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The Mole and
Chemical Compounds
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.
The Mole Concept and Chemical Compounds
Molecular mass
Molecular mass is the mass of a molecule in atomic mass units.
= 2(1.00794 u) + 15.9994 u
molecular mass H2O = 2(atomic mass H) + (atomic mass O)
= 18.0153 u
formula mass Mg(NO3)2 = atomic mass Mg + 2[atomic mass N + 3(atomic mass O)]
= 24.3050 u + 2[14.0067 u + 3(15.9994 u)]
= 148.3148 u
One mole of a substance is equal to 6.022 × 1023 units of that substance (such as atoms,
molecules, or ions). The number 6.022 × 1023 is known as Avogadro's number or
Avogadro's constant.
The mole can be used to convert between atomic mass units and grams
1 mol H2O = 18.0153 g H2O = 6.02214 x 10 23 H2O molecules
1 mol MgCl2 = 95.211 g MgCl2 = 6.02214 x 10 23 MgCl2 formula unit
1 mol Mg(NO3)2 = 148.3148 g Mg(NO3)2
= 6.02214 x 10 23 Mg(NO3)2 formula units
EXAMPLE 3-1
An analytical balance can detect a mass of 0.1 mg. How many ions are present in this
minimally detectable quantity of MgCl2 ?
EXAMPLE 3-2
The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances
known. It is sometimes added to natural gas to make gas leaks detectable. How many
C2H6S molecules are contained in a sample? The density of liquid ethyl mercaptan is 0.84
g/mL
Break the calculation into three steps:
(1) a conversion from volume to mass,
(2) a conversion from mass to amount in moles, and
(3) a conversion from amount in moles to molecules
Convert from volume to mass
Convert from mass to amount in moles
Convert from moles to molecules
PRACTICE EXAMPLE A:
Gold has a density of 19.32 g/cm3. A piece of gold foil is 2.50 cm on each side and
0.100 mm thick. How many atoms of gold are in this piece of gold foil?
PRACTICE EXAMPLE B:
If the1.0 µL sample of liquid ethyl mercaptan from Example 3-2 is allowed to
evaporate and distribute itself throughout a chemistry lecture room with dimensions
62 ft x 35 ft x 14 ft will the odor of the vapor be detectable in the room? The limit of
detectability is 9 x 10-4 µmol/m3 .
Composition of Chemical Compounds
MC2H Br Cl F3 = 2MC + MH + MBr + MCl + 3MF
= [12 x 12.01072) + 1.00794 + 79.904 + 35.453 + 13 * 18.99842] g/mol
= 197.382 g/mol
Problem statement
The colorless, volatile liquid halothane has been used as a fire extinguisher
and also as an inhalation anesthetic. Both its empirical and molecular formulas
are C2HBr ClF3 its molecular mass is 197.382 u, How it was calculated?
Example
How many C atoms are present per mole of
halothane? In this case, the factor needed is 2 mol C / mol C2HBr ClF3. That is,
EXAMPLE 3-3
How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g/mL)
PRACTICE EXAMPLE A:
How many grams of Br are contained in 25.00 mL of halothane
(d = 1.871 g/mL)
PRACTICE EXAMPLE B:
How many milliliters of halothane would contain 1.00 x 1024
Br atoms?
Calculating % Composition from a Chemical Formula
The mass composition of a compound is the collection of mass percentages of the
individual elements in the compound.
1. Determine the molar mass of the compound. This is the denominator
2. Determine the contribution of the given element to the molar mass. This product of the
formula unit and the molar mass of the element appears in the numerator
3. Formulate the ratio of the mass of the given element to the mass of the compound as a
whole. This is the ratio of the numerator from step 2 to the denominator from step 1.
4. Multiply this ratio by 100% to obtain the mass percent of the element
Example 3-4
What is the mass percent composition of halothane, C2HBrClF3?
