Percentage composition and Molecular formula Questions in English

(C) The mineral Beryl has the chemical formula $Be_3 Al_2 Si_6 O_{18}$.
To determine $n$ using the principle of charge neutrality (sum of oxidation states $= 0$):
$Be = +2$,$Al = +3$,$Si = +4$,$O = -2$
$n(+2) + 2(+3) + 6(+4) + 18(-2) = 0$
$2n + 6 + 24 - 36 = 0$
$2n - 6 = 0$
$2n = 6$
$n = 3$

(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$.
Moles of $H$ atoms $= 0.05 \ mol \times 2 = 0.1 \ mol$.
Mass of $H$ atoms in $0.01 \ mol$ of $(X) = 0.1 \ mol \times 1 \ g \ mol^{-1} = 0.1 \ g$.
Mass of $H$ atoms in $1 \ mol$ of $(X) = \frac{0.1 \ g}{0.01 \ mol} = 10 \ g \ mol^{-1}$.
Given that the compound contains $10\%$ hydrogen by mass,we have:
$10\% = \frac{\text{Mass of } H \text{ in } 1 \ mol \text{ of } (X)}{\text{Molar mass of } (X)} \times 100$.
$10 = \frac{10}{M} \times 100$.
$M = 100 \ g \ mol^{-1}$.

(C) Step $1$: Calculate the moles of each element in $100 \ g$ of the compound:
$n_C = \frac{54.2}{12} = 4.516$
$n_H = \frac{9.2}{1} = 9.2$
$n_O = \frac{36.6}{16} = 2.287$
Step $2$: Determine the simplest molar ratio by dividing by the smallest value $(2.287)$:
$C = \frac{4.516}{2.287} \approx 2$
$H = \frac{9.2}{2.287} \approx 4$
$O = \frac{2.287}{2.287} = 1$
Step $3$: The empirical formula is $C_2H_4O$.
Step $4$: Calculate the empirical formula mass:
$E.F. \text{ mass} = (2 \times 12) + (4 \times 1) + (1 \times 16) = 24 + 4 + 16 = 44 \ g \ mol^{-1}$.
Step $5$: Calculate the value of $n$ (where $n = \frac{\text{Molar mass}}{\text{E.F. mass}}$):
$n = \frac{132}{44} = 3$.
Step $6$: Determine the molecular formula:
$\text{Molecular formula} = (C_2H_4O)_3 = C_6H_{12}O_3$.

(B) Total percentage of $C, H, Cl = 14.5 + 1.8 + 64.46 = 80.76\%$.
Percentage of Oxygen $(O) = 100 - 80.76 = 19.24\%$.
Relative number of moles:
$n_C = \frac{14.5}{12} = 1.208$
$n_H = \frac{1.8}{1} = 1.8$
$n_{Cl} = \frac{64.46}{35.5} = 1.815$
$n_O = \frac{19.24}{16} = 1.2025$
Simplest molar ratio (divide by $1.2025$):
$C: \frac{1.208}{1.2025} \approx 1$
$H: \frac{1.8}{1.2025} \approx 1.5$
$Cl: \frac{1.815}{1.2025} \approx 1.5$
$O: \frac{1.2025}{1.2025} = 1$
Simplest whole number ratio $(\times 2)$: $C: 2, H: 3, Cl: 3, O: 2$.
Empirical formula: $C_2H_3Cl_3O_2$.
Empirical formula mass $= (2 \times 12) + (3 \times 1) + (3 \times 35.5) + (2 \times 16) = 24 + 3 + 106.5 + 32 = 165.5 \ g \ mol^{-1}$.
Value $= 1655 \times 10^{-1}$.

