Chapter 6 -

PARAMOUNT SCHOOL SYSTEM

Subject: Chemistry – I

Unit 6: Stoichiometry

CONCEPT ASSESSMENT EXERCISE 6.1

Write the empirical formulas for the compound containing carbon to hydrogen in the following ratios.

1 : 4 CH₄
2 : 6 CH₃
2 : 2 CH
6 : 6 CH

CONCEPT ASSESSMENT EXERCISE 6.2

1. Aspirin is used as a mild pain killer. There are nine carbon atoms, eight hydrogen atoms and four oxygen atoms, in this compound. Write its empirical and molecular formulas.

Molecular formula of Aspirin is C₉H₈O₄

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula for Aspirin is: C₉H₈O₄

2. Vinegar is 5% acetic acid. This contains 2 carbon atoms, four hydrogen atoms and 2 oxygen atoms. Write its empirical and molecular formulas.

Molecular formula of Vinegar is C₂H₄O₂

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula for Vinegar is: CH₂O

3. Caffeine (C₈H₁₀N₄O₂) is found in tea and coffee. Write the empirical formula for caffeine.

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula for Caffeine is: C₄H₅N₂O

CONCEPT ASSESSMENT EXERCISE 6.3

Potassium Chlorate (KClO₃) is used commonly for the laboratory preparation of oxygen gas. Calculate its formula mass.

Formula mass of KClO₃ =39 + 35.5 + 3×16

= 39 + 35.5 + 48

= 122.5 amu

2. When baking soda, NaHCO₃ is heated it releases carbon dioxide, which is responsible for the rising of cookies and bread. Determine the formula mass of baking soda and carbon dioxide.

Formula mass of NaHCO₃ =23+1+12+3×16

= 23+1+12+48

= 84 amu

Formula mass of CO₂ = 12+2×16

= 12+32

= 44 amu

3. Following compounds are used as fertilizers. Determine their formula masses.

(i) Urea, (NH₂)₂ CO

Formula mass of (NH₂)₂ CO = 2×14 + 2×2 + 12 + 16

= 28 + 4 + 12 + 16

= 60 amu

Ammonium nitrate, NH₄NO₃

Formula mass of NH₄NO₃ = 14 + 1×14 + 14 + 3×16

= 14 + 4 +14 + 48

= 80 amu

CONCEPT ASSESSMENT EXERCISE 6.4

1. Calculate the one mole of

(a) Copper

1 mole of Copper (Cu) = 63.5 g

(b) Iodine

Molecular mass of Iodine (I₂) = 2×127

1 mole of I₂ = 254 g

(c) Potassium

1 mole of K = 39g

(d) Oxygen (O₂)

Molecular mass of Oxygen (O₂) = 2 x 16

1 mole of O = 32g

2. Differentiate between gram formula mass and gram molecular mass.

Gram Formula Mass	Gram Molecular Mass
Represents one mole of ionic formula units of a compound.	Represents one mole of molecules of a compound or element.
Contains 6.022 × 10²³ formula units.	Contains 6.022 × 10²³ molecules.
Used for ionic compounds.	Used for molecular compounds or elements in the molecular state.
Example: NaCl	Example: H₂O, O₂

CONCEPT ASSESSMENT EXERCISE 6.5

1. The molecular formula of a compound used for bleaching hair is Hydrogen Peroxide (H₂O₂₎. Calculate:

(a) Mass of this compound that would contain 2.5 moles.

Solution:

Number of moles = 2.5 moles

Molar mass of (H₂O₂) = 2 x 1 + 2 x 16

= 34g

Mass (in grams) = ?

= 2.5 x 34g

= 85g

(b) Number of moles of this compound that would exactly weight 30 g.

Solution:

Mass of (H₂O₂) = 30g

Molar mass of (H₂O₂) = 34g

Number of moles = ?

= 0.88 moles

2. A spoon of table salt NaCl contains 12.5 grams of this salt. Calculate the number of moles it contains.

Solution:

Mass of Sodium Chloride (NaCl) = 12.5g

Molar mass of NaCl = 23+35.5

= 58.5g

Number of moles = ?

