Question 15 Marks
The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon $X$ is purified by fractional distillation.
$0.145 g$ of $X$ was heated with dry copper (II) oxide and $224 cm ^3$ of carbon dioxide was collected at S.T.P.
(a) Which elements does X contain?
(b) What was the purpose of copper (II) oxide?
(c ) Calculate the empirical formula of $X$ by the following steps:
(i) Calculate the number of moles of carbon dioxide gas.
(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample $X$.
(iii) Calculate the mass of hydrogen in sample X.
(iv) Deduce the ratio of atoms of each element in $X$ (empirical formula).
Answer$22400 cm ^3 CO _2$ has mass $=44 g$
so, $224 cm ^3 CO _2$ will have mass $=0.44 g$
Now since $CO _2$ is being formed and $X$ is a hydrocarbon so it contains $C$ and $H$.
In $0.44 g CO _2$, mass of carbon $=0.44-0.32=0.12 g =0.01 g$ atom
So, mass of Hydrogen in $X=0.145-0.12=0.025 g$
$=0.025 g$ atom
Now the ratio of $C : H$ is $C =1: H =2.5$ or $C =2: H =5$
i.e. the formula of hydrocarbon is $C _2 H _5$
(a) $C$ and $H$
(b) Copper (II) oxide was used for reduction of the hydrocarbon.
(c)
(i) no. of moles of $CO _2=0.44 / 44=0.01$ moles
(ii) mass of $C =0.12 g$
(iii) mass of $H =0.025 g$
(iv) The empirical formula of $X= C _2 H _5$
View full question & answer→Question 25 Marks
Find the molecular formula of a hydrocarbon having vapour density $15,$ which contains $20\%$ of Hydrogen.
Answer$ \% \text { of hydrogen }=20 \%$
$\% \text { of carbon }=100-20=80 \% $
| |
$\%$ Weight |
Atomic Weight |
Relative No. of Moles |
Simplest Ratio |
| $C$ |
$80$ |
$12$ |
$80/12 = 6.667$ |
$6.667/6.667 = 1$ |
| $H$ |
$20$ |
$1$ |
$20/1 = 20$ |
$20/6.667 = 2.99 \approx 3$ |
$\text { Empirical formula }= CH _3$
$\text { Empirical formula weight }=1 \times 12+1 \times 3=12+3=15$
$\text { Vapour Density }=15$
$\text { Relative molecular mass }=15 \times 2=30$
$N =\frac{\text { Relative molecular mass }}{\text { Empirical Weight }}=\frac{30}{15}=2$
$\text { Molecular formula }= n \times \text { empirical formula }$
$=2 \times CH _3$
$= C _2 H _6$ View full question & answer→Question 35 Marks
Water decomposes to $O _2$ and $H _2$ under suitable conditions as represented by the equation below:
$2 H _2 O \rightarrow 2 H _2+ O _2$
(a) If $2500 cm ^3$ of $H _2$ is produced, what volume of $O _2$ is liberated at the same time and under the same conditions of temperature and pressure?
(b) The $2500 cm ^3$ of $H _2$ is subjected to $2$ times an increase in pressure (temp. remaining constant). What volume of $H _2$ will now occupy?
(c) Taking the value of $H _2$ calculated in $5$(b), what changes must be made in Kelvin (absolute) temperature to return the volume to $2500 cm ^3$ pressure remaining constant.
Answer$2 H _2 O \rightarrow 2 H _2+ O _2$
$2 V 2 V 1 V$
(a) From equation, $2 V$ of water gives $2 V$ of $H _2$ and $1 V$ of $O _2$ where $2 V =2500 cm ^3$
so, volume of $O _2$ liberated $=2 V / V =1250 cm ^3$
$\text { (b) } \frac{ P _1 V _1}{ T _1}=\frac{ P _2 V _2}{ T _2} $
$\frac{ P _1 V _1}{ T _1}=\frac{7 P _1 \times V _2}{2 \times T _1} $
$V _2=\frac{2500 \times 2}{7} $
$V _2=\frac{5000}{7} cm ^3$
$\text { (c) } \frac{ V _1}{ V _2}=\frac{ T _1}{ T _2} $
$=\frac{5000}{7 \times 2500}=\frac{ T _1}{ T _2} $
$T _2=3.5 T _1$
i.e. temperature should be increased by $3.5$ times.
