Form 4 Chemistry Chapter 3 Chemical Formula and Equation
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The document provides information about chemical formulas and equations. It defines empirical and molecular formulas, and explains how to determine them through calculation of moles and mass ratios of elements in a compound. It also describes writing and balancing chemical equations, naming ionic compounds based on their constituent ions, and using formulas and equations to solve stoichiometric problems. Key topics covered include determining formulas from experimental data, relating formulas to molecular structure and mass, and representing chemical reactions systematically.
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Determining empirical formula
Example
1.08g of aluminium powder combines completely with 0.96g of oxygen
to produce an oxide compound. What is the empirical formula of the
oxide? [Relative atomic mass: O, 16; Al, 27]
Element Aluminium, Al Oxygen, O
Mass (g) 1.08 0.96
Number of
moles of atoms
1.08
27
= 0.04
0.96
16
= 0.06
Ratio of
number moles
0.04
0.04
= 1
0.06
0.04
= 1.5
Simplest ratio
of number of
moles
2 3
Solution
Al2O3](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-6-320.jpg)
![8
Check Point
1. A compound contains 3.06% of hydrogen, 31.63% of phosphorus and 65.31% of
oxygen. What is the empirical formula of the compound?
[RAM: O, 16; H, 1; P, 31]
Element Hydrogen, H Phosphorus, P Oxygen, O
Mass 3.06 31.63 65.31
Number of
moles of atoms
3.06
1
= 3.06
31.63
31
=1.02
65.31
16
=4.08
Ratio of
number moles
4.08
1.02
= 4
3.06
1.02
= 3
1.02
1.02
= 1
Simplest ratio
of number of
moles
3 1 4
H3PO4](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-8-320.jpg)
![10
Determining molecular formula
Example
A compound has an empirical formula of CH2 and a relative
molecular mass of 70. What is the compound’s molecular formula?
[Relative atomic mass: H,1; C, 12]
Solution
Let the molecular formula of the compound to be (CH2)n.
(CH2)n = n[12 + 2(1)]
= 14n
So, 14n = 70
n =
70
14
= 5
Therefore, the molecular formula of the compound is (CH2) 5 , which is C5H10.](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-10-320.jpg)
![11
Check Point
(a) What is meant by empirical formula? [1 mark]
(b) A carbon compound contains 92.3% of carbon and 7.7% of hydrogen by
mass. The relative molecular mass of this compound is 78. Find the
molecular formula of this compound. [RAM: C = 12; H =1] [5 marks]
(c) Describe how you could determine the empirical formula of magnesium
oxide in the laboratory. Your description should include
- the procedure of the experiment
- tabulation of the data
- the calculation of the results
[10 marks]
Empirical formula is a chemical formula that shows the simplest ratio of the
number of atoms for each element in the compound.
Element C H
Mass
No. of moles
Ratio of moles
Empirical formula = CH
Relative molecular mass = (CH)n
(CH)n = 78
(12+1)n = 78
13n = 78
n = 6 Molecular formula: C6H6
92.3% 7.7%
92.3/12 = 7.7 7.7/1 =7.7
1 1](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-11-320.jpg)
![12
Calculation involving empirical
and molecular formula
• Empirical formula and molecular formula can help us to solve
calculation problems related to the composition of compounds.
Example
What is the mass of zinc that is needed to combine with 0.5 mole of chlorine
atoms to form zinc chloride, ZnCl2 ? [Relative atomic mass: Zn, 65]
Solution
Number of moles of Cl = 0.5
mass?
Zn + Cl2 ZnCl2
0.5 mole
Therefore, number of moles of Zn = 0.5÷ 2
= 0.25
Number of moles =
𝒎𝒂𝒔𝒔
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
0.25 =
𝒎𝒂𝒔𝒔
𝟔𝟓
mass = 0.25 x 65
= 16.25g](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-12-320.jpg)
![13
Calculation involving empirical
and molecular formula
The percentage composition based on mass can be calculated as follows:
Percentage of an element by mass in a compound = Mass of the element in 1
mole of the compound
Mass of 1 mole of the
compound
X 100%
Example
Calculate the percentage by mass of carbon in octane, C8H18 [RAM: H, 1; C, 12]
Solution
The mass of 1 mole of octane = 8(12) +18(1)
=114
The mass of carbon in 1 mole of octane = 8 x 12
= 96
So, the percentage by mass of carbon =
96
114
x 100% = 84.21%](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-13-320.jpg)
![22
Chemical Equations
Example
What is the mass of sodium needed to react with 0.5 mole of chlorine?
