1. Houston Community College System Chemistry 1405 Chapter 4 Chemical Calculations By Mounia Elamrani Blei / Odian ’ s General, Organic, and Biochemistry

2. Mounia ElamraniChapter 4 –Chemical Calculations2 Objectives Formula mass of compounds What is a “ mole ” Formula mass as a conversion factor Avogadro ’ s number and atoms in a sample Empirical and molecular formulas Chemical reactions and chemical equations Balancing chemical equations Predicting the amount of compound produced or consumed in a reaction

3. Mounia ElamraniChapter 4 –Chemical Calculations3 4.1 Chemical Formulas and Formula Masses The formula mass allows counting of molecules by mass The molecular mass is the sum of atomic masses of the atoms in a covalent compound ’ s formula For example the molar mass of water, H 2 O, is: 2 x mass of H (1.008) + 1 x mass of O (15.999) = 18.015 amu Strictly speaking, ionic compounds do not have a “ molecular mass ” because they don ’ t contain molecules, they have a formula mass For example the formula mass of calcium oxide, CaO, is: 1 x mass of Ca (40.08) + 1 x mass of O (15.999) = 56.08amu

4. Mounia ElamraniChapter 4 –Chemical Calculations4 Exercises Problem 4.1 Calculate the formula masses of: Zn(HPO 4 ) Fe 3 O 4 C 3 H 6 O 2 N

5. Mounia ElamraniChapter 4 –Chemical Calculations5 4.2-4.3 The Mole and Avogrado ’ s Number 1 trio= 3 singers 1 six-pack Cola=6 cans Cola drink 1 dozen donuts=12 donuts 1 gross of pencils=144 pencils 1 mole of pencils=6.02 x 10 23 pencils

6. Mounia ElamraniChapter 4 –Chemical Calculations6 4.2-4.3 The Mole and Avogrado ’ s Number The mole (mol) is a counting unit for objects - atoms, ions, molecules, etc. A mole represents 6.02x10 23 objects. This is known as Avogadro ’ s number. Avogadro ’ s number is chosen so that 1 mole of 12 C atoms has a mass of exactly 12 grams. Since 6.02x10 23 x 12 amu = 12 g, 6.02x10 23 amu = 1 gor1amu = 1.66054x10 -24 g

7. Mounia ElamraniChapter 4 –Chemical Calculations7 4.2-4.3 The Mole and Avogrado ’ s Number Counting formula units by moles is no different than counting eggs by the dozen (12 eggs) or pens by the gross (144 pens) Avogadro ’ s number is huge because atoms and molecules are so small: a huge number of them are needed to make a lab-sized sample Avogadro ’ s number links moles and atoms or molecules and provides an easy way to link mass and number of atoms or molecules

8. Mounia ElamraniChapter 4 –Chemical Calculations8 4.2-4.3 The Mole and Avogrado ’ s Number The molar mass of a substance is the mass of one mole in grams. It is numerically equal to the formula mass in amu. Mass of 1 C atom = 12.011 amu Mass of 1 mol of C = 12.011 g Mass of 1 H 2 O molecule = 18.015 amu Mass of 1 mol of H 2 O = 18.015 g Mass of 1 MgO unit = 40.304 amu Mass of 1 mol of MgO = 40.304 g

9. Mounia ElamraniChapter 4 –Chemical Calculations9 Exercises Problem 4.3 Calculate the number of moles of CaCl 2 in 138.8g of CaCl 2 Problem 4.4 How many moles are in 0.3600 of glucose (1mol glucose = 270.0g) Problem 4.5 How many moles of each atom are there in 1 mol of the following substances: ZnHPO 4 H 2 SO 4 Al 2 O 3 Ca 3 P 2

10. Mounia ElamraniChapter 4 –Chemical Calculations10 4.4-4.5 Empirical and Molecular Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical MolecularName CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide CH 2 OC 5 H 10 O 5 ribose An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound. Empirical and molecular formulas are related through their formula masses

11. Mounia ElamraniChapter 4 –Chemical Calculations11 Exercise A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3 D. If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN2) SN 4 3) S 4 N 4

12. Mounia ElamraniChapter 4 –Chemical Calculations12 Empirical and Molecular Formulas molar mass = a whole number = n simplest mass n = 1 molar mass = 1 x empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula n=?molecular formula = n x empirical formula Example: A compound has a molar mass of 176.0 g and an empirical formula of C 3 H 4 O 3. What is the molecular formula?

