1. Chemical Quantities The Mole
CHAPTER 7 NOTES
Chemical Quantities
The Mole

2. THE MOLE
One mole of substance represents 6.02  representative particles of that substance.
This number has been experimentally determined.
This number is called Avogadro’s number after Amedeo Avogadro di Quarenga.

3. The term representative particle usually refers to atoms, molecules, formula units, or ions.
BrINClHOF elements are all diatomic, that is they exist in the elemental state as two atoms sharing electrons. (Br2)

4. To convert atoms to moles:
x Atoms 1 mol substance
 1023 atoms
Remember! Putting the “known” over 1 is done simply to keep track of the numerator and denominator

5. To convert molecules to moles:
x molecules 1 mol substance
 1023 molecules
Molecules represent molecular compounds.

6. To convert ions to moles
x ions 1 mol substance
 1023 ions
Ions are atoms that have gained (anions) or lost (cations) electrons. Polyatomic ions are groups of covalently bonded atoms with a charge.

7. To Convert Formula Units to Moles
x formula units 1 mol substance
 1023 formula units
Formula units represent ionic compounds.

8. To convert moles to atoms, molecules or ions:
x Moles  1023 molecules (ions,atoms)
1                        1 mol substance

9. GRAM FORMULA, ATOMIC, AND MOLECULAR MASS
The gram atomic mass is the atomic mass of an element expressed in grams (gram formula mass for C is 12.0g) (Atoms only! Not molecules or ions)
The gram atomic mass of any two elements must contain the same number of atoms. (16.0 g O and 12.0 g C each contain 6.02  1023 atoms

10. The gram molecular mass is the mass of one mole of that compound
The gram molecular mass is the mass of one mole of that compound. (H2O has a gram molecular mass of 18g.)(Molecular compounds only! Not atoms or ions.)
The gram formula mass is the mass of one mole of an ionic compound. (NaCl has a gram formula mass of 58.1 g) (Ionic compounds only!)

11. To calculate GAM,GFM,GMM
First, you must find each elements atomic mass.
EX: Iron Fe
55.9 g
EX: Sodium Chloride NaCl
Na = 23.0 g
Cl = 35.5 g
NaCl = 58.5 g

12. EX: H2O 2 H at 1.0 g each = 2.0 g
1 O at 16.0 g = 16.0 g
H2O = 18.0 g

13. 7.2 MOLAR MASS
The molar mass may be used in place of gam, gfm, and gmm.
Molar mass is expressed in grams per mol (g/mol)
Complete the calculations as in the previous examples, but report the results in g/mol.

14. EX: How many grams are in 9.45 mol of dinitrogen trioxide?
First, write the formula: N2O3
Second, calculate the molar mass.
2 N = 28.0 g
3 O = 48.0 g
N2O3 = 76.0 g/mol

15. Third, perform the conversion/calculation
9.45 mol  76.0 g N2O3 = g N2O3
mol N2O3
Three significant figures are allowed in the result:
718 g N2O3

16. To convert from grams to moles, multiply the grams of substance A by the molar mass of A
  To find the number of moles in 92.2 g Fe2O3:
The molar mass of Fe2O3 is:
(2)(55.8 g/mol) + (3)(16 g/mol) = g/mol
92.2 g Fe2O3 1 mol Fe2O3 = mol Fe2O3
g Fe2O3

17. THE VOLUME OF A MOLE OF GAS AT STP
Standard pressure is kPa, 1 atm, or 760 mm Hg (all the same)
Standard temperature is 0C (273 K)
At standard temperature and pressure (STP) the molar volume of any gas is 22.4 L.

18. To convert from moles of gas at STP to liters:
0.600 mol SO2 STP 22.4 L = 13.4 L SO2
1                          1.0 mol

19. GAS DENSITY Density of a gas is usually measured in g/L.
Several calculations are associated with gas density   

20. An example of using density to determine molar mass
A compound consisting of C and O has a density at STP of g.L. Determine the molar mass and formula (CO or CO2)
Since at STP one mole of gas occupies 22.4L, we can determine the number of grams per mole( molar mass) as follows:

21. 22.4 L g = 44.0 g
1 mol L mol
CO has a molar mass of 28 g/mol. CO2 has a molar mass of 44 g/mol. The gas is carbon dioxide.

22. The Mole Road Map
On p. 186 of your text you have a copy of a “road map” that can assist you as you work on the problems in chapter 7.
Keep track of your numerator and denominator!
Make sure to use the EE or Exp key! Do not use the ^ key unless you are an expert at parenthesis!

23. PERCENT COMPOSITION
The relative amounts of elements in a compound are expressed as percent composition.
To calculate percent composition, find the molar mass, divide each element’s individual mass by the total molar mass, multiply by 100.

24. C3H8 has a molar mass of : 3X12g/mol= 36 g/mol
  36 g/mol 100 = 81.8% ; 8 g/mol 100=18.2% H
44 g/mol g/mol

25. `
An 8.20 g piece of Mg combines completely with 5.40 g O to form a compound. What is the percent composition of this compound?
8.20 g g = g

26. %Mg = 8.20 g 100 = 60.3% Mg
13.60 g
%O= 5.40 g 100 = 39.7% O

27. If you need to find the mass of an element in a substance, first find the percent composition, then multiply the mass given by the percent of the compound in question.

28. EMPIRICAL FORMULAS
Empirical formulas are formulas expressed as the lowest whole number ratio of atoms in the compound.
C2H6 is not an empirical formula CH3 is the empirical formula. (LCD)
To find empirical formulas, you need to know percent composition and molar masses of the elements.

29. What is the empirical formula of a compound that is 25
What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?  
Change the ratio of masses to a ratio of moles by converting grams to moles. Since you are given percentages assume a 100 gram sample (25.9 g N and 74.1 g O)

30. N: 25.9 g N 1 mol N = 1.85 mol N
1                 14.0 g N
O: 74.1 g O 1 mol O = 4.63 mol O
1              16.0 g O

31. The empirical formula must be whole numbers. So far we have this:. N1
The empirical formula must be whole numbers. So far we have this: N1.85O4.63
Not whole numbers! Divide each subscript by the smallest number.
N1.85O4.63 = N1O2.5 Almost there.

32. Multiply each subscript by 2 to get whole numbers: N2O5

33. MOLECULAR FORMULAS
Molecular formulas may or may not be reported to the smallest whole number ratio.
Molecular formulas represent the actual composition of molecular substances.

34. When given molar mass and empirical formulas:
Molar mass = 60 g
Empirical formula = CH4N
Molar mass of CH4N = 30 g/mol
Multiply each subscript by 2: C2H8N2