MOLE CONCEPT AND STOICHIOMETRY

Multiple Choice Questions

Q1. The percentage composition of oxygen in calcium sulphate (CaSO₄) is

A. 18.18%

B. 16%

C. 11.76%

D. 47.05%

Q2. Dhan Bdr has a balloon that contains 5.22 moles of helium, how many grams of helium is present in the balloon?

A. 21g

B. 20g

C. 10g

D. 4g

Q3. A compound has an empirical formula of C₂H₄O with molar mass of 132g. What is the molecular formula of the compound?

A. C₁₀H₁₂

B. C₆H₁₂O₃

C. C₄H₄O₅

D. C₂H₄

Q4. The molar volume of an ideal gas at STP is

A. 21000 mL

B. 22000 mL

C. 23000 mL

D. 22400 mL

Q5. Which of the following molecule contains the maximum number of moles?

A. 10g of Nitrogen

B. 10g of Carbon dioxide

C. 10g of Sulphur dioxide

D. 10g of Ammonia

Q6. The equation for the combustion of butane is shown below.  How many molecules of water are formed when one molecule of butane burns completely?

2C₄H₁₀  +  13O₂  →   8CO₂  +  10H₂O

A. 10 molecules

B. 8 molecules

C. 5 molecules

D. 4 molecules

Q7. The volume occupied by one mole of carbon dioxide at STP is

A. 44.8 L

B. 22.4 L

C. 11.2 L

D. 5.6 L

Q8. The percentage composition of oxygen in H₂SO₄ is

A. 16.3%

B. 32.6%

C. 48.9%

D. 65.3%

Q9. What would be the relative molecular mass of CuSO₄? [At.wt of Cu=64]

A. 162

B. 161

C. 160

D. 159

Q10. The relative molecular mass of glucose (C₆H₁₂O₆) is

A. 120

B. 160

C. 180

D. 200

Q11. 1.5 moles of oxygen at STP occupies a volume of

A. 11.2 L

B. 22.4 L

C. 33.6 L

D. 44.8 L

Q12. The number of molecules present in one mole of sulphur dioxide is

A. 6.023 × 10²¹ molecules

B. 6.023 × 10²² molecules

C. 6.023 × 10²³ molecules

D. 6.023 × 10²⁴ molecules

Q13. Which one of the following weighs the least?

A. 1 mole of NH₃

B. 1 mole of H₂O

C. 1 mole of CO₂

D. 1 mole of SO₂

Short Questions

Q1. Fill in the blanks

i. Number of mole(s) present in 100g of CaCO₃ is ________________.

Ans.: 1

ii. The gram molecular mass of Ca(OH)₂ is  ________________.  

Ans.: 74g

iii. The formula which can give the simple whole-number ratio of an atom is

_________________________.

Ans.: Empirical formula

iv. The vapour density of SO₂ is ________________.

Ans.: 32 (As vapour density = 64/2 = 32)

v. Fill the blanks in the following table.

Mole(s)	Molecules	Volume at S.T.P (L)
2 moles	12.046 × 1023	A……………
1 mole	B……………..	22.4
C……………	9.034 × 1023	33.6

Ans.:

A. 44.8L              

B. 6.023 ×10²³         

C. 1.5 mole

Q2. Differentiate the following pairs

i. Nitrogen and oxygen (gram molecular mass)

Ans.:

Nitrogen	Oxygen
28 grams	32 grams

ii. Mole and stoichiometry (define)

Ans.:

Mole	Stoichiometry
An amount of substance containing 6.023 × 1023particles.	It is the quantitative relation between the number of moles of various products and reactants in a chemical reaction.

Q3. Write the relationship between empirical formula and molecular formula.

Ans.: Molecular formula= n × empirical formula

Q4. What is the empirical formula of hydrogen peroxide (H₂O₂)?

Ans.: HO

Q5. 1 mole of any substance contains 22.4 L at STP and 6.023 × 10²³ molecules. Explain.

Ans.: The volume occupied by one mole of any substance at standard temperature and pressure is called molar volume which is equal to 22.4 L. And the number of molecules present in 22.4 L of substance is equal to 6.023 × 10²³.

Q6. Masses of atoms are expressed relative to the mass of a carbon-12. Give TWO reasons.

Ans.:

• Atoms are too small to be weighed directly
• Inconvenient to express masses of atoms in Kg or gram
• Carbon-12 is stable and light.

Q7. How many gram(s) of sodium makes one mole?

Ans.: Since 1 mole of any atom = Gram atomic mass

                    Gram atomic mass = Relative atomic mass   (except the unit changes to gram)

       So, one mole of sodium is equal to 23 grams.

Q8. The volume occupied by 1 mole of helium is 22.4 L at STP. What would be the volume occupied by 1 mole of oxygen at STP?

Ans.:

1 mole of any gas occupies 22.4 L

So, 1 mole of oxygen occupies 22.4 L

Numerical Questions

Q1. Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. The relative molecular mass of the compound is 162.

Determine the empirical formula and molecular formula of the compound.

