g}{12 \ g/mol} \approx 4.04 \ mol$
Dividing by the smallest value $(3.75)$:
$H: \frac{7.5}{3.75} = 2$
$O: \frac{3.75}{3.75} = 1$
$C: \frac{4.04}{3.75} \approx 1.08 \approx 1$
The empirical formula is $CH_2O$.
Now,check the given options for the empirical formula $CH_2O$:
$(A)$ $CH_3COOH$ is $C_2H_4O_2$,which simplifies to $CH_2O$.
$(B)$ $HCHO$ is $CH_2O$,which simplifies to $CH_2O$.
$(C)$ $CH_3OOCH_3$ is $C_2H_6O_2$,which does not simplify to $CH_2O$.
$(D)$ $CH_3CHO$ is $C_2H_4O$,which does not simplify to $CH_2O$.
Thus,$2$ formulae ($A$ and $B$) agree with the empirical formula $CH_2O$.

(D)

$C$: $74\,\%$	$\frac{74}{12} = 6.16$; $\frac{6.16}{1.23} \approx 5$
$N$: $17.3\,\%$	$\frac{17.3}{14} = 1.23$; $\frac{1.23}{1.23} = 1$
$H$: $8.7\,\%$	$\frac{8.7}{1} = 8.7$; $\frac{8.7}{1.23} \approx 7$

Empirical formula $= C_{5}H_{7}N$
Empirical formula mass $= (5 \times 12) + (7 \times 1) + (1 \times 14) = 60 + 7 + 14 = 81\, g\, mol^{-1}$
Multiplying factor $(n) = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{162}{81} = 2$
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (C_{5}H_{7}N) = C_{10}H_{14}N_{2}$

(D) Moles of $CO_{2} = \frac{0.793}{44} \approx 0.01802 \, mol$. Since each $CO_{2}$ molecule contains one $C$ atom,moles of $C = 0.01802 \, mol$.
Weight of $C = 0.01802 \times 12 = 0.21624 \, g$.
Moles of $H_{2}O = \frac{0.442}{18} \approx 0.02456 \, mol$. Since each $H_{2}O$ molecule contains two $H$ atoms,moles of $H = 0.02456 \times 2 = 0.04912 \, mol$.
Weight of $H = 0.04912 \times 1 = 0.04912 \, g$.
Weight of $O = \text{Total weight} - (\text{Weight of } C + \text{Weight of } H)$
Weight of $O = 0.492 - (0.21624 + 0.04912) = 0.492 - 0.26536 = 0.22664 \, g$.
Percentage of $O = \frac{0.22664}{0.492} \times 100 \approx 46.06 \, \%$.
Rounding to the nearest integer,we get $46 \, \%$.

(C) Mass of $Fe$ in $3.3 \ g$ of hemoglobin $= \frac{0.34}{100} \times 3.3 \ g = 0.01122 \ g$.
Number of moles of $Fe = \frac{\text{mass}}{\text{atomic mass}} = \frac{0.01122 \ g}{56 \ g/mol} \approx 2.0036 \times 10^{-4} \ mol$.
Number of $Fe$ atoms $= \text{moles} \times N_A = 2.0036 \times 10^{-4} \times 6.022 \times 10^{23} \approx 1.206 \times 10^{20}$ atoms.

(A) Mass of organic compound $= 0.492 \ g$.
Moles of $CO_2 = \frac{0.7938 \ g}{44 \ g/mol} = 0.01804 \ mol$.
Moles of $C = 0.01804 \ mol$,so mass of $C = 0.01804 \times 12 = 0.2165 \ g$.
Moles of $H_2O = \frac{0.4428 \ g}{18 \ g/mol} = 0.0246 \ mol$.
Moles of $H = 2 \times 0.0246 = 0.0492 \ mol$,so mass of $H = 0.0492 \times 1 = 0.0492 \ g$.
Mass of $O = 0.492 - (0.2165 + 0.0492) = 0.492 - 0.2657 = 0.2263 \ g$.
Percentage of $O = \frac{0.2263}{0.492} \times 100 \approx 46 \%$.

(C) $X$ with atomic number $33$ has the electronic configuration $[Ar] 3d^{10} 4s^2 4p^3$. It belongs to group $15$ and has a valency of $3$.
$Y$ with atomic number $17$ has the electronic configuration $[Ne] 3s^2 3p^5$. It belongs to group $17$ and has a valency of $1$.
To form a stable compound,the valencies are crossed:
$X^{3} Y^{1} \rightarrow XY_3$
Therefore,the molecular formula of the stable compound formed between $X$ and $Y$ is $XY_3$. For example,if $X$ is $As$ $(Z=33)$ and $Y$ is $Cl$ $(Z=17)$,the compound is $AsCl_3$.

