ch 11: the mole. section 11.1 measuring matter dozen= 12 ream= 500 pair= 2 gross= 144

24
Ch 11: The Mole

Upload: polly-warren

Post on 21-Dec-2015

222 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

Ch 11: The Mole

Page 2: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

Section 11.1 Measuring Matter

• Dozen=• 12

• Ream=• 500

• Pair=• 2

• Gross=• 144

Page 3: 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)

Page 4: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• 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

Page 5: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Review Scientific Notation

• Review Rounding

Page 6: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Example:• 3.50 mol sucrose has how many molecules?

• Work practice problems 1-3

• Example:• How many moles are in 3.58 X 1020 atoms of Ca?

• Work practice problems 4-7

Page 7: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

closure

• How is a mole similar to a dozen?• What is the relationship between avagadr

o’s number & one mole?• Why do chemists use moles?

• **worksheet: The mole & Avogadro’s number

• STOP

Page 8: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

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.

Page 9: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• 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

Page 10: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Example:

• 1 mol Zn =

65.4 g/mol

• 1 mol O2 =

32.0 g/mol

Page 11: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Practice

• 3 mol Zinc =

196 g/mol Zn

• 1 mol H2O =

18.0g/mol H2O

• 1 mol sulfuric acid =

98.1 g/mol H2SO4

Page 12: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

Mass Mole conversions

• Example:

• Calculate the mass in grams of 0.0450 mol Cr.

• Work Practice problems 1-4

Page 13: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Hydrate: CuSO4·5H2O

4) Calculate the mass in grams of 2.45 mol of CaCl2·2H2O

Page 14: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

Mole Mass conversions

• Example:

• Determine the number of moles for 25.5 g Ag. (mass mol)

• Work practice problems 5-7.

**worksheet: moles & mass

Page 15: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

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

Page 16: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

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.

Page 17: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

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.

Page 18: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

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?

Page 19: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• 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

Page 20: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• Ex. Determine the percent composition of H2O.

• (If you had 350. g of water, then how much is oxygen?)

• Practice problems 1-3.

Page 21: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• 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.

Page 22: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

POEM:

1. Percent to Mass

2. Mass to Mole

3. Divide by small

4. Multiply til whole.

Page 23: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• 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

Page 24: Ch 11: The Mole. Section 11.1 Measuring Matter Dozen= 12 Ream= 500 Pair= 2 Gross= 144

• **worksheet: determining empirical formulas