(A) Moles of $C = n_{CO_2} = \frac{1.46}{44} \approx 0.03318 \ mol$.
Mass of $C = 0.03318 \times 12 = 0.398 \ g$.
Moles of $H = 2 \times n_{H_2O} = 2 \times \frac{0.567}{18} = 0.063 \ mol$.
Mass of $H = 0.063 \times 1 = 0.063 \ g$.
Mass of $O = 1.0 - (0.398 + 0.063) = 0.539 \ g$.
Moles of $O = \frac{0.539}{16} \approx 0.0337 \ mol$.
Ratio of $C:H:O = 0.03318 : 0.063 : 0.0337 \approx 1:2:1$.
Empirical formula $= CH_2O$.
Empirical formula mass $= 12 + (2 \times 1) + 16 = 30 \ g \ mol^{-1}$.

(B) The percentage of carbon in an organic compound is calculated using the formula: $\% \text{ of } C = \frac{12}{44} \times \frac{\text{mass of } CO_2}{\text{mass of compound}} \times 100$.
Substituting the given values: $\% \text{ of } C = \frac{12}{44} \times \frac{1.46}{0.5} \times 100$.
Calculating the value: $\% \text{ of } C = 0.2727 \times 2.92 \times 100 = 79.636\%$.
Rounding to the nearest integer,we get $80\%$.

(A) The combustion reaction is: $\text{Organic compound} \xrightarrow{\Delta, CuO} CO_2 + H_2O$
$1.$ Calculate the moles of $CO_2$ produced:
$n_{CO_2} = \frac{220 \times 10^{-3} \ g}{44 \ g \ mol^{-1}} = 5 \times 10^{-3} \ mol$
$2.$ Calculate the mass of carbon in $CO_2$:
Since $1 \ mol$ of $CO_2$ contains $1 \ mol$ of $C$,the moles of $C$ are $5 \times 10^{-3} \ mol$.
$m_{C} = 5 \times 10^{-3} \ mol \times 12 \ g \ mol^{-1} = 60 \times 10^{-3} \ g = 60 \ mg$
$3.$ Calculate the percentage of carbon:
$\% \ C = \frac{\text{mass of } C}{\text{mass of compound}} \times 100 = \frac{60 \ mg}{500 \ mg} \times 100 = 12 \%$

(B) In the combustion of an organic compound,all $C$ in $CO_2$ and all $H$ in $H_2O$ comes from the organic compound.
Mass of $C$ in $CO_2 = \frac{12}{44} \times 0.307 \ g = 0.0837 \ g$.
Mass of $H$ in $H_2O = \frac{2}{18} \times 0.127 \ g = 0.0141 \ g$.
Percentage of $H = \frac{0.0141}{0.210} \times 100 = 6.714 \% \approx 6.72 \%$.
Mass of $O$ in the compound = Total mass - (Mass of $C$ + Mass of $H$) = $0.210 - (0.0837 + 0.0141) = 0.1122 \ g$.
Percentage of $O = \frac{0.1122}{0.210} \times 100 = 53.428 \% \approx 53.41 \%$.
Therefore,the percentages of $H$ and $O$ are $6.72 \%$ and $53.41 \%$ respectively.

(C) To determine the least expensive fertilizer,we calculate the percentage of nitrogen $(N)$ in each compound:
$1$. Urea,$(NH_2)_2CO$: Molar mass = $60 \ g/mol$. $\% \ N = (28/60) \times 100 = 46.7 \%$.
$2$. Ammonia,$NH_3$: Molar mass = $17 \ g/mol$. $\% \ N = (14/17) \times 100 = 82.3 \%$.
$3$. Ammonium nitrate,$NH_4NO_3$: Molar mass = $80 \ g/mol$. $\% \ N = (28/80) \times 100 = 35 \%$.
$4$. Guanidine,$HNC(NH_2)_2$: Molar mass = $59 \ g/mol$. $\% \ N = (42/59) \times 100 = 71.1 \%$.
Since the price is based on nitrogen content,the fertilizer with the lowest percentage of nitrogen will be the least expensive per $50 \ kg$ bag. Therefore,ammonium nitrate is the least expensive.

