2. Chapter 3 Stoichiometry

3. The study of the quantities of materials produced and consumed in chemical reactions Relative Atomic Mass- atoms are small and as chemists we deal with samples of matter that contain huge numbers of atoms The number of atoms in a sample is determined by its average atomic mass

4. Atomic Masses The work of Dalton, Gay-Lussac, Lavoisier, Avogadro and Berzelius produced information about atomic masses Modern system of atomic mass instituted 1961 – Based on Carbon-12 – It is assigned a mass of exactly 12 amu – Masses of all other atoms are relative to carbon- 12 standard

5. Atomic Masses Mass Spectrometer- most accurate method to compare the masses of atoms An analytical technique that produces spectra of the masses of the molecules in sample Spectra are used to determine the elemental composition of the matter Samples can be solids, liquids or gases

6. Atomic Masses

7. Mass Spectrometer Ionization- atoms/molecules are passed through a beam of high speed electrons- creates cations Acceleration-Electric field is applied to cations Deflection- applied magnetic field changes path of ion depending on its mass (most massive deflected the least) Detection-Ions hit the detector plate, gives accurate values of their relative masses

8. Atomic Masses Atomic Mass of carbon is 12.01 amu Natural carbon is a mixture of isotopes 12 C, 13 C, and 14 C Natural carbon is composed of 98.89% 12 C and 1.11% 13 C, the amount of 14 C is negligible – 12 C has a mass of 12 amu and 13 C has a mass of 13.0034 amu Calculate the average atomic mass

9. The Mole

10. The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C One mole of something consists of 6.02 x 10 23 units of that substance (Avogadro’s number) The mass of 1 mole of an element is equal to its atomic mass in grams

11. Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li6.94 g Li 1 mol Li = 45.1 g Li

12. Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li 6.94 g Li = 2.62 mol Li

13. Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol 6.02 x 10 23 atoms 1 mol =2.07 x 10 24 atoms

14. Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.94 g Li 6.02 x 10 23 atoms Li 1 mol Li = 1.58 x 10 24 atoms Li

15. Molar Mass Mass in grams of one mole of the compound Also referred to as molecular weight Calculate the molar mass of methane CH 4

16. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO 3. Calculate the formula mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

17. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

18. Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH

19. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

20. Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

21. Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

22. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

23. Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

24. Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

25. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

26. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

27. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

28. Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactantsproducts 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water When the equation is balanced it has quantitative significance:

29. Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to mass or volume, if necessary. 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to mass or volume, if necessary.

30. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

31. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =12.3 g Al 2 O 3

32. Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.