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Chapter 9

Chapter 9. The Mole concept and Chemical Formulas. Law of definite proportions. In a pure compound elements are always present in the same definite proportion by mass. Formula Mass.

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Chapter 9

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  1. Chapter 9 The Mole concept and Chemical Formulas

  2. Law of definite proportions • In a pure compound elements are always present in the same definite proportion by mass.

  3. Formula Mass • Sum of atomic masses of all the atoms present in one formula unit of a substance expressed in atomic mass units. • Formula mass is also called as Molar mass.

  4. Problems • 1.Calculate the molar mass of LiClO4. • 2.Calculate the molar mass of (NH4)3PO4 • 3.Calculate the molar mass of C2H6O • Practice Ex: 9.2 • Use sig figures like addition rules. • Practice Ex: 9.3

  5. Percent composition: • Percent composition of a compound is the mass percent of each element in the compound. • If the formula of a compound is known a two-step process is needed to calculate the percent composition. • Step 1 Calculate the molar mass of the formula. • Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

  6. Percent composition:

  7. Percent Composition From Experimental Data • Percent composition can be calculated from experimental data without knowing the composition of the compound. • Step 1 Calculate the mass of the compound formed. • Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100. • 5. A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. • Practice Ex: 9.4,9.5

  8. The Mole: The Mole • The mass of a single atom is too small to measure on a balance. • mass of hydrogen atom = 1.673 x 10-24 g • 1 mole = 6.022 x 1023 objects • This number is called as Avogadro’s Number • 1 mole of any element contains 6.022 x 1023 particles of that substance. • The atomic mass in grams of any element2contains 1 mole of atoms. • This is the same number of particles as there are in exactly 12 grams of

  9. 1 mole = 6.022 x 1023 objects

  10. 6.022 x 1023 is a very LARGE number

  11. 6.022 x 1023 is Avogadro’s Number number

  12. If 10,000 people started to count Avogardro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

  13. 1 mole of any element contains 6.022 x 1023 particles of that substance.

  14. The atomic mass in grams of any element23 contains 1 mole of atoms.

  15. This is the same number of particles6.022 x 1023as there are in exactly 12 grams of

  16. H SpeciesQuantity Number of H atoms 1 mole 6.022 x 1023

  17. Fe SpeciesQuantity Number of Fe atoms 1 mole 6.022 x 1023

  18. C6H6 SpeciesQuantity Number of C6H6molecules 1 mole 6.022 x 1023

  19. The molar mass of an element is its atomic mass in grams. • It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.

  20. 1 mol of atoms = 6.022 x 1023 atoms 1 mol of molecules = 6.022 x 1023 molecules 1 mol of ions = 6.022 x 1023 ions

  21. Avogadro’s Number of Particles 6 x 1023 Particles 1 MOLE Molar Mass

  22. Avogadro’s Number ofH2O molecules 6 x 1023H2O molecules 1 MOLE H2O 18.02 g H2O

  23. These relationships are present when hydrogen combines with chlorine.

  24. In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.

  25. Problems: • 6. Convert 1.20 moles of CO to molecules. • 7. Convert 2.53 moles of Ag to Ag atoms. • 8. Convert 0.025 mole of magnesium sulfate to formula units. • Do Practice : Ex: 9.6

  26. The Mass of a Mole: • The molar mass of an element is its atomic mass in grams. • •It contains 6.022 x 1023 atoms (Avogadro’s number) of the element. • Moles=grams/molar mass. • 9. Convert 1.50 moles of CH4molecule to mass in grams. • 10. Convert 2.50 moles of NaCl formula units to mass in grams. • 11. Convert 1.68 moles of N2 molecule to mass in grams. • 12. Convert 1.68 moles of N atoms to mass in grams. • Do Practice Ex: 9.7. • 13. What is the mass in grams of a molecule whose mass on the amu scale is 104.00 amu?

  27. Problems • Do Practice Ex: 9.8 • Do Practice Ex: 9.9 • Do Practice Ex: 9.10 • 14. 56.04 g of N2 contains how many N2 molecules? • 15. How many moles of benzene, C6H6, are present in 390.0 grams of benzene? • 16.How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4? • Do Practice Ex: 9.11,9.12.

  28. Problems • 17. Calculate the mass in grams for : • a) a single atom of Cu. • b) a single molecule of CO2 • 18. Caffeine has the formula C8H10N4O2 How many grams of N are present in a 50.0g sample of caffeine. • Do Practice Ex: 9.14.

  29. Problems • 19. The compound cholesterol has a formula C27H46O. How many H atoms are present in 2.000 g sample of cholesterol. • Percent purity: Percent by mass of a specified substance in an impure sample of the substance. • 20. A 32.00 g sample of HNO3 has a purity of 96.20 % by mass. Calculate the following for this sample of nitric acid. • a) The mass in grams of HNO3 present,b) the mass in grams of impurities present. • Do Practice Ex: 9.16. • 21. How many Fe atoms are present in 1 25.00 g sample of Fe-ore that has a purity of 87.70 % by mass Fe? Assume that there are no Fe containing impurities present.

  30. Empirical Formula versus Molecular Formula • The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound. • The molecularformula is the true formula of a compound. • Two compounds can have identical empirical formulas and different molecular formulas.

  31. Empirical Formula CH2 Smallest Whole Number Ratio C:H 1:2 Molecular Formula C2H4

  32. Smallest Whole Number Ratio C:H 1:1 Molecular Formula C6H6 Empirical Formula CH

  33. Calculating Empirical Formulas • Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if if the actual amount is not given, and express the mass of each element in grams. • Step 2 Convert the grams of each element into moles of each element using each element’s molar mass.

  34. Calculating Empirical Formulas • Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value. • – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. • – If the numbers obtained are not whole numbers, go on to step 4. • Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers • Use these whole numbers as the subscripts in the empirical formula.

  35. Resulting WholeNumber 1 1 2 1 3 Some Common Fractions and Their Decimal Equivalents DecimalEquivalent Common Fraction 0.25 0.333… Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number. 0.666… 0.5 0.75

  36. Problems: • 22.The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. • 23. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. • Do Practice Ex: 9.18, 9.19, 9.20

  37. Calculating the Molecular Formula from the Empirical Formula •The molecular formula can be calculated from the empirical formula if the molar mass is known. •The molecular formula will be equal to the empirical formula or some multiple, n, of it. •To determine the molecular formula evaluate n. •n is the number of units of the empirical formula contained in the molecular formula

  38. Problems: • 24. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g? • Do Practice EX: 9.23, 9.24

  39. Combustion analysis • A method used to measure the amounts of C and H present in a combustible compound that contains these two elements when burned in O2. • 25. Ethylene is burned in combustion analysis apparatus and 3.14 g of CO2 and 1.29 g of H2O are produced. What is the formula of ethylene? • Do Practice Ex: 9.21.

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