[Relative atomic mass: Na, 23]
Solution
Na + Cl2 NaCl22
0.5molemass ?
So, number of mole of Na = 1 mole
Number of moles =
𝑚𝑎𝑠𝑠
𝑅𝐴𝑀
1 =
𝑚𝑎𝑠𝑠
23
mass = 23 x 1
= 23g](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-22-320.jpg)
![23
SPM Target
Question 1
Sodium thiosulphate has the formula of Na2S2O3. [RAM: Na, 23; S, 32; O, 16]
(a) What is the mass of sulphur in 0.2 mole of sodium thiosulphate?
(b) How many moles of oxygen atoms are present in 79g of sodium
thiosulphate?
Number of moles =
𝒎𝒂𝒔𝒔(𝒈)
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
0.2 =
𝒈
𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔)
0.2 =
𝒈
𝟏𝟓𝟖
g = 31.6
158 31.6g
2(32) x g
2022.4 = 158 x
Let x be the mass of sulphur
X = 12.8g
Number of moles of sodium thiosulphate =
𝒎𝒂𝒔𝒔(𝒈)
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
=
𝟕𝟗
𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔)
= 0.5
1 mol of Na2S2O3 can produce 3 atoms of
oxygen.
Therefore, number of moles of oxygen atoms
= 3 x 0.5
= 1.5 moles](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-23-320.jpg)
![25
SPM Target
Equation for the reaction between potassium and oxygen:
K + O2 K2O
Question 3
24
What is the maximum mass of potassium oxide that is formed when 19.5g
potassium is burned completely in excess oxygen? [RAM: O, 16; K, 39]
19.5g Mass?
No.of moles of potassium = 19.5
39
=0.5mol
No.of moles of potassium oxide
= 0.5/2
=0.25mol
0.25 = mass
2(39) +16
Mass = 0.25 x 94
= 23.5g](https://image.slidesharecdn.com/f4-chem-chapter3-chemicalformulaandequation-200625142536/85/Form-4-Chemistry-Chapter-3-Chemical-Formula-and-Equation-25-320.jpg)
Form 4 Chemistry Chapter 3 Chemical Formula and Equation
- 1. Chemistry Chapter 3 Chemical Formula & Equations -Ms Cindy-
- 3. What is chemical formula? 3 Representation of a chemical substance using letters for atoms and subscript numbers to show the number of each type of atoms that are present in the substance.
- 4. 4 What is chemical formula? Molecular formula Empirical formula Formula that show the actual number of atoms of each element present in one molecule of the compound. Formula that shows the simplest whole number ratio of atoms of each element present in the compound. Compound Molecular formula Simplest whole number ratio Empirical formula Water H20 H:O = 2:1 H20 Benzene C6H6 C:H = 1:1 CH Vitamin C C6H8O6 C:H:O = 3:4:3 C3H4O3
- 5. 5 Determining empirical formula Determining the mass of each element in the compound. Converting the mass of each element into the number of moles of atoms. Determining the simplest ratio of number of moles of atoms of each element. 1 2 3 𝒎𝒂𝒔𝒔(𝒈) 𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
- 6. 6 Determining empirical formula Example 1.08g of aluminium powder combines completely with 0.96g of oxygen to produce an oxide compound. What is the empirical formula of the oxide? [Relative atomic mass: O, 16; Al, 27] Element Aluminium, Al Oxygen, O Mass (g) 1.08 0.96 Number of moles of atoms 1.08 27 = 0.04 0.96 16 = 0.06 Ratio of number moles 0.04 0.04 = 1 0.06 0.04 = 1.5 Simplest ratio of number of moles 2 3 Solution Al2O3
- 7. 7 Determining empirical formula Procedure: 1. A crucible with its lid is weighed. 2. Mg ribbon is cleaned with sand paper to remove its oxide layer. 3. The ribbon is put into the crucible covered with lid, and weighed. 4. When the burning is complete, the lid is removed and the crucible is heated strongly for 1 to 2 minutes. 5. After cool down, the crucible, its lid and content are weighed again. Results: Items Mass Crucible + lid X Crucible + lid + mg ribbon Y Crucible + lid + magnesium oxide z Element Mg O Mass Number of moles of atoms Simplest ratio of number of moles of atom Empirical formula = MgpOq
