APPLICATIONS OF THE MOLE Molar Mass of a

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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. • 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.

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.

Molar Mass of a Compound • In the case of NH 3, the molar mass equals the mass of one mole of nitrogen atoms plus the mass of three moles of hydrogen atoms.

Molar Mass of a Compound Molar mass of NH 3 = molar mass of N + 3(molar mass of H) Molar mass of NH 3 = 14. 0 g + 3(1. 0 g) = 17. 0 g/mol

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.

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)

Molar Mass Conversion 2. How many moles are in 146 grams of NH 3? (Answer: 8. 59 mol)

Molar Mass Conversion 3. How many moles are in 295 grams of Cr(OH)3? (Answer: 2. 86 mol)

Molar Mass Conversion 4. How many moles are in 22. 5 grams of HCl? (Answer: 0. 616 mol)

Molar Mass Conversion 5. How many grams of sodium chloride in 3. 45 moles of Na. Cl? 3. 45 moles Na. Cl 58. 5 g Na. Cl __ __ 1 mole Na. Cl (202 g)

Molar Mass Conversion 6. How many grams are in 0. 120 moles of Al. F 3? (Answer: 10. 1 g)

Molar Mass Conversion 7. How many grams are in 13. 0 moles of H 2 SO 4? (Answer: 1280 g)

Molar Mass Conversion 8. How many grams are in 1. 6 moles of K 2 Cr. O 4? (Answer: 250 g)

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.

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

Percent Composition

Example • Determine the percent composition of chlorine in calcium chloride (Ca. Cl 2). • 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.

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

Calculating Percent Composition • To the mass of one mole of Ca. Cl 2, 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 Ca. Cl 2.

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 Ca. Cl 2 can be calculated as follows.

Calculating Percent Composition Ca. Cl 2 %X= (molar mass of X) (# X ions) molar mass of compound % Cl = ( 35. 5 g) ( 2 Cl ions) 111. 1 g x 100 X 100 % Cl in Ca. Cl 2 = 63. 9%

Problem 9. Determine the percent composition of carbon in sodium acetate (Na. C 2 H 3 O 2). %X= (molar mass of X) (# X ions) molar mass of compound x 100

Calculating Percent Composition Na. C 2 H 3 O 2 %C= ( 12. 0 g) ( 2 C’s) 82. 0 g X 100 (% C = 29. 3%)

Problem 10. Calculate the percent composition aluminum of aluminum oxide (Al 2 O 3). 52. 9% Al

Problem 11. Determine the percent composition of oxygen in magnesium nitrate, Mg(NO 3)2. 64. 7% O

Problem 12. Determine the percent composition of sulfur in aluminum sulfate, which has the formula Al 2(SO 4)3. 28. 1% S

Problem 13. Determine the percent composition of oxygen in zinc nitrite, which has the formula Zn(NO 2)2. 40. 7% O

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.

Percent Water in a Hydrate • Gypsum is a hydrate with two water molecules present for every formula unit of Ca. SO 4. The chemical formula for gypsum is Ca. SO 4 • 2 H 2 O and the chemical name is calcium sulfate dihydrate. Note that the dot in the formula (or multiplication sign) indicates that the waters are there.

Percent Water in a Hydrate • Other examples of hydrates are: lithium perchlorate trihydrate Li. Cl. O 4 • 3 H 2 O; Ø magnesium carbonate pentahydrate Mg. CO 3 • 5 H 2 O; Ø and copper (II) sulfate pentahydrate Cu. SO 4 • 5 H 2 O. Ø

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.

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.

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.

Problem 14. Determine the percent water in Cu. SO 4 • 5 H 2 O (s).

