1. Atomic Mass Unit: amu (atomic mass unit) amu is defined as a mass exactly equal to on-twelfth the mass of Carbon-12 atom amu = 1/12 of carbon-12 Hydrogen atom is 8.400% of Carbon-12 atom Therefore the atomic mass of hydrogen is x 12 = amu

2. Average Atomic Mass Carbon-12: 98. 90% (atomic mass = 12) Carbon-13: 1
Average Atomic Mass Carbon-12: 98.90% (atomic mass = 12) Carbon-13: 1.10% (atomic mass = The average atomic mass of Carbon is (98.90/100 x 12 amu) + (1.10/100 x amu) = 12.01amu Average atomic mass: The average of atomic mass of naturally occurring isotopes 6 C
12.01

3. Avogadro number and the Molar mass of an element “Atoms and molecules are measured in moles in Si system” Mol (mole) is the amount of a substance that contains as many elementary entities (atoms and molecules) 1 mole of carbon-12 = 12g (0.012kg) In 12g of Carbon-12 the number of atoms: NA = x or x This number is called is Avagadro’s number Avagadro’s Number: Number of atoms in 1 mole

4. Molar mass and Atomic mass unit Carbon-12 molar mass = 12g Mass of one carbon-12 atom is x 10-23g Mass of one Carbon-12 atom is 12 amu 12 amu = x g 1 g = 12 amu/1.993 x = x 1023 amu 1 amu = 1g/6.022 x = x 10-24g

5. Conversion factors for moles & number of atoms 1 mol/molar mass 1mol/Avogadro’s number How many moles of Helium atoms in 6.46g of Helium? 6.46g x 1mol/4.003g = 1.61 mol

6. How may atoms are in 16. 3g of sulfur
How may atoms are in 16.3g of sulfur ? First find the moles 1 mol of S = 32.07g No. of moles = 16.3g/32.07 = mol Now calculate the number of atoms x 1 mol/avogadro’s number = x 6.22 x = 3.06 x 1023 atoms or 16.3g x 1mol/32.07 x x 1023/1mol = x 1023 atoms “Calculate number of atoms in g of potassium”

7. Molecular Mass “Molecular mass or weight is the sum of the atomic masses of a molecule” Molecular mass of water atomic mass of 2H + atomic mass of one O 2 x = amu Calculate moles in 6.07g of CH4, and molecular mass of 1mol of sulfur dioxide, 1mole of caffeine (C8H10N4O2)

8. Mass Spectrometer

9. Percent Composition of Compounds “The percent composition by mass is the percent by mass of each element in compound” Percent composition = n x molar mass of element x 100% of an element molar mass of compound n = number of moles Calculate of H% and O% in hydrogen peroxide and water (2 moles of H and 2 mole of O)

10. Determination of Empirical formula Combustion of 11
Determination of Empirical formula Combustion of 11.5g ethanol produced 22.0g of CO2 and 13.5g of water Oxygen Unused oxygen heat H2O CO2 Determine the empirical formula

11. Mass of C = 22. 0g of CO2 x 1mol CO2 x 1mol C x 12g C 44
Mass of C = 22.0g of CO2 x 1mol CO2 x 1mol C x g C g CO mol CO mol C = 6.00 g Mass of H = 13.5g of H2O x 1mol H2O x 2 mol H x g H g H2O mol H2O mol H H = 1.51g Mass of O = Mass sample – (Mass of C + Mass of H) – (6.00 g g) = 4.0g Mole of C = 6.00g C x 1 mol C/12.01 g C = mol C Mole of H = 1.51 g H x 1 mol H/1.008 g H = 1.50 mol H Moles of O = 4.0 g O x 1 mol O/16.00 g O = 0.25 mol O = C2H6O

12. Determination of Molecular Formula Empirical formula is based on the smallest ration of the whole numbers. To calculate the molecular formula, the approximate molar mass should be known Calculate MF of percent of Nitrogen and 69.4 percent of Oxygen. Molar mass is to be between 90 g and 95g

13. Chemical Reactions and Chemical Equations Chemical reaction is a process in which a substance (or substances) is changed to one or more new substances Chemical equation is a symbolic expression of a chemical reaction Ex: H2 + O H2O Balanced equation is 2H2 + O H2O

14. Chemical Equation 2H O H2O 2 Molecules One molecules two molecules 2 moles One mole two moles 2(2.02g) = 4.04g g g g reactants g product

15. Chemical Equations NaCl(s) H2O NaCl(aq) KBr(aq) + AgNO3(aq) KNO3(aq) + AgBr(s)

16. Balancing Chemical equation 1) Identify all the reactants and products with correct formulas 2) Try to change the coefficients but not subscripts 3) Look for an element that appears once on each side but unequal number of atoms. Balance such element first. KClO KCl + O2 C2H O CO2 + H2O

17. Stoichiometry of the reaction Stoichiometry is the quantitative study of the reactants and products in a chemical reaction

18. Amount of Reactants and Products N2(g) H2(g) > NH3 1 Molecule molecules molecules x 1023 molecules x x 1023 molecules x x 1023 molecules 1 mole 3 moles moles mole H2 ~ 2 mole NH3 Conversion factors For H2 required in moles for given moles of NH3: 3 mole H2/2 mol NH3 For NH3 produced from given H2 (moles): 2 mol NH3/3 mol H2 Calculate the amount of NH3 produced when 16g H2 was consumed in the reaction with N2

19. Summary of Stoichiometry 1) Write the balanced equation for the reaction 2) Convert the given amount of the reactant into moles 3) Use the mole ratio from the balanced equation to calculate the number of moles of product formed 4) Convert the moles of product into grams

20. Limiting reagents and excess reagents 1) The reagent used up first in the reaction is called Limiting reagent 2) The reactant present in quantities greater than necessary to react with the quantity of limiting reagent is called excess reagent Determine the limiting reagent when 4 mols CO and 6 mols H2 used to produce methanol CO (g) + 2H2 (g) CH3OH (g)

21. Reaction yield Based on balanced equation, the yield of product can be determined. This is called theoretical yield The actual yield is the amount of the product actually obtained from a reaction %yield = actual yield/theoretical yield x 100