Like this presentation? Why not share!

# Chemical reactions

## by UITM Shah Alam on Oct 28, 2011

• 476 views

### Views

Total Views
476
Views on SlideShare
476
Embed Views
0

Likes
0
13
0

No embeds

### Categories

Uploaded via SlideShare as Microsoft PowerPoint

## Chemical reactionsPresentation Transcript

• CHEMICAL REACTIONS Dr Sharipah Ruzaina Syed Aris Empirical and Molecular Formula Balancing Chemical Reactions Types of Chemical Reactions
• Empirical and Molecular Formulas Empirical Formula - Molecular Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. The formula of the compound as it exists, it may be a multiple of the empirical formula.
• The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula.
• The Molecular Mass (formula mass, formula weight or molecular weight) of a compound is a multiple of the empirical formula mass.     MM = n x empirical formula mass
• Empirical Formula can be calculated from the percentage (or percent) composition of a compound.
• mass(g) of each element Sample Problem 3.4 Determining the Empirical Formula from Masses of Elements PROBLEM: PLAN: SOLUTION: amount(mol) of each element empirical formula Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? preliminary formula change to integer subscripts use # of moles as subscripts divide by M (g/mol) Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). 2.82 g Na = 0.123 mol Na 4.35 g Cl = 0.123 mol Cl 7.83 g O = 0.489 mol O Na 1 Cl 1 O 3.98 NaClO 4 Na 1 Cl 1 O 3.98 mol Na 22.99 g Na mol Cl 35.45 g Cl mol O 16.00 g O NaClO 4 NaClO 4 is sodium perchlorate.
• assume 100g lactic acid and find the mass of each element Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: PLAN: amount(mol) of each element During physical activity. lactic acid ( M =90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. preliminary formula empirical formula divide each mass by mol mass( M ) molecular formula use # mols as subscripts convert to integer subscripts divide mol mass by mass of empirical formula to get a multiplier
• Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass continued SOLUTION: Assuming there are 100. g of lactic acid, the constituents are 40.0 g C 6.71 g H 53.3 g O 3.33 mol C 6.66 mol H 3.33 mol O C 3.33 H 6.66 O 3.33 CH 2 O empirical formula 3 mol C 12.01g C mol H 1.008 g H mol O 16.00 g O 3.33 3.33 3.33 mass of CH 2 O molar mass of lactate 90.08 g 30.03 g C 3 H 6 O 3 is the molecular formula
• EXERCISES
• What is the empirical formula and molecular formula of a compound with a molar mass of 245.8g The composition is 19.53 C 2.44 H 13.02 O and 65.01 Br?
• Calculate the simplest formulas for the compounds whose compositions are listed:
• a) carbon, 15.8%; sulfur, 84.2%
• b) silver,70.1%; nitrogen,9.1%; oxygen,20.8%
• c) K, 26.6%; Cr, 35.4%, O, 38.0%
• The simplest formula for glucose is CH 2 O and its molar mass is 180 g/mol. What is its molecular formula?
• specify states of matter Balancing Chemical Equations translate the statement balance the atoms adjust the coefficients check the atom balance
• SOME PRACTICE PROBLEMS:
• 1. __NaCl + __BeF 2 -> __NaF + __BeCl 2
• 2. __FeCl 3 + __Be 3 (PO 4 ) 2 -> __BeCl 2 +__FePO 4
• 3. __AgNO 3 + __LiOH -> __AgOH + __LiNO 3
• 4. __CH 4 + __O 2 -> __CO 2 + __H 2 O
• 5. __Mg + __Mn 2 O 3 -> __MgO + __Mn
• TYPES OF CHEMICAL REACTIONS
• DECOMPOSITION REACTIONS
• In a decomposition reaction a single substance is broken down to form two or more simpler substrances.
reactant -------> product + product Exercise: CaCO 3 (s) ->
• COMBINATION REACTIONS
• Also called synthesis or addition reactions.
• Two or more substances react to form a single substance.
• Exp: 2Mg (s) + O 2 (g) -> 2MgO (s)
• Exercises:
• SO 3 (g) + H 2 O (l) ->
• PCl 3 (l) + Cl 2 (g) ->
• 2Cu + O 2 ->
• DOUBLE REPLACEMENT REACTIONS
• Two compounds switch places to form two new compounds. Two reactants yield two products.
• For example when silver nitrate combines with sodium chloride, two new compounds--silver chloride and sodium nitrate are formed because the sodium and silver switched places.
Exercise: HCl (aq) + NaOH (aq) ->
• COMBUSTION REACTIONS
• Combustion reactions are different in that they are characterized by the fact that one of the reactants is always oxygen.
• A combustion reaction is a reaction of a substance with oxygen, usually with the rapid release of heat to produce a flame.
• Organic compounds burn in oxygen to produce carbon dioxide and water vapor.
• Exp: butane burning in air to produce carbon dioxide and water vapor.
• 2C 4 H 10 (g) + 13O 2 (g) -> 8CO 2 (g) + 10H 2 O(g)
• FERMENTATION REACTION
• Ethanol fermentation is a biological process in which sugars such as glucose , fructose , and sucrose are converted into cellular energy and thereby produce ethanol and carbon dioxide as metabolic waste products.
• The chemical equation below summarizes the fermentation of glucose. One glucose molecule is converted into two ethanol molecules and two carbon dioxide molecules:
• C 6 H 12 O 6 -> 2C 2 H 5 OH + 2CO 2
• all metal oxides soluble in water gives alkalies(bases)
• CaO (s) + H 2 O (l) ↔ Ca(OH) 2
• Na 2 O + H 2 O -> 2 NaOH
• (MgO + H 2 O -> Mg(OH) 2 ),
• All nonmetals form covalent oxides with oxygen, which react with water to form acids
• N 2 O 5 + H 2 O -> 2HNO 3 .
• Metal oxide reaction with water
• Non-metal oxide reaction with water