Tag: percent composition

Questions Related to percent composition

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

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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%