1. Unit 6 The Mole: % Composition and Emperical Formula
PLEASE GRAB A CALCULATOR FROM THE BACK
CHECK YOUR BOX

2. Formula Mass
Molecular mass – the sum of the average atomic masses of all the atoms in a molecule
***Remember***
A molecule is a neutral group of atoms that are held together by covalent bonds!!!

3. The Mole and Molar Mass
Molar Mass
The atomic molar mass is the mass of one mole of atoms. This is the value found on your Periodic Table.
All values are rounded to three decimal places.
The term molar mass is a general term referring to the mass of one mole of a compound. Also called molecular mass, formula mass, molar mass.
Add the molar mass together for each atom bonded together

4. Molar Mass of Compounds
The molar mass (mm) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound)
Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.08g
Avg. Atomic mass of Chlorine = 35.45g
Molar Mass of calcium chloride = g/mol Ca + (2 X 35.45) g/mol Cl  g/mol CaCl2 20
Ca 
17 Cl
35.45

5. The Mole Mole: 6.02 x 1023 items. Example:
Atoms, ions and molecules are so small it is impractical to refer to 1, 2 or 3 of these particles.
Mole: 6.02 x 1023 items.
This is the number of atoms required to produce the atomic mass of an element when that mass is expressed in grams.
Example:
1 atom Mg = amu (4.04 x g)
6.02 x 1023 atoms Mg = g

6. Avogadro’s Number
Avogadro’s Number (symbol N) is the number of atoms in grams of carbon.
Its numerical value is 6.02 × 1023.
Therefore, a g sample of carbon contains 6.02 × 1023 carbon atoms.

7. The Mole Reading Balanced Equations 2 Mg(s) + O2(g)  2 MgO(s)
2 Mg atoms + 1 O2 molecule produces 2 MgO formula units.
12.04 x 1023 Mg atoms x 1023 O2 molecules produces x 1023 MgO formula units.
2 mol Mg atoms + 1 mol O2 molecules produces mol MgO formula units.
The terms atoms, molecules and formula units are usually omitted:
Two moles magnesium react with one mole oxygen to produce two moles magnesium oxide.

8. The Mole One mole of copper metal
Other Examples: Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
One mole of copper metal
reacts with two moles of silver nitrate solution
to produce two moles of silver metal
and one mole of copper(II) nitrate solution.

9. How Big Is a Mole?
The volume occupied by one mole of softballs would be about the size of the Earth.
One mole of Olympic shot put balls has about the same mass as the Earth.

10. Molar Mass What is the molar mass of KCl?
What is the molar mass of Ba(NO3)2?

11. Molar Mass
What is the molar mass of CuSO4•5H2O?

12. Calculations Involving Mass and Moles
Mass to Moles Calculations: OR
Given number of grams
Calculated number of moles = x

13. Calculations Involving Mass and Moles
Example: g NaCl is how many moles?
1 Na 1 x =
1 Cl 1 x =
58.44 g/mol

14. Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)

15. Everything must go through Moles!!!
Calculations
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!

16. Calculations Involving Mass and Moles
Moles-to-Mass Calculations:
# of grams = moles x molar mass m = n M OR
Calculated molar mass
Given number of moles
Calculated number of grams = x

17. Calculations Involving Mass and Moles
Example: 0.21 mol NaCl is how many grams?
1 Na 1 x =
1 Cl 1 x =
58.44 g/mol
m = n M = 0.21 mol x g/mol = 12 g NaCl

18. Percent Composition
Law of Definite Proportions:
Elements in a compound combine in a constant proportion by mass.
For example, 1 g hydrogen combines with 8 g oxygen to produce 9 g water.
If atoms of an element have a specific mass, it must be true that atoms combine in a specific ratio.
Commonly expressed as percent composition

19. Percent Composition
Example (using experimental data): g of magnesium are burned to produce g of magnesium oxide. Determine the mass percent of magnesium in the compound.

20. Percent Composition
Example (using molar mass calculations): What is the percent by mass of each of the elements in magnesium sulfate, MgSO4?
1 Mg 1 x =
1 S x =
4 O x =
g/mol

21. 4. Combustion analysis of 0. 500 g of an unknown hydrocarbon yielded 1
4. Combustion analysis of g of an unknown hydrocarbon yielded g of CO2 and g H2O.
a) What is the percent by mass of carbon in the compound?

22. Empirical/Molecular Formulas and REVIEW

23. EMPERICAL FORMULA
The Empirical Formula is the simplest integer ratio of the elements in a compound.
For example: C4H8 is NOT an empirical formula. CH2 IS.

24. EMPERICAL FORMULA
Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).
Consider the amounts you are given as being in units of grams.
Convert the grams to moles for each element.
Find the smallest whole number ratio of moles for each element.

25. EMPIRICAL FORMULA EXAMPLE:
Find the empirical formula for a compound consisting of 63% Mn and 37% O

26. EMPIRICAL FORMULA
Assuming 100 g of the compound, there would be 63 g Mn and 37 g O
Look up the number of grams per mole for each element using the Periodic Table. There are grams in each mole of manganese and grams in a mole of oxygen.
63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn
37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O

27. EMPIRICAL FORMULA
Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. In this case there is less Mn than O, so divide by the number of moles of Mn:
1.1 mol Mn/1.1 = 1 mol Mn
2.3 mol O/1.1 = 2.1 mol O
The best ratio is Mn:O of 1:2 and the formula is MnO2
The empirical formula is MnO2

28. Molecular Formula
Molecular Formula is the exact ratio of the elements in the compound.
You can find the Molecular Formula of a compound by using the empirical formula and the molar mass.
The mass of the molecular formula is called the molecular mass (in the same way that the mass of the empirical formula is called the formula mass)

29. Molecular Formula
1) Calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula in order to find the ratio between the molecular formula and the empirical formula.
2)Then multiply all the atoms by this ratio to find the molecular formula!

30. Molecular Formula EXAMPLE:
What is the molecular formula of a compound that has a molecular weight of 240g? A 5.00g sample contains 2.00g of Carbon, 0.34g of Hydrogen and 2.69g of Oxygen.