- 8. 8 Check Point 1. A compound contains 3.06% of hydrogen, 31.63% of phosphorus and 65.31% of oxygen. What is the empirical formula of the compound? [RAM: O, 16; H, 1; P, 31] Element Hydrogen, H Phosphorus, P Oxygen, O Mass 3.06 31.63 65.31 Number of moles of atoms 3.06 1 = 3.06 31.63 31 =1.02 65.31 16 =4.08 Ratio of number moles 4.08 1.02 = 4 3.06 1.02 = 3 1.02 1.02 = 1 Simplest ratio of number of moles 3 1 4 H3PO4
- 9. 9 Determining molecular formula • Molecular formula is a multiple of empirical formula, where n is an integer. Molecular formula = (Empirical formula)n Compound Empirical Formula n Molecular formula Water H20 1 (H20)1 = H20 Ethane CH3 2 (CH3) 2 = C2H6 Glucose CH2O 6 (CH2O) 6 = C6H12O6 • The determination of a molecular formula requires the following information: Its empirical formula Its relative molecular mass or its molar mass
- 10. 10 Determining molecular formula Example A compound has an empirical formula of CH2 and a relative molecular mass of 70. What is the compound’s molecular formula? [Relative atomic mass: H,1; C, 12] Solution Let the molecular formula of the compound to be (CH2)n. (CH2)n = n[12 + 2(1)] = 14n So, 14n = 70 n = 70 14 = 5 Therefore, the molecular formula of the compound is (CH2) 5 , which is C5H10.
- 11. 11 Check Point (a) What is meant by empirical formula? [1 mark] (b) A carbon compound contains 92.3% of carbon and 7.7% of hydrogen by mass. The relative molecular mass of this compound is 78. Find the molecular formula of this compound. [RAM: C = 12; H =1] [5 marks] (c) Describe how you could determine the empirical formula of magnesium oxide in the laboratory. Your description should include - the procedure of the experiment - tabulation of the data - the calculation of the results [10 marks] Empirical formula is a chemical formula that shows the simplest ratio of the number of atoms for each element in the compound. Element C H Mass No. of moles Ratio of moles Empirical formula = CH Relative molecular mass = (CH)n (CH)n = 78 (12+1)n = 78 13n = 78 n = 6 Molecular formula: C6H6 92.3% 7.7% 92.3/12 = 7.7 7.7/1 =7.7 1 1
- 12. 12 Calculation involving empirical and molecular formula • Empirical formula and molecular formula can help us to solve calculation problems related to the composition of compounds. Example What is the mass of zinc that is needed to combine with 0.5 mole of chlorine atoms to form zinc chloride, ZnCl2 ? [Relative atomic mass: Zn, 65] Solution Number of moles of Cl = 0.5 mass? Zn + Cl2 ZnCl2 0.5 mole Therefore, number of moles of Zn = 0.5÷ 2 = 0.25 Number of moles = 𝒎𝒂𝒔𝒔 𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀 0.25 = 𝒎𝒂𝒔𝒔 𝟔𝟓 mass = 0.25 x 65 = 16.25g
- 13. 13 Calculation involving empirical and molecular formula The percentage composition based on mass can be calculated as follows: Percentage of an element by mass in a compound = Mass of the element in 1 mole of the compound Mass of 1 mole of the compound X 100% Example Calculate the percentage by mass of carbon in octane, C8H18 [RAM: H, 1; C, 12] Solution The mass of 1 mole of octane = 8(12) +18(1) =114 The mass of carbon in 1 mole of octane = 8 x 12 = 96 So, the percentage by mass of carbon = 96 114 x 100% = 84.21%
- 14. 14 Chemical formula of ionic compound Cation Anion+ Formula of ionic compound: Sodium ion Na+ Magnesium ion Mg2+ Aluminium ion Al3+ Lead(IV) ion Pb4+ Chloride ion Cl- Hydroxide ion OH- Oxide ion O2- Phosphate ion PO4 3-
- 15. 15 Chemical formula of ionic compound List of Cations and Anions
- 16. 16 Chemical formula of ionic compound • The formula of an ionic compound combines the formula of its cation (positive ion) and the formula of its anion (negative ion). • However, the formula of an ionic compound is neutral because the total positive charges equal the total negative charges. Example Iron (III) chloride Fe3+ Cl- FeCl3
- 17. 17 Chemical formula of ionic compound • When polyatomic ions such as SO4 2-, OH-, NO3 - , NH4 + are involved, brackets are used to show the number of ions in the formula.