Calculating Percent Composition • Mass of Cu. SO 4 = 63. 6 + 32. 1 + 16. 0 (4) = 159. 7 • Mass of 5 H 2 O = 1. 0 (2) + 16. 0 = 18. 0 x 5 = 90. 0 • Mass of Cu. SO 4. 5 H 2 O = 159. 7 + 90. 0 = 249. 7

Calculating Percent Composition % H 2 O = ( 18. 0 g) ( 5 H 2 O’s) 249. 7 g X 100 % H 2 O = 36. 0%

Problem 15. Determine the percent water in Mg. CO 3 • 5 H 2 O (s). 51. 6% H 2 O

Problem 16. Determine the percent water in Li. Cl. O 4 • 3 H 2 O (s). 33. 7% H 2 O

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.

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 Na 1 Cl 1, or Na. Cl, which is the true formula for the compound.

Empirical Formula • The formula for glucose is C 6 H 12 O 6. • The coefficients in glucose are all divisible by 6. • The empirical formula of glucose is CH 2 O.

Problem 17. Determine the empirical formula for Tl 2 C 4 H 4 O 6. Tl. C 2 H 2 O 3

Problem 18. Determine the empirical formula for N 2 O 4. NO 2

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.

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

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 54. 9 g Mn = 0. 7000 mol Mn

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 12. 0 g C = 1. 400 mol C

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 16. 0 g O = 2. 798 mol O

Empirical Formula from Percent Composition • The results show the following relationship. mol Mn : mol C : mol O 0. 7000 : 1. 400 : 2. 798

Empirical Formula from Percent Composition • To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles. 0. 7000 mol Mn 0. 7000 = 1 mol Mn

Empirical Formula from Percent Composition 1. 400 mol C 0. 7000 2. 798 mol O 0. 7000 = 2 mol C = 4 mol O

Empirical Formula from Percent Composition • The mole ratio for the compound is: 1 Mn : 2 C : 4 O • The empirical formula for the compound is Mn. C 2 O 4.

Problem 19. Determine the empirical formula of the following compound. 31. 9 g Mg, 27. 1 g P

Problem • The number of moles of magnesium may be calculated as follows. Given: 31. 9 g Mg 1 mol Mg 24. 3 g Mg = 1. 31 mol Mg

Problem • The number of moles phosphorous may be calculated as follows. Given: 27. 1 g P 1 mol P 31. 0 g P = 0. 874 mol P

Empirical Formula from Percent Composition • The results show the following relationship. mol Mg : mol P 1. 31 : 0. 874

Empirical Formula from Percent Composition • To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles. 0. 874 mol P 0. 874 = 1 mol P

Empirical Formula from Percent Composition 1. 31 mol Mg 0. 874 = 1. 5 mol Mg

Empirical Formula from Percent Composition • The mole ratio for the compound is: 1. 5 Mg : 1 P • You cannot have a fraction, so multiply both numbers by 2. 3 Mg : 2 P

Empirical Formula from Percent Composition • The empirical formula for the compound is Mg 3 P 2.

Problem 20. The composition of an unknown acid is 40. 00% carbon, 6. 71% hydrogen, and 53. 29% oxygen. Calculate the empirical formula for the acid. CH 20

Problem 21. The composition of an unknown ionic compound is 60. 7% nickel and 39. 3% fluorine. Calculate the empirical formula for the ionic compound. Ni. F 2

Problem 22. The composition of a compound is 6. 27 g calcium and 1. 46 g nitrogen. Calculate the empirical formula for the compound. Ca 3 N 2

Problem 23. Find the empirical formula for a compound consisting of 63. 0% Mn and 37. 0% O. Mn. O 2

Molecular Formulas • For many compounds, the empirical formula is not the true formula. • A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound.

Molecular Formulas • The molecular formula for a compound is either the same as the empirical formula or a whole-number multiple of the empirical formula.

Molecular Formulas • In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula.

Molecular Formulas • Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula as shown in the following example problem.

Problem 24. The molecular mass of benzene is 78 g/mol and its empirical formula is CH. What is the molecular formula for benzene?

Determining a Molecular Formula • Calculate the molar mass represented by the formula CH. • Here, the molar mass is the sum of the masses of one mole of each element. Molar mass CH = 12. 0 g + 1. 0 g Molar mass CH = 13. 0 g/mol

Problem • As stated in the problem, the molecular mass of benzene is known to be 78. 0 g/mol.

