Chapter 11 : Matter Notes

Mole (mol) is equal to 6.02x10 23 The mole was named in honor of Amedeo Avogadro. He determined the volume of one mole of gas.

Molar Mass: mass in grams of 1 mole of any pure substance. Equal to the atomic mass for an element. Unit is g/mol. Converting moles to Mass: Number of moles number of grams = mass(g) 1 mole

Converting moles to particles. Number of moles_ 6.02x10 23 _ particles = particle 1 mole Example: How many molecules are in 3.50 moles of sucrose? 3.50 mol sucrose 6.20 x molecules of sucrose 1 mole sucrose = 2.11 x10 23 molecules of sucrose

Converting Particles to Moles: Number of particles 1 mole = number of moles 6.02 x10 23 particles Example: How many moles are in 4.5 x atoms of Zinc? 4.5 x atoms Zn 1 mole Zn 6.02 x10 23 atoms Zn = 7.48 moles Zn

Converting Moles to Mass: Element: How many grams are in 3 moles of Manganese? 3.00 mol Mn54.9g Mn= 165g Mn 1 mol Mn

Converting Moles to Mass: Compound: How many grams are in 3 moles of K 2 CrO 4 ? 1. Find molar mass of compound. 2 mol K g K = g K 1 mol K 1mol Cr 52.00g Cr = g Cr 1 mol Cr 4 mol O g O = g O 1mol O g K 2 CrO 4

Converting Moles to Mass: Compound: 1. Multiply moles by molar mass: 3 moles K 2 CrO g K 2 CrO 4 mol K 2 CrO 4 = g K 2 CrO 4

Converting Mass to Moles: Number of grams1 mole = moles mass (g) Element: How many moles are in 525g of Ca? 525g Ca1 mol Ca = 13.1 mol Ca 40.06g Ca

Converting Mass to Moles: Compound: How many moles are in 325g of Calcium Hydroxide [Ca(OH) 2 ]? 1. Find the molar mass of compound: 1mol Ca 40.08g Ca = g 1mol Ca 2mol O 16.00g O = g 1mol O 2mol H 1.008g H = 2.026g 1mol H g Ca(OH) 2

Converting Mass to Moles: Compound: 1. Convert grams to moles: 325g Ca(OH) 2 1mol Ca(OH) g Ca(OH) 2 = 4.39 mol Ca(OH) 2

Converting Mass to Particles: Number of grams 1 mol 6.02 x particles mass(g) 1 mol = particles Element: How many atoms are in 25 g of Gold? 25g Au 1 mol Au 6.02 x atoms Au g Au 1 mol Au = 7.65 x atoms Au

Converting Mass to Particles: Compound: How many formula units are in 35.6 g of aluminum chloride? [AlCl 3 ] 1. Find the molar mass of compound: 1 mol Al g Al = g 1mol Al 3 mol Cl 35.45g Cl= g 1 mol Cl g AlCl 3

Converting Mass to Particles: Compounds: 1. Convert grams to formula units: 35.6 g AlCl 3 1 mol AlCl x formula units g AlCl 3 1 mol AlCl 3 = 1.61 x10 23 formula units AlCl 3

Chapter 11 Matter Notes Section 11.4

Percent Composition: Percent composition is the percent by mass of any element in a compound. Mass of element X 100% = percent composition Mass of compound

Percent Composition: Example: Calculate the percent composition of water. H 2 O 2 mol H 1.01 g H =2.02 g H 1 mol H g H 2 O 1 mol O g O = g O 1 mol O g H 2 O

Percent Composition: con’t g H H X 100% = 11.2% H 18.02g H 2 O g O 88.8% O O X 100% = 88.8% O g H 2 O

Empirical Formula Empirical formula is a formula with the smallest whole-number ratio of the element. Example: Determine the empirical formula for methyl acetate which has 48.64% Carbon, 8.16% Hydrogen, % Oxygen.

Empirical Formula: con’t. 1. Find the number of moles of each element 48.64g C 1 mol C = 4.05 mol C g C 8.16 g H 1 mol H = 8.10 mol H g H g O 1 mol O = 2.7 mol O g O

Empirical Formula: con’t. 1. Divide each mole by the smallest number: 4.05 mol C = 1.5 mol C mol H = 3 mol H mol O = 1 mol O 2.7

Empirical Formula: con’t. 1. Multiply each mole by the smallest number that will produce whole numbers: 1.5 mol C x 2 = 3 C 3 mol H x 2 = 6 H 1 mol O x 2 = 2 O 1. Use the new whole numbers as subscripts for each element in the compound. C 3 H 6 O 2

Molecular Formula: Experimental molar mass = n empirical formula mass Determine the molecular formula for a compound composed of 40.68% carbon, 5.08 % Hydrogen, and % Oxygen which has a molar mass of g/mol.

Molecular Formula: 1. Find the empirical formula: g C 1 mol C = mol C g C 5.08 g H 1 mol H = 5.04 mol H g H g O 1 mol O = 3.39 mol O g O

Molecular Formula: ( finding the empirical formula con’t.) mol C = 1 mol C x 2 = mol H = 1.5 mol H x 2 = mol O = 1 mol O x 2 = empirical formula: C 2 H 3 O 2

Molecular Formula: 1. Find the molar mass of the empirical formula: 2 mol C g C = 24.2 g 1 mol C 3 mol H g H = g 1 mol H 2 mol O g O = 32.00g 1 mol O g C 2 H 3 O 2

Molecular Formula: 1. Divide the molar mass given in the problem by the molar mass of the empirical formula to find n: n = g 59.04g n = 2

Molecular Formula: 1. Multiply all subscripts of the empirical formula by n: (C 2 H 3 O 2 ) 2 molecular formula: C 4 H 6 O 4