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Ch. 6 Notes -- Chemical Composition What is a mole? Mole is a unit of quantity. Like a _____________. 1 mole = 6.02x10 23 atoms or molecules = “X” grams.

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Presentation on theme: "Ch. 6 Notes -- Chemical Composition What is a mole? Mole is a unit of quantity. Like a _____________. 1 mole = 6.02x10 23 atoms or molecules = “X” grams."— Presentation transcript:

1 Ch. 6 Notes -- Chemical Composition What is a mole? Mole is a unit of quantity. Like a _____________. 1 mole = 6.02x10 23 atoms or molecules = “X” grams = 22.4 L gas Dozen

2 The Mole!!! A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion… (602,000,000,000,000,000,000,000) ___________ (in scientific notation) Amedeo (1776 – 1856)This number is named in honor of Amedeo _________ (1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present…6.02 x 10 23 Avogadro 6.02 x 10 23

3 Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of un- popped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

4 1 dozen cookies = ___ cookies 1 mole of cookies = ___________ cookies 1 dozen cars = ___ cars 1 mole of cars = __________ cars 1 dozen Al atoms = ___ Al atoms 1 mole of Al atoms = __________ atoms Note that the NUMBER is always the same, but the ______ is very different! Mole is abbreviated ______. The Mole 12 6.02 X 10 23 12 6.02 X 10 23 12 6.02 X 10 23 MASS mol

5 The Mole and Mass Mass in grams of 1 mole equal to __________ of the atomic masses Practice problem: Calculate the mass of 1 mole of CaCl 2 Ca = 1 x ________ g/mol = 40.1 g/mol Cl = 2 x ________ g/mol = 71.0 g/mol 40.1 g/mol + 71.0 g/mol = __________ g/mol CaCl 2 1 mole of CaCl 2 = 111.1 g/mol the sum 40.1 35.5 111.1

6 Ch. 7 Notes -- Chemical Quantities Practice Problems: (1) How many atoms of hydrogen are there in each compound? a) Ca(OH) 2 ___ b) C 3 H 8 O___ c) (NH 4 ) 2 HPO 4 ___ d) HC 2 H 3 O 2 ___ (2) Calculate the formula mass of each compound. (Add up all the atomic masses for each atom from the Periodic Table.) a) CaCO 3 b) (NH 4 ) 2 SO 4 c) C 3 H 6 Od) Br 2 2894 Ca = 40.1 C = 12.0 3 O’s =3 x 16.0 = 48.0 Add them up! 100.1 g/mol 2 N’s = 2 x 14.0 = 28.0 8 H’s = 8 x 1.0 = 8.0 S = 32.1 4 O’s = 4 x 16.0 = 64.0 Add them up! 132.1 g/mol C = 3 x 12.0 = 36.0 H = 6 x 1.0 = 6.0 O =16.0 Add them up! 58.0 g/mol 2 Br’s = 2 x 79.9 = 159.8 g/mol

7 3) Convert 835 grams of SO 3 to moles. 4) How many molecules of CH 4 are there in 18 moles? 5) How many grams of helium are there in 5.6 x 10 23 atoms of helium? 6) How many molecules are there in 3.7 grams of H 2 O? 80.1 g SO 3 1 mole SO 3 835 g SO 3 x= 10.4 moles of SO 3 1 mole CH 4 6.02 x 10 23 molecules CH 4 18 moles CH 4 x= 4.0 grams He 5.6 x 10 23 atoms He x 3.72 grams He 18.0 grams H 2 O 3.7 grams H 2 O x= 1.23 x 10 23 molecules H 2 O 1.08 x 10 25 molecules CH 4 6.02 x 10 23 atoms He = 6.02 x 10 23 molecules H 2 O

8 Calculating Percent Composition by Mass Step 1: Find the formula mass of the compound by adding the individual masses of the elements together. Step 2: Divide each of the individual masses of the elements by the formula mass of the compound. Step 3: Convert the decimal to a % by multiplying by 100. Practice Problems: (1) Find the % composition of the elements in each compound. a) Na 3 PO 4 b) SnCl 4 3 Na’s = 3 x 23.0 = 69.0 P = 31.0 4 O’s = 4 x 16.0 = 64.0 + 164 ÷ 164 = 0.421 = 42.1% = 0.189 = 18.9% = 0.390 = 39.0% Sn = 118.7 4 Cl’s = 4 x 35.5 = 142.0 + 260.7 ÷ 260.7 = 45.5% = 54.5%

9 Elements in the Universe: % Composition by Mass

10 Earth’s Crust: % Composition by Mass

11 Entire Earth (Including Atmosphere): % Composition by Mass

12 Human Body: % Composition by Mass

13 Meet The Elements

14 Determining the Empirical Formula for a Compound The empirical formula for a compound is the simplest __________ number __________ of the atoms in the compound. Examples: H 2 O is the empirical formula for water. _______ is the empirical formula for glucose, C 6 H 12 O 6. Practice Problems: What is the empirical formula for the following compounds? a) C 6 H 6 = ________ b) C 8 H 14 O 2 = ________ c) C 10 H 14 O 2 = _________ d) Ca 5 Br 10 = ________ e) N 3 O 9 = ________ whole ratio C1H2O1C1H2O1 CH C4H7OC4H7O C5H7OC5H7O CaBr 2 NO 3

15 Determining the Molecular Formula for a Compound The molecular formula for a compound is either the same as the empirical formula ratio or it is a “_________ _________ of this ratio. It represents the true # of atoms in the molecule. Examples: 1) H 2 O is the empirical & molecular formula for water. 2) CH 2 O is the empirical formula for sugar, ethanoic acid, and methanol. The molecular formula for glucose is C 6 H 12 O 6, (___times the empirical ratio!) Practice Problems: (1) If the empirical formula for a compound is CH 2, which of the following is a possible molecular formula for the compound? a) C 8 H 16 b) C 8 H 8 c) C 4 H 2 d) C 3 H 9 (2) If the empirical formula for a compound is C 2 H 3, which of the following is a possible molecular formula for the compound? a) C 2 H 6 b) C 10 H 15 c) C 6 H 12 d) C 8 H 14 whole # multiple 6


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