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Percentage composition and empirical formula - class-XII

Description: percentage composition and empirical formula
Number of Questions: 21
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Tags: introduction to analytical chemistry stoichiometric relationships stoichiometry atoms and molecules basic concepts of chemistry chemistry mole concept and stoichiometry the language of chemistry some basic concepts of chemistry moles and equations
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What changes does not occur in presence of organic solvent when acidified $K _{2}Cr _{2}O _{7}$ reacts with $H _{2}O _{2}$ solutions

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. Orange colour of solution turns blue

  2. $O.S$ if $Cr$ atom decreases

  3. $O.S$ of $Cr$ atom remains constant

  4. Unpaired electron remain same

Show answer
Correct Option: A

From $\ 2 \ mg$ calcium $1.2\times 10^{19}$ atoms are removed. The number of $g$ - atoms of calcium left is $(Ca=40)$:

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. $5\times 10^{-5}$

  2. $2\times 10^{-5}$

  3. $3\times 10^{-5}$

  4. $5\times 10^{-6}$

Show answer
Correct Option: C
Explanation:

$2\ mg$ Calcium moles $=\dfrac{2\times 10^{-3}}{40}=5\times 10^{-5}$


$1$ mole $=6.022\times 10^{23}$ atoms

$x=\dfrac{1.2\times 10^{19}}{6.022\times 10^{23}}$

$=1.99\times 10^{-5}$ moles removed
$\therefore$ Remaining moles $=5\times 10^{-5}-1.99\times 10^{-5}$
$=3.00\times 10^{-5}$ moles
$1\ g$ atom $=1$ mole
$\therefore$ Option $C$ correct.

Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 30, 20, 45, 5

  2. 28, 1, 14, 57

  3. 24, 23, 12, 1

  4. None of above

Show answer
Correct Option: B
Explanation:

22.99 g (1 mol) of Na
12.01          (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO _3$ is $ 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$%
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$%
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$%
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%

Determine the percentage composition of $K$ in $KMnO _4$.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. $31\%$

  2. $43.58\%$

  3. $25\%$

  4. $55\%$

Show answer
Correct Option: C
Explanation:
To find:- $\%$ composition of K in $KMnO _4$
A/c
Molar mass of $K=39$g
Molar mass of $KMnO _4=158$gm
$\%$ of k in $KMnO _4=\dfrac{39}{158}\times 100=24.68\%$
$\approx 25\%$.
Option C is correct.

Given:
Mass of beaker is X g
Mass of beaker + mixture is Y g
Mass of washed and dried sand is Z g
Find the percentage of sand in the mixture.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. $\frac{Y-X}{100}\times Z$

  2. $\frac{Y-X}{Z}\times 100$

  3. $\frac{Z}{Y-X}\times 100$

  4. $\frac{100}{Y-X}\times Z$

Show answer
Correct Option: C
Explanation:

Mass of mixture=$Y-X$

Percentage of sand= $\frac{weight   of   sand}{weight   of   mixture}\times 100$
                                 =$\frac{Z}{Y-X}\times100$

If the % composition of a 'x' component is 35%, find the mass of the dried 'x' component in 100g? mixture + beaker = 50 g, mass of beaker = 23 g

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 15.5 g

  2. 9.45 g

  3. 35.5 g

  4. 13.4 g

Show answer
Correct Option: B
Explanation:

mass of beaker $= 23g$
mass of mix + beaker$ = 50g$
mass of mixture $= 50 - 23 = 27g$
$\%$ composition of mixture $= 35\%$
Mass of the x component $= 35 \times  \dfrac{27}{100} = 9.45g$

The mass of a sand and powdered mixture along with a beaker is 56 g. If the mass of the dried mixture is 20 g, find the % composition of the mixture in 100 g?
(weight of beaker = 20 g).

