1 dozen =12 1 gross =144 1 ream =500 1 mole = 6.02 x (Avogadro’s Number) A mole is a Very large Number

Molar Mass: -The mass of 1 mole of an element or compound -Atomic mass tells us…. -Atomic mass units per atom (amu) -Grams per mole (g/mol) -Round to 2 decimal places

Examples Carbon - C g/mol Aluminum -Al g/mol Zinc - Zn g/mol Mercury - Hg g/mol

H20H20 2 (1.01) = g/mol NaCl = g/mol NaHCO (16.00) = g/mol C 6 H 12 O 6 6(12.01) + 12(1.01) + 6(16.00) = = = g/mol

Molar Conversions: Mole Atoms or molecules Atoms or molecules Liters Grams 6.02 x L MolarMassMolarMass x ÷

Examples How many moles of Carbon in 26 g of Carbon? 26 g C1 mole g C = 2.2 moles

1.53 moles of LiCl contains how many atoms? 1.53 mol LiCl 1 mol LiCl 6.02 x atoms = 9.21 x atoms LiCl

How many grams of methane gas (CH 4 ) are in 45 L? 45 L CH 4 1 mole 22.4 L1 mole g CH 4 = 32.24g CH 4

Courtesy Percent Composition:

Used to find the makeup of any compound or mixture- %Composition = Mass of desired component Mass of total compound X 100 Examples: % of H in H 2 O?H= 1.01(2) = 2.02 O= = 18.02g = 2.02g H 18.02g H 2 O X100 =11.21 %

Calculate the percentage composition of magnesium carbonate, MgCO 3. Formula mass of magnesium carbonate: g g + 3(16.00 g) = g

Do Now Q: How much copper can be found in a 205g sample of Copper (II) Sulfate? A: Molar mass of CuSO 4 : = 160 g/mol 64 g 160 g X 100 = 40 % 40 % of 205 g = 82 g

Formulas  molecular formula = (empirical formula) n  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

Empirical Formula Determination 1.Base calculation on 100 grams of compound. Determine moles of each element in compound. 2.Divide each value of moles by the smallest of the values. 3.If necessary, Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles

Empirical Formula Determination Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g Divide the molecular mass by the mass given by the emipirical formula.

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

CH 2