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A sample problem

A sample problem. 3.43)

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A sample problem

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  1. A sample problem 3.43) Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%, H: 6.21%, S: 39.5%, O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g? empirical formula = molecular formula = C6H10S2O *Note that we arbitrarily chose a 100-g sample to make our math easier.

  2. Another sample problem 3.52) The empirical formula of a compound is CH. If the molar mass of this compound is about 78g, what is its molecular formula? • Figure out molar mass of the empirical formula. • Divide the molar mass of the compound by the molar mass of the empirical formula. • Multiply each subscript on the empirical formula by this quotient to get the molecular formula. 78/13.01=6 C6H6

  3. Yet another sample problem Practice exercise, p. 71) A sample of a compound containing boron and hydrogen contains 6.444 g of boron and 1.803 g of hydrogen. The molar mass of the compound is about 30 g. What is its molecular formula? • Figure out molar mass of the empirical formula. • Divide the molar mass of the compound by the molar mass of the empirical formula. • Multiply each subscript in the empirical formula by this whole number to get the molecular formula. B2H6

  4. 3 ways of representing the reaction of H2 with O2 to form H2O: reactants products A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction

  5. How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO 2 grams Mg + 1 gram O2 makes 2 g MgO *Note: Stoichiometric coefficients = atoms or moles… NOT grams!

  6. Balancing Chemical Equations C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane (C2H6)reacts with oxygen to form carbon dioxide & water • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

  7. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

  8. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2 3.7

  9. reactants products C 4 C 4 H 12 H 12 O 14 O 14 4CO2 + 6H2O 2C2H6 + 7O2 Balancing Chemical Equations • 5. Check your balanced eqn to make sure you have the same number of each type of atom on both sides of the eqn arrow. final tally

  10. Mass Changes in Chemical Reactions • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units

  11. Balance a combustion reaction Provide a balanced equation for the combustion of methanol (CH3OH). CH4O + O2 CO2 + H2O

  12. Balance a combustion reaction Provide a balanced equation for the combustion of methanol (CH3OH). 2CH4O + 3O2 2CO2 + 4H2O

  13. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18 g H2O 1 mol CH3OH = x x x 2 mol CH3OH 1 mol H2O 32 g CH3OH Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH molar mass H2O coefficients chemical equation 209 g CH3OH 235 g H2O

  14. Limiting Reagents When a chemist carries out a reaction, the reactants are usually not present in stoichiometric amounts, that is, in the proportions indicated by the balanced equation. It is common to use the cheaper reactant in excess to ensure that all of the expensive reagent is consumed. The more expensive reagent is the limiting reagent and is often completely consumed in a reaction.

  15. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160 g Fe2O3 1 mol Al = x x x 27 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3. Which reactant is the limiting reagent? 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

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