View full question & answer→Question 45 Marks
The volume of gases $A, B, C$, and $D$ are in the ratio, $1: 2: 2: 4$ under the same conditions of temperature and pressure.
(i) Which sample of gas contains the maximum number of molecules?
(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of $A$ when the number of molecules is doubled?
(iii)If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
(iv)If the volume of A is actually $5.6 dm ^3$ at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro's number is $6 \times 10^{23}$ ).
(v) Using your answer from (iv), state the mass of $D$ if the gas is dinitrogen oxide $\left( N _2 O \right)$
Answer(i) D contains the maximum number of molecules because the volume is directly proportional to the number of molecules.
(ii) The volume will become double because the volume is directly proportional to the no. of molecules at constant temperature and pressure.
$ V _1 / V _2= n _1 / n _2$
$V _1 / V _2= n _1 / 2 n _1 $
So, $V _2=2 V _1$
(iii) Gay lussac's law of combining volume is being observed.
(iv) The volume of $D=5.6 \times 4=22.4 dm ^3$, so the number of molecules $=6 \times 10^{23}$ because according to mole concept $22.4$ litre volume at STP has $=6 \times 10^{23}$ molecules
(v) No. of moles of $D=1$ because volume is $22.4$ litre
$\text { so, mass of } N _2 O =1 \times 44=44 g$
View full question & answer→Question 55 Marks
The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide, and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?
AnswerAccording to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure, contain the equal number of molecules.
So, 1 mole of each gas contains $=6.02 \times 10^{23}$ molecules
Mol. Mass of $H _2(2), O _2(32), CO _2(44), SO _2(64), Cl _2(71)$
(1) Now $2 g$ of hydrogen contains molecules $=6.02 \times 10^{23}$
So, $8 g$ of hydrogen contains molecules $=8 / 2 \times 6.02 \times 10^{23}$
$=4 \times 6.02 \times 10^{23}=4 M$ molecules
(2) $32 g$ of oxygen contains molecules $=8 / 32 \times 6.02 \times 10^{23}= M / 4$
(3) $44 g$ of carbon dioxide contains molecules $=8 / 44 \times 6.02 \times 10^{23}=2 M / 11$
(4) $64 g$ of sulphur dioxide contains molecules $=6.02 \times 10^{23}$
So, $8 g$ of sulphur dioxide molecules $=8 / 64 \times 6.02 \times 10^{23}= M / 8$
(5) $71 g$ of chlorine contains molecules $=6.02 \times 10^{23}$
So, $8 g$ of chlorine molecules $=8 / 72 \times 6.02 \times 10^{23}=8 M / 71$
Since $8 M / 71<M / 8<2 M / 11<M / 4<4 M$
Thus $Cl _2< SO _2< CO _2< O _2< H _2$
(i) Least number of molecules in $Cl _2$
(ii)Most number of molecules in $H _2$
View full question & answer→Question 65 Marks
Solid ammonium dichromate decomposes as:
$\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
If $63 g$ of ammonium dichromate decomposes. Calculate what will be the loss of mass?
AnswerSolid ammonium dichromate decomposes as:
$\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
$\text { Number of moles }=\frac{\text { Weight }}{\text { Molecular weight }} $
$=\frac{63}{252} $
$=0.25 \text { moles } $
$\qquad\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
$\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
$0.25$ moles of ammonium dichromate gives $0.25$ moles of $N _2=7 g$
$1$ mole of $H _2 O =18 g$
Therefore, total loss of mass $=7+18$
$=25 g$
View full question & answer→Question 75 Marks
i. A compound has the following percentage composition by mass: carbon $14.4 \%$, hydrogen $1.2 \%$, and chlorine $84.5 \%$. Determine the empirical formula of this compound. Work correctly to $1$ decimal place.
$( H =1 ; C =12 ; Cl =35.5)$
ii. The relative molecular mass of this compound is $168,$ so what is its molecular formula?
