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

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 Avogadro 6.02 x 10 23

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.

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 X X X MASS mol

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 g/mol = __________ g/mol CaCl 2 1 mole of CaCl 2 = g/mol the sum

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 Ca = 40.1 C = O’s =3 x 16.0 = 48.0 Add them up! g/mol 2 N’s = 2 x 14.0 = H’s = 8 x 1.0 = 8.0 S = O’s = 4 x 16.0 = 64.0 Add them up! 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 = g/mol

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 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 g SO 3 x= 10.4 moles of SO 3 1 mole CH x molecules CH 4 18 moles CH 4 x= 108 x molecules CH grams He 5.6 x atoms He x 3.72 grams He 18.0 grams H 2 O 3.7 grams H 2 O x= 1.23 x molecules H 2 O or 1.08 x molecules CH x atoms He = 6.02 x molecules H 2 O

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 = O’s = 4 x 16.0 = ÷ 164 = = 42.1% = = 18.9% = = 39.0% Sn = Cl’s = 4 x 35.5 = ÷ = 45.5% = 54.5%

Elements in the Universe: % Composition by Mass

Earth’s Crust: % Composition by Mass

Entire Earth (Including Atmosphere): % Composition by Mass

Human Body: % Composition by Mass

Meet The Elements

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. Step 1: Divide the % composition data by the atomic mass of the element. This will give you a ratio of the # of atoms in the formula. Step 2: Divide each of these answers by the smallest ratio. Step 3: If there is still a decimal, multiply each answer by the denominator of the “freak”, (i.e. -- multiply all the ratios by the denominator of the ratio that is still a decimal.) [1/2= 0.5 1/3≈ /3≈ /4= 0.75] whole ratio Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole. CH 2 O

Practice Problems: 1) An unknown compound is composed of 81.8% carbon and 18.2% hydrogen. Determine the empirical formula for the compound. 2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound. C = 81.8% = 81.8 g H = 18.2% = 18.2 g (mass to moles) 81.8 g C ÷ 12.0 = 6.82 moles 18.2 g H ÷ 1.0 = 18.2 moles (÷ by small) C 6.82 H C 1 H 2.67 (x ‘til whole) x 3 C3H8C3H8 C = 42.9% = 42.9 g O = 57.1% = 57.1 g (mass to moles) 42.9 g C ÷ 12.0 = moles 57.1 g O ÷ 16.0 = moles (÷ by small) C O C 1.0 O 1.0 CO

2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound. C = 42.9% = 42.9 g O = 57.1% = 57.1 g (mass to moles) 42.9 g C ÷ 12.0 = moles C 57.1 g O ÷ 16.0 = moles O (÷ by small) C O C 1.0 O 1.0 CO To check your answer, you can simply find the % composition by mass of your formula… % C = 12.0/28 = 42.9% %O = 16.0/28 = 57.1% CO = 28 g/mole That matches up with the original problem!!

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!) Step 1: Determine the empirical formula for the compound. (See the previous steps in the notes.) Step 2: Calculate the empirical formula mass of the compound. Step 3: Determine the “whole # multiple” by dividing the molecular formula mass (given in the problem) by the empirical formula mass. Multiply each of the empirical ratios by this whole number. whole # multiple 6

Practice Problems: 1) An unknown compound has an empirical formula of C 5 H 10 O 2. The molecular formula mass is 204 g. Determine the molecular formula for the compound. Our formula mass = 102 g (Compare) (204 ÷ 102 = 2) [C 5 H 10 O 2 ] x 2 = C 10 H 20 O 4