(B) $1$. Calculate the percentage of oxygen: $O = 100 - (57.82 + 3.6) = 38.58 \%$.
$2$. Determine the empirical formula:
$C: \frac{57.82}{12} = 4.818 \Rightarrow \frac{4.818}{2.41} = 2$
$H: \frac{3.6}{1} = 3.6 \Rightarrow \frac{3.6}{2.41} = 1.5 = \frac{3}{2}$
$O: \frac{38.58}{16} = 2.41 \Rightarrow \frac{2.41}{2.41} = 1$
Multiplying by $2$ to get whole numbers: $C = 4, H = 3, O = 2$.
Empirical Formula ($E$.$F$.) = $C_4H_3O_2$.
$3$. Calculate Empirical Formula Mass ($E$.$F$.$M$.): $(4 \times 12) + (3 \times 1) + (2 \times 16) = 48 + 3 + 32 = 83$.
$4$. Calculate Molecular Formula Mass ($M$.$F$.$M$.): $2 \times \text{Vapour Density} = 2 \times 83 = 166$.
$5$. Calculate $n$: $n = \frac{M.F.M.}{E.F.M.} = \frac{166}{83} = 2$.
$6$. Molecular Formula = $n \times E.F. = 2 \times C_4H_3O_2 = C_8H_6O_4$.

(C) $1$. Calculate the percentage of oxygen: $\% O = 100 - (38.71 + 9.67) = 51.62 \%$.
$2$. Calculate the relative number of moles for each element:
$C = \frac{38.71}{12} = 3.226$
$H = \frac{9.67}{1} = 9.67$
$O = \frac{51.62}{16} = 3.226$
$3$. Determine the simplest molar ratio by dividing by the smallest value $(3.226)$:
$C = \frac{3.226}{3.226} = 1$
$H = \frac{9.67}{3.226} \approx 3$
$O = \frac{3.226}{3.226} = 1$
$4$. The empirical formula is $CH_3O$.

(A) At $\text{STP}$,$224 \ mL$ of the gas weighs $0.44 \ g$.
Since $22400 \ mL$ $(1 \ mol)$ of any gas at $\text{STP}$ weighs equal to its molar mass,
Molar mass $= \frac{0.44 \ g}{224 \ mL} \times 22400 \ mL = 44 \ g/mol$.
The molar mass of $N_2O$ is $(2 \times 14) + 16 = 44 \ g/mol$.
Therefore,the compound is $N_2O$.

(D) The chemical formula of ethanol is $C_2H_5OH$.
The molar mass of $C_2H_5OH = (2 \times 12.01) + (6 \times 1.008) + (1 \times 16.00) = 46.07 \text{ g/mol}$.
The mass of oxygen in $1 \text{ mole}$ of $C_2H_5OH$ is $16.00 \text{ g}$.
The mass percent of oxygen $= \frac{\text{Mass of Oxygen}}{\text{Molar mass of } C_2H_5OH} \times 100$.
Mass percent $= \frac{16.00}{46.07} \times 100 \approx 34.73 \% $.

(C) Mass of iron in one molecule of haemoglobin $= \frac{0.33}{100} \times 67200 = 221.76 \ g/mol$.
Number of iron atoms $= \frac{\text{Mass of iron in one molecule}}{\text{Atomic mass of } Fe} = \frac{221.76}{56} = 3.96$.
Since the number of atoms must be an integer,we round $3.96$ to $4$.

(C) The relationship between molecular mass,empirical formula mass,and the integer $n$ is given by the formula:
$\text{Molecular mass} = n \times \text{Empirical formula mass}$
Given that the molecular mass is $129$ and $n = 2$:
$\text{Empirical formula mass} = \frac{\text{Molecular mass}}{n} = \frac{129}{2} = 64.5$.

(D) First,calculate the percentage of oxygen: $\% O = 100 - (40.0 + 6.66) = 53.34 \%$.
Construct the table for empirical formula calculation:

Element	% Composition	Atomic Mass	Mole Ratio	Simple Ratio
$C$	$40.0$	$12$	$40/12 = 3.33$	$1$
$H$	$6.66$	$1$	$6.66/1 = 6.66$	$2$
$O$	$53.34$	$16$	$53.34/16 = 3.33$	$1$

Empirical formula $= CH_{2} O$.
Empirical formula mass $= 12 + (2 \times 1) + 16 = 30$.
Calculate $n = \text{Molar mass} / \text{Empirical formula mass} = 180 / 30 = 6$.
Molecular formula $= n \times (\text{Empirical formula}) = 6 \times (CH_{2} O) = C_{6} H_{12} O_{6}$.

