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Chapter 3: Calculations with Chemical Formulas and Equations

Chapter 3: Calculations with Chemical Formulas and Equations. MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT Molecular weight : (MW) sum of the atomic weights of all the atoms in a molecule of the substance.

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Chapter 3: Calculations with Chemical Formulas and Equations

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  1. Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT • Molecularweight: (MW) sum of the atomic weights of all the atoms in a molecule of the substance. • H2O = 18.0 amu (2 x 1.0 amu for two H atoms plus 16.0 amu from one O atom)

  2. Formulaweight (FW) – sum of the atomic weights of all atoms in a formula unit of a compound. Whether molecular or not. Sodium chloride, NaCl (formula unit) formula weight of 58.44 amu (22.99 amu from Na plus 35.45 amu from Cl). NaCl is ionic (no molecular weight)

  3. 3.2 THE MOLE CONCEPT • Developed to deal with the enormous numbers of molecules or ions in samples of substances. • Mole: the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12.

  4. The number of atoms in a 12-g sample of carbon-12 is called Avogadro’sNumber (NA). = 6.02 x 10^23 A mole of a substance contains Avogadro’s number of molecules (6.02 x 10^23) Mole (much like dozen or gross) refers to a particular number of molecules. 1 mole of ethanol = 6.02 x 10^23 ethanol molecules When talking about ionic substances we are referring to the number of formula units not molecules.

  5. One mole of sodium carbonate Na2CO3 Contains: 6.02 x 10^23 Na2CO3 units Which contains: 2 x 6.02 x 10^23 Na+ ions 1 x 6.02 x 10^23 CO32- ions

  6. Specify Mole of oxygen atoms means 6.02 x 10^23 O atoms Mole of oxygen molecules means 6.02 x 10^23 O2 molecules (2 x 6.02 x 10^23 O atoms)

  7. Molar mass: mass of one mole of the substance. Carbon-12 has a molar mass of exactly 12g/mol Ethanol, molecular formula C2H5OH MW = 46.1 amu Molar mass = 46.1 g/mol For all substances, the molar mass in grams per mole is numerically equal to the formula weight in atomic mass units.

  8. MOLE CALCULATIONS • Conversion from mol to g • Ethanol is 46.1 g/mol 1 mol C2H5OH = 46.1 g Cr2H5OH To convert grams to moles

  9. You need to prepare acetic acid from 10.0g of ethanol, C2H5OH. Convert 10.0g C2H5OH to moles C2H5OH by multiplying by the appropriate conversion factor.

  10. 3.3 MASS PERCENTAGES FROM THE FORMULA • Mass percentage of A as the parts of A per hundred parts of the total, by mass.

  11. 3.4 ELEMENTAL ANALYSIS: PERCENTAGES OF CARBON, HYDROGEN, AND OXYGEN • To determine the formula of a new compound is to obtain its percentage composition. • Ex. Burn a sample of the compound of known mass and get CO2 and H2O. • Next relate the masses of CO2 and H2O to the masses of carbon and hydrogen. • Then calculate the mass percentages of C and H. • Determine the mass percentage of O by difference.

  12. 3.5 DETERMINING FORMULAS • Empirical formula (simplest formula) – formula of a substance written with the smallest integer subscripts. • For most ionic substances, the empirical formula is the formula of the compound. • This is often not the case for molecular substances. • Hydrogen peroxide: molecular formula is H2O2 • The empirical formula (just tells you the ratio of numbers of atoms in the compound) HO

  13. Why do ionic compounds usually have the same formula and empirical formula? • No multiple proportions like molecules.

  14. Compounds with different molecular formulas can have the same empirical formula. • They will also have the same percentage composition. • Acetylene C2H2 • Benzene C6H6 • Same empirical formula CO, different molecular formulas, different chemical structures.

  15. To obtain the molecular formula of a substance, you need to know: • The percentage composition, from which the empirical formula can be determined • Molecular weight

  16. EMPIRICAL FORMULA FROM THE COMPOSITION • You can find the empirical formula from the composition of the compound by converting masses of the elements to moles.

  17. Ex. 3.10

  18. MOLECULAR FORMULA FROM EMPIRICAL FORMULA • Molecular formula of a compound is a multiple of its empirical formula. • Acetylene C2H2 is equivalent to (CH)2 • Benzene C6H6 is equivalent to (CH)6 Molecular weight = nx empirical formula weight

  19. N = number of empirical formula units in the molecule. • Molecular formula is obtained by multiplying the subscripts of the empirical formula by n n = molecular weight/empirical formula weight

  20. After you find the empirical formula, calculate its empirical formula weight. • From an experimental determination of its molecular weight, you can calculate n and then the molecular formula.

  21. Ex. 3.11

  22. STOICHIOMETRY: QUANTITATIVE RELATIONS IN CHEMICAL REACTIONS Stoichiometry – the calculation of the quantities of reactants and products involved in a chemical reaction. Relationship between mass and moles.

  23. 3.6 MOLAR INTERPRETATION OF A CHEMICAL EQUATION N2 + 3H2 = 2NH3 How much hydrogen is required to give a particular quantity of ammonia? One N2 One mole of N2 reacts with Three H2 three moles of H2 to give two Two NH3 moles of NH3.

  24. Write pg. 81

  25. 3.7 AMOUNTS OF SUBSTANCES IN A CHEMICAL REACTION -

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