Ans.:

Element	% by weight	Atomic weight	Atomic ratio	Simplest ratio
C	74.02	12	74.02/12= 6.16	6.16/1.23= 5
H	8.710	1	8.71/1=8.71	8.71/1.23=7
N	17.27	14	17.27/14=1.23	1.23/1.23=1

Empirical formula: C₅H₇N₁

Empirical formula mass= 5 × 12 + 7 × 1 + 1 ×14 = 81

n= 162/81

n= 2

Molecular formula= n × Empirical formula

Molecular formula = 2 × C₅H₇N₁

Molecular formula = C₁₀H₁₄N₂

Q2.  Calculate the following based on the equation given:

Fe₂O₃(s) + 3CO → 2Fe(s) + 3CO₂ (g)

(Atomic weight of Fe=56)

Determine the volume of CO₂ that will be produced from 80g of Fe₂O₃ at STP.

Ans.: Molecular mass of Fe₂O₃ = 56×2 + 3×16 = 160g

160g of Fe₂O₃ produces 3 moles of CO₂ = 3 x 22.4 L volume at STP

=67.2L

80g of Fe₂O₃ = 80/160 x 67.2

= 33.6 L of CO₂ produced at STP

Q3. Calculate the percentage composition of nitrogen in NH₄NO₃.

Ans.: Relative molecular mass = 80

Mass of nitrogen = 28g

% composition = 28/80 x 100 = 35%

Q4. If 16.4 grams of Calcium nitrate is heated as shown in the reaction:

2Ca(NO₃)₂ →   2CaO  +  4NO₂  +  O₂

i. Calculate the volume of nitrogen dioxide produced at STP.

Ans.:

2Ca(NO₃)₂ → 2CaO  +  4NO₂  +  O₂

2Ca(NO₃)₂ = 4 x NO₂

328g = 4 x 22.4

16.4g = X

X= 4×22.4/328 x 16.4 = 4.48L

ii. Calculate the weight of calcium oxide obtained.

Ans.:

328g of Ca(NO₃)₂ gives →  112g of CaO

16.4  → 112×16.4/328

=5.6g of CaO

Q5. Silver nitrate and Aluminum chloride react according to the following equation. Calculate the amount of silver chloride formed from 200g of Aluminum chloride. (Atomic mass of Ag=108)

3AgNO₃ + AlCl₃ →   3AgCl  + Al(NO₃)₃

Ans.:

AlCl₃      →         3AgCl

133.5g    →          430.5g

200g       →         x

X=430.5g x 200g/133.5g

=644g

Q6. Calculate the mass of 60 mL of chlorine at STP.

Ans.:

Molecular mass of chlorine (Cl₂) = 35.5 x 2= 71g

1 mole (1 gram molecular mass) of any gas at STP occupies 22.4 l or 22400 mL

Mass of 22400 mL of Cl₂ at STP = 71 g

Mass of 60 mL of Cl₂ at STP =  71/22400 x 60 =0.19g

Hence, mass of 60 mL of chlorine at STP = 0.19 g

Q7. An acid of phosphorus has the following percentage composition:

       2.47% H, 38.27% P, 59.26% O. The molar mass of acid is 162.

a) Determine its empirical formula.

Ans.:

Element	% composition	At. Weight	Relative no. of atoms	Simplest Ratio
H	2.47	1	2.47/1=2.47	2.47/1.23= 2
P	38.27	31	38.27/31=1.23	1.23/1.23= 1
O	59.26	16	59.26/16=3.70	3.70/1.23= 3

Empirical formula = H₂PO₃

b) Calculate the molecular formula of the compound.

Ans.: Empirical formula mass = 2 + 31 + 48 = 81

n = 162/81= 2

Molecular formula = n × Emp. Formula

                                 = 2 ×  H₂PO₃ =  H₄P₂O₆

Molecular formula = H₄P₂O₆

Q8. A compound of carbon, hydrogen, and oxygen is found to contain 40% of carbon, 6.7 of hydrogen, and 53.3% of oxygen. If its vapour density is 30, find:

i. Empirical formula

Ans.:

Element	% by weight	Atomic weight	Atomic ratio	Simplest ratio
C	40	12	40/12=3.33	3.33/3.33 =1
H	6.7	1	6.7/1=6.7	6.7/3.33=2
O	53.3	16	53.3/16=3.33	3.33/3.33=1

Empirical formula= CH₂O

ii. Molecular formula

Ans.:    Empirical formula mass = 30

Molecular mass= 2 × VD = 2 × 30= 60

n= 60/30 =2

Molecular formula = n × empirical formula

Molecular formula = 2 × CH₂O = C₂H₄O₂

Molecular formula = C₂H₄O₂

Q9. Urea is a chemical fertilizer that is used by farmers to enhance the nitrogen content in the soil. If the farmer uses 1 mole of urea (COCNH₂)₂, calculate the percentage of nitrogen that the farmer is able to increase?

Ans.:

Gram molecular mass or mass of 1 mole of urea = 112g

The mass of nitrogen in 1 mole of urea = 28g

The % composition of nitrogen in urea= mass of the nitrogen / GMM of urea x 100

= 28/112 x 100 = 25

Compiled from BCSE 2017-2021