1. Calculations with Chemical Formulas and Equations Chapter 3 Dr. Victor Vilchiz

2. What is a mole? A mole is a unit of measurement used to specified amounts of chemical substances. A mole is a unit of measurement used to specified amounts of chemical substances. It is not a unit of mass.It is not a unit of mass. It is similar to “a dozen”It is similar to “a dozen” A dozen eggs is not the same as a dozen cars but they are still both a dozen. A dozen eggs is not the same as a dozen cars but they are still both a dozen.

3. What is a mole? A mole is defined as the number of atoms of carbon in 12 g of Carbon-12. (examples) A mole is defined as the number of atoms of carbon in 12 g of Carbon-12. (examples)(examples) 1mole=6.022x10 23 atoms and can be applied to any moiety 1mole=6.022x10 23 atoms and can be applied to any moiety 6.022x10 23 is also known as Avogadro’s Number (N A )6.022x10 23 is also known as Avogadro’s Number (N A ) 1mol of Carbon=12g Carbon = 6.022x10 23 C atoms 1mol of Carbon=12g Carbon = 6.022x10 23 C atoms 1mol of water= 18g H 2 O =6.022x10 23 water molecules 1mol of water= 18g H 2 O =6.022x10 23 water molecules

4. Why the mole? The mole helps determine amounts of substances and allows for conversion between species. The mole helps determine amounts of substances and allows for conversion between species. CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + H 2 CO 3 (aq)CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + H 2 CO 3 (aq) From this we cannot say 1g of CaCO 3 will react with 2 grams of HCl; however, we can say 1mol of CaCO 3 reacts with 2 moles of HCl. From this we cannot say 1g of CaCO 3 will react with 2 grams of HCl; however, we can say 1mol of CaCO 3 reacts with 2 moles of HCl. 1 mole of CaCO 3 is not the same as 1 mole of HCl mass wise, but both have 6.022x10 23 molecules.1 mole of CaCO 3 is not the same as 1 mole of HCl mass wise, but both have 6.022x10 23 molecules.

5. Masses Isotopic Mass is the mass of an elemental isotope (given in amu). Isotopic Mass is the mass of an elemental isotope (given in amu). Atomic Weight is the weighted average of all natural occurring isotopes of an element (amu). Atomic Weight is the weighted average of all natural occurring isotopes of an element (amu). Molecular Mass is the sum of the atomic masses of the atoms in a molecule (amu). Molecular Mass is the sum of the atomic masses of the atoms in a molecule (amu). Molar Weight is the weight of one mole of chemical moieties (g/mol). Molar Weight is the weight of one mole of chemical moieties (g/mol).

6. Formula Weight The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu.

7. Molar Mass Examples Molar Mass of Ca(C 2 H 3 O 2 ) 2, Calcium Acetate. Molar Mass of Ca(C 2 H 3 O 2 ) 2, Calcium Acetate. 2x(40.1)+4(12.0)+6(1.01)+4(16.0)=198.3 g/mol Molar Mass of Ethylene Glycol, C 2 H 4 O 2. Molar Mass of Ethylene Glycol, C 2 H 4 O 2.2x(12.0)+4(1.01)+2(16.0)=60.0g/mol Molar Mass of Ammonium Oxalate, (NH 4 ) 2 C 2 O 4. Molar Mass of Ammonium Oxalate, (NH 4 ) 2 C 2 O 4. 2x(14.0)+8x(1.01)+2(12.0)+4(16.0)=124. 1g/mol

8. Stoichiometry.- conversion from mass to moles to mass Stoichiometry is the study of chemical quantities and their chemical reaction relationship. Stoichiometry is the study of chemical quantities and their chemical reaction relationship. Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations.Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations.

9. Mass and Moles of a Substance Mole calculations Mole calculations Example: 2.934g of NaCl = ? Moles NaCl 1.321moles of CuSO 4 = ? g CuSO 4 Chart

10. Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. The percent composition of a compound is the mass percentage of each element in the compound. We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is,

11. Mass Percentages from Formulas Let’s calculate the percent composition of butane, C 4 H 10. Let’s calculate the percent composition of butane, C 4 H 10. First, we need the molecular mass of C 4 H 10. Now, we can calculate the percents.

