Unit 4 COUNTING PARTICLES Counting By Weighing We

Unit 4 COUNTING PARTICLES

Counting By Weighing We can weigh a large number of the objects and find the average mass. Once we know the average mass we can equate that to any number of the objects. EXAMPLE: ◦ The average mass of a book is 40. 0 grams. ◦ How many books are present in a sample with a mass of 2000. 0 grams? ◦ 2000. 0 g/40. 0 g = 50. 0 books 6 -2

Counting By Weighing When we know the average mass of the atoms of an element, as that element occurs in nature, we can calculate the number of atoms in any given sample of that element by weighing the sample. The atomic mass of an element, as found on the periodic table, allows us to count by weighing. 6 -3

Avogadro’s Number is the number of atoms in 12. 01 grams of carbon. Its numerical value is 6. 022 × 1023. Therefore, a 12. 01 g sample of carbon contains 6. 022 × 1023 carbon atoms.

The Mole The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro’s number of particles, that is 6. 022 × 1023 particles. 1 mol = Avogadro’s Number = 6. 022 × 1023 units We can use the mole relationship to convert between the number of particles and the mass of a substance.

Section 6. 3: The Mole The mole (mol) is known as the “chemists dozen” and represents Not this type of mole! 6. 022 x 1023 things (atoms, particles, molecules, etc). • A mole (from the Latin - “lump of stuff”), was originally defined as 1. 0 g of the lightest element known – hydrogen. 6 -6

How Big Is a Mole? The volume occupied by one mole of softballs would be about the size of the Earth. One mole of Olympic shot put balls has about the same mass as the Earth.

Molar Mass The atomic mass of any substance expressed in grams is the molar mass of that substance. The atomic mass of iron is 55. 85 amu. Therefore, the molar mass of iron is 55. 85 g/mol (grams in a mole) Since oxygen occurs naturally as a diatomic, O 2, the molar mass of oxygen gas is 2 times 16. 00 g or 32. 00 g/mol.

Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of magnesium nitrate, Mg(NO 3)2? The sum of the atomic masses is: 24. 31 + 2(14. 01 + 16. 00) = 24. 31 + 2(62. 01) = 148. 33 amu The molar mass for Mg(NO 3)2 is 148. 33 g/mol.

Molar Mass (g/mol) • The molar mass of any substance is the mass (in grams) of one mole of the substance. • The molar mass of a compound is obtained by summing the masses of ALL component atoms. 6 -10

Molar Mass: Practice 1) Calculate the following molar masses: a) Water – H 2 O b) Ammonia – NH 3 c) Propane – C 3 H 8 d) Glucose – C 6 H 12 O 6 6 -11

Mole Calculations We will be using Dimensional analysis again. Recall: ◦ First we write down the unit asked for ◦ Second we write down the given value ◦ Third we apply unit factor(s) to convert the given units to the desired units

Mass-Mole Calculations What is the mass of 1. 33 moles of titanium, Ti? We want grams, we have 1. 33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = 47. 88 g Ti 1. 33 mole Ti × = 63. 7 g Ti 1 mole Ti

Mole Calculations Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance. 6. 02 × 1023 particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles and than we can convert moles to mass.

Mole Calculations How many sodium atoms are in 0. 120 mol Na? ◦ Step 1: we want atoms of Na ◦ Step 2: we have 0. 120 mol Na ◦ Step 3: 1 mole Na = 6. 02 × 1023 atoms Na 0. 120 mol Na × 6. 02 × 1023 atoms Na 1 mol Na = 7. 22 × 1022 atoms Na

Mole Calculations I How many moles of potassium are in 1. 25 × 1021 atoms K? ◦ Step 1: we want moles K ◦ Step 2: we have 1. 25 × 1021 atoms K ◦ Step 3: 1 mole K = 6. 02 × 1023 atoms K 1. 25 × 1021 atoms K × 1 mol K 6. 02 × 1023 atoms K = 2. 08 × 10 -3 mol K

Mole Calculations What is the mass of 2. 55 × 1023 atoms of lead? We want grams, we have atoms of lead. Use Avogadro’s number and the molar mass of Pb 2. 55 × 1023 1 mol Pb 207. 2 g Pb atoms Pb × × 23 1 mole Pb 6. 02× 10 atoms Pb = 87. 8 g Pb