- 18. 18 Naming of chemical compounds • Chemical compounds are named systematically according to the guidelines given by the International Union of Pure and Applied Chemistry (IUPAC). Name of cation Name of anion Magnesium sulphate Chemical formula Name CO Carbon monoxide CO2 Carbon dioxide SO3 Sulphur trioxide PCl5 Phosphorus pentachloride MgSO4 Chemical formula Name Ca(OH) Calcium hydroxide KCl Potassium chloride FeCl3 Iron (III) chloride NaBr Sodium bromide
- 19. 19 Check Point Name each of the following ionic compound. (a) Al(OH) 3 (b) FeSO4 (c) NH4Cl (d) Ca(NO3) 2 (e) K2CO3 (f) ZnS Aluminium hydroxide Iron(II) sulphate Ammonium chloride Calcium nitrate Potassium carbonate Zinc sulphide
- 20. 20 Chemical Equations • Describe a chemical reaction. • Can be written in words or chemical formulae. 2 H2 (g) + O2 (g) 2 H2O (l) reactants products • The state symbols (s), (l) and (g) represent the solid, liquid and gaseous states respectively. The symbol (aq) represents the aqueous solution.
- 21. 21 Chemical Equations • According to the law of conservation of mass, a matter can neither be created nor destroyed. So, the numbers of atoms before and after a chemical reaction are the same. Therefore, a chemical equation must be balanced. Example Iron fillings react with copper(II) chloride solution to form iron (III) chloride and copper metal. Fe CuCl2+ FeCl3 + Cu3 22 3 (2mol) (3mol) (2 mol) (3 mol)
- 22. 22 Chemical Equations Example What is the mass of sodium needed to react with 0.5 mole of chlorine? [Relative atomic mass: Na, 23] Solution Na + Cl2 NaCl22 0.5molemass ? So, number of mole of Na = 1 mole Number of moles = 𝑚𝑎𝑠𝑠 𝑅𝐴𝑀 1 = 𝑚𝑎𝑠𝑠 23 mass = 23 x 1 = 23g
- 23. 23 SPM Target Question 1 Sodium thiosulphate has the formula of Na2S2O3. [RAM: Na, 23; S, 32; O, 16] (a) What is the mass of sulphur in 0.2 mole of sodium thiosulphate? (b) How many moles of oxygen atoms are present in 79g of sodium thiosulphate? Number of moles = 𝒎𝒂𝒔𝒔(𝒈) 𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀 0.2 = 𝒈 𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔) 0.2 = 𝒈 𝟏𝟓𝟖 g = 31.6 158 31.6g 2(32) x g 2022.4 = 158 x Let x be the mass of sulphur X = 12.8g Number of moles of sodium thiosulphate = 𝒎𝒂𝒔𝒔(𝒈) 𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀 = 𝟕𝟗 𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔) = 0.5 1 mol of Na2S2O3 can produce 3 atoms of oxygen. Therefore, number of moles of oxygen atoms = 3 x 0.5 = 1.5 moles
- 24. SPM Target 24 Diagram shows the structural formula of vitamin C. (a) What is the molecular formula of vitamin C? (b) What is the empirical formula of vitamin C? (c) State the molar mass of vitamin C. (d) Calculate the percentage of oxygen by mass in a molecule of vitamin C. Question 2 C6H8O6 C3H4O3 6(12) + 8(1) + 6(16) = 176g/mol 𝟔(𝟏𝟔) 𝟏𝟕𝟔 x 100% = 54.55%
- 25. 25 SPM Target Equation for the reaction between potassium and oxygen: K + O2 K2O Question 3 24 What is the maximum mass of potassium oxide that is formed when 19.5g potassium is burned completely in excess oxygen? [RAM: O, 16; K, 39] 19.5g Mass? No.of moles of potassium = 19.5 39 =0.5mol No.of moles of potassium oxide = 0.5/2 =0.25mol 0.25 = mass 2(39) +16 Mass = 0.25 x 94 = 23.5g
- 26. THANKS! Any questions? You can find me at ✘ Fb: https://www.facebook.com/brilliantA.edu/ ✘ IG: instagram.com/mybrilliant_a ✘ Tel: 011-15093232 26
- 27. 27 Form 4: Chapter 6 Acid Bases and Salts Form 5: Chapter 3 Oxidation and Reduction See you!