(C) $1$. Calculate the percentage of oxygen: $\% O = 100 - (18.6 + 1.55 + 55.04) = 24.81 \%$.
$2$. Calculate the moles of each element in $100 \ g$ of the compound:
$n_C = \frac{18.6}{12} = 1.55$,
$n_H = \frac{1.55}{1} = 1.55$,
$n_{Cl} = \frac{55.04}{35.5} = 1.55$,
$n_O = \frac{24.81}{16} = 1.55$.
$3$. Determine the simplest molar ratio: $1.55 : 1.55 : 1.55 : 1.55 = 1 : 1 : 1 : 1$.
$4$. The empirical formula is $CHClO$.

(A) The chemical formula of urea is $NH_2CONH_2$.
The molar mass of urea is $(2 \times 14) + (4 \times 1) + 12 + 16 = 60 \ g/mol$.
The number of carbon atoms in one molecule of urea is $1$.
The percentage of carbon by mass is calculated as: $\frac{\text{Mass of Carbon}}{\text{Molar mass of Urea}} \times 100 = \frac{12}{60} \times 100 = 20 \%$.

(C) The molar mass of $NaOH$ is calculated as: $23 + 16 + 1 = 40 \ u$.
The percentage by mass of oxygen is given by the formula: $\frac{\text{Atomic mass of } O}{\text{Molar mass of } NaOH} \times 100$.
Substituting the values: $\frac{16}{40} \times 100 = 40 \%$.

(B) The molecular formula of urea is $NH_2CONH_2$.
We know that the mass percentage of an element in a compound is given by:
$\text{Mass percentage} = \frac{\text{Total mass of the element in one mole of compound}}{\text{Molar mass of the compound}} \times 100$
In one mole of urea $(NH_2CONH_2)$,there is $1$ mole of carbon atom.
The molar mass of carbon is $12 \ g \ mol^{-1}$.
The molar mass of urea is given as $60 \ g \ mol^{-1}$.
Therefore,the mass percentage of carbon in urea $= \frac{12 \ g \ mol^{-1}}{60 \ g \ mol^{-1}} \times 100 = 20 \%$.
Hence,option $B$ is correct.

(B) The percentage of oxygen $= 100 - (52 + 13) = 35 \%$.
Calculating the empirical formula:
$C: 52/12 = 4.33 \rightarrow 2$
$H: 13/1 = 13.00 \rightarrow 6$
$O: 35/16 = 2.18 \rightarrow 1$
Empirical formula $= C_2H_6O$.
Empirical formula weight $= 2 \times 12 + 6 \times 1 + 16 = 46$.
Molecular weight $= 2 \times \text{vapour density} = 2 \times 23 = 46$.
Since molecular weight equals empirical formula weight,the molecular formula is $C_2H_6O$.
Since the compound reacts with sodium metal to liberate $H_2$ gas,it must be an alcohol,which is $C_2H_5OH$ (ethanol).
The functional isomer of ethanol $(C_2H_5OH)$ is methoxy methane $(CH_3OCH_3)$.

(D) Moles of carbon $= \frac{2.4 \ g}{12 \ g/mol} = 0.2 \ mol$
Moles of $H = \frac{1.2 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.2 \ mol$
Moles of oxygen atoms $= 0.2 \ mol$
The molar ratio of $C:H:O$ is $0.2:0.2:0.2$,which simplifies to $1:1:1$.
Therefore,the empirical formula is $CHO$.

