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APPLICATIONS OF THE MOLE

APPLICATIONS OF THE MOLE. Molar Mass of a Compound. The molar mass of a compound is the mass of a mole of the representative particles of the compound.

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APPLICATIONS OF THE MOLE

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  1. APPLICATIONS OF THE MOLE

  2. Molar Mass of a Compound • The molar mass of a compound is the mass of a mole of the representative particles of the compound. • Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle.

  3. Molar Mass of an Element • To determine the molar mass of an element, find the element’s symbol on the periodic table and round the mass so there is one digit beyond the decimal. • The molar mass of carbon (C) is 12.0 g/mol, of chlorine (Cl) is 35.5 g/mol and of iron (Fe) is 55.8 g/mol.

  4. Molar Mass of a Compound • In the case of NH3, the molar mass equals the mass of one mole of nitrogen atoms plus the mass of three moles of hydrogen atoms.

  5. Molar Mass of a Compound Molar mass of NH3= molar mass of N + 3(molar mass of H) Molar mass of NH3= 14.0 g + 3(1.0 g) = 17.0 g/mol

  6. Molar Mass of a Compound • You can use the molar mass of a compound to convert between mass and moles, just as you used the molar mass of elements to make these conversions.

  7. Molar Mass Conversion 1.How many moles of magnesium in 56.3 g of Mg? 56.3 g Mg __ 1 mole Mg __ 24.3 g Mg (2.32)

  8. Molar Mass Conversion 2.How many moles are in 146 grams of NH3? (Answer: 8.59 mol)

  9. Molar Mass Conversion 3.How many moles are in 295 grams of Cr(OH)3? (Answer: 2.86 mol)

  10. Molar Mass Conversion 4.How many moles are in 22.5 grams of HCl? (Answer: 0.616 mol)

  11. Molar Mass Conversion 5.How many grams of sodium chloride in 3.45 moles of NaCl? 3.45 moles NaCl 58.5 __ g NaCl __ 1 mole NaCl (202 g)

  12. Molar Mass Conversion 6.How many grams are in 0.120 moles of AlF3? (Answer: 10.1 g)

  13. Molar Mass Conversion 7.How many grams are in 13.0 moles of H2SO4? (Answer: 1280 g)

  14. Molar Mass Conversion 8. How many grams are in 1.6 moles of K2CrO4? (Answer: 250 g)

  15. Percent Composition • Recall that every chemical compound has a definite composition - a composition that is always the same wherever that compound is found. • The composition of a compound is usually stated as the percent by mass of each element in the compound.

  16. (molar mass of X) (# X ‘s) x 100 % X = molar mass of compound Percent Composition • The percent of an element (X) in a compound can be found in the following way.

  17. Percent Composition

  18. Example • Determine the percent composition of chlorine in calcium chloride (CaCl2). • First, analyze the information available from the formula. • A mole of calcium chloride consists of one mole of calcium ions and two moles of chloride ions.

  19. Calculating Percent Composition • Next, gather molar mass information from the atomic masses on the periodic table.

  20. Calculating Percent Composition • To the mass of one mole of CaCl2, a mole of calcium ions contributes 40.1 g, and two moles of chloride ions contribute 2 x 35.5 g = 71.0 g for a total molar mass of 111.1 g/mol for CaCl2.

  21. Calculating Percent Composition • Finally, use the data to set up a calculation to determine the percent by mass of an element in the compound. • The percent by mass of chlorine in CaCl2 can be calculated as follows.

  22. (molar mass of X) (# X ions) x 100 % X = molar mass of compound Calculating Percent Composition CaCl2 ( g) ( Cl ions) 35.5 2 % Cl = X 100 111.1 g % Cl in CaCl2 = 63.9%

  23. (molar mass of X) (# X ions) x 100 % X = molar mass of compound Problem 9.Determine the percent composition of carbon in sodium acetate (NaC2H3O2).