Answeri.
| Element |
% Weight |
Atomic Weight |
Atomic Ratio |
Simplest Ratio |
| $C$ |
$14.4$ |
$12$ |
$14.4/12 = 1.2$ |
$1.2/1.2=1$ |
| $H$ |
$1.2$ |
$1$ |
$1.2/1 =1.2$ |
$1.2/1.2=1$ |
| $Cl$ |
$84.5$ |
$35.5$ |
$84.5/35.5=2.3$ |
$2.3/1.2=1.9=2$ |
Empirical formula $= CHCl _2$
ii.
Empirical formula $= CHCl _2$
Empirical formula weight $=1 \times 12+1 \times 1+(2 \times 35.5)$
$ =12+1+70$
$=83 $
Relative molecular mass $=168$
$N =\frac{\text { Relative molecular mass }}{\text { Empirical Weight }}=\frac{168}{83}=2.02 \approx 2$
Molecular formula $= n x$ empirical formula
$ =2 \times CHCl _2$
$= C _2 H _2 Cl _4 $$ View full question & answer→Question 85 Marks
Calculate the mass of substance 'A' which in gaseous form occupies $10$ litres at $27 {}^\circ C$ and $700\ mm$ pressure. The molecular mass of 'A' is $60.$
Answer$ V _1=10 \text { litres } V _2=?$
$T _1=27+273=300 K T _2=273 K$
$P _1=700\ mm P _2=760\ mm $
Using the gas equation
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$
$V_2=\frac{P_1 V_1 T_2}{T_1 P_2}=\frac{700 \times 10 \times 273}{300 \times 760}$
Molecular weight $A=60$
So, weight of $22.4$ litres of $A$ at STP $=60 g$
weight of litres of A at STP $=\frac{700 \times 10 \times 273}{300 \times 760}$
$=\frac{60}{22.4} \times \frac{700 \times 10 \times 273}{300 \times 760} \text { g or } 22.45 g$
View full question & answer→Question 95 Marks
$1.56 g$ of sodium peroxide reacts with water according to the following equation:
$2 Na _2 O _2+2 H _2 O 4 NaOH + O _2$
Calculate:
(a) mass of sodium hydroxide formed,
(b) The volume of oxygen liberated at S.T.P.
(c) Mass of oxygen liberated.
Answer$2 Na _2 O _2+2 H _2 O 4 NaOH + O _2$
$2 V 4 V 1 V$
(a) Mol. Mass of $Na _2 O _2=223+216=78 g$
Mass of $2 Na _2 O _2=156 g$
$156 g Na _2 O _2$ gives $=160 g$ of $NaOH (440 g )$
So, $1.56 Na _2 O _2$ will give $=160$ 1.56/156 $=1.6 g$
(b) $156 g Na _2 O _2$ gives $=22.4$ litres of oxygen
So, $1.56 g$ will give $=22.41 .56 / 156=0.224$ litres $=224 cm ^3$
(c) $156 g Na _2 O _2$ gives $=32 g O _2$
So, $1.56 g Na _2 O _2$ will give $=32$ 1.56/156 $=32 / 100=0.32 g$
View full question & answer→Question 105 Marks
Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide were collected at $27^{\circ} C$ and normal pressure.
$
CaCO _3+2 HCl \rightarrow CaCl _2+ H _2 O + CO _2
$
Calculate:
(a) The mass of salt required.
(b) The mass of the acid required to prepare the 2 litres of $CO _2$ at $27 C$ and normal pressure.
Answer(a) The mass of salt required is, $8 . 1 2$ grams
(b) The mass of the acid required is, $5 . 9 3$ grams
Explanation: Given,
The volume of carbon dioxide gas $=2 L$
The temperature of carbon dioxide gas $=27^{\circ} C =273+27=300 K$
The pressure of carbon dioxide gas $=1 atm$
First, we have to calculate the moles of carbon dioxide gas by using the ideal gas equation.
$
\text { PV }= n R T
$
where,
$P=$ pressure of carbon dioxide gas
$V=$ volume of carbon dioxide gas
$T$ = temperature of carbon dioxide gas
$n =$ number of moles of carbon dioxide gas
$R =$ gas constant $=0.0821 L$.atm $/ mole . K$
$(1 atm ) \times(2 L )= n \times(0.0821 L . atm / mole . K ) \times(300 K )$
$n =0.0812$ mole
(a) Now we have to calculate the mass of salt required.