Mole Calculations How many O 2 molecules are present in 0. 470 g of oxygen gas? We want molecules O 2, we have grams O 2. Use Avogadro’s number and the molar mass of O 2 1 mol O 2 6. 02× 1023 molecules O 2 0. 470 g O 2 × × 1 mole O 2 32. 00 g O 2 8. 84 × 1021 molecules O 2

How many Oxygen atoms in 8. 84 × 1021 molecules O 2? 8. 84 × 1021 molecules O 2 x 2 oxygen atoms = molecule

Mole Video

6. 022 X 1023 pennies: would make at least 7 stacks that would reach the moon. 6. 02 X 1023 blood cells: would be more than the total number of blood cells found in every human on earth. A mol is A LOT of particles: 602, 200, 000, 000

The Mole • It is VERY important to understand that the value listed as the mass number on the periodic table really tells us TWO things: – The mass of one atom in units of amu’s. – The mass of one mole of atoms in grams. – One mol of Al = 26. 98 g & one atom = 26. 98 amu – One mol of Au = 196. 97 g & one atom = 196. 97 amu – One mol of B = 10. 81 g & one atom = 10. 81 amu –Remember one mol = 6. 022 X 1023 atoms (or particles or molecules or donuts) 6 -22

• Remember: can talk about one mole of atoms or Molar we Mass one mole of molecules. • One mole of oxygen atoms (O) weighs 16. 00 g. • One mole of oxygen molecules (O 2) weighs 32. 00 g. • Two moles of O atoms weigh 32. 00 g. • Two moles of O 2 molecules weigh 64. 00 g. • And so on. . . 6 -23

The Mole • To summarize: a sample of any element that weighs a number of grams equal to the molar mass (from the periodic table) contains 6. 022 x 1023 atoms (1 mol) of that element. • Therefore, atomic weight of an element = #g in one mol of that element • We write that as g/mol. 6 -24

The Mole: Practice 1) Calculate the number of moles in a 25. 0 g sample of calcium. How many atoms are there? 2) Calculate the number of moles and the number of atoms in a 57. 7 g sample of sulfur. 3) Calculate the number of atoms in a 23. 6 g sample of zinc. 4) Calculate the number of molecules in a 12. 83 g sample of calcium carbonate. 6 -25

1. . 624 mol and 3. 76 x 1023 atoms 2. 1. 80 mol and 1. 08 x 1024 atoms 3. 3. 61 x 10 -4 mol and 2. 17 x 1020 atoms 4. 1. 189 x 10 -3 mol and 7. 160 x 1020 atoms

Molar Mass: Practice 1. Calculate the number of moles of H 2 CO 3 present in a 7. 55 g sample. 2. How many water molecules are present in 10. 0 g of water? (Hint: find moles first) 3. How many molecules of sucrose (C 12 H 22 O 11) are present in a 5. 0 kg bag of sugar? (Hint: find moles first) 6 -27

Representative Particles= atoms, molecules, Items, etc

Percent Composition of Compounds • Sometimes it is not enough to know a compound’s composition in terms of numbers of atoms; it may also be useful to know its composition in terms of the masses of its elements. • We can calculate the mass fraction by dividing the mass of a given element in one mole of a compound by the mass of one mole of the compound. 6 -29

Percent Composition of Compounds • Once we know the mass fraction we can multiply by 100 to get the percent. • Remember: percent = part/whole X 100. 6 -30

In one mole of methane (CH 4), there is one mole of C and four moles of H: 1 mol C = 12. 01 g 4 mol H = 4(1. 01) = 4. 04 g 1 mol CH 4 = 12. 01 g + 4. 04 g = 16. 05 g %C= mass of 1 mol C x 100 mass of 1 mol CH 4 %C = 12. 01 X 100 = 74. 83 %C 16. 05 %H= mass of 4 mol H x 100 mass of 1 mol CH 4 %H = 4. 04 16. 05 X 100 = 25. 17 %H

Percent Composition Practice Determine the mass percent of each element in the following: ◦ H 2 SO 4 (sulfuric acid) ◦ C 3 H 7 OH (isopropyl alcohol) ◦ C 6 H 12 O 6 (glucose) 6 -32

Formulas of Compounds • The object of this section is to do the opposite of the previous section. Instead of getting the mass from the formula, we will determine the formula from the mass. • To do this, the mass must be converted to moles using each element’s mass number. How can we convert mass to moles? 6 -33