Solve
The molar mass of halothane is 197.38 g/mol. The mass % are
Thus, the percent composition of halothane is
12.17% C, 0.51% H, 40.48% Br, 17.96% Cl, and 28.88% F
PRACTICE EXAMPLE A:
Adenosine triphosphate (ATP) is the main energy-storage molecule in cells. Its chemical
formula is C10H16N5P3O13. What is its mass percent composition?
PRACTICE EXAMPLE B:
Without doing detailed calculations, determine which two compounds from the following
list have the same percent oxygen by mass: (a) CO; (b) CH3COOH(c) C2O3(d) N2O(e)
C6H12O6(f ) HOCH2CH2OH.
EXAMPLE 3-5
Determining the empirical and molecular formulas of a compound from its mass percent
composition
Dibutyl succinate is an insect repellent used against household ants and roaches. Its
composition is 62.58% C, 9.63% H, and 27.79% O. Its experimentally determined
molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl
succinate?
Determine the mass of each element in a sample of 100.00 g,
62.58 g C, 9.63 g H, 27.79 g O
Convert each of these masses to an amount in moles
Write a tentative formula based on these numbers of moles
C5.21 H9.55 O1.74
Divide each of the subscripts of the tentative formula by the smallest subscript (1.74),
round off any subscripts that differ only slightly from whole numbers; that is, round 2.99 to 3.
C3H5.49O
Multiply all subscripts by a small whole number to make them integral (here by the factor
2), and write the empirical formula. C2x3 H2x5.49 O2x1 = C6 H10.98 O2
Empirical formula: C6H11O2
To establish the molecular formula, first determine the empirical formula mass
[(6 x 12.0) + (11 x 1.0) + (2 x 16.0)] u = 115 u
Since the experimentally determined formula mass (230 u) is twice the empirical formula
mass, the molecular formula is twice the empirical formula.
Molecular formula: C12H22O4

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The mole and chemical compounds

  • 1. The Mole and Chemical Compounds Dr. K. Shahzad Baig Memorial University of Newfoundland (MUN) Canada Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario. Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.
  • 2. The Mole Concept and Chemical Compounds Molecular mass Molecular mass is the mass of a molecule in atomic mass units. = 2(1.00794 u) + 15.9994 u molecular mass H2O = 2(atomic mass H) + (atomic mass O) = 18.0153 u formula mass Mg(NO3)2 = atomic mass Mg + 2[atomic mass N + 3(atomic mass O)] = 24.3050 u + 2[14.0067 u + 3(15.9994 u)] = 148.3148 u
  • 3. One mole of a substance is equal to 6.022 × 1023 units of that substance (such as atoms, molecules, or ions). The number 6.022 × 1023 is known as Avogadro's number or Avogadro's constant. The mole can be used to convert between atomic mass units and grams 1 mol H2O = 18.0153 g H2O = 6.02214 x 10 23 H2O molecules 1 mol MgCl2 = 95.211 g MgCl2 = 6.02214 x 10 23 MgCl2 formula unit 1 mol Mg(NO3)2 = 148.3148 g Mg(NO3)2 = 6.02214 x 10 23 Mg(NO3)2 formula units
  • 4.
  • 5. EXAMPLE 3-1 An analytical balance can detect a mass of 0.1 mg. How many ions are present in this minimally detectable quantity of MgCl2 ?
  • 6. EXAMPLE 3-2 The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances known. It is sometimes added to natural gas to make gas leaks detectable. How many C2H6S molecules are contained in a sample? The density of liquid ethyl mercaptan is 0.84 g/mL Break the calculation into three steps: (1) a conversion from volume to mass, (2) a conversion from mass to amount in moles, and (3) a conversion from amount in moles to molecules
  • 7. Convert from volume to mass Convert from mass to amount in moles Convert from moles to molecules
  • 8. PRACTICE EXAMPLE A: Gold has a density of 19.32 g/cm3. A piece of gold foil is 2.50 cm on each side and 0.100 mm thick. How many atoms of gold are in this piece of gold foil? PRACTICE EXAMPLE B: If the1.0 µL sample of liquid ethyl mercaptan from Example 3-2 is allowed to evaporate and distribute itself throughout a chemistry lecture room with dimensions 62 ft x 35 ft x 14 ft will the odor of the vapor be detectable in the room? The limit of detectability is 9 x 10-4 µmol/m3 .