(C) Given,
Mass of organic compound $= 0.48 \ g$
Mass of $CO_2$ produced $= 0.22 \ g$
$\% C = ?$
Using the formula:
$\% C = \frac{12}{44} \times \frac{\text{mass of } CO_2}{\text{mass of organic compound}} \times 100$
$= \frac{12}{44} \times \frac{0.22}{0.48} \times 100$
$= \frac{1}{44} \times \frac{0.22}{0.04} \times 100$
$= 12.5 \%$

(C) To find the empirical formula,we calculate the relative number of moles of each element:

Element	Percentage (%) / Atomic Mass	Relative Moles	Simplest Ratio
$C$	$40 / 12$	$3.33$	$3.33 / 3.33 = 1$
$H$	$13.33 / 1$	$13.33$	$13.33 / 3.33 \approx 4$
$N$	$46.67 / 14$	$3.33$	$3.33 / 3.33 = 1$

The empirical formula is $CH_{4}N$.

(D) $1$. Calculate the mass of $C$ and $H$ in the compound:
Mass of $C = \frac{12}{44} \times 0.44 \ g = 0.12 \ g$
Mass of $H = \frac{2}{18} \times 0.18 \ g = 0.02 \ g$
Mass of $O = 0.30 - (0.12 + 0.02) = 0.16 \ g$
$2$. Calculate the molar ratio:
Moles of $C = \frac{0.12}{12} = 0.01$
Moles of $H = \frac{0.02}{1} = 0.02$
Moles of $O = \frac{0.16}{16} = 0.01$
$3$. Determine the empirical formula:
The ratio $C:H:O = 0.01:0.02:0.01 = 1:2:1$. Thus,the empirical formula is $CH_2O$.
$4$. Determine the molecular formula:
Empirical formula mass $= 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2$.
Molecular formula $= 2 \times (CH_2O) = C_2H_4O_2$.

(A) The empirical formula of the compound is $CH_2O$.
Empirical formula mass $= (1 \times 12) + (2 \times 1) + (1 \times 16) = 30 \ g/mol$.
Molecular mass $= 90 \ g/mol$.
Calculate the value of $n$ using the formula: $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{90}{30} = 3$.
The molecular formula is given by $n \times (\text{Empirical formula}) = 3 \times CH_2O = C_3H_6O_3$.

(C) Potash alum is a double salt with the chemical formula $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$.
This can also be written as $2KAl(SO_4)_2 \cdot 12H_2O$.
Summing the atoms: $K_2$,$Al_2$,$S_4$,$O_{40}$,$H_{48}$.
Thus,the molecular formula is $K_2Al_2S_4H_{48}O_{40}$.

(B) Moles of metal $(M) = \frac{3.2 \ g}{64 \ g/mol} = 0.05 \ mol$.
Weight of oxygen $= 3.6 \ g - 3.2 \ g = 0.4 \ g$.
Moles of oxygen $= \frac{0.4 \ g}{16 \ g/mol} = 0.025 \ mol$.
Ratio of moles of $M : O = 0.05 : 0.025 = 2 : 1$.
Therefore,the formula of the oxide is $M_2O$.
Hence,the correct option is $(B)$.

(D) Molecular mass of $C_{57}H_{110}O_6 = (12 \times 57) + (110 \times 1) + (6 \times 16) = 684 + 110 + 96 = 890$.
Mass $\%$ of $C = \frac{\text{Mass of } C}{\text{Molecular mass of compound}} \times 100 = \frac{684}{890} \times 100 = 76.85 \%$

(C) Let the mass of $C$ be $4x$ and the mass of $H$ be $1x$.
Number of moles of $C = \frac{4x}{12} = \frac{x}{3}$.
Number of moles of $H = \frac{1x}{1} = x$.
Ratio of moles of $C:H = \frac{x}{3} : x = \frac{1}{3} : 1 = 1 : 3$.
Therefore,the empirical formula is $CH_3$.

(C) The chemical formula of Calgon is $Na_2[Na_4(PO_3)_6]$.
Calgon is the trade name for sodium hexametaphosphate,which has the molecular formula $(NaPO_3)_6$.
An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound.
For $(NaPO_3)_6$,the ratio of $Na:P:O$ is $6:6:18$,which simplifies to $1:1:3$.
Therefore,the empirical formula is $NaPO_3$.

