Chemical reactions


Published on

Published in: Technology, Business
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Chemical reactions

  1. 1. CHEMICAL REACTIONS Dr Sharipah Ruzaina Syed Aris Empirical and Molecular Formula Balancing Chemical Reactions Types of Chemical Reactions
  2. 2. 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.
  3. 3. <ul><li>The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula. </li></ul><ul><li>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 </li></ul><ul><li>Empirical Formula can be calculated from the percentage (or percent) composition of a compound. </li></ul>
  4. 4. 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.
  5. 5. 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
  6. 6. 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
  7. 7. EXERCISES <ul><li>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? </li></ul><ul><li>Calculate the simplest formulas for the compounds whose compositions are listed: </li></ul><ul><li>a) carbon, 15.8%; sulfur, 84.2% </li></ul><ul><li>b) silver,70.1%; nitrogen,9.1%; oxygen,20.8% </li></ul><ul><li>c) K, 26.6%; Cr, 35.4%, O, 38.0% </li></ul><ul><li>The simplest formula for glucose is CH 2 O and its molar mass is 180 g/mol. What is its molecular formula? </li></ul>
  8. 8. specify states of matter Balancing Chemical Equations translate the statement balance the atoms adjust the coefficients check the atom balance
  9. 9. SOME PRACTICE PROBLEMS: <ul><li>1. __NaCl + __BeF 2 -> __NaF + __BeCl 2 </li></ul><ul><li>2. __FeCl 3 + __Be 3 (PO 4 ) 2 -> __BeCl 2 +__FePO 4 </li></ul><ul><li>3. __AgNO 3 + __LiOH -> __AgOH + __LiNO 3 </li></ul><ul><li>4. __CH 4 + __O 2 -> __CO 2 + __H 2 O </li></ul><ul><li>5. __Mg + __Mn 2 O 3 -> __MgO + __Mn </li></ul>
  11. 11. DECOMPOSITION REACTIONS <ul><li>In a decomposition reaction a single substance is broken down to form two or more simpler substrances. </li></ul>reactant -------> product + product Exercise: CaCO 3 (s) ->
  12. 12. COMBINATION REACTIONS <ul><li>Also called synthesis or addition reactions. </li></ul><ul><li>Two or more substances react to form a single substance. </li></ul><ul><li>Exp: 2Mg (s) + O 2 (g) -> 2MgO (s) </li></ul><ul><li>Exercises: </li></ul><ul><ul><li>SO 3 (g) + H 2 O (l) -> </li></ul></ul><ul><ul><li>PCl 3 (l) + Cl 2 (g) -> </li></ul></ul><ul><ul><li>2Cu + O 2 -> </li></ul></ul>
  13. 13. DOUBLE REPLACEMENT REACTIONS <ul><li>Two compounds switch places to form two new compounds. Two reactants yield two products. </li></ul><ul><li>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. </li></ul>Exercise: HCl (aq) + NaOH (aq) ->
  14. 14. COMBUSTION REACTIONS <ul><li>Combustion reactions are different in that they are characterized by the fact that one of the reactants is always oxygen. </li></ul><ul><li>A combustion reaction is a reaction of a substance with oxygen, usually with the rapid release of heat to produce a flame. </li></ul><ul><li>Organic compounds burn in oxygen to produce carbon dioxide and water vapor. </li></ul><ul><li>Exp: butane burning in air to produce carbon dioxide and water vapor. </li></ul><ul><li>2C 4 H 10 (g) + 13O 2 (g) -> 8CO 2 (g) + 10H 2 O(g) </li></ul>
  15. 15. FERMENTATION REACTION <ul><li>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. </li></ul><ul><li>The chemical equation below summarizes the fermentation of glucose. One glucose molecule is converted into two ethanol molecules and two carbon dioxide molecules: </li></ul><ul><li>C 6 H 12 O 6 -> 2C 2 H 5 OH + 2CO 2 </li></ul>
  16. 16. <ul><li>all metal oxides soluble in water gives alkalies(bases) </li></ul><ul><li>CaO (s) + H 2 O (l) ↔ Ca(OH) 2 </li></ul><ul><li>Na 2 O + H 2 O -> 2 NaOH </li></ul><ul><li>(MgO + H 2 O -> Mg(OH) 2 ), </li></ul><ul><li>All nonmetals form covalent oxides with oxygen, which react with water to form acids </li></ul><ul><li>N 2 O 5 + H 2 O -> 2HNO 3 . </li></ul><ul><li>Metal oxide reaction with water </li></ul><ul><li>Non-metal oxide reaction with water </li></ul>