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 20%

  2. 36%

  3. 55%

  4. 60%

Show answer
Correct Option: C
Explanation:

Mass of beaker = 20 g
Mass of mixture + beaker = 56 g
Mass of mixture = 56 - 20 = 36 g
Mass of washed and dried sand = 20 g
100 g of mixture contains $= \dfrac{20}{36} \times 100 = 55$% of sand

Calculate the % composition of Carbon in $CO _2$. Molar mass is 44.01.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 40%

  2. 67%

  3. 30%

  4. 27%

Show answer
Correct Option: D
Explanation:

Molar mass of compound:
Mass due to carbon:   12.01 g/mol
Total molar mass $= 12.01 + 2(16.00) = 44.01g/mol$             ($CO _2$ has two $O _2$ atoms)
Percent composition of carbon: $12.01/44.01 \times 100 = 27.28$%

If the % composition of Cl in HCl is 46%, what is the mass of Cl in HCl?

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 13.3 g

  2. 66 g

  3. 25 g

  4. 16.76 g

Show answer
Correct Option: D
Explanation:

$\%$ $ composition =\cfrac{Mass \quad of\quad an \quad element.}{Total \quad mass \quad of\quad compound.}$

$Total \quad mass\quad of\quad HCl=1+35.5 g=36.5g$
$\therefore$ $\cfrac{46}{100}=\cfrac{mass\quad of\quad Cl}{36.5}\ mass\quad of\quad Cl=16.76g$

Mass of beaker=20 g
Mass of beaker+mixture=50 g
Mass of washed and dried sand=5 g
Find the percentage of sand in the mixture.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 16.67%

  2. 20%

  3. 25%

  4. 30%

Show answer
Correct Option: A
Explanation:

Mass of mixture=$50 g-20 g=30 g$

Mass of washed and dried sand in the mixture=$5g$
Percentage=$\frac{5}{30}\times 100$
                   =16.67%


A mixture contains $2.5$g $CaCO _3$ and $3.0$g $NaCl$. What is the percent by mass of $CaCO _3?$

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. $45.45$

  2. $75$

  3. $53$

  4. $60$

Show answer
Correct Option: A
Explanation:

Total mass$=2.5+3.0g=5.5g$


$\%$ by mass $CaCO _3=\dfrac{2.5}{5.5}\times 100 = 45.45\%$

Mass of beaker=10g
Mass of mixture+beaker=25g
Mixture contains 30% sand. Find the weight of sand.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

  1. 5 g

  2. 4.5 g

  3. 3 g

  4. 3.5 g

Show answer
Correct Option: B
Explanation:

Mass of mixture=$25g-10g=15g$

Percentage of sand=30%=$\frac{Mass  of  sand}{Mass  of  mixture}$
 Mass of sand =$0.3\times15g=4.5 g$

Which of the following compound represents an alkane?

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. ${C} _{5}{H} _{8}$

  2. ${C} _{7}{H} _{16}$

  3. ${C} _{8}{H} _{6}$

  4. ${C} _{9}{H} _{10}$

Show answer
Correct Option: B
Explanation:

Alkanes have general molecular formula is $C _nH _{2n+2}$


$\therefore C _7H _{16}$ is Alkane.

Hence, the correct option is $\text{B}$

The number of $g$ molecules of oxygen in $6.023\times 10^{24} \ CO$ molecules is ________________.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $1\ g$ molecule

  2. $0.5\ g$ molecule

  3. $5\ g$ molecules

  4. $10\ g$ molecules

Show answer
Correct Option: C
Explanation:

$1$ mole of $CO=6.023\times 10^{23}$ molecules

$x$ mole of $CO=6.023\times 10^{24}$ molecules
$\therefore x=10$ moles $=10\ g$ molecules
$CO=1$ oxygen atom
$\therefore$ Oxygen molecules $=\dfrac{10}{2}$
$=5\ g$ molecules 
Option $C$ is correct.