  24. Calculating Percent Composition NaC2H3O2 ( g) ( C’s) 12.0 2 % C = X 100 82.0 g (% C = 29.3%)

  25. Problem 10. Calculate the percent composition aluminum of aluminum oxide (Al2O3). 52.9% Al

  26. Problem 11. Determine the percent composition of oxygen in magnesium nitrate, Mg(NO3)2. 64.7% O

  27. Problem 12. Determine the percent composition of sulfur in aluminum sulfate, which has the formula Al2(SO4)3. 28.1% S

  28. Problem 13. Determine the percent composition of oxygen in zinc nitrite, which has the formula Zn(NO2)2. 40.7% O

  29. Percent Water in a Hydrate • Hydrates are compounds that incorporate water molecules into their fundamental solid structure. In a hydrate (which usually has a specific crystalline form), a defined number of water molecules are associated with each formula unit of the primary material.

  30. Percent Water in a Hydrate • Gypsum is a hydrate with two water molecules present for every formula unit of CaSO4. The chemical formula for gypsum is CaSO4 • 2 H2O and the chemical name is calcium sulfate dihydrate. Note that the dot in the formula (or multiplication sign) indicates that the waters are there.

  31. Percent Water in a Hydrate • Other examples of hydrates are: • lithium perchlorate trihydrate - LiClO4 • 3 H2O; • magnesium carbonate pentahydrate - MgCO3 • 5 H2O; • and copper (II) sulfate pentahydrate - CuSO4 • 5 H2O.

  32. Percent Water in a Hydrate • The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.

  33. Percent Water in a Hydrate • Experimentally measuring the percent water in a hydrate involves first heating a known mass of the hydrate to remove the waters of hydration and then measuring the mass of the anhydrate remaining. The difference between the two masses is the mass of water lost.

  34. Percent Water in a Hydrate • Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. Multiplying this fraction by 100 gives the percent water.

  35. Problem 14. Determine the percent water in CuSO4 • 5 H2O (s).

  36. Calculating Percent Composition • Mass of CuSO4 = 63.6 + 32.1 + 16.0 (4) = 159.7 • Mass of 5 H2O = 1.0 (2) + 16.0 = 18.0 x 5 = 90.0 • Mass of CuSO4 . 5 H2O = 159.7 + 90.0 = 249.7

  37. Calculating Percent Composition ( g) ( H2O’s) 18.0 5 % H2O = X 100 249.7 g % H2O = 36.0%

  38. Problem 15. Determine the percent water in MgCO3 • 5 H2O (s). 51.6% H2O

  39. Problem 16. Determine the percent water in LiClO4 • 3 H2O (s). 33.7% H2O

  40. Empirical Formula • You can use percent composition data to help identify an unknown compound by determining its empirical formula. • The empirical formula is the simplest whole-number ratio of atoms of elements in the compound. In many cases, the empirical formula is the actual formula for the compound.

  41. Empirical Formula • For example, the simplest ratio of atoms of sodium to atoms of chlorine in sodium chloride is 1 atom Na : 1 atom Cl. • So, the empirical formula of sodium chloride is Na1Cl1, or NaCl, which is the true formula for the compound.

  42. Empirical Formula • The formula for glucose is C6H12O6. • The coefficients in glucose are all divisible by 6. • The empirical formula of glucose is CH2O.

  43. Problem 17. Determine the empirical formula for Tl2C4H4O6. TlC2H2O3

  44. Problem 18. Determine the empirical formula for N2O4. NO2

  45. Empirical Formula from Percent Composition • The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. • Because percent means “parts per hundred parts,” assume that you have 100 g of the compound.

  46. Empirical Formula from Percent Composition • Then calculate the number of moles of each element in the 100 g of compound.

  47. Empirical Formula from Percent Composition • The number of moles of manganese may be calculated as follows. Given: 38.43% Mn 38.48 g Mn 1 mol Mn = 0.7000 mol Mn 54.9 g Mn

  48. Empirical Formula from Percent Composition • The number of moles of carbon may be calculated as follows. Given: 16.80% C 16.80 g C 1 mol C = 1.400 mol C 12.0 g C

  49. Empirical Formula from Percent Composition • The number of moles of carbon may be calculated as follows. Given: 44.77% O 44.77 g O 1 mol O = 2.798 mol O 16.0 g O

  50. Empirical Formula from Percent Composition • The results show the following relationship. mol Mn : mol C : mol O 0.7000 : 1.400 : 2.798

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