(C) In the first oxide $XO_2$,the percentage of oxygen is $50 \%$. This means the percentage of metal $X$ is also $50 \%$.
Since the mass of $2$ oxygen atoms is $2 \times 16 = 32 \ g$,and this corresponds to $50 \%$ of the total mass,the mass of $X$ must also be $32 \ g$.
In the second oxide,the percentage of oxygen is $40 \%$,which implies the percentage of metal $X$ is $60 \%$.
Since the mass of $X$ remains $32 \ g$,we have $60 \% \equiv 32 \ g$.
Therefore,$40 \% \equiv \frac{32}{60} \times 40 = \frac{64}{3} \approx 21.33 \ g$.
Wait,let us re-evaluate: If $X$ is $60 \%$ and $O$ is $40 \%$,then $\frac{\text{mass of } X}{\text{mass of } O} = \frac{60}{40} = 1.5$.
Given mass of $X = 32 \ g$,mass of oxygen $= \frac{32}{1.5} = 21.33 \ g$.
Number of oxygen atoms $= \frac{21.33}{16} = 1.33$. This does not yield an integer.
Let us re-read: 'Two oxides of $X$ contain $50 \%$ and $40 \%$ of non-metal'. If $X$ is the non-metal,then in $XO_2$,$X$ is $50 \%$.
Mass of $X = 32 \ g$. In the second oxide,$X$ is $40 \%$,so oxygen is $60 \%$.
Ratio $\frac{O}{X} = \frac{60}{40} = 1.5$.
Mass of $O = 1.5 \times 32 = 48 \ g$.
Number of $O$ atoms $= \frac{48}{16} = 3$.
Thus,the formula is $XO_3$.

(C) To find the empirical formula,calculate the mole ratio of each element:
Carbon: $\frac{12.8}{12.0} = 1.066$
Hydrogen: $\frac{2.1}{1.008} = 2.083$
Bromine: $\frac{85.1}{79.9} = 1.065$
Dividing by the smallest value $(1.065)$:
Carbon: $\frac{1.066}{1.065} \approx 1$
Hydrogen: $\frac{2.083}{1.065} \approx 2$
Bromine: $\frac{1.065}{1.065} = 1$
Thus,the empirical formula is $CH_2Br$.
Empirical formula mass $= 12.0 + 2(1.008) + 79.9 = 93.996 \approx 94$.
Calculate the factor $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{187.9}{94} = 2$.
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (CH_2Br) = C_2H_4Br_2$.

(B) The mass percentage of an element in a compound is calculated as: $\text{Mass } \% = \frac{\text{Total mass of the element}}{\text{Molar mass of the compound}} \times 100$.
In $CCl_4$,the molar mass is $12.01 + 4 \times 35.45 = 153.81 \approx 154 \ g/mol$.
Mass percentage of $C = \frac{12.01}{153.81} \times 100 \approx 7.81 \%$.
Mass percentage of $Cl = \frac{4 \times 35.45}{153.81} \times 100 \approx 92.19 \%$.
Given the options provided,the closest values are $7.84$ and $92.80$.

(A) Given,mass of $CO_2 = 0.535 \ g$ and mass of $H_2O = 0.13 \ g$.
Mass of compound $= 0.765 \ g$.
Percentage of $C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 \approx 19.07 \%$.
Percentage of $H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 \approx 1.88 \% \approx 2.00 \%$.
Ratio of percentage of $C$ and $H = 19.07 : 1.88 \approx 19 : 2$.

(D) The reaction sequence is the Kolbe-Schmitt reaction followed by acetylation to form aspirin $(C_9H_8O_4)$.
$1$. Phenol reacts with $NaOH$, $CO_2$, and $H^+$ to form salicylic acid.
$2$. Salicylic acid reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of $H^+$ to form aspirin $(Z)$ and acetic acid $(CH_3COOH)$.
$3$. The molecular formula of aspirin $(Z)$ is $C_9H_8O_4$.
$4$. The molar mass of $C_9H_8O_4 = (9 \times 12) + (8 \times 1) + (4 \times 16) = 108 + 8 + 64 = 180 \ g/mol$.
$5$. The mass of carbon in one mole of $C_9H_8O_4 = 9 \times 12 = 108 \ g$.
$6$. Percentage of carbon = $\frac{\text{Mass of Carbon}}{\text{Molar Mass of } Z} \times 100 = \frac{108}{180} \times 100 = 60 \%$.

