602X ,000,000,000, 000,000,000, X Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements
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1 Pre- AP Chemistry Chemical Quan44es: The Mole Mole SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Representa9ve par9cles can be atoms, molecules, formula units, or ions. For example Atoms: Fe, Al, Mg, C, etc. Molecules: H 2 0, CO, SO 2, etc. Formula units: NaCl, KBr, FeSO 4, etc. Ions: Na +, Cl -, SO 4 2-, etc. Diatomic Elements Certain elements are only stable in pairs or with other elements in a compound. These are known as the diatomic elements: Hydrogen (H 2 ); Bromine (Br 2 ); Oxygen: (O 2 ); Nitrogen ( ); Chlorine (Cl 2 ); Iodine (I 2 ); and Fluorine (F 2 ) A mole of substance represents 6.02 x representa9ve par9cles of a par9cular substance. The experimentally determined 6.02 x is called Avogadro s number. 6.02X X ,000,000,000, 000,000,000,000 1
2 **Conversion factor between representa9ve par9cles and moles.** 1 mol = 6.02 x representa9ve par9cles Example: How many atoms are in mol of aluminum? mol Al 6.02x10 23 Al atoms = 1 mol Al 2.23X10 23 Al atoms How many moles are 9.3 x atoms of lead? 9.3 x Pb atoms 1 mol Pb = 6.02x10 23 Pb atoms 1.5x10-8 Pb mol How many molecules are in 4.50 moles of water? 4.50 mol H 2 O 6.02x10 23 H 2 O molec. = 1 mol H 2 O 2.71x10 24 H 2 O molecules How many moles are in 8.5 x molecules of carbon dioxide? 8.5 x molec. CO 2 1 mol CO 2 = 6.02x10 23 CO 2 molec. 140 CO 2 mole How many chloride ions are in mol calcium chloride? mol CaCl 2 2 mol Cl x10 23 Cl - = 9.0 x Cl - ions 1 mol CaCl 2 1 mol Cl - 2
3 Gram Atomic Mass (gam) atomic mass of an element expressed in grams. n The gam is the mass of 1 mole of atoms of a monatomic element n use the periodic table and take masses to 0.1g For example Mg = 24.3 g = mass of 6.02 x atoms 24.3 g/mol is the gram atomic mass of Mg Gram Molecular Mass (gmm) the mass of 1 mol of a molecule. 1 mol CO 2 : For example Sum of the atomic masses of each atom making up the molecule. 1 mol C = 1 x 12.0 g C / mol = 12.0 g 2 mol O = 2 x 16.0 g O/ mol = 32.0 g 44.0 g CO 2 Example: What is the gram molecular mass of ace9c acid, CH 3 COOH? 1 mol CH 3 COOH : 2 mol C = 2 x 12.0 g C / mol = 24.0 g 4 mol H = 4 x 1.0 g H/ mol = 4.0 g 2 mol O = 2 x 16.0 g O/mol = 32.0g 60.0g CH 3 COOH Gram Formula Mass (gfm) mass of one mole of an ionic compound. Sum of the atomic masses of each atom making up the ionic compound 3
4 Example: Recap: What is the gram formula mass of ferric sulfate? Fe 2 (SO 4 ) 3 2 mol Fe 2 x 55.9 g = g 3 mol S 3 x 32.1 g = 96.3 g 12 mol O 12 x 16.0 g = g g ** gam of an element contains 1 mole of atoms ** gmm of a molecular compound contains 1 mole of molecules ** gfm of an ionic compound contains 1 mole of formula units The term molar mass or molecular weight can be used to refer to a mole of an element, molecular compound, formula unit, or ion. Gram formula mass can also be used for either ionic or molecular compounds. Molar Mass of any substance is the mass (in grams) or 1 mole of substance. Example