6 02 X 1023 Avogadros Number 6 02

6. 02 X 1023 Avogadro’s Number

6. 02 X 1023 of that thing 1 Mole of anything

Can be: - atoms - molecules - ions Particles

3. 01 X 1023 0. 5 Mole

1. 50 X 1023 0. 25 Mole

12. 04 X 1023 Or 1. 204 X 1024 2. 0 Mole

22. 4 Liters 1. 0 Mole of any gas at STP

Standard Temperature = 0 C Standard Pressure = 1 atm STP

11. 2 Liters 0. 5 mole of any gas

44. 8 Liters 2. 0 mole of any gas

67. 2 Liters 3. 0 mole of any gas

5. 6 Liters 0. 25 mole of any gas

Sum of the masses of the elements in the compound. Expressed in atomic mass units Formula Mass

2 X H = 2 X 1. 0 = 2. 0 1 X O = 1 X 16. 0 = 16. 0 Sum = 18. 0 amu’s Formula Mass of H 2 O

3 X H = 3 X 1. 0 = 3. 0 1 X N = 1 X 14. 0 = 14. 0 Sum = 17. 0 amu Formula Mass of NH 3

1 X C = 1 X 12. 0 = 12. 0 2 X O = 2 X 16. 0 = 32. 0 Sum = 44. 0 amu Formula Mass of CO 2

Formula mass expressed in grams. Equals the molar mass of the compound. Gram Formula Mass

Mass of one mole of the substance. Molar Mass

H 2 O = NH 3 = CO 2 = 18. 0 grams 17. 0 grams 44. 0 grams Molar Mass

For Paren: Sub. Inside X Suboutside N: 2 H: 8 S: 1 O: 4 Count up the atoms in (NH 4)2 SO 4

For Paren: Sub. Inside X Suboutside Coefficients X subs in formula Mg: 6 P: 4 O: 16 The “ 2” applies to every element in the formula. Count up the atoms in 2 Mg 3(PO 4)2

# of Grams by formula mass by 6. 02 X 1023 X Formula Mass # of Moles X 6. 02 X 1023 # of Particles X 22. 4 L/mole by 22. 4 # of Liters (gas)

Part X 100% Whole Percent

Part X 100% Whole = 2 X 100% 18 Percent H in H 2 O

Part X 100% Whole = 16 X 100% 18 Percent O in H 2 O

smallest whole number ratio of the elements in a compound Empirical Formula

Gives exact composition of molecule Molecular Formula

Formula contains all nonmetals Covalent Compound

Molecular (vast majority) & Network (Si. O 2, Si. C, Cdia, Cgraph) Types of covalent substances

Formula contains metal plus nonmetal Ionic Compound

Ionic compound that has H 2 O molecules incorporated into its structure. Hydrate

Substance that remains after the water is removed from a hydrate. Anhydrate

Formula of a hydrated salt. • means “is associated with. ” H 2 O molecules are stuffed in the empty spaces. Cu. SO 4 • 5 H 2 O

Cu. SO 4 + 5 H 2 O anhydrate Evaporates into air Cu. SO 4· 5 H 2 O ? hydrate Heat to constant mass

Mass of Cu. SO 4 plus mass of 5 water molecules. 249. 6 grams/mole Formula mass of Cu. SO 4 • 5 H 2 O

Part X 100% Whole = 90 X 100% 249. 6 Percent H 2 O in Cu. SO 4 • 5 H 2 O (from the formula)

All elements to the left of the staircase except H Metals

All elements to the right of the staircase plus H Nonmetals

Compound made from 2 elements Binary Compound

H 2 O 2 C 6 H 12 O 6 KCl CH 4 C 2 H 6 P 4 O 10 Ca. F 2 Which formulas are empirical?

Crystal Lattices Ionic Compounds, Metals, & Network Solids make….

Smallest repetitive unit in a crystal lattice Formula Unit

Have distinctly different properties than molecular substances. Substances with crystal lattices…

Types of Substances Ionic • May be Binary • Or may have > 2 elements • May contain a transition metal Covalent Metallic • Cu. Al 2 • May be Molecular • or Network (Si. O 2, Si. C, Cdia, Cgraph)

Ionic, Metallic, and Network solids What kind of substances have Crystal Lattices?

Empirical formulas only Substances that make crystal lattices have

Have both empirical & molecular formulas. The molecular formula is a wholenumber multiple of the empirical formula. Molecular Covalent Substances

1) Find empirical mass 2) Divide formula mass/empirical mass 3) Multiply subscripts in empirical formula by answer in step 2 Given empirical formula & Formula Mass, find Molecular Formula

1) Empirical mass = 13 2) Divide formula mass/empirical mass = 78/13 = 6 3) Multiply subscripts: C 6 H 6 Empirical formula = CH & Formula Mass = 78, find Molecular Formula

1) Mass of H 2 O = 12 – 8 = 4 g 2) Percent H 2 O = 4/12 X 100% 3) Percent salt = 8/12 X 100% Percent water in hydrate from experimental data. 12 grams of hydrated salt is heated to constant mass. After heating the mass is 8. 0 grams. What is the percent salt & the percent H 2 O?

1 atom of He or 1 mole of He 1 atom per molecule He

1 molecule of O 2 or 1 mole of O 2 2 atoms per molecule O 2

1 molecule of O 3 or 1 mole of O 3 3 atoms per molecule O 3

1. Convert to mass. 2. Convert to moles. 3. Divide by small. 4. Multiply ‘til whole. Percent composition to empirical formula

1) 2) 3) 4) 45. 27 g C, 9. 50 g H, and 45. 23 g O 3. 773 mol C, 9. 50 mol H, and 2. 827 mol O Divide by 2. 827: C 1. 33 H 3. 36 O 1 Multiply by 3: C 4 H 10 O 3 Find the empirical formula: 45. 27% C, 9. 50% H, 45. 23% O