  • 9. Composition of Chemical Compounds MC2H Br Cl F3 = 2MC + MH + MBr + MCl + 3MF = [12 x 12.01072) + 1.00794 + 79.904 + 35.453 + 13 * 18.99842] g/mol = 197.382 g/mol Problem statement The colorless, volatile liquid halothane has been used as a fire extinguisher and also as an inhalation anesthetic. Both its empirical and molecular formulas are C2HBr ClF3 its molecular mass is 197.382 u, How it was calculated?
  • 10. Example How many C atoms are present per mole of halothane? In this case, the factor needed is 2 mol C / mol C2HBr ClF3. That is,
  • 11. EXAMPLE 3-3 How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g/mL)
  • 12. PRACTICE EXAMPLE A: How many grams of Br are contained in 25.00 mL of halothane (d = 1.871 g/mL) PRACTICE EXAMPLE B: How many milliliters of halothane would contain 1.00 x 1024 Br atoms?
  • 13. Calculating % Composition from a Chemical Formula The mass composition of a compound is the collection of mass percentages of the individual elements in the compound. 1. Determine the molar mass of the compound. This is the denominator 2. Determine the contribution of the given element to the molar mass. This product of the formula unit and the molar mass of the element appears in the numerator 3. Formulate the ratio of the mass of the given element to the mass of the compound as a whole. This is the ratio of the numerator from step 2 to the denominator from step 1. 4. Multiply this ratio by 100% to obtain the mass percent of the element
  • 14. Example 3-4 What is the mass percent composition of halothane, C2HBrClF3? Solve The molar mass of halothane is 197.38 g/mol. The mass % are
  • 15. Thus, the percent composition of halothane is 12.17% C, 0.51% H, 40.48% Br, 17.96% Cl, and 28.88% F
  • 16. PRACTICE EXAMPLE A: Adenosine triphosphate (ATP) is the main energy-storage molecule in cells. Its chemical formula is C10H16N5P3O13. What is its mass percent composition? PRACTICE EXAMPLE B: Without doing detailed calculations, determine which two compounds from the following list have the same percent oxygen by mass: (a) CO; (b) CH3COOH(c) C2O3(d) N2O(e) C6H12O6(f ) HOCH2CH2OH.
  • 17. EXAMPLE 3-5 Determining the empirical and molecular formulas of a compound from its mass percent composition Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H, and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Determine the mass of each element in a sample of 100.00 g, 62.58 g C, 9.63 g H, 27.79 g O
  • 18. Convert each of these masses to an amount in moles Write a tentative formula based on these numbers of moles C5.21 H9.55 O1.74 Divide each of the subscripts of the tentative formula by the smallest subscript (1.74),
  • 19. round off any subscripts that differ only slightly from whole numbers; that is, round 2.99 to 3. C3H5.49O Multiply all subscripts by a small whole number to make them integral (here by the factor 2), and write the empirical formula. C2x3 H2x5.49 O2x1 = C6 H10.98 O2 Empirical formula: C6H11O2 To establish the molecular formula, first determine the empirical formula mass [(6 x 12.0) + (11 x 1.0) + (2 x 16.0)] u = 115 u Since the experimentally determined formula mass (230 u) is twice the empirical formula mass, the molecular formula is twice the empirical formula. Molecular formula: C12H22O4

Editor's Notes

  1. The final factor we need is based on the fact that there are three ions (one and two ) per formula unit (fu) of MgCl2
  2. The molecular formula of C2HBr ClF3 tells us that per mole of halothane there are two moles of C atoms, one mole each of H, Br, and Cl atoms, and three moles of F atoms.