Honors Chemistry Chapter 8 The Mole

The Mole

!   Symbol of an elem represents 1 atom of that elem

!   Formula represents 1 molec. or formula unit of the compound !   These may also represent a group of atoms or formula

units

!   Atoms are too small to deal with, so chemists deal w/ large groups of atoms ! A mole – contains a specific # of atoms or formula units

8.1 Molecular & Formula Mass

!   H atom’s mass = 1.67 x 10-24g

!   O atom’s mass = 16 x H or 2.66 x 10-23g

!   C atom’s mass = 12 x H or 2.00 x 10-23g

!   Since these #’s are so small, we use a mass scale defined on the atomic level !   Masses of atoms are compared by using the atomic mass

scale !   Standard is the atomic mass unit (u or amu)

!   Based on 1/12 of a C-12 atom

!   H is 1u, O is 16u, ∴ H2O = 18u

8.1 Molecular & Formula Mass

! Molecular Mass – sum of the atomic masses of all the atoms in a molecule !   This term is incorrect for ionic compounds

! Formula Mass – sum of atomic masses of atoms in a formula unit

8.2 The Avogadro Constant

!   Since chemists can’t count out atoms or molecs, they obtain mass quantities of a subst. !   ∴ It’s important to obtain a relationship betw mass & #

of particles

! Molec & formula masses are in amu’s – too small to meas by ordinary means !   Need larger unit such as grams

8.2 The Avogadro Constant

!   Need to choose a # of atoms that would have a mass in grams = to the mass of 1 atom in amu’s

!   H 1.0079g 1u 1 atom H = ?

!   1.66 x 10-24g 1.0079u

!   O 15.9994g 1u 1 atom O = ?

!   1.66 x 10-24g 15.9994u

!   Answer for each is 6.02 x 1023 atoms

8.2 The Avogadro Constant

!   6.02 x 1023 is called the Avogadro Constant in honor of Amedeo Avogadro, an Italian scientist.

8.3 The Mole !   The Avogadro Constant is an SI standard

!   Symbol is NA

!   6.02 x 1023 units of anything is call one mole (mol)

!   1 mole of particles (atoms, ions, molecs, etc) has a mass in grams = to that of one particle in atomic mass units.

!   ∴ if 1 mole of a certain type of particle has a mass of 3.01g, then a single particle has a mass of 3.01u.

8.3 The Mole

!   1 mole of molecules = 6.02 x 1023 molecs

!   1 mole of atoms = 6.02 x 1023 atoms

!   1 mole of formula units = 6.02 x 1023 formula units

!   1 mole of ions = 6.02 x 1023 ions !   ∴ NA can have any of the following units:

!   molecs/mole, atoms/mole, ions/mole, or

!   formula units/mole

8.3 The Mole

! Molar Mass – mass of 1 mole of particles

!   (Save room for examples)

8.4 Moles in Solution

!   Many methods of expressing the relationship betw dissolved subst & the soln. Most common – ! Molarity (M) – ratio betw the moles of dissoved subst &

the vol of soln in dm3

!   M = moles of solute

dm3 of soln

∴  a one-molar (1M) soln of salt water contains 1 mole of NaCl in 1 dm3 of soln ∴  A .35M soln of KNO3 contains .35 moles of KNO3 in 1

dm3 of soln

8.4 Moles in Solution

!   ***Note – molarity is expressed in moles per dm3 of solution, not solvent. !   ∴ to make a 1M soln of NaCl, you do NOT add 1dm3

H2O to 58.5 g NaCl. The final volume must be 1dm3. !   Dissolve the salt in water, then add enough water to make

1dm3 of solution.

!   (Leave room for examples)

8.5 Percent Composition !   Percent composition of a comp is a statement of the

relative mass each elem conributes to the mass of the compound as a whole !   % composition of Cu = 100% bec it’s an elem

!   %comp of NaCl=

!   Mass Na x 100% = 23.0u x100% = 39.3% Na

!   Mass NaCl 58.5u

!   Mass Cl x 100% = 35.5u x 100% = 60.7% Cl

!   Mass NaCl 58.5u

!   All % should add up to 100% !   (leave room for examples)

8.6 Empirical Formula

!   - Simplest ratio of the elements in that compound

!   Empirical formulas can be found using experimental data !   Need only the mass of ea elem in the sample

! Elems in a compound combine in simple, whole-number ratios !   ∴ if atoms of elem are present in simple ratios, then the

moles of atoms for ea elem in the comp will also be in small, whole-number ratios.

8.6 Empirical Formula

!   Ex) We have a 2.50 sample of a comp which contains 0.900g Ca and 1.60g Cl. Find the empirical formula !   Step 1 – find the # of moles (leave room in notes)

!   Step 2 – Find the ratio of moles of Ca to moles of Cl !   Divide each by the smallest # of moles

!   Leave room in notes

!   Step 3 – Find the empirical formula

8.6 Empirical Formula

!   Empirical formulas can also be found from % composition !   Assume a 100g sample & change all % to g

!   Ex) A sample has a % comp of 40.0% C, 6.71% H, and 53.3% O. Find the empirical formula !   Leave room in notes

8.6 Empirical Formula

!   Sometimes dividing by the smallest # of moles doesn’t yield a ratio close to a whole # !   Must multiply all ratios by a whole # to make them

whole #’s

!   Ex) a sample of a compound is 66% Ca and 34% P. Find the empirical formula !   (leave room in notes)

8.7 Molecular Formulas

!   To calc molecular formulas, you need the empirical formula and the molar (or molecular) mass !   Molecular formula is a multiple of the empirical form

!   Ex) Emp. form = CH2O, molec. Mass = 180u !   Mass of CH2O = 12.0u + 2.02u + 16.0u = 30.0u

!   30 x ? = 180 ? = 6

!   ∴ multiply all subscripts by 6 !   C6H12O6

8.8 Hydrates

!   Some comps crystallize from a water soln w/ water molecs adhering to ions or molecs & become part of the crystal !   Hydrates – crystals that contain water molecs

!   Can be dried by heating & driving off the water !   Then meas how much mass is lost

!   NiSO3 � 6H2O !   The � means that 6 molecs of water adhere to 1

formula unit.

8.8 Hydrates

!   To calculate the formula mass of a hydrate, add the mass of the salt (NiSO3) to the mass of the water !   NiSO3 = 138.8u

6 H2O = 108.0u

!   246.8u = NiSO3 � 6H2O

8.8 Hydrates

!   From sample problem on p. 215 or your packet: !   10.407g hydrate

!   − 9.520g dry (anhydrous) BaI2

!   0.887g water

!   Convert mass of dry BaI2 and mass of water to moles !   (leave room in notes)

!   Find the simplest ratio (divide by smallest #) !   ∴ BaI2 � 2H2O