(B) The atomic masses of elements are: $Na = 23 \ g/mol$,$S = 32 \ g/mol$,$O = 16 \ g/mol$,$H = 1 \ g/mol$.
The molecular mass of $Na_2SO_4 \cdot 10H_2O = (2 \times 23) + 32 + (4 \times 16) + (10 \times 18) = 46 + 32 + 64 + 180 = 322 \ g/mol$.
Given mass of $Na_2SO_4 \cdot 10H_2O = 32.2 \ g$.
Moles of $Na_2SO_4 \cdot 10H_2O = \frac{32.2 \ g}{322 \ g/mol} = 0.1 \ mol$.
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.
Moles of oxygen atoms $= 0.1 \ mol \times 14 = 1.4 \ mol$.
Mass of oxygen $= 1.4 \ mol \times 16 \ g/mol = 22.4 \ g$.

(A) To find the empirical formula,we calculate the molar ratio of the elements:
$1$. Moles of $C = \frac{60}{12} = 5.0$
$2$. Moles of $H = \frac{4.48}{1} = 4.48$
$3$. Moles of $O = \frac{35.5}{16} \approx 2.22$
Divide each by the smallest value $(2.22)$:
$C: \frac{5.0}{2.22} \approx 2.25$
$H: \frac{4.48}{2.22} \approx 2.0$
$O: \frac{2.22}{2.22} = 1.0$
Multiplying by $4$ to get whole numbers: $C = 9, H = 8, O = 4$.
Thus,the empirical formula is $C_9H_8O_4$.

(C) Given,empirical formula of compound $= C_2H_5O$
Vapour density $= 45$
Molecular weight $= 2 \times \text{vapour density} = 2 \times 45 = 90$
Empirical formula mass $= (2 \times 12) + (5 \times 1) + (1 \times 16) = 24 + 5 + 16 = 45$
Calculate the value of $n$ using the formula: $n = \frac{\text{Molecular weight}}{\text{Empirical formula mass}} = \frac{90}{45} = 2$
Molecular formula $= n \times (\text{empirical formula}) = 2 \times (C_2H_5O) = C_4H_{10}O_2$

(C) To determine the empirical formula,we calculate the molar ratio of the elements:
$C: \frac{52.14}{12} = 4.345 \implies \frac{4.345}{2.17} \approx 2$
$H: \frac{13.13}{1} = 13.13 \implies \frac{13.13}{2.17} \approx 6$
$O: \frac{34.73}{16} = 2.17 \implies \frac{2.17}{2.17} = 1$
Thus,the empirical formula is $C_2H_6O$.
Empirical formula mass $= 2(12) + 6(1) + 1(16) = 24 + 6 + 16 = 46 \ g/mol$.
Given molar mass $= 46.068 \ g/mol$.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{46.068}{46} \approx 1$.
Therefore,the molecular formula is $(C_2H_6O)_1 = C_2H_6O$.

(A) The percentage of $C$ is $92.3 \%$,so the percentage of $H$ is $100 - 92.3 = 7.7 \%$.

Element	Atomic Ratio (Percentage / Atomic Mass)	Simple Molar Ratio
$C$	$92.3 / 12 = 7.69$	$7.69 / 7.69 = 1$
$H$	$7.7 / 1 = 7.70$	$7.70 / 7.69 \approx 1$

Thus,the empirical formula is $CH$.