13. Mounia ElamraniChapter 4 –Chemical Calculations13 Finding the Empirical and Molecular Formula A compound contains Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol. 1. State mass percents as grams in a 100.00-g sample of the compound Cl 71.65 gC 24.27 g H 4.07 g

14. Mounia ElamraniChapter 4 –Chemical Calculations14 Finding the Empirical and Molecular Formula 2. Calculate the number of moles of each element. 71.65 g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl 24.27 g C x 1 mol C = 2.02 mol C 12.0 g C 4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

15. Mounia ElamraniChapter 4 –Chemical Calculations15 Finding the Empirical and Molecular Formula 3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 = 1 Cl 2.02 C: 2.02 = 1 C 2.02 H: 4.04 = 2 H 2.02 4. Write the simplest or empirical formula CH 2 Cl

16. Mounia ElamraniChapter 4 –Chemical Calculations16 Finding the Empirical and Molecular Formula 5. EM (empirical mass) of CH 2 Cl = 1(C) + 2(H) + 1(Cl) = 49.5 g 6. n = molar mass/empirical mass Molar mass = 99.0 g = n = 2 Empirical mass 49.5 g 7.Molecular formula (C 1 H 2 Cl 1 ) x 2 = C 2 H 4 Cl 2

17. Mounia ElamraniChapter 4 –Chemical Calculations17 Finding Subscripts Sometimes the numbers obtained cannot be rounded: A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for the subscripts: (1/2) 0.5 x 2 = 1 (1/3)0.333 x 3 = 1 (1/4)0.25 x 4 = 1 (3/4)0.75 x 4 = 3

18. Mounia ElamraniChapter 4 –Chemical Calculations18 Exercise Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its empirical formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

19. Mounia ElamraniChapter 4 –Chemical Calculations19 Chemical Change and Chemical Reactions Atoms in the reactants are rearranged to form one or more different substance Old bonds are broken; new bonds form: Fe and O 2 form rust (Fe 2 O 3 ) Ag and S form tarnish (Ag 2 S) A Chemical Reaction is a process in which at least one new substance is produced as a result of chemical change

20. Mounia ElamraniChapter 4 –Chemical Calculations20 A Chemical Reaction Reactants Products

21. Mounia ElamraniChapter 4 –Chemical Calculations21 Writing a Chemical Equation Chemical symbols give a “ before-and-after ” picture of a chemical reaction ReactantsProducts 1) MgO + CCO + Mg magnesium oxide to form carbon monoxide reacts with carbon and magnesium 2) 4 NH 3 + 5 O 2 4 NO + 6 H 2 O 4 molecules of NH 3 react with 5 molecules of O 2 to produce 4 molecules of NO and 6 molecules of H 2 O or 4 moles of NH 3 react with 5 moles of O 2 to produce 4 moles of NO and 6 moles of H 2 O

22. Mounia ElamraniChapter 4 –Chemical Calculations22 A Balanced Chemical Equation The same numbers of each type of atom need to appear on each side of the equation. This illustrates the law of conservation of mass: 1)Al + S Al 2 S 3 Not Balanced 2Al + 3S Al 2 S 3 Balanced 2)H 2 + Cl 2 2 HCl + + Total atoms=Total atoms 2 H, 2 Cl2H, 2 Cl Total Mass=Total Mass 2(1.0) + 2(35.5) =73.0g=2(36.5)

23. Mounia ElamraniChapter 4 –Chemical Calculations23 Balance Equations with Coefficients Coefficients in front of formulas balance each type of atom 4NH 3 + 5O 2 4NO + 6H 2 O 4 N = 4 N 12 H =12 H 10 O =10 O

24. Mounia ElamraniChapter 4 –Chemical Calculations24 Tips on Balancing Chemical Equations Always balance by adjusting coefficients, and never the charge or the subscripts in formulas Balance elements other than hydrogen and oxygen first Balance elements hydrogen and oxygen next Check that you have conservation of matter (same number of atoms) and mass (same molar mass)

25. Mounia ElamraniChapter 4 –Chemical Calculations25 Exercise Fe 3 O 4 + 4 H 2 3 Fe + 4 H 2 O A. Number of H atoms in 4 H 2 O 1) 22) 43) 8 B. Number of O atoms in 4 H 2 O 1) 22) 43) 8 C. Number of Fe atoms in Fe 3 O 4 1) 12) 33) 4

26. Mounia ElamraniChapter 4 –Chemical Calculations26 Exercise Balance each equation. The coefficients for each equation are read from left to right A. Mg + N 2 Mg 3 N 2 B. Al + Cl 2 AlCl 3 C. Fe 2 O 3 + C Fe + CO 2 D. Al + FeO Fe + Al 2 O 3 E. Al + H 2 SO 4 Al 2 (SO 4 ) 3 + H 2