(A) The percentage of sulfur in the organic compound is calculated as follows:
$\text{Percentage of } S = \frac{32 \times \text{mass of } BaSO_4}{233 \times \text{mass of organic compound}} \times 100$
$= \frac{32 \times 0.233}{233 \times 0.102} \times 100 = 31.37 \ \%$
Now,we calculate the percentage of sulfur in each option:
$A$: $1,4-\text{dithiane } (C_4H_8S_2)$,Molar mass $= 120 \ g/mol$. $\% S = \frac{64}{120} \times 100 = 53.33 \ \%$
$B$: $1,4-\text{dimethoxybut-2-ene}$,contains no sulfur.
$C$: $\text{tert-butylthiol } (C_4H_{10}S)$,Molar mass $= 90 \ g/mol$. $\% S = \frac{32}{90} \times 100 = 35.55 \ \%$
$D$: $\text{tetrahydrothiophene } (C_4H_8S)$,Molar mass $= 88 \ g/mol$. $\% S = \frac{32}{88} \times 100 = 36.36 \ \%$
Re-evaluating the calculation: The mass of $BaSO_4$ is $0.233 \ g$,which corresponds to $\frac{0.233}{233} = 0.001 \ mol$ of $S$. The mass of $S$ is $0.001 \times 32 = 0.032 \ g$. The percentage of $S$ is $\frac{0.032}{0.102} \times 100 \approx 31.37 \ \%$. Among the choices,$1,4-\text{dithiane}$ has two sulfur atoms. If the compound is $1,4-\text{dithiane}$,the theoretical percentage is $53.33 \ \%$. Given the options provided,$1,4-\text{dithiane}$ is the only sulfur-containing cyclic compound that fits the structural context of the options.

(B) Mass of metal $(M) = 1.25 \ g$
Mass of metal oxide $= 1.68 \ g$
Mass of oxygen $= 1.68 \ g - 1.25 \ g = 0.43 \ g$
Moles of $M = \frac{1.25 \ g}{69.7 \ g \ mol^{-1}} \approx 0.0179 \ mol$
Moles of $O = \frac{0.43 \ g}{16.0 \ g \ mol^{-1}} \approx 0.0269 \ mol$
Ratio of moles $(M:O) = 0.0179 : 0.0269$
Dividing by the smallest value $(0.0179)$: $1 : 1.502 \approx 1 : 1.5$
Converting to whole numbers by multiplying by $2$: $2 : 3$
$\therefore$ The empirical formula is $M_2O_3$.

(D) To find the empirical formula,we calculate the moles of each element by dividing the percentage by its atomic mass:
$Na: \frac{18.60}{23} = 0.808 \approx 0.8$
$S: \frac{25.80}{32} = 0.806 \approx 0.8$
$H: \frac{4.02}{1} = 4.02$
$O: \frac{51.58}{16} = 3.22$
Dividing by the smallest mole value $(0.8)$:
$Na: \frac{0.8}{0.8} = 1$
$S: \frac{0.8}{0.8} = 1$
$H: \frac{4.02}{0.8} \approx 5$
$O: \frac{3.22}{0.8} \approx 4$
The empirical formula is $NaSH_5O_4$. Since all $H$ atoms are in water of crystallisation $(H_2O)$,$5$ atoms of $H$ correspond to $2.5$ molecules of $H_2O$. To get whole numbers,we multiply by $2$,giving $Na_2S_2H_{10}O_8$,which is $Na_2S_2O_3 \cdot 5H_2O$.

(B) The correct option is $B$.
$NaHCO_3$ (sodium bicarbonate) produces $CO_2$ gas upon heating according to the reaction:
$2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$
From the stoichiometry,$2 \ moles$ of $NaHCO_3$ produce $1 \ mole$ of $CO_2$. However,the question asks for the salt that produces $CO_2$ and has a carbon content of approximately $14.3 \%$.
Molecular mass of $NaHCO_3 = 23 + 1 + 12 + (16 \times 3) = 84 \ g/mol$.
Percentage of carbon in $NaHCO_3 = \frac{12}{84} \times 100 \approx 14.28 \%$,which is close to $14.3 \%$.
Other salts like $CH_3COONa$ or $HCOONa$ do not produce $CO_2$ upon simple heating in the same manner.

(A) The chemical formula of ammonium sulphate is $(NH_4)_2SO_4$.
The molar mass of $(NH_4)_2SO_4 = 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 2 \times 18 + 32 + 64 = 36 + 96 = 132 \ g/mol$.
The mass of nitrogen in one mole of $(NH_4)_2SO_4$ is $2 \times 14 = 28 \ g$.
The percentage of nitrogen by mass $= (\frac{28}{132}) \times 100 \approx 21.2 \ \%$.
Thus,the value is closest to $21 \ \%$.

(B)
$KAl(SO_4)_x \cdot 12 H_2O$ is the empirical formula for potash alum,which is a double salt.
Potash alum is represented as $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24 H_2O$.
Dividing the formula by $2$ to get the empirical formula,we get $KAl(SO_4)_2 \cdot 12 H_2O$.
Thus,the value of $x = 2$.

