The Mathematics of Chemistry Mole Concept Kenneth E. Schnobrich

Counting Things It is relatively easy to count the number of things in: A pair of soxs A dozen oranges A gross of paper clips Atoms, molecules, and ions are extremely small units of matter and visually it would be impossible to count them in any sample we are given. Experimentally, we can determine the number of atoms in a given sample indirectly

Avogadro’s Number It has been determined that the number of atoms contained in a mass equivalent to the atomic mass of an element, expressed in grams = 6.02 x 1023 atoms of that element. The term used for 6.02 x 1023 “things” is a MOLE (or MOL). It is also referred to as Avogadro’s Number (it is unitless).

An Example There are 1 x 1028 atoms in the average sized person (an Octillion) If the atom were the size of a pea, an Octillion peas would cover the complete surface of 250,000 planets, each the size of Earth, to a depth of 4 feet The atom is an incredibly small unit of matter.

Some Mole Relationships AMOUNT # OF MOLES 7 grams of Li 1 mole of Li 28 grams of Fe 0.5 moles of Fe 8 grams of Mg 0.25 moles of Mg 1.204 x 1024 Na atoms 2 moles of Na 3.01 x 1023 S atoms 0.5 moles of S

Doing the Conversion Always start with what you are given: 28 g Fe 1 mole Fe 56 g Fe 0.5 mole Fe = 1.204 x 1024 atoms Na 6.02 x 1023 atoms Na 1 mole Na = 2 mole Na

Using Table T* On Table T a formula is given for calculating the number of moles - let’s use it for a calculation: # of moles (n) = given mass/gram formula mass* # of moles (n) = given mass/gram atomic mass** *Gram formula mass - mass of the compound in grams **Gram atomic mass - mass of the element in grams Given 8.0 grams of Mg how many moles do you have? #moles(n) = 8.0 grams/24.0 grams #moles(n) = 0.25 *NYS Reference Tables for Chemistry

Determining Formula Mass Element Atomic Mass # atoms Total Mass Na 23 1 H S 32 O4 16 4 64 NaHSO4 Total Mass (1 mole) 120

Determining Formula Mass Element Atomic Mass # atoms Total Mass C 12 6 72 H 1 O 16 96 C6H12O6 Total Mass (1 mole) 180

Using Formula Mass Given 90. grams of C6H12O6 how many moles in the sample? #moles (n) = given mass/ gram formula mass #moles (n) = 90. grams/ 180 grams/mole = 0.50 moles

Mixing It Up A Bit Substance Mass (g) #moles CaCO3 300. (NH4)2SO4 0.50 Cr(NO3)3 150. CuSO4•5H2O 2.0 Sr(OH)2 50.

At STP one mol of any gas occupies a volume of 22.4 L Mols and Gases At STP one mol of any gas occupies a volume of 22.4 L So, if we had 2 mols of H2 gas at STP it would occupy 44.8 Liters PROBLEM: If a sample of N2 gas at STP occupies a volume of 67.2 Liters, how many moles of the gas do you have in the sample? # mols = volume given/volume of one mol at STP # mols = 67.2 Liters/22.4 Liters/mol = 3 mols PROBLEM: How many grams of N2 gas are there in the sample? #g = #mols x #g/one mol # g = 3 mol (28 g/mol) = 84 grams

*NYS Reference Tables for Chemistry Percent Composition These problems are worked like any percentage - if you miss 25 out of 50 questions you have a 50%. % = (mass of part/mass of whole) x 100 This equation can be found on Table T* If I wanted to know the percent by mass of hydrogen in water %H = (2.00 g H/18.0 g H2O) x 100 = 11.2% *NYS Reference Tables for Chemistry

% Composition Examples Compound %C %H %O C6H12O6 CH4 No Oxygen C2H5OH CH3COOH

% H2O in a Hydrate Certain substances have water molecules associated with their structure. Many salts fall into this category and we refer to them as hydrated salts. Different salts have different numbers of of water molecules - BaCl2•2H2O; CuSO4•5H2O. PROBLEM: Calculate the % by mass of water in one mole of BaCl2•2H2O %H2O = (mass of H2O in one Mol/mass of BaCl2•2H2O) x 100 %H2O = (36 grams/243 grams) x 100 = 14.8% H2O

Empirical & Molecular Formulas Empirical formulas represent the smallest whole number ratio of atoms in a compound (cannot be reduced any further) Molecular (or True) formula represents the actual whole number ratio of atoms in a compound (it will be a small whole number multiple of the Empirical formula)

Empirical & Molecular Formulas CH2O would be the empirical formula of a typical monosaccharide C6H12O6 would represent the molecular or true formula of the monosaccharide 6 x (CH2O) - each of the subscripts would be multiplied by 6

Simple EF/MF Problems Vitamin C has an empirical formula of C3H4O3 and a molecular mass of 176 amu. What is the molecular formula? First determine the mass of the empirical formula: (3 x 12) + (4 x 1) + (3 x 16) = 88 Now divide the molecular mass by the empirical mass: 176/88 = 2 Now multiply each of the subscripts in the empirical formula by 2: C6H8O6

Some Problems Empirical Formula Molecular Mass (amu) Molecular Formula CH2 70 CH 78 28 CH3 30

Moles in Equations In chemical reactions it is important that equations be balanced and the mole ratio taken into account. For this unit we will provide the balanced equation 2C2H6 + 7O2 -> 4CO2 + 6H2O The coefficients represent the # of moles of each substance.

Moles in Equations Regardless of the number of moles of any reactant or product, the ratio must always remain - 2:7:4:6 for this reaction 2C2H6 + 7O2 -> 4CO2 + 6H2O Problem: Starting with 1 mole of C2H6 how many moles of H2O will be produced? Divide the coefficients by 2 The new ratio is - 1:7/2:2:3 Therefore we will produce 3 moles of H2O starting with 1 mole of C2H6

Other Mole Relationships Let’s use another reaction and take a slightly different approach: 2KClO3 = 2KCl + 3O2 Problem: 3.5 moles of KClO3 will produce how many moles of O2? Place the given # of moles over the substance Place an X over the substance in question Below the substance given indicate the # of moles from the balanced equation Below the substance you are looking for place the number of moles from the balanced equation

Mole Relationships(cont.) 3.5 moles X 2KClO3 = 2KCl + 3O2 2 moles 3 moles Now setup a proportion 3.5 moles 2 moles 3 moles X = X = 5.25 moles O2

Mole Relationships(cont.) 35 grams X 2KClO3 = 2KCl + 3O2 2 (122.5 grams) 3 moles K = 39 Cl = 35.5 O = 16 (3) = 48 Now setup a proportion 35 grams 245 grams 3 moles X = 122.5 X = 0.43 moles O2

Mole Relationships(cont.) 35 grams X 2KClO3 = 2KCl + 3O2 2 (122.5 grams) 3 (32 grams) K = 39 Cl = 35.5 O = 16 (3) = 48 Now setup a proportion 35 grams 245 grams 48 grams X = 122.5 X = 6.86 grams O2

Same Mole Problem Using The Factor Label Method 2KClO3 = 2KCl + 3O2 35 g KClO3 mol KClO3 3 mol O2 32 g O2 1 122.5 g 2 mol KClO3 mol O2 KClO3 Answer = 6.86 gram of O2