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atoms. At $0^o C$ and $1\ atm$, its density is $1.25\ g$ per litre. The formula of the gas would be ______________.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $CH _2$

  2. $C _2H _4$

  3. $C _2H _6$

  4. $C _4H _8$

Show answer
Correct Option: B
Explanation:

$PV=nRT$


$P=\dfrac {dRT}{M}$

$M=\dfrac {1.25\times 0.082\times 273}{1}$

$=27.98$

Molar mass $\equiv 28\ gram/mole$

$\therefore \ $ as empirical formula $=CH _2$

$\therefore \ $ emprical mass $=12+2=14$

$n=\dfrac {28}{14}=2$

$\therefore \ $ Molecular formula $=(CH _2) _2$

$=C _2H _4$

Hence, the correct option is $\text{B}$

When burnt in air, $14.0\ g$ mixture of carbon and sulpher gives a mixture of $CO 2$ and $SO _2$ in the volume ratio of $2:1$, volume being measured at the same conditions of temperature and pressure. Moles of carbon in the mixture is _________.

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $0.25$

  2. $0.40$

  3. $0.5$

  4. $0.75$

Show answer
Correct Option: C
Explanation:

Consider the mass of $C=x$


mass of $S=14-x$

moles of $C=\dfrac {x}{12};$ moles of $\dfrac {14-x}{32}$

$C+O _2\to CO _2;\quad S+O _2\to SO _2$

moles of $C=$ moles of $CO _2=\dfrac {x}{12}\quad $ moles of $S=$ moles of $SO _2=\dfrac {14-x}{32}$

Given,

$\dfrac {V _c}{V _s}=\dfrac {2}{1}$

$V\alpha $ no. of mole

$\dfrac {\dfrac {x}{12}}{\dfrac {14-x}{32}}=\dfrac {2}{1}$

$\dfrac {x}{12}\times \dfrac {32}{14-x}=\dfrac {2}{1}$

$\dfrac {4x}{42-3x}=1$

$4x=42-3x$

$7x=42$ 

$\boxed {x=6}$

$\therefore \ $ moles of $C$ in mixture $=\dfrac {w}{M}=\dfrac {6}{12}=0.5$

Answer is option $C$

If air is pumped slowly but continuously into a metallic cylinder of strong wall, what would happen to the air inside the cylinder?

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. Temperature of air would increase

  2. Pressure of air would increase

  3. Pressure of air would decrease

  4. Temperature and pressure of air would increase

Show answer
Correct Option: B
Explanation:

Due to continuous pumping number of moles will increase
We know,
$n\propto P$
$n\propto V$
$n\propto \dfrac{1}{T}$
$\therefore$ Pressure will increase.
Option B.

For $10$ min each, at $27^o$C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of $3$l capacity. The resulting pressure is $4.18$ bar and the mixture contains $0.4$ mole of nitrogen. The molar mass of the unknown gas is?

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $112$g $mol^{-1}$

  2. $242$g $mol^{-1}$

  3. $224$g $mol^{-1}$

  4. $422$g $mol^{-1}$

Show answer
Correct Option: C

What is the molecular formula of arsenious chloride?

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $ AsCI _{3}$

  2. $ As _{2}CI _{6}$

  3. $ As _{2}CI _{5}$

  4. $ AsCI _{5}$

Show answer
Correct Option: B

If two compounds have the same empirical formula but different molecular formula, they must have :

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. different percentage composition

  2. different molecular weights

  3. same vapour density

  4. none of these

Show answer
Correct Option: B
Explanation:

If two compounds have the same empirical formula but different molecular formula, they must have different molecular weights.


For example, $CH _2O$ and $C _6H _{12}O _6$ have the same empirical formula but different molecular formula, they have different molecular weights.

Option B is correct.

Calculate the number of molecules present in $0.5$ moles of magnesium oxide $\left( MgO \right) $. 

[Atomic weights :  $Mg=24, O=16$]

chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

  1. $14.09\times { 10 }^{ 23 }$ molecules

  2. $3.0115\times { 10 }^{ 23 }$ molecules

  3. $30.12\times { 10 }^{ 23 }$ molecules

  4. $14.09\times { 10 }^{ -23 }$ molecules

Show answer
Correct Option: B
Explanation:
1 mole of MgO contains Avogadro's number of molecules which is $ \displaystyle 6.023 \times 10^{23}$ molecules.

0.5 mole of MgO will contain $ \displaystyle 0.5 \times 6.023 \times 10^{23}=3.0115 \times 10^{23}$ molecules.
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