(B) Mass of $C = \frac{12}{44} \times 12.571 \ g = 3.428 \ g$
Mass of $H = \frac{2}{18} \times 5.143 \ g = 0.571 \ g$
Moles of $C = \frac{3.428}{12} = 0.2857$
Moles of $H = \frac{0.571}{1} = 0.571$
Ratio of $C:H = 0.2857 : 0.571 \approx 1 : 2$

Element	Mass $(g)$	Moles	Simple Ratio
$C$	$3.428$	$0.2857$	$1$
$H$	$0.571$	$0.571$	$2$

$\therefore$ Empirical formula $= CH_2$

(C) The molar mass of the crystalline solid $MSO_{4} \cdot nH_{2}O$ is $250 \ g \ mol^{-1}$.
Given that the percentage of anhydrous salt $(MSO_{4})$ is $64\%$,the percentage of water $(H_{2}O)$ is $100\% - 64\% = 36\%$.
Mass of $H_{2}O = \frac{36}{100} \times 250 = 90 \ g$.
Moles of $H_{2}O = \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.
Since the formula is $MSO_{4} \cdot nH_{2}O$,the value of $n$ corresponds to the number of moles of water,which is $5$.
Hence,option $(C)$ is the correct answer.

(A) The atomic weight of nitrogen is $14 \ g/mol$. In an organic molecule,there must be at least one nitrogen atom present.
Given that the compound contains $20 \%$ nitrogen by mass.
Let the molecular weight be $M$.
Then,$\frac{14}{M} \times 100 = 20$.
Solving for $M$: $M = \frac{14 \times 100}{20} = 70 \ g/mol$.
Thus,the molecular weight is $70$.

(A) The molar mass of magnesium pyrophosphate $(Mg_2P_2O_7)$ is: $2 \times 24 + 2 \times 31 + 7 \times 16 = 48 + 62 + 112 = 222 \ g \ mol^{-1}$.
The moles of $Mg_2P_2O_7$ formed = $\frac{1.79 \ g}{222 \ g \ mol^{-1}} \approx 0.008063 \ mol$.
Since each mole of $Mg_2P_2O_7$ contains $2 \ mol$ of phosphorus atoms,the moles of $P$ = $2 \times 0.008063 = 0.016126 \ mol$.
Mass of phosphorus = $0.016126 \ mol \times 31 \ g \ mol^{-1} \approx 0.4999 \ g$.
Percentage of phosphorus = $\frac{\text{Mass of } P}{\text{Mass of compound } X} \times 100 = \frac{0.4999}{1.00} \times 100 = 49.99\%$.
Rounding to the nearest integer,we get $50\%$.

(A) Mass of $CO_2 = 0.25$ g. Moles of $CO_2 = \frac{0.25}{44} \approx 0.00568$ mol.
Mass of $C = \text{Moles of } CO_2 \times 12 = 0.00568 \times 12 = 0.06818$ g.
Given percentage of $C = 25\%$.
So,$0.06818 = 0.25 \times X \implies X = \frac{0.06818}{0.25} = 0.2727$ g.
Similarly for $H$,Mass of $H_2O = 0.12$ g.
Moles of $H_2O = \frac{0.12}{18} \approx 0.00666$ mol.
Mass of $H = 0.00666 \times 2 = 0.01333$ g.
Given percentage of $H = 4.89\%$.
So,$0.01333 = 0.0489 \times X \implies X = \frac{0.01333}{0.0489} = 0.2726$ g.
Thus,$X \approx 0.273$ g.
Expressing as $273 \times 10^{-3}$ g.

(B) $1$. Percentage of $Fe = 69.9\%$. Therefore,percentage of $O = 100\% - 69.9\% = 30.1\%$.
$2$. Calculate moles of each element:
Moles of $Fe = \frac{69.9}{56} = 1.248 \text{ mol}$.
Moles of $O = \frac{30.1}{16} = 1.881 \text{ mol}$.
$3$. Determine the simplest molar ratio:
Divide by the smallest value $(1.248)$:
$Fe = \frac{1.248}{1.248} = 1$.
$O = \frac{1.881}{1.248} \approx 1.5$.
$4$. Convert to whole numbers:
Multiply by $2$ to get $Fe : O = 2 : 3$.
$5$. Thus,the empirical formula is $Fe_2O_3$.