The balanced chemical reaction is,
$
CaCO _3( s )+2 HCl ( l ) \longrightarrow CaCl _2( s )+ H _2 O ( l )+ CO _2( g )
$
From the balanced reaction, we conclude that,
As, 1 mole of $CO _2$ obtained from 1 mole of $CaCO _3$
So, 0.0812 moles of $CO _2$ obtained from 0.0812 moles of $CaCO _3$
Mass of $CaCO _3=$ Moles of $CaCO _3 \times$ Molar mass of $CaCO _3=0.0812$ mole $\times 100 g / mole =8.12 g$
Therefore, the mass of the salt required is, $8 . 1 2$ grams
(b) Now we have to calculate the mass of acid required.
The balanced chemical reaction is,
$
CaCO _3( s )+2 HCl ( l ) \longrightarrow CaCl _2( s )+ H _2 O ( l )+ CO _2( g )
$
From the balanced reaction we conclude that,
As, 1 mole of $CO _2$ obtained from 2 mole of $HCl$
So, 0.0812 mole of $CO _2$ obtained from 0.1624 mole of $HCl$
Mass of $HCl =$ Moles of $HCl \times$ Molar mass of $HCl =0.1624$ mole $\times 36.5 g / mole =5.93 g$
Therefore, the mass of the acid required is, 5.93 grams
View full question & answer→Question 115 Marks
If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula.
AnswerFor acetylene, molecular mass $=2 \times V . D =2 \times 13=26 g$
The empirical mass $=12( C )+1( H )=13 g$
$
n =\frac{\text { Molecular formula mass }}{\text { Empirical formula weight }}=\frac{26}{13}=2
$
Molecular formula of acetylene $=2 \times$ Empirical formula $= C _2 H _2$
Similarly, for benzene molecular mass $=2 \times$ V.D $=2 \times 39=78$
$
n =78 / 13=6
$
So, the molecular formula $= C _6 H _6$
View full question & answer→Question 125 Marks
$10.47\ g$ of a compound contained $6.25\ g$ of metal A and rest non-metal B. Calculate the empirical formula of the compound [At. wt of $A = 207, B = 35.5$]
AnswerGiven:
Wt. of the compound: $10.47\ g$
Wt. of metal A: $6.25\ g$
Wt. of non-metal B: $10.47 – 6.25 = 4.22\ g$
| Element |
mass |
At. Wt. |
Relative no. of atoms |
Simplest ratio |
| A |
$6.25g$ |
$207$ |
$6.25/207=0.03$ |
$0.03/0.03=1$ |
| B |
$4.22g$ |
$35.5$ |
$4.26/35.5=0.12$ |
$0.12/0.03=4$ |
Hence, the empirical formula is $AB_4$ View full question & answer→Question 135 Marks
A gaseous hydrocarbon contains $82.76\%$ of carbon. Given that its vapour density is $29$, find its molecular formula.
Answer$\%$ of carbon $= 82.76\%$
$\%$ of hydrogen $= 100 - 82.76 = 17.24\%$
| Element |
% Weight |
Atomic Weight |
Relative No. of Moles |
Simplest Ratio |
| $C$ |
$82.76$ |
$12$ |
$82.76/12 = 6.89$ |
$6.89/6.89 = 1 x 2 = 2$ |
| $H$ |
$17.24$ |
$1$ |
$17.24/1 = 17.24$ |
$17.24/6.89 = 2.5 x 2 = 5$ |
Empirical formula $= C_2H_5$
Empirical formula weight $= 2 x 12 + 1 x 5 = 24 + 5 = 29$
Vapour Density $= 29$
Relative molecular mass $= 29 x 2 = 58$
N = `"Relative molecular mass"/"Empirical weight" $= 58/29 = 2`$
Molecular formula = n × empirical formula
$= 2 x C_2H_5$
$= C_4H_{10}$ View full question & answer→Question 145 Marks
An unknown gas shows a density of $3 g$ per litre at $273^{\circ} \mathrm{C}$ and $1140 mm$ Hg pressure. What is the gram molecular mass of this gas?
