Tag: percentage composition and empirical formula

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

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

  1. 5 g

  2. 4.5 g

  3. 3 g

  4. 3.5 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

Which of the following compound represents an alkane?

  1. ${C} _{5}{H} _{8}$
  2. ${C} _{7}{H} _{16}$
  3. ${C} _{8}{H} _{6}$
  4. ${C} _{9}{H} _{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


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

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

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

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

  1. $1\ g$ molecule
  2. $0.5\ g$ molecule
  3. $5\ g$ molecules
  4. $10\ g$ molecules
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 ______________.

  1. $CH _2$
  2. $C _2H _4$
  3. $C _2H _6$
  4. $C _4H _8$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\text{B}$

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

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 _________.

  1. $0.25$
  2. $0.40$
  3. $0.5$
  4. $0.75$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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?

  1. $112$g $mol^{-1}$
  2. $242$g $mol^{-1}$
  3. $224$g $mol^{-1}$
  4. $422$g $mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Graham's Law of Effusion, the rate of effusion is inversely proportional to the square root of molar mass. Given the conditions and pressure, the molar mass of the unknown gas calculates to 224 g/mol.

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

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

  1. different percentage composition

  2. different molecular weights

  3. same vapour density

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.