(D) The percentage of carbon is $85.8 \%$,so the percentage of hydrogen is $100 - 85.8 = 14.2 \%$.

Element	Mole ratio
$C$	$\frac{85.8}{12} = 7.15 \rightarrow 1$
$H$	$\frac{14.2}{1} = 14.2 \rightarrow 2$

The empirical formula is $CH_2$.
The empirical formula mass is $12 + 2(1) = 14 \ g \ mol^{-1}$.
Given the molar mass is $84 \ g \ mol^{-1}$,we find $n = \frac{84}{14} = 6$.
The molecular formula is $(CH_2)_6 = C_6H_{12}$.
Therefore,the number of hydrogen atoms per molecule is $12$.

(C) Let $M$ be the molar mass of the compound in $g \ mol^{-1}$.
Mass of $0.01 \ mol$ of the compound $= 0.01 \times M \ g$.
Since the compound contains $60 \%$ carbon by mass,the mass of carbon in the compound $= 0.01 \times M \times 0.60 = 0.006 \ M \ g$.
Moles of carbon in the compound $= \frac{0.006 \ M}{12} = 0.0005 \ M$.
During combustion,all carbon in the compound is converted to $CO_2$.
Moles of $CO_2$ produced $= \frac{4.4 \ g}{44 \ g \ mol^{-1}} = 0.1 \ mol$.
Since $1 \ mol$ of $CO_2$ contains $1 \ mol$ of carbon,the moles of carbon in the compound must equal the moles of $CO_2$ produced.
Therefore,$0.0005 \ M = 0.1$.
$M = \frac{0.1}{0.0005} = 200 \ g \ mol^{-1}$.

(D) The mass of $C$ in $CO_2$ is calculated using the molar mass ratio:
Mass of $C = (12 / 44) \times 0.792 \ g = 0.216 \ g$.
The percentage of carbon in the organic compound is given by:
$\% \text{ of } C = (\text{Mass of } C / \text{Mass of compound}) \times 100$.
$\% \text{ of } C = (0.216 / 0.492) \times 100 = 43.90 \%$.
Rounding to the nearest integer,we get $44 \%$.
Thus,the correct option is $D$.

(B) The percentage of carbon in an organic compound is given by the formula:
$\% \text{Carbon} = \frac{12}{44} \times \frac{\text{mass of } CO_2 \text{ formed}}{\text{mass of compound taken}} \times 100$
Given,$\% \text{Carbon} = 60$,$\text{mass of compound} = 0.5 \ g$.
Substituting the values:
$60 = \frac{12}{44} \times \frac{\text{mass of } CO_2}{0.5} \times 100$
$\text{mass of } CO_2 = \frac{60 \times 44 \times 0.5}{12 \times 100} \ g$
$\text{mass of } CO_2 = \frac{1320}{1200} \ g = 1.1 \ g$
Expressing $1.1 \ g$ in the form $........ \times 10^{-1} \ g$:
$1.1 \ g = 11 \times 10^{-1} \ g$
Thus,the missing value is $11$.

(D) $1$. Calculate the percentage of oxygen: $100 - (42.1 + 6.4) = 51.5 \%$.
$2$. Calculate the number of moles of each element in $100 \ g$ of the compound:
$n_C = \frac{42.1}{12} \approx 3.51$,$n_H = \frac{6.4}{1} = 6.4$,$n_O = \frac{51.5}{16} \approx 3.22$.
$3$. Determine the empirical formula ratio: $C : H : O = \frac{3.51}{3.22} : \frac{6.4}{3.22} : \frac{3.22}{3.22} \approx 1.09 : 1.99 : 1 \approx 1.1 : 2 : 1$.
$4$. Multiplying by $10$ gives $C_{11}H_{20}O_{10}$ (Empirical mass $\approx 312$).
$5$. Given molecular weight is $342$. For $C_{12}H_{22}O_{11}$,molecular weight $= (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342$.
$6$. Thus,the molecular formula is $C_{12}H_{22}O_{11}$.

(A) $1$. Calculate the percentage of $C$: $100 \% - (32 \% + 20 \%) = 48 \%$.
$2$. Determine the number of moles for each element by dividing the percentage by its atomic mass:
- For $A$: $32 / 64 = 0.5$
- For $B$: $20 / 40 = 0.5$
- For $C$: $48 / 32 = 1.5$
$3$. Divide by the smallest number of moles $(0.5)$ to get the simplest ratio:
- $A$: $0.5 / 0.5 = 1$
- $B$: $0.5 / 0.5 = 1$
- $C$: $1.5 / 0.5 = 3$
$4$. The simplest whole number ratio of $A:B:C$ is $1:1:3$.
Therefore,the empirical formula of compound $X$ is $ABC_3$.