27. Mounia ElamraniChapter 4 –Chemical Calculations27 4.7 Oxidation-Reduction Reactions Oxidation-Reduction (or redox) reactions involve a loss/gain of electrons: Oxidation is the loss of electrons (LEO) ZnZn 2+ + 2e - The substance that loses electrons is being oxidized Reduction is the gain of electrons (GER) Cu 2+ + 2e - Cu The substance that gains electrons is being reduced For a redox reaction to occur, something must accept the electrons that are lost by another substance

28. Mounia ElamraniChapter 4 –Chemical Calculations28 Balancing Redox Equations Examples of redox reactions include:  Combustion reactions  Food metabolism  Batteries  Rusting of metals Combine the oxidation and reduction reactions to make: Loss of electrons = Gain of electrons Zn + Cu 2+ + 2e-Zn 2+ + 2e- + Cu Zn + Cu 2+ Zn 2+ + Cu

29. Mounia ElamraniChapter 4 –Chemical Calculations29 Exercise Identify the following as oxidation or reduction: __A. SnSn 4+ + 4e- __B. Fe 3+ + 1e - Fe 2+ __C. Cl 2 + 2e - 2Cl - From the examples above, combine an oxidation reaction with a reduction reaction and balance the equation.

30. Mounia ElamraniChapter 4 –Chemical Calculations30 Redox Reactions with Oxygen In organic and biological reactions Loss of Hydrogen = Oxidation = Gain of Oxygen Gain of Hydrogen = Reduction = Loss of Oxygen Metals can gain oxygen to form oxides: 2 Mg + O 2  2 MgO 4 Fe + 3 O 2  2 Fe 2 O 3 (iron rust) In this processes, the metals are oxidized while the oxygen is reduced

31. Mounia ElamraniChapter 4 –Chemical Calculations31 Redox Reactions with Oxygen Metal oxides from ores can lose oxygen to recover the metal in elemental form: 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2 Other compounds can react with oxygen producing carbon dioxide and water as in: combustion reactions like burning octane: C 8 H 8 + 10 O 2  8CO 2 + 4H 2 O + flame Food metabolism like digesting glucose: C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O Exercise: Write the balanced combustion reactions of C 4 H 10 and of lactic acid (C 3 H 6 O 3 )

32. Mounia ElamraniChapter 4 –Chemical Calculations32 Redox Reactions with Hydrogen In organic and biological reactions Loss of Hydrogen = Oxidation = Gain of Oxygen Gain of Hydrogen = Reduction = Loss of Oxygen Hydrogen loss happens in biological systems to extract energy from food: C 4 H 6 O 5 + NAD +  C 4 H 4 O 5 + NADH + H +

33. Mounia ElamraniChapter 4 –Chemical Calculations33 4.8 Stoichiometry Stoichiometry is the study of the relationship between the mass of the different substances involved in a chemical reaction. Stoichiometry is based on balanced equations The stoichiometric coefficients help write conversion factors, in terms of moles, between the different substances This will help answer stoichiometry questions: How much product can be made How much reactant is required

34. Mounia ElamraniChapter 4 –Chemical Calculations34 Conversion factors from balanced equations Problem 4.15: Derive all possible unit-conversion factors: Mg(OH) 2 + FeCl 3  Fe(OH) 3 + MgCl 2 (unbalanced) How many moles of aluminum hydroxide are required to produce 3.200 moles of Al 2 (SO 4 ) 3 ? Al(OH) 3 + H 2 SO 4  Al 2 (SO 4 ) 3 + H 2 O (unbalanced) How many moles of ammonia will be formed from 0.500 mol of N 2 and 1.500 mol of H 2 ?

35. Mounia ElamraniChapter 4 –Chemical Calculations35 Solving Stoichiometry Problems

36. Mounia ElamraniChapter 4 –Chemical Calculations36 Calculating the mass relations in a chemical reaction 1. Balance the equation 2. Convert the mass of substance A to moles of A 3. Use a conversion factor to calculate moles of B from moles of A 4. Convert the mass of substance B to moles of B Problem 4.17: Calculate the grams of BaCl 2 needed to react with 42.4 g of K 3 PO 4 and the grams of Ba 3 (PO 4 ) 2 produced. K 3 PO 4 + BaCl 2  Ba 3 (PO 4 ) 2 + KCl

37. Mounia ElamraniChapter 4 –Chemical Calculations37 End of Chapter Problems 4.2 b, 4.4 e 4.6, 4.8 4.13 4.16a, 4.18b 4.19, 4.20, 4.23 4.34