Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

1. Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition 3.6 Chemical Formulas

2. AtomicMass

3. Atomic Mass • Atoms are so small, it is difficult to discuss how much they weigh in common units. • The decimal numbers on the table are atomic masses in amu. • Atomic masses are not decimals because they are based on averages of atoms & of isotopes. • The average atomic mass is from the mass of the isotopes and their relative abundance.

4. Atomic Mass Examples • There are two isotopes of carbon 12C with a mass of 12.00000 amu(98.892%), and 13C with a mass of 13.00335 amu (1.108%). • There are two isotopes of nitrogen, one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each?

5. Moles

6. Avogadro’s Number • The mole is a number. • A very large number, but still, just a number. • 6.022 ´ 1023 of anything is a mole • A large dozen.

7. Avogadro’s Number 1 mole of 12C has 6.02 ´ 1023 atoms and a mass of 12 g

8. Molar Mass • By definition, these are the mass of 1 mol of a substance (i.e., g/mol). • The molar mass of an element is the mass number for the element that we find on the periodic table • Often called molecular weight or formula weight (in g/mol)

9. Using Moles Moles provide a bridge from the molecular scale to the real-world scale.

10. Mole Relationships • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. • One mole of moleculesorformula unitscontains Avogadro’s number times the number of atoms or ions of each element in the compound.

11. Percent Composition

12. mass of element in sample total mass of sample % element = ´ 100 Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

13. (2)(12.0 amu) %C = (30.0 amu) x 100 24.0 amu = 30.0 amu Percent Composition So the percentage of carbon in ethane, C2H6, is… = 80.0%

14. Empirical Formulas

15. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

16. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

17. Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g × = 5.105 mol C H: 5.14 g × = 5.09 mol H N: 10.21 g × = 0.7288 mol N O: 23.33 g × = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 16.00 g 1 mol 1.01 g Calculating Empirical Formulas

18. Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 5.105 mol 0.7288 mol 1.458 mol 0.7288 mol Calculating Empirical Formulas

19. Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2

20. Combustion Analysis • Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like this • C is determined from the mass of CO2 produced. • H is determined from the mass of H2O produced. • O is determined by difference after the C and H have been determined.

21. Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

22. Molecular Formulas

23. Empirical To Molecular Formulas • Empirical is lowest ratio. • Molecular is actual molecule. • Need Molar mass. • Ratio of empirical to molar mass will tell you the molecular formula. • Must be a whole number because...

24. Example • A compound is made of only sulfur and oxygen. It is 50.0% S by mass. Its molar mass is 192 g/mol. What is its formula?

25. Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition 3.6 Chemical Formulas