= 0.21 moles

3. Before the digestive systems x-rayed, people are required to swallow suspensions of barium sulphate (BaSO₄). Calculate mass of one mole of BaSO₄ .

Solution:

Molar Mass of BaSO₄= 137+32+4×16

= 137+32+64

= 233g

Therefore,

Mass of 1 mole of BaSO₄= 233g

CONCEPT ASSESSMENT EXERCISE 6.6

Aspirin is compound that contains carbon, hydrogen and oxygen. It is used as a painkiller. An aspirin tablet contains 1.25 x 10³⁰ molecules. How many moles of this compound are present in the tablet?

Solution:

Number of molecules = 1.25×10³⁰ molecules

Avogadro’s number = N_A = 6.022×10²³

Number of moles = ?

= 2.076 x 10⁶ moles

A method used to prevent rusting in ships and underground pipelines involves connecting the iron to a block of a more active metal such as magnesium. This method is called cathodic protection. How many moles of magnesium are present in 1 billion (1 x 10⁹) atoms of magnesium.

Solution:

Number of atoms = 1×10⁹ atoms

Avogadro’s number = N_A = 6.022×10²³

Number of moles = ?

= 1.66 x10^-15 moles

CONCEPT ASSESSMENT EXERCISE 6.7

Represent the following chemical reactions by chemical equations.

1. Burning of hydrogen (H₂) to produce water.

Hydrogen gas + Oxygen → Water

H₂ + O₂ → H₂O

2. Burning of magnesium (Mg) to produce magnesium oxide (MgO).

Magnesium metal + Oxygen → Magnesium Oxide

2Mg + O₂ → 2MgO

CONCEPT ASSESSMENT EXERCISE 6.8

Transform the following chemical equations into ionic equations.

1. AgNO₃ + NaCl → AgCl + NaNO₃

Write the substances that are soluble in water in their dissociated form. AgCl (silver chloride) is insoluble in water and precipitates out.

Ag⁺ + NO₃⁻ + Na⁺ + Cl⁻ → AgCl + Na⁺ + NO₃⁻

Remove common ions from both sides,

Ag⁺ + Cl⁻ → AgCl

2. Zn + 2HCl → ZnCl₂ + H₂

Write the substances that are soluble in water in their dissociated form.

Zn + 2H⁺+ 2Cl⁻→ Zn⁺²+ 2Cl⁻ + H₂

Remove common ions from both sides,

Zn + 2H⁺→ Zn⁺²+ H₂

CONCEPT ASSESSMENT EXERCISE 6.9

Write the molecular formulae of the following compounds.

1. CH₃—CH₂— OH

C₂H₆O

2. CH₃— CH₂— NH₂

C₂H₇N

3. CH₃– CO – CH₃

C₃H₆O

REVIEW QUESTIONS

1. Encircle the correct answer.

What is the formula mass of CuSO₄.5H₂O. (Atomic masses: Cu-3.5, S=32, H= 1)

a. 159.5 b. 185.5 c. 249.5 d. 149.5

A compound with chemical formula Na₂CX₃ has formula mass 106amu: Atomic mass of the element X is:

a. 106 b. 23 c. 12 d. 16

How many moles of molecules are there in 16g oxygen?

a. 1 b. 0.5 c. 0.1 d. 0.05

What is the mass of 4 moles of hydrogen gas?

a. 8.064g b. 4.032g c. 1g d.1.008g

What is the mass of carbon present in 44g of carbon dioxide?

a. 12g b. 6g c. 24g d. 44g

Which term is the same for one mole of oxygen and one mole of water?

a. volume b. mass c. atoms d. molecules

If one mole of carbon contains x atoms, what is the number of atoms contained in 12g of Mg.

a. x b. 0.5x c. 2x d. 1.5x

2. Give short answer.

i. What is mole?

A mole is an amount of a substance that contains 6.022×10²³ particles of that substance. This experimentally determined number is known as Avogadro’s number. It is represented by N_A.

ii. Differentiate between empirical formula and molecular formula.