12. Determining Chemical Formulas Determining both empirical and molecular formulas of a compound from the percent composition. Determining both empirical and molecular formulas of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts.

13. Determining Chemical Formulas Determining the empirical formula from the percent composition. Determining the empirical formula from the percent composition. Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest whole- number ratio of the subscripts in the formula Cx HyOzCx HyOz

14. Determining Chemical Formulas Determining the empirical formula from the percent composition. Determining the empirical formula from the percent composition. For the purposes of this calculation and making calculations simpler, we will assume we have 100.0 grams of sample benzoic acid. Then the percentage of each element equals the mass of each element in the sample. Since x, y, and z in our formula represent mole-mole ratios, we must first convert these masses to moles.

15. Determining Chemical Formulas This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain:

16. Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: now it’s not too difficult to see that the smallest whole number ratio is 7:6:2. The empirical formula is C 7 H 6 O 2.

17. Determining Chemical Formulas Determining the “true” molecular formula from the empirical formula. Determining the “true” molecular formula from the empirical formula. An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The “true” molecular formula could be a multiple of the empirical formula (since both would have the same percent composition). To determine the “true” molecular formula, we must know the “true” molecular weight of the compound.

18. Determining Chemical Formulas Determining the “true” molecular formula from the empirical formula. Determining the “true” molecular formula from the empirical formula. For example, suppose the empirical formula of a compound is CH 2 O and its “true” molecular weight is 60.0 g/mol. The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g/mol. This would imply that the “true” molecular formula is actually the empirical formula doubled 2(CH 2 O) or C2H4O2C2H4O2

19. Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry.

20. Molar Interpretation of a Chemical Equation Balancing equations is part of Stoichiometry. Balancing equations is part of Stoichiometry. Balancing equations Balancing equations A balanced chemical equation:A balanced chemical equation: 2H 2 +1O 2  2H 2 O can be interpreted to read 2 moles of Hydrogen react with one mole of oxygen to produce 2 moles of water. In the balanced equation the 2, 1, and 2 are known as the stoichiometric coefficients.In the balanced equation the 2, 1, and 2 are known as the stoichiometric coefficients. At the molecular level they refer to the number of molecules reacting. At the molecular level they refer to the number of molecules reacting.

21. Molar Interpretation of a Chemical Equation Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. 2H 2 +1O 2  2H 2 O 2(2.02g)H 2 react with 1(32.0g) O 2 to yield 2(18.0g)H 2 O 4.04g H 2 react with 32.0g O 2 to yield 36.0g H 2 O

22. Stoichiometric Calculations Suppose we wished to determine the number of moles of NH 3 we could obtain from 4.8 mol H 2. Suppose we wished to determine the number of moles of NH 3 we could obtain from 4.8 mol H 2. Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple.

23. Limiting Reagent In the previous example we assumed that we had enough N 2 to react with the 4.8 moles of H 2. However it is possible that we could had ran out of nitrogen before consuming the hydrogen in that case the N 2 will be labeled as the limiting reactant. In the previous example we assumed that we had enough N 2 to react with the 4.8 moles of H 2. However it is possible that we could had ran out of nitrogen before consuming the hydrogen in that case the N 2 will be labeled as the limiting reactant. The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion.

24. Limiting Reactant The limiting reagent ultimately determines how much product can be obtained. For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made. Pictorial Example

25. Stoichiometric Calculations Amounts of substances in a chemical reaction by mass. Amounts of substances in a chemical reaction by mass. How many grams of HCl are required to react with 5.00 grams manganese(IV) oxide, according to the following chemical equation?

26. First, you write what is given (5.00 g MnO 2 ) and convert this to moles. Then convert to moles of what is desired, mol HCl, using the stoichiometric coefficients. Finally, you convert this to mass of HCl, (g HCl). Stoichiometric Calculations

27. Limiting Reagent Example Zinc metal reacts with hydrochloric acid by the following reaction. Zinc metal reacts with hydrochloric acid by the following reaction. If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H 2 are produced?

