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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1Topic
    |
    1 Quiz
  4. Separation Techniques II | Week 4
    5Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5Topics
    |
    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
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    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
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    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
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    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3Topics
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    1 Quiz
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The molecular formula of an element or compound is the formula which shows the exact number of atoms present in one molecule of the element or compound.

Relationship between Empirical and Molecular Formula

(Empirical formula) n   =  Relative molecular mass

n = The actual number of atoms present in the element or compound.

Relative molecular mass of Empirical formula = Relative molecular mass of the compound

Molecular formula  = n x (Empirical formula)

Example I

A hydrocarbon contains 80% carbon and 20% hydrogen. 

(a) Find the Empirical Formula of the compound. (b) If the relative molecular mass is 30, find its molecular formula. (H =1, C = 12)

Solution:

(a)

CH
Divide by RAM\( \frac{80}{12} \)\( \frac{20}{1} \)
Divide by the Smallest Value\( \frac{6.67}{6.67} \)\( \frac{20}{6.67} \)
13

Empirical Formula   =    CH3

(b) \( \scriptsize (CH_3)_n = 30 \)

\( \scriptsize (12 \: + \: 1 \: \times \: 3)_n = 30 \)

\( \scriptsize 15_n = 30 \)

\( \scriptsize n = \normalsize \frac{30}{15} \)

\( \scriptsize n = 2 \)

Molecular Formula  = (Empirical Formula)n

Molecular Formula   = (CH3)2

Molecular Formula   =  C2H6

Example II

Find the Empirical formula of a compound which on analysis yields the following as the reacting masses. 2.0g of carbon, 0.34g of hydrogen, and 2.67g of oxygen. From your results, find the molecular formula of the compound, if its relative molecular mass is 60. (C = 12, H = 1, O = 16)

Solution:

CarbonHydrogenOxygen
Divide by the RAM\( \frac{2.0}{12} \\ =\scriptsize 0.17\)\( \frac{0.34}{1} \\ = \scriptsize 0.34 \)\( \frac{2.67}{16} \\ =\scriptsize 0.17\)
Divide by the smallest value\( \frac{0.17}{0.17} \\ \)\( \frac{0.34}{0.17} \\ \)\( \frac{0.17}{0.17} \\ \)
121

Empirical Formula = CH2O

\( \scriptsize (CH_2O)_n = 60 \)

\( \scriptsize (12 \: + \: 1 \: \times \: 2 \: + \: 16)_n = 60 \)

\( \scriptsize (30)_n = 60 \)

\( \scriptsize n = \normalsize \frac{60}{30} \)

\( \scriptsize n = 2 \)

Molecular Formula =  (CH2O)2

Molecular Formula = C2H4O2   or   CH3COOH

Theory Questions

1. Atoms of the same element are alike in every aspect but differ from atoms of other elements. This is one of the ideals of Dalton’s atomic theory:

 (a) Which modern discovery rendered the ideal invalid 

 (b) How has the ideal been modified by the modern atomic theory

View Answer

2. (a) An element T exists in two Isotopic forms in ratio 2:3. If the relative masses of the Isotopes are 15 and 17 respectively. Calculate the relative atomic mass of the Element T.

(b) Calculate the relative molecular mass (RMM) of CaSO4 [Ca=40, S=32, O=16]

View Answer

3. Copy and complete the table below

ElementAtomic NumberMass NumberNo of ProtonNo of ElectronElectronic Configuration
A2714
B1818
C1152,3
View Answer

4. (a) Define the relative atomic mass of an element 

(b) Calculate the relative molecular mass of 

(i) Slaked lime Ca(OH)2                     

(ii) Sodium trioxonitrate(V), NaNO3  

(iii) Copper(II)tetraoxosulphate. CuSO   

(iv) Pentahydrate,CuSO4.5H2O     

[Ca=40, O=16, H=1, Na=23, N=14, Cu=635,S=32]

View Answer

5. a (i) Which instrument can be used to measure the mass of an atom of an element 

(ii)Who invented the instrument mentioned in 5a (i) above.

b (i) What are Isotopes?

   (ii) Name two elements that exhibits Isotopy, and give their Isotopic forms.

View Answer

Theory Question 1

1. Atoms of the same element are alike in every aspect but differ from atoms of other elements. This is one of the ideals of Dalton’s atomic theory:

 (a) Which modern discovery rendered the ideal invalid 

Answer: The discovery of isotopes.

 (b) How has the ideal been modified by the modern atomic theory

Answer: Through the discovery of isotopes. Atoms of the same element are all alike but have different masses. These atoms are called isotopes.

Theory Question 2

2. (a) An element T exists in two Isotopic forms in ratio 2:3. If the relative masses of the Isotopes are 15 and 17 respectively. Calculate the relative atomic mass of the Element T.

Solution:

Total ratio = 5

:> \( \frac {2}{5} \; \times \; \scriptsize 15 \\ = \scriptsize 2 \; \times \; 3 \\ = \scriptsize 6 \)

:> \( \frac {3}{5} \; \times \;\scriptsize 17 \\ = \scriptsize 10.2 \)

Total = 6 + 10.2 = 16.2

R.A.M = 16.2

(b) Calculate the relative molecular mass (RMM) of CaSO4 [Ca=40, S=32, O=16]

Solution

CaSO4; Ca = 40; S = 32; O4 = 16 x 4 = 64

R.M.M = 40 + 32 + 64

= 136

Theory Question 3

Copy and complete the table below

Element Atomic Number Mass Number No of Proton No of Electron Electronic Configuration
A 27 14
B 18 18
C 11 5 2,3

Solution:

Element

Atomic Number

Mass Number

No of Proton

No of Electron

Electronic Configuration

A

14

27

14

14

2, 8, 4

B

18

18

C

5

11

5

5

2,3

Theory Question 4

4. (a) Define the relative atomic mass of an element 

Answer:

The relative atomic mass of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth the mass of one atom of carbon 12.

(b) Calculate the relative molecular mass of 

(i) Slaked lime Ca(OH)2     

Answer:

Ca(OH)2   = 40 + (16+1)2

= 40 + 34

= 84g/mol

 

(ii) Sodium trioxonitrate(V), NaNO3  

Answer:

NaNO3 = 23 + 14 + (16 x 3)

= 23 + 14 + 48

= 85g/mol

 

(iii) Copper(II)tetraoxosulphate, CuSO   

Answer:

= 63 + 32 + (16×4)

= 63 + 32 + 64

= 159

 

(iv) Pentahydrate,CuSO4.5H2O  

Answer:

CuSO4.5H2O

= 63 + 32 + (16×4) + 5(1×2) + 16×5

= 63 + 32 + 64 + 10 + 80

= 249g/mol

 

[Ca=40, O=16, H=1, Na=23, N=14, Cu=635,S=32]

Theory Question 5

a (i) Which instrument can be used to measure the mass of an atom of an element

Answer:  mass spectrometer

 

(ii)Who invented the instrument mentioned in 5a (i) above.

Answer: F.W. Aston

 

b (i) What are Isotopes?

Answer: Isotopes are atoms of the same element with the same atomic number but different mass number.

 

(ii) Name two elements that exhibit Isotopy, and give their Isotopic forms.

Answer:  Oxygen and Carbon

:- \( \scriptsize _{8} ^{16} \textrm {0}, \; \scriptsize _{8} ^{17} \textrm {0}, \; \scriptsize _{8} ^{18} \textrm {0}\)

:- \( \scriptsize _{6} ^{12} \textrm {C}, \; \scriptsize _{6} ^{13} \textrm {C} \)

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