Ch 11: The Mole

Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

What is a mole? It is the SI base unit used to measure the amount of a substance Abbreviated mol Represents particles Is also called Avogadro’s number 6.02 x 1023 (3 significant figures)

Remember conversion factors? 12 roses = 1 dozen So, 3.5 dozen= ? Roses = 42 Well, 6.02 x 1023 atoms (or any representative paticles) = 1 mol

Review Scientific Notation Review Rounding

Work practice problems 1-3 Example: 3.50 mol sucrose has how many molecules? Work practice problems 1-3 How many moles are in 3.58 X 1020 atoms of Ca? Work practice problems 4-7

closure How is a mole similar to a dozen? What is the relationship between avagadro’s number & one mole? Why do chemists use moles? **worksheet: The mole & Avogadro’s number STOP

Section 11.2 Mass & the Mole Different particles (atoms) have different masses. Remember atomic mass. Each element has its own specific mass. Therefore each compound has its own specific mass.

Molar mass (g/mol)-mass in grams of any pure substance The molar mass of any element is numerically equal to its atomic mass. Thus…1 mol Mn = 54.94 g/mol Mn = 6.02 x 1023 atoms Mn

Example: 1 mol Zn = 65.4 g/mol 1 mol O2 = 32.0 g/mol

Practice 3 mol Zinc = 196 g/mol Zn 1 mol H2O = 18.0g/mol H2O 1 mol sulfuric acid = 98.1 g/mol H2SO4

Mass  Mole conversions Example: Calculate the mass in grams of 0.0450 mol Cr. Work Practice problems 1-4

Hydrate: CuSO4·5H2O 4) Calculate the mass in grams of 2.45 mol of CaCl2·2H2O

Mole Mass conversions Example: Determine the number of moles for 25.5 g Ag. (mass mol) Work practice problems 5-7. **worksheet: moles & mass

Mass  Atom Conversion Example: How many atoms of gold (Au) are in a pure nugget having the mass of 25.0g? Practice problems 1-5

Atom  Mass Conversion Example A party balloon contains 5.50 x 1022 atoms of helium (He) gas. What is the mass in grams of the helium? Practice problems 6-10.

Mole  Mole Example According to the following balanced equation, how many moles of O2 is produced from 3.00 moles of CuO? 2CuO  2Cu + O2 Practice problems 1 & 2.

Section 11.4 Empirical & Molecular Formula Percent Composition is the percent by mass of each element in a compound. Example: If we had 100 g of a sample of some new compound contains 55g of element X & 45 g of element Y. What is the % of element X & Y?

Mass of element in 1 mol of compound X 100 If we already know the _chemical formula for a compound, you can calculate its percent composition. % by mass= Mass of element in 1 mol of compound X 100 Molar mass of compound

Ex. Determine the percent composition of H2O. (If you had 350. g of water, then how much is oxygen?) Practice problems 1-3. 311g

Empirical Formula is the smallest whole number ratio of the elements. Calculating Empirical formula from percent composition: Directions: The percent should be assumed to be 100 g & converted to moles. Then we figure out the mole ratio by dividing each by the smallest mole.

POEM: Percent to Mass Mass to Mole Divide by small Multiply til whole.

Example 1 The percent composition of an oxide of sulfur is 40.05% S & 59.95% O. Example 2 Determine the empirical formula for methyl acetate which has the following chemical analysis:48.64% C, 8.16% H, & 43.20% O. Practice Problems

**worksheet: determining empirical formulas