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Quantitative Composition of Compounds Chapter 7

Quantitative Composition of Compounds Chapter 7. Hein and Arena. Version 1.1. Chapter Outline. 7.1 The Mole. 7.4 Empirical Formula versus Molecular Formula. 7.2 Molar Mass of Compounds. 7.5 Calculating Empirical Formulas. 7.3 Percent Composition of Compounds.

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Quantitative Composition of Compounds Chapter 7

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  1. Quantitative Composition of CompoundsChapter 7 Hein and Arena Version 1.1

  2. Chapter Outline 7.1 The Mole 7.4 Empirical Formula versusMolecular Formula 7.2 Molar Mass ofCompounds 7.5 Calculating EmpiricalFormulas 7.3 Percent Composition ofCompounds 7.6 Calculating the MolecularFormula from the EmpiricalFormula

  3. 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

  4. This is an infinitesimal mass 1.673 x 10-24 g

  5. 1 mole = 6.022 x 1023 objects

  6. 6.022 x 1023 is a very LARGE number

  7. 6.022 x 1023 is Avogadro’s Number number

  8. 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.

  9. 1 mole of any element contains 6.022 x 1023 particles of that substance.

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

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

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

  13. H2 SpeciesQuantity Number of H2molecules 1 mole 6.022 x 1023

  14. Na SpeciesQuantity Number of Na atoms 1 mole 6.022 x 1023

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

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

  17. 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

  18. 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.

  19. How many moles of iron does 25.0 g of iron represent? Conversion sequence: grams Fe → moles Fe Atomic mass iron = 55.85 Set up the calculation using a conversion factor between moles and grams.

  20. How many iron atoms are contained in 40.0 grams of iron? Atomic mass iron = 55.85 Conversion sequence: grams Fe → atoms Fe Set up the calculation using a conversion factor between atoms and grams.

  21. What is the mass of 3.01 x 1023 atoms of sodium (Na)? Molar mass Na = 22.99 g Conversion sequence: atoms Na → grams Na Set up the calculation using a conversion factor between grams and atoms.

  22. What is the mass of 0.365 moles of tin? Atomic mass tin = 118.7 Conversion sequence: moles Sn → grams Sn Set up the calculation using a conversion factor between grams and atoms.

  23. How many oxygen atoms are present in 2.00 mol of oxygen molecules? Two conversion factors are needed: Conversion sequence: moles O2→ molecules O→ atoms O

  24. Molar Mass of Compounds The molar mass of a compound can be determined by adding the molar masses of all of the atoms in its formula.

  25. Calculate the molar mass of C2H6O. 2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g

  26. Calculate the molar mass of LiClO4. 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = 35.45 g 4 O = 4(16.00 g) = 64.00 g 106.39 g

  27. Calculate the molar mass of (NH4)3PO4 . 3 N = 3(14.01 g) = 42.03 g 12 H = 12(1.01 g) = 12.12 g 1 P = 1(30.97 g) = 30.97 g 4 O = 4(16.00 g) = 64.00 g 149.12 g

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

  29. Avogadro’s Number ofCa atoms 6 x 1023 Ca atoms 1 MOLE Ca 40.078 g Ca

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

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

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

  33. Calculate the molar mass of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the molar mass of 1 mole of H2molecules. 2 H = 2(1.01 g) = 2.02 g

  34. How many moles of benzene, C6H6, are present in 390.0 grams of benzene? The molar mass of C6H6 is 78.12 g. Conversion sequence: grams C6H6→ moles C6H6

  35. How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4? The molar mass of (NH4)3PO4 is 149.12 g. Conversion sequence: moles (NH4)3PO4→ grams (NH4)3PO4

  36. 56.04 g of N2 contains how many N2 molecules? The molar mass of N2 is 28.02 g. Conversion sequence: g N2 → moles N2→ molecules N2 Use the conversion factors

  37. 56.04 g of N2 contains how many N2 atoms? The molar mass of N2 is 28.02 g. Conversion sequence: g N2 → moles N2→ molecules N2 → atoms N Use the conversion factors

  38. Percent Compositionof Compounds H2O 11.19% H by mass 88.79% O by mass Percent composition of a compound is the mass percent of each element in the compound.

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

  40. Calculate the percent composition of hydrosulfuric acid H2S. Step 1Calculate the molar mass of H2S. 2 H = 2 x 1.01g = 2.02 g 1 S = 1 x 32.07 g = 32.07 g 34.09 g

  41. Calculate the percent composition of hydrosulfuric acid H2S. S94.07% H5.93% Step 2Divide the mass of each element by the molar mass and multiply by 100.

  42. Percent Composition From Formula Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1Calculate the mass of the compound formed. Step 2Divide the mass of each element by the total mass of the compound and multiply by 100.

  43. 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. Step 1Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of product

  44. 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. N30.5% O69.5% Step 2Divide the mass of each element by the total mass of the compound formed.

  45. 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 empirical formula gives the relative number of atoms of each element present in the compound.

  46. The molecularformula is the true formula of a compound. • The molecularformula represents the total number of atoms of each element present in one molecule of a compound.

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

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

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