Tag: chemistry

Questions Related to chemistry

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.