Empirical Formula	Molecular Formula
Which gives the simplest whole number ratio of atoms of each element of a compound	Which shows the actual number of atoms of each element present in a compound.
It does not show the structure of compound	It shows the structure of compound
Two or more compounds can have same empirical formula	Two or more compounds cannot have same molecular formula.
Example:CH₂O, CH are empirical formula of glucose and benzene	Example: C₆H₁₂O₆, C₆ H₆ are molecular formula of glucose and benzene

iii. What is the number of molecules in 9.0g of steam?

Solution:

Mass in grams = 9.0g

Molar mass of steam (H₂O) = 2 x 1 + 16 = 18g

Avogadro’s number = N_A = 6.022×10²³

Number of molecules = ?

= 6.022×10²³

= 0.5 x 6.022 x 10²³

= 3.011 x 10²³ Molecules

iv. What are the molar masses of Uranium^-238 and

Uranium^-235?

Molar mass = Atomic mass

Molar mass of uranium-238 = 238g

Molar mass of uranium-235 = 235g

v. Why one mole of hydrogen molecules and one mole of H atoms have different mases?

One mole of hydrogen molecule (H₂) contains two hydrogen atoms, its molar mass is 1×2=2g. One mole of H-atom contains only one hydrogen atom, its molar mass is 1×1=1g. That’s why one mole of hydrogen molecules and one mole of H-atoms have different masses.

3. Define:

Ion:

Ion is a charged specie formed from an atom or chemically bonded group of atoms by adding or removing electrons.

Ion may have positive or negative charge.

Molecular Ion:

When a molecule losses or gains electrons, the resulting species is called molecular ion. These are short lived species and only exist at high temperature. Molecular ions do not form ionic compounds.

Formula Unit:

The simplest unit which represents an ionic compound is called formula unit.

Free radical:

Free radical is an atom or group of atoms that contains an unpaired electron. Free radical bear no charge

Example ,

Atomic number:

The number of protons in the nucleus of an atom is known as atomic number.

Mass number:

The total number of protons and neutrons in an atom is known as its mass number.

Atomic mass unit:

One atomic mass unit (amu) is defined as a mass exactly equal to one twelfth the mass of one C-12 atom.

Mass of one C-12 atom = 12 amu

4. Describe how Avogadro’s number is related to a mole of any substance.

Avogadro’s number is related to a mole of any substance by a relation.

5. Calculate the number of moles of each substance in samples with the following masses.

(a) 2.4g of He:

Mass of Helium (He) = 2.4g

Molar mass of Helium(He)= 4g

= 0.6 moles

(b) 250 mg of Carbon:

Mass of carbon (C) = 250mg = 250g/1000= 0.25g

Molar mass of carbon = 12g

= 0.021 moles

(c) 15g of sodium chloride:

Mass of sodium chloride (NaCl) = 15g

Molar mass of Sodium Chloride = 23 + 35.5 = 58.5g

= 0.256 moles

(d) 40 g of Sulphur:

Mass of Sulphur (S) = 40 g

Molar mass of Sulphur = 32g

= 1.25 moles

(e) 1.5 kg of MgO: (1kg = 1000 g)

Mass of MgO = 1.5 Kg = 1.5 x 1000 = 1500 g

Molar mass of MgO = 24 + 16 = 40g

= 37.5 moles

6. Calculate the mass in grams of each of the following samples:

(a) 1.2 moles of K

Number of moles of Potassium (K) = 1.2 moles

Molar mass of Potassium = 39g

Mass in gram =Number of moles × Molar mass

Mass in grams = 1.2 x 39 = 46.8 g

(b) 75 moles of H₂

Number of moles of hydrogen(H₂) = 75 moles

Molar mass of hydrogen(H₂)= 2 x 1 = 2g

Mass in gram =Number of moles × Molar mass

Mass in grams = 75 x 2 = 150g

(c) 0.25 moles of steam

Number of moles of steam (H₂O) = 0.25 moles

Molar mass of steam (H₂O) = 2 x 1+16 = 18g

Mass in gram = Number of moles × Molar mass

Mass in grams = 0.25 x 18 = 4.5g

(d) 1.05 moles of CuSO₄.5H₂O

Number of moles of Copper Sulphate Penta Hydrate= 1.05 moles

Molar mass of CuSo₄ .5H₂O = 63.5+32+4×16+5(2×1+16)