AnswerGiven:
$ P =1140 mm Hg$
$\text { Density }= D =2.4 g / L$
$T=273{ }^{\circ} C =273+273=546 K$
$M =? $
We know that, at STP, the volume of one mole of any gas is $22.4 L$
Hence we have to find out the volume of the unknown gas at STP.
First, apply Charle's law.
We have to find out the volume of one litre of unknown gas at standard temperature $273 K$.
$ V _1=1 L T _1=546 K$
$V _2=? T _2=273 K$
$V _1 / T _1= V _2 / T _2$
$V _2=\left( V _1 \times T _2\right) / T _1$
$=(1 L \times 273 K ) / 546 K$
$=0.5 L $
We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.
Apply Boyle's law.
$ P_1=1140 mm Hg V_1=0.5 L$
$P_2=760 mm Hg V_2=\text { ? }$
$P_1 \times V_1=P_2 \times V_2$
$V_2=\left(P_1 \times V_1\right) / P_2$
$ =(1140 mm Hg \times 0.5 L ) / 760 mm Hg$
$=0.75 L $
Now, 22.4 $L$ is the volume of 1 mole of any gas at STP, then $0.75 L$ is the volume of $X$ moles at STP
$ X \text { moles }=0.75 L / 22.4 L$
$=0.0335 \text { moles } $
The original mass is $2.4 g$
$ n = m / M$
$0.0335 \text { moles }=2.4 g / M$
$M =2.4 g / 0.0335 \text { moles }$
$M =71.6 g / \text { mole } $
Hence, the gram molecular mass of the unknown gas is $71.6 g$
View full question & answer→Question 155 Marks
$112 cm^3$ of $H_2S(g)$ is mixed with $120 cm^3$ of $Cl_2(g)$ at STP to produce $HCl(g)$ and sulphur(s). Write a balanced equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.
AnswerBalanced chemical equation:
$
\underset{1 \text { mole, } 112 cm ^3}{ H _2 S _{( g )}}+\underset{1 \text { mole, } 120 cm ^3}{ Cl _{2( g )}} \longrightarrow \underset{2 \text { moles }}{2 HCl _{( g )}}+\underset{1 mol }{ S _{( s )}}
$
(i) At STP, 1 mole gas occupies $22.4 L$.
As 1 mole $H _2 S$ gas produces 2 moles $HCl$ gas,
22.4 L $H _2 S$ gas produces $22.4 \times 2=44.8 L HCl$ gas.
Hence, $112 cm ^3 H _2 S$ gas will produce $112 \times 2=224 cm ^3 HCl$ gas.
(ii) 1 mole $H _2 S$ gas consumes 1 mole $Cl _2$ gas.
This means $22.4 L H _2 S$ gas consumes $22.4 L Cl _2$ gas at STP.
Hence, $112 cm ^3 H _2 S$ gas consumes $112 cm ^3 Cl _2$ gas.
$120 cm ^3-112 cm ^3=8 cm ^3 Cl _2$ gas remains unreacted.
Thus, the composition of the resulting mixture is $\underline{224 cm ^3 HCl }$ gas $+8 cm ^3 Cl _2$ gas.
View full question & answer→Question 165 Marks
What volume of air (containing $20 \% O _2$ by volume) will be required to burn completely $10 cm ^3$ each of methane and acetylene?
$CH _4+2 O _2 \rightarrow CO _2+2 H _2 O$
$2 C _2 H _2+5 O _2 \rightarrow 4 CO _2+2 H _2 O$
Answer$CH _4+2 O _2 \rightarrow CO _2+2 H _2 O $
$1 V 2 V 1 V $
$2 C _2 H _2+5 O _2 \rightarrow 4 CO _2+2 H _2 O $
$2 V 5 V 4 V$
From the equations, we can see that
$1 V CH _4 \text { requires oxygen }=2 V O _2$
So, $10 cm ^3 CH _4$ will require $=20 cm ^3 O _2$
Similarly $2 V C _2 H _2$ requires $=5 V O _2$
So, $10 cm ^3 C _2 H _2$ will require $=25 cm ^3 O _2$
Now, $20 V O _2$ will be present in $100 V$ air and $25 V O _2$ will be present in $125 V$ air , so the volume of air required is $225 cm ^3$
View full question & answer→