1. Chapter 3 Molecules, Compounds, and Chemical Equations

2. 3.7 Formula Mass versus Molar mass Formula mass ◦ The average mass of a molecule or formula unit in amu ◦ also known as molecular mass or molecular weight (MW) ◦ whole = sum of the parts ! Molar mass ◦ Total mass of a compound in gram per 1 mol of its molecules or formula unit

3. Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H 2 O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H 2 O = 2(1.01 g H) + 16.00 g O = 18.02 g so the Molar Mass of H 2 O is 18.02 g/mole 3 Why multiplying 2?

4. Molar Mass of Na 2 SO 4 Calculate the molar mass of Na 2 SO 4. 4 ElementNumber of Moles Atomic MassTotal Mass in each element Na S O Total mass in 1 mol Na 2 SO 4

5. 3.8 Mass Percent Composition Percentage of each element in a compound ◦ By mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding 5

6. Example - Mass Percent as a Conversion Factor Calculate the mass percent of Na in NaCl Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

7. Chemical Formulas and Elemental Composition chemical formulas have inherent in them relationships between numbers of atoms and molecules ◦ or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules ◦ like percent composition Vol A  Grams A  Moles A  Moles B  Grams B Grams A  Moles A  Moles B  Grams B 7

8. Example Butane (C 4 H 10 ) is the liquid fuel in lighters. a.Determine the number of atoms ratio between carbon and 1 molecule of C 4 H 10 b.Determine the number moles ratio between C and 1 mol of C 4 H 10 c.How many grams of carbon are present within a lighter containing 7.5 mL of butane? The density of quid butane is 0.601 g/mL

9. Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analysis ◦ masses of elements formed when decompose or react compound  combustion analysis ◦ percent composition 9

10. Steps in determine the Empirical formula Step 1: Obtain the mass of each element (in grams) E.G 100% = 100g therefore mass percent is the same numerical value in grams Step 2: Determine the numbers moles of each atom present ◦ Use molar mass of each element Step 3: Divide the smallest moles by numbers moles of each atom to obtain the closet integer as possible. ◦ if result is within 0.1 of whole number, round to whole number Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible. E.g 1.5 x 2 = 3.0 atoms 1.33 x 3 = 3. 99 = 4 atoms Step 5: Write the result (number atoms) from step 4 as a subscript for the appropriate element.

11. Example Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70g/mol) and the rest fluorine (19.00 g/mol) An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?

12. Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound 12 Multiple (n) = molecular mass empirical formula mass Molecular formula = empirical formula x n where n = 1, 2, 3, 4

13. Example Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula and molecular formula C = 60.00% H = 4.48% O = 35.53% 13

14. Determining Empirical Formulas: Elemental Analysis Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O 2 (g) xCO 2 (g) + yH 2 O(g) carbon hydrogen

15. Combustion Analysis Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal) gCO 2  moles CO 2  moles C  gC gH 2 O  moles H 2 O  moles H  gH g O = g sample – (g H + g C) ◦ gO  moles O Follow steps in determine the empirical formula and molecular formula

16. Example Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO 2 = 2.445 g H 2 O = 0.6003 g Determine the empirical formula of the compound 16

17. Example Upon combustion, a compound containing only carbon and hydrogen produced 1.60g CO 2 and 0.819g H 2 O. Find the empirical formula 17

18. Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules ◦ Elements are not transmuted during a reaction 18

19. Chemical Equations Chemical Equations A chemical equation gives the formulas of the reactants on the left of the arrow. the formulas of the products on the right of the arrow. ReactantsProduct 19 C(s) O 2 (g) CO 2 (g)

20. Symbols Used in Equations Symbols in chemical equations show the states of the reactants. the states of the products. the reaction conditions. 20 TABLE

21. Chemical Equations are Balanced In a balanced chemical reaction no atoms are lost or gained. the number of reacting atoms is equal to the number of product atoms. 21

22. Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical. 2NaCl(s)2Na(s) + Cl 2 (g) right side: 2 Na 2 Cl left side: 2 Na 2 Cl

23. Balancing Chemical Equations 2.Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation. 2H 2 O(l)2H 2 (g) + O 2 (g) 1.Write the unbalanced equation using the correct chemical formula for each reactant and product. H 2 O(l)H 2 (g) + O 2 (g) 3.Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator. 2H 2 O(l)2H 2 (g) + O 2 (g)

24. Balancing Chemical Equations 4.Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. 2H 2 O(l)2H 2 (g) + O 2 (g) right side: 4 H 2 O left side: 4 H 2 O

25. Balancing Chemical Equations Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients. H 2 O(l)H 2 (g) + O 2 (g) unbalanced 2H 2 O(l)2H 2 (g) + O 2 (g) Chemical equation changed! H2O2(l)H2O2(l)H 2 (g) + O 2 (g) Balanced properly

26. Examples Balance the coefficients from reactants to products. A. __N 2 (g) + __H 2 (g) __ NH 3 (g) A. B. __Co 2 O 3 (s) + __ C(s) __Co(s) + __CO 2 (g) Write a balanced equation for the reaction between a. carbon dioxide gas and aqueous potassium hydroxide to form potassium carbonate and water. b. The combustion of gaseous ethane (C 2 H 6 ) 26