= 63.5 +32 +64 +5(18)

= 249.5g

Mass in gram =Number of moles × Molar mass

Mass in grams = 1.05 x 249.5 = 261.96g

(e) 0.15 moles of H₂SO₄

Number of moles of Sulphuric Acid (H₂SO₄) = 0.15 moles

Molar mass of H₂SO₄ = 2 x 1 + 32 + 4×16 = 98g

Mass in gram =Number of moles × Molar mass

Mass in grams = 0.15 x 98 = 14.7g

7. Calculate the number of molecules present in each of the following samples

(a) Number of moles = 2.5 moles

Avogadro’s number = N_A = 6.022×10²³

Number of molecules = ?

Number of molecules = Number of Moles x N_A

Number of molecules = 2.5 x 6.022 x 10²³

= 1.505 x 10²⁴ molecules

(b) 3.4 moles of ammonia, NH₃

Number of moles = 3.4 moles

Avogadro’s number = N_A = 6.022×10²³

Number of molecules = ?

Number of molecules = Number of Moles x N_A

Number of molecules = 3.4 x 6.022 x 10²³

= 2.05 x 10²⁴ molecules

(c) 1.09 moles of benzene, C₆H₆

Number of moles = 1.09 moles

Avogadro’s number = N_A = 6.022×10²³

Number of molecules = ?

Number of molecules = Number of Moles x N_A

Number of molecules = 1.09 x 6.022 x 10²³

= 6.56 x 10²³ molecules

(d) 0.01 moles of acetic acid, CH₃COOH

Number of moles = 0.01 moles

Avogadro’s number = N_A = 6.022×10²³

Number of molecules = ?

Number of molecules = Number of Moles x N_A

Number of molecules = 0.01 x 6.022 x 10²³

= 6.02 x 10²¹ molecules

8. Decide whether or not each of the following is an example of empirical formula

(a) Al₂ Cl₆

This formula is not in its simplest form, so this is not an empirical formula.

(b) Hg₂ Cl₂

This formula is not in its simplest form, so this is not an empirical formula.

(c) NaCl

This formula is in its simplest form, so this is an empirical formula.

(d) C₂H₆O

This formula is in its simplest form, so this is an empirical formula.

9. TNT or trinitrotoluene is an explosive compound used in bombs. It contains 7 C-atoms, 6 H-atoms, 5 N-atoms and 6-O atoms. Write its empirical formula

There are:

7 carbon (C) atoms
6 hydrogen (H) atoms
5 nitrogen (N) atoms
6 oxygen (O) atoms

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula for trinitrotoluene (TNT) is: C₇ H₆ N₅ O₆

10. A molecule contains four Phosphorus atoms and ten Oxygen atoms. Write the empirical formula of this compound. Also determine the molar mass of this molecule

There are:

4 phosphorus (P) atoms
10 oxygen (O) atoms

Molecular formula = P₄O₁₀

Molar mass of P₄O₁₀ = 4×31 + 10×16

= 124 + 160 = 284g

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula is: P₂O₅

11. Indigo (C₁₆H₁₀N₂O₂) the dye used to colour blue jeans is derived from a compound known as indoxyl (C₈H₇ON). Calculate the molar masses of these compounds. Also write their empirical formulas

Indigo (C₁₆H₁₀N₂O₂)

Molar mass = 16×12 + 10 x 1 + 2×14 + 2×16

= 192 + 10 + 28 + 32

= 262g

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula is: C₈H₅NO

Indoxyl (C₈H₇ON)

Molar mass = 8 x12 + 7×1 + 16 + 14

= 96 + 7 + 16 + 14

= 133g

The empirical formula is the simplest whole-number ratio. Therefore, the empirical formula is: C₈H₇ON