Formulas of Compounds ex) An unknown compound with a mass of 0. 2015 g is is found to contain: ◦ 0. 0806 g C ◦ 0. 01353 g H ◦ 0. 1074 g O It must contain: 0. 0806 g (1 mol C/12. 01 g C) = 0. 00671 mol C 0. 01353 g(1 mol H/1. 008 g H) = 0. 01342 mol H 0. 1074 g(1 mol/16. 00 g O) = 0. 00671 mol O 6 -34

The numbers from the previous slide allow us to determine the C: H: O ratio. 0. 00671 (C) : 0. 01342 (H) : 0. 00671 (O) If we divide each number by the smallest number we get 1: 2: 1 for the C: H: O ratio. This leads us to a formula of C 1 H 2 O 1 or CH 2 O. This is not necessarily the TRUE formula of the compound, but represents the RELATIVE numbers of atoms. 6 -35

This represents the lowest whole number ratio of the compound. The actual formula could be CH 2 O, C 2 H 4 O 2, C 3 H 6 O 3, C 4 H 8 O 4, C 5 H 10 O 5, C 6 H 12 O 6, . . . Any formula with a C: H: O ratio of 1: 2: 1 is possible (in theory, an infinite number). C 1 H 2 O 1 represents the simplest possible formula or the EMPIRICAL FORMULA. The multiples represent possible MOLECULAR FORMULAS. 6 -36

Formulas of Compounds: Practice Determine the empirical formula from each of the following molecular formulas: ◦ H 2 O 2 (hydrogen peroxide) ◦ C 4 H 10 (butane) ◦ CCl 4 (name? ) ◦ HC 2 H 3 O 2 (acetic acid) ◦ C 6 H 12 O 6 (glucose) 6 -37

Calculation of Empirical Formulas There are four steps to determine the empirical formula of a compound: 1. Obtain the mass of each element present (in grams). 2. Determine the number of moles of each type of atom present (use atomic mass). 3. Divide each number by the smallest number. 4. Multiply all numbers by the smallest integer that will make them all integers 6 -38

Empirical Formulas: Practice 1) A 1. 500 g sample of a compound containing only carbon and hydrogen is found to contain 1. 198 g of carbon. Determine the empirical formula for this compound. 2) A 3. 450 g sample of nitrogen reacts with 1. 970 g of oxygen. Determine the empirical formula for this compound. 3) When a 2. 000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2. 573 g. Determine the empirical formula for this compound. 6 -39

If the relative amounts of elements are presented as percentages, assume we are starting with a 100 g sample (100%). Then each percentage simply becomes a mass (in grams). For example if 15% of a compound is carbon, we just assume it is 15 g of a 100 g sample; from there we convert to moles. 6 -40

Empirical Formulas: More Practice 1) The simplest amino acid, glycine, has the following mass percents: 32. 00% carbon, 6. 714% hydrogen, 42. 63% oxygen, and 18. 66% nitrogen. Determine the empirical formula for glycine. 2) A compound has been analyzed and found to have the following mass percent composition: 66. 75% copper, 10. 84% phosphorous, and 22. 41% oxygen. Determine the empirical formula for this compound. 6 -41

Calculation of Molecular Formulas If we know the empirical formula AND the molar mass, we can calculate a compound’s molecular formula. Note: without the molar mass, the best you can find is the empirical formula. Once the molar mass is known, one must ALWAYS find the empirical formula before one can calculate the molecular formula. It is impossible to do the reverse. 6 -42

Calculation of Molecular Formulas It is also important to note that the molecular formula is ALWAYS an integer multiple of the empirical formula. We can represent the molecular formula in terms of the empirical formula: (Empirical Formula)n = molecular formula It should also be noted when n = 1, the empirical and molecular formulas are identical to each other. 6 -43

Molecular Formulas: Practice 1) A compound containing carbon, hydrogen, and oxygen is found to be 40. 00% carbon and 6. 700% hydrogen by mass. The molar mass of this compound is between 115 and 125 g/mol. Determine the molecular formula for this compound. 2) Caffeine is composed of 49. 47% C, 5. 191% H, 28. 86% N, and 16. 48% O. The molar mass is about 194 g/mol. Determine the molecular formula for caffeine. 6 -44