28. Limiting Reagent Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent. Since HCl is the limiting reagent, produces the least H 2, the amount of H 2 produced must be 0.26 mol.

29. Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. The percent yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). The percent yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated).

30. Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of H 2 in the previous example was 0.26 mol (or 0.52 g) H 2. To illustrate the calculation of percentage yield, recall that the theoretical yield of H 2 in the previous example was 0.26 mol (or 0.52 g) H 2. If the actual yield of the reaction had been 0.22 g H 2, then

31. Reactions in a sequence In the case when you have reactions in a sequence: In the case when you have reactions in a sequence: balance all reactions in the sequence.balance all reactions in the sequence. add all reactions (both sides).add all reactions (both sides). Cancel species found on both sides of the overall reaction (make sure they in the same physical state before canceling).Cancel species found on both sides of the overall reaction (make sure they in the same physical state before canceling). The left over is the net equation.The left over is the net equation.

32. Solution Stoichiometry Molarity is the measurement of the concentration of a chemical in solution. Molarity is the measurement of the concentration of a chemical in solution. The unit of molarity is the Molar (M).The unit of molarity is the Molar (M). Example: Calculate the molarity of a solution made by dissolving 12.94g of Ca(OH) 2 in enough water to make 1.23L of solution.

33. Molarity Example: how many grams of ammonium nitrate are in a 172.7mL sample of 1.21M NH 4 NO 3 solution?

34. Diluting Solutions When diluting a solution the number of moles is constant. When diluting a solution the number of moles is constant. (Molarity)(Volume) =moles M 1 xV 1 = n = M 2 xV 2 So if we change the volume the new molarity is obtained by: So if we change the volume the new molarity is obtained by: M 2 =M 1 xV 1 /V 2

35. Operational Skills Calculating the formula weight from a formula. Calculating the formula weight from a formula. Converting moles of substance to grams and vice versa. Converting moles of substance to grams and vice versa. Calculating the number of molecules in a given mass. Calculating the number of molecules in a given mass. Calculating the percentage composition from the formula. Calculating the percentage composition from the formula. Calculating the mass of an element in a given mass of compound. Calculating the mass of an element in a given mass of compound. Determining the empirical formula from percentage composition. Determining the empirical formula from percentage composition. Determining the true molecular formula. Determining the true molecular formula. Relating quantities in a chemical equation. Relating quantities in a chemical equation. Calculating with a limiting reagent. Calculating with a limiting reagent.

36. One mole each of various substances. Photo courtesy of American Color.

37. Mass Difference of one mole of substance Return to lecture

38. Stoichiometric Calculations Road Map Back to Lecture

39. Balancing Chemical Equations There are some steps that can make balancing equations easier: There are some steps that can make balancing equations easier: Translate the problem from words to a preliminary equation.Translate the problem from words to a preliminary equation. Balance the atoms using the stoichiometric coefficients.Balance the atoms using the stoichiometric coefficients. Start with substance with most atoms or different types of atoms Start with substance with most atoms or different types of atoms Leave simple atoms for last specially H and O Leave simple atoms for last specially H and O Adjust coefficients so no fractions are present.Adjust coefficients so no fractions are present. Note: Remember that coefficients affect all atoms in the formula, subscripts only affect the atom or ion preceding it.

40. Balancing Chemical Equations Example: Example: When propane burns in excess oxygen both water vapor and carbon dioxide are produced.When propane burns in excess oxygen both water vapor and carbon dioxide are produced. Translate Translate C 3 H 8 (g) + O 2 (g)  H 2 O(g) +CO 2 (g) C 3 H 8 (g) + O 2 (g)  H 2 O(g) +CO 2 (g) Balance the atoms using the stoichiometric coefficients. Balance the atoms using the stoichiometric coefficients. C 3 H 8 (g) + O 2 (g)  4H 2 O(g) +3CO 2 (g) Both C and H are balanced but not the oxygen, we have 4(1)+3(2)=10 Oxygen atoms on the right so we need a 5 in front of the O 2 on the left to balance the equation. C 3 H 8 (g) + 5O 2 (g)  4H 2 O(g) +3CO 2 (g) Back to Lecture

41. Limiting Reactant Back to Lecture