12. Identify the substance that has formula mass of 133.5 amu

(a) MgCl₂

Formula mass = 24 + 2×35.5

= 24 + 71

= 95 amu

(b) S₂Cl₂

Formula mass = 2×32 + 2×35.5

= 64 + 71

= 135 amu

(c) BCl₃

Formula mass = 11 + 3×35.5

= 11 + 106.5

= 117.5 amu

(d) AlCl₃

Formula mass = 27 + 3×35.5

= 27 + 106.5

= 133.5 amu

The substance with a formula mass of 133.5amu is AlCl₃

13. Calculate the number of atoms in each of the following samples.

(a) 3.4 moles of nitrogen atoms

Number of moles = 3.4 moles

Number of atoms = No. of moles x N_A

= 3.4 x 6.022×10²³

=2.05 x10²⁴ atoms

(b) 23g of Na

Mass in grams = 23 g

Molar mass of Sodium (Na) = 23 g

= 6.022×10²³

= 6.022 x 10²³

(c) 5 g of H atoms

Mass in grams = 5g

Molar mass of hydrogen(H)= 1g

= 6.022×10²³

= 3.01 x 10²⁴ atoms

14. Calculate the mass of the following

(a) 3.24 x 10¹⁸ atoms of iron

Number of atoms = 3.24 x 10¹⁸ atoms

= 3.01 x 10^-4 g

(b) 2 x 10¹⁰ molecules of nitrogen gas

Number of molecules = 2 x 10¹⁰ molecules

Molar mass of nitrogen gas (N₂) = 2 x 14 = 28g

= 9.3 x 10^-13 g

(c) 1 x 10 ²⁵ molecules water

Number of molecules = 1 x 10²⁵ atoms

Molar mass of water (H₂O)= 1×2 + 16 = 18g

= 2.99 x 10²g

(d) 3 x 10⁶ atoms of Al

Number of atoms = 3 x 10⁶ atoms

Molar mass of Aluminum(Al)= 27 g

= 1.346 x 10^-16g

15. Balance the following chemical equations.

a. Na (s) + H₂O (l) → NaOH (aq) + H₂ (g)

Balanced equation:

2Na (s) + 2 H₂O (l) → 2 NaOH (aq) + H₂ (g)

b. NH₃ → N₂ + H₂

Balanced equation:

2NH₃ → N₂ + 3 H₂

16. Potassium is Group 1 element. It is silvery white metal. It burns in air and forms both potassium oxide and potassium nitride. The nitride ion is N^3-.

(a) Predict the formula of potassium oxide and potassium nitride.

Potassium oxide (K₂O): Potassium (K) has a +1 charge, and oxygen (O) has a -2 charge. To balance the charges, the formula is K₂O.

Potassium nitride (K₃N): Potassium (K) has a +1 charge, and the nitride ion (N³⁻) has a -3 charge. To balance the charges, the formula is K₃N.

(b) A 0.5g sample of K was added in 100cm³ of water.

K _(s) + H₂O _(l) → KOH _(aq) + H_2(g)

Show that 1.28 x10^-2 mole of K were added to the water.

Mass of K = 0.5 g

Molar mass of K = 39 g/mol

= 1.28 x10^-2 mole

2K _(s) + 2H₂O _(l) → 2KOH _(aq) + H_2(g)

(d) Transform above chemical equation into ionic equation.

Write the substances that are soluble in water in their dissociated form

K _(s) + H₂O _(l) → K⁺ _(aq) + OH⁻ _(aq) + H₂ _(g)

(e) Calculate the number of atoms present in the sample of K.

Number of moles of K = 1.28×10⁻²moles

Avogadro’s number N_A = 6.022×10²³

Number of atoms = No. of moles x N_A

= 1.28×10⁻² x 6.022×10²³

= 7.7 × 10²¹ atoms

(f) Predict period number of potassium in the periodic table.

Potassium (K) is in Period 4 of the periodic table.