Determining a Molecular Formula • To determine the molecular formula for benzene, calculate the whole number multiple, n, to apply to its empirical formula. n= 78. 0 g/mol 13. 0 g/mol =6

Determining a Molecular Formula • This calculation shows that the molar mass of benzene is six times the molar mass of its empirical formula CH.

Determining a Molecular Formula • Therefore, the molecular formula must have six times as many atoms of each element as the empirical formula. • Thus, the molecular formula is (CH)6 = C 6 H 6

Problem 25. The simplest formula for butane is C 2 H 5 and its molecular mass is about 60. 0 g/mol. What is the molecular formula of butane?

Determining a Molecular Formula • To determine the molecular formula for butane, calculate the whole number multiple, n, to apply to its empirical formula. n= 60. 0 g/mol 29. 0 g/mol =2

Determining a Molecular Formula • Therefore, the molecular formula must have two times as many atoms of each element as the empirical formula. • Thus, the molecular formula is (C 2 H 5)2 = C 4 H 10

Problem 26. What is its molecular formula of cyanuric chloride, if the empirical formula is CCl. N and the molecular mass is 184. 5 g/mol? C 3 Cl 3 N 3

Problem 27. The simplest formula for vitamin C is C 3 H 4 O 3. Experimental data indicates that the molecular mass of vitamin C is about 180. What is the molecular formula of vitamin C? C 6 H 8 O 6

Problem 28. Maleic acid is a compound that is widely used in the plastics and textiles industries. The composition of maleic acid is 41. 39% carbon, 3. 47% hydrogen, and 55. 14% oxygen. Its molar mass is 116. 1 g/mol. Calculate the molecular formula for maleic acid.

Determining a Molecular Formula • Start by determining the empirical formula for the compound.

Problem • The number of moles of carbon may be calculated as follows. Given: 41. 39% C 41. 39 g C 1 mol C 12. 0 g C = 3. 449 mol C

Problem • The number of moles of hydrogen may be calculated as follows. Given: 3. 47% H 3. 47 g H 1 mol H 1. 0 g H = 3. 47 mol H

Problem • The number of moles of oxygen may be calculated as follows. Given: 55. 14% O 55. 14 g O 1 mol O 16. 0 g O = 3. 446 mol O

Determining a Molecular Formula • The numbers of moles of C, H, and O are nearly equal, so it is not necessary to divide through by the smallest value. • You can see by inspection that the smallest whole-number ratio is 1 C : 1 H : 1 O, and the empirical formula is CHO.

Determining a Molecular Formula • Next, calculate the molar mass represented by the formula CHO. • Here, the molar mass is the sum of the masses of one mole of each element. Molar mass CHO = 12. 0 g + 16. 0 g Molar mass CHO = 29. 0 g/mol

Problem • As stated in the problem, the molar mass of maleic acid is known to be 116. 1 g/mol.

Determining a Molecular Formula • To determine the molecular formula for maleic acid, calculate the whole number multiple, n, to apply to its empirical formula. n= 116. 1 g/mol 29. 0 g/mol =4

Determining a Molecular Formula • This calculation shows that the molar mass of maleic acid is four times the molar mass of its empirical formula CHO.

Determining a Molecular Formula • Therefore, the molecular formula must have four times as many atoms of each element as the empirical formula. • Thus, the molecular formula is (CHO)4 = C 4 H 4 O 4

Determining a Molecular Formula

Determining a Molecular Formula

Problem 29. The composition of silver oxalate is 71. 02% silver, 7. 91% carbon, and 21. 07% oxygen. If the molar mass of silver oxalate is 303. 8 g/mol, what is its molecular formula? Ag 2 C 2 O 4

Problem 30. The composition of a compound is 85. 6% carbon and 14. 4% hydrogen. If the molar mass of the compound is 42. 1 g/mol, what is its molecular formula? C 3 H 6