Ch. 6 – Chemical Quantities The Mole What is a mole? It is a unit for _________in chemistry. It is similar to a dozen, except instead of 12 things, it’s 602 billion trillion things… (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 equal volumes of gases under the same conditions of temperature and pressure contained the same number of particles. Avogadro 6.02 x counting

A mole is a term for a certain ______________ of objects. 1 mole = 6.02 x objects *Other words that represent quantities: 1 pair = __ objects; 1 dozen = __ objects 1 gross = ____ objects; 1 _____ = 24 cans of soda Since this value is so huge, it is used to measure very small objects like ___________ and _______________. The Mole Concept number case atomsmolecules

Just How Big is a Mole? It’s 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 Mole is abbreviated ______. The Mole x x x mol

Gram-Formula Mass The # of grams that 6.02 x particles, (or ___ mole), weighs is called the gram-formula mass. The mass is found from the atomic masses of the elements on the ____________ __________. This quantity has other nicknames too! ________ ______or _________ ________ or Formula Weight 1 periodic table Practice Problems: Calculate the gram-formula mass of each compound. a) CaCO 3 b) (NH 4 ) 2 SO 4 c) N 2 O 5 Molar MassFormula Mass 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 2 N’s = 2 x 14.0 = O’s = 5 x 16.0 = 80.0 Add them up! 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= 1.24 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 molar 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 molar 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

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. C1H2O1C1H2O1

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 is composed of 58.8% carbon and 9.8% hydrogen and 31.4% oxygen. The molecular formula mass is 204 g. Determine the molecular formula for the compound. C = 58.8% = 58.8 g H = 9.8% = 9.8 g (mass to moles) 58.8 g C ÷ 12.0 = 4.9 moles C 9.8 g H ÷ 1.0 = 9.8 moles H (÷ by small) C 4.9 H 9.8 O C 2.5 H 5 O 1 (x ‘til whole) x 2 x 2 x 2 C 5 H 10 O 2 O = 31.4% = 31.4 g31.4 g O ÷ 16.0 = 1.96 moles O 1.96 Our formula mass = 102 g (Compare) (204 ÷ 102 = 2) [C 5 H 10 O 2 ] x 2 = C 10 H 20 O 4

Practice Problems: 2) An unknown compound is composed of 40% carbon, 6.6% hydrogen, and 53.4% oxygen. Determine the molecular formula for the compound if the mass of one mole of the compound is 120 g. C = 40% = 40 g H = 6.6% = 6.6 g (mass to moles) 40 g C ÷ 12.0 = 3.3 moles C 6.6 g H ÷ 1.0 = 6.6 moles H (÷ by small) C 3.3 H 6.6 O C1H2O1C1H2O1 O = 53.4% = 53.4 g53.4 g O ÷ 16.0 = 3.3 moles O 3.3 Our formula mass = 30 g (Compare) (120 ÷ 30 = 4) [C 1 H 2 O 1 ] x 4 = C4H8O4C4H8O4