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# Lecture08221

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### Lecture08221

1. 1. Stoichiometry: Writing Reactions, Calculating Amounts Lecture 8
2. 2. Think in moles. It is truly chemical.
3. 3. Which information tells us more? <ul><li>2.016 g of H 2 and 38.00 g of F 2 react to form 40.02 g of HF </li></ul><ul><li>1 mol of H 2 and 1 mol of F 2 react to form 2 mol of HF </li></ul>
4. 4. Macroscopic change corresponds to submicroscopic change. We see it when express reactions in terms of moles.
5. 5. Chemical reaction is a statement in formulas that expresses the identities and quantities of the substances involved in a chemical or physical change.
6. 6. Chemical reaction must be balanced to depict all amounts accurately. The same number of each type of atoms must appear on both sides of the equation.
7. 7. Converting chemical statements into balanced chemical equations: <ul><li>Translating the Statement </li></ul><ul><li>Balancing the Atoms </li></ul><ul><li>Adjusting the Coefficients </li></ul><ul><li>Checking </li></ul><ul><li>Specifying the States of Matter. </li></ul>
8. 8. A sample problem on balancing chemical equations.
9. 9. Remember: <ul><li>A coefficient operates on all the atoms in the formula that follows it: 2MgO means 2 x (MgO), or 2 Mg atoms and 2 O atoms; 2Ca(NO 3 ) 2 means 2 x [Ca(NO 3 ) 2 ], or 2Ca atoms, 4 N atoms, and 12 O atoms. </li></ul>
10. 10. Remember: <ul><li>In balancing an equation, chemical formulas cannot be altered: </li></ul><ul><li>In Mg + O 2  MgO we cannot balance the O atoms by changing MgO to MgO 2 because MgO 2 has a different elemental composition and thus is a different compound. </li></ul>
11. 11. Remember: <ul><li>We cannot add other reactants or products to balance the equation because it would represent a different chemical reaction. </li></ul><ul><li>In Mg + O 2  MgO we cannot balance the O atoms by changing O 2 to O or by adding one O atom to the products, because the chemical statement does not say that the reaction involves O atoms. </li></ul>
12. 12. Remember: <ul><li>A balanced equation remains balanced even if you multiply all the coefficients by the same number . </li></ul><ul><li>2Mg (s) + O 2(g)  2MgO (s) </li></ul><ul><li>4Mg (s) + 2O 2(g)  4MgO (s) </li></ul><ul><li>6Mg (s) + 3O 2(g)  6MgO (s) </li></ul><ul><li>1024Mg (s) + 512O 2(g)  1024MgO (s) </li></ul>
13. 13. If we know the amount of one substance in the reaction, the balanced equation tells us the amounts of all the others in the reaction.
14. 14. The amount of one substance in the reaction is stoichiometrically equivalent to the amounts of any other substance in the reaction.
15. 15. Stoichiometrically equivalent molar ratios are used as conversion factors to calculate amounts of substances.
16. 16. General approach for solving any stoichiometry problem: <ul><li>Write a balanced equation for the reaction. </li></ul><ul><li>Convert the given mass (or number of entities) of the first substance to amount (mol). </li></ul><ul><li>Use the appropriate molar ratio from the balanced equation to calculate the amount (mol) of the second substance. </li></ul><ul><li>Convert the amount of the second substance to the desired mass (or number of entities). </li></ul>
17. 17. A sample problem on calculating amounts of reactants and products.
18. 18. General approach for writing overall equations: <ul><li>Write the sequence of balanced equations. </li></ul><ul><li>Adjust the equations arithmetically to cancel out the common substance. </li></ul><ul><li>Add the adjusted equations together to obtain the overall balanced equations. </li></ul>
19. 19. A sample problem on writing an overall reaction for a reaction sequence.
20. 20. A sample limiting-reactant problem on using molecular depictions.
21. 21. A sample problem on calculating amounts of reactant and product in reactions involving a limiting reagent.
22. 22. Theoretical yield is amount of substance indicated by the stoichiometrically equivalent molar ratio in the balanced equation.
23. 23. Side reactions is one reason the theoretical yield is never obtained is that other reactions lead some of the reactants along side paths to form undesired products.
24. 24. Actual yield is amount of substance that we actually obtain.
25. 25. Percent yield is the actual yield expressed as a percentage of the theoretical yield: %yield= (actual / theoretical) x 100%
26. 26. A sample problem on calculating percent yield.
27. 27. Overall percent yield in the multistep syntheses equals to the product of all the step yields expressed as fractions and multipled by 100%
28. 28. THE END