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What is Avogadro’s Number

What is Avogadro’s Number. Avogadro’s Number is the number of “things” in 1 mole 6.022 X 10 23. Calculating Molar Mass. Molar mass is the mass of 1 mole of a compound. Calculate the molar mass of magnesium carbonate. MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) =. 84.32 g/mol. Practice.

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What is Avogadro’s Number

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  1. What is Avogadro’s Number Avogadro’s Number is the number of “things” in 1 mole 6.022 X 1023

  2. Calculating Molar Mass Molar mass is the mass of 1 mole of a compound. Calculate the molar mass of magnesium carbonate. MgCO3 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

  3. Practice • Al2S3 • NaNO3 • Ba(OH)2 • Ba(NO3)2 150.17 g/mol 85.00 g/mol 171.35 g/mol 261.35 g/mol

  4. 84.32 g 1 mol 1 mol 84.32 g Gram/Mole Conversions Molar masses are used as conversion factors. MgCO3 = 84.32 g/mol or Conversion factors

  5. Mole to Mass Conversions How many grams are in 2.36 mol of MgCO3?

  6. More Practice! What is the mass of 1.36 mol of H2O? What is the mass of 4.36 mol of Ba(OH)2?

  7. Mass to Mole Conversions How many moles are in 0.37 g of MgCO3?

  8. More Practice! How many moles are in 4.50 g of H2O? How many moles are in 471.6 g of Ba(OH)2?

  9. Pause for a Cause #8 Calculate the number of moles in each of the following masses: a. 45.0 grams of acetic acid (CH3COOH) b. 7.04 grams of lead II nitrate (Pb(NO3)2 c. 5000 kg of iron III oxide (Fe2O3) Calculate the mass of each of the following amounts: a. 3.00 mol of selenium oxybromide ( SeOBr2) b. 488 mol of calcium carbonate (CaCO3) c. 0.0091 mol of retonic acid (C20H28O2)

  10. Pause for a Cause #8 • 45.0 grams of acetic acid (CH3COOH) • 7.04 grams of lead II nitrate Pb(NO3)2 • 5000 kg of iron III oxide (Fe2O3) 0.749 mol(CH3COOH) 0.0213 mol (Pb(NO3)2 3 .14 X 104 mol (Fe2O3) • 3.00 mol of selenium oxybromide (SeOBr2) • 488 mol of calcium carbonate (CaCO3) • 0.0091 mol of retonic acid (C20H28O2) 764.88g (SeOBr2) 4.88 X 104 g (CaCO3) 2.73g (C20H28O2)

  11. 1 mole = molar mass in grams = 6.022 X1023 molecules Calculate the number of molecules/ formula units a. 4.27 moles of tungsten oxide (WO3) b. 0.003 00 moles of strontium nitrate Sr(NO3)2 Calculate the number of molecules/ formula units a. 285 grams of iron III phosphate Fe(PO4)3 b. 0.0084 grams of C5H5N Calculate the mass of each of the following quantities a. 8.39 X 1023 molecules of F2 (fluorine) b. 6.82 X 1024 formula units of beryllium sulfate BeSO4

  12. Pause for a Cause #8 • Calculate the number of molecules/ formula units • a. 4.27 moles of tungsten oxide (WO3) • b. 0.003 00 moles of strontium nitrate Sr(NO3)2 • Calculate the number of molecules/ formula units • 285 grams of iron III phosphate Fe(PO4)3 • 0.0084 grams of C5H5N • Calculate the mass of each of the following quantities • a. 8.39 X 1023 molecules of F2 (fluorine) • b. 6.82 X 1024 formula units of beryllium sulfate BeSO4 2.57 X 1024 Molecules 1.81 X 1021 Molecules 5.04 X 1023 Molecules 6.4 X 1019 Molecules 52.9g F2 1.19 X 103 Molecules

  13. Molarity Molarity is the concentration of mole per liter of solution M = mole/L

  14. What is the mass of 6.022 X 1023 atoms of Iron Oxide What is the mass of 6.022 X 1024 atoms of Iron II Oxide 6.022 X 1024 atoms FeO2 * ____1 mole______ * 72g FeO = 720g FeO 6.022 X 1023 atoms 1 mole

  15. Calculating Percentage Composition Page 241

  16. Calculating Percentage Composition Calculate the percentage composition of barium sulfate. Step 1: Determine the molar mass. Step 2: Divide the mass of the element in the compound by the molar mass, then multiply by 100%. Page 241

  17. Let’s Practice! #10 • Determine the percent composition of NaCl. • Na- 1 x 23.0=23.0g Cl- 1 x 35.5=35.5g23.0g + 35.5g = 58.5 (58.5 is total) • Determine the percent composition of Ba3(PO4)2 (23.0g/58.5g) x 100 = 39.3% Na (35.5g/58.5g) x 100 = 60.7% Cl

  18. Formulas Empirical formula:the lowest whole number ratio of atoms in a compound. (found experimentally by comparing ratio) Molecular formula:the true number of atoms of each element in the formula of a compound. (actual compound) • molecular formula = C6H6 • empirical formula = CH Khan Academy Video

  19. Formulas (continued) Formulas forionic compoundsare ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  20. Formulas(continued) Formulas formolecular compoundsMIGHT be empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

  21. Empirical Formula Determination • Base calculation on 100 grams of compound. • Determine moles of each element in 100 grams of compound. • Divide each value of moles by the smallest of the values. • Convert decimals to fractions and determine the least common denominator.

  22. C?H?O? 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? C?O?H?

  23. Carbon: Oxygen: Hydrogen Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 C3H5O2 Empirical formula:

  24. Pause for a Cause #11 A compound is found to contain 36.48% Na, 25.41% S, and 38.11% O. Find its empirical formula. Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound?

  25. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? Step 1: Find the molar mass of the empirical formula. 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 C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? Step 2: Divide the molar mass of the compound by the molar mass given by the empirical 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 C3H5O2. The molar 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 (C3H5O2) x 2 = C6H10O4

  28. Pause for a Cause #12 What is the molecular formula of a compound that has an empirical formula of CH2 and a molar mass of 28 g/mol? A compound is 81.06% boron and 18.94% hydrogen. The molar mass was experimentally determined to be 54 g/mol. What is the molecular formula of the compound?

  29. The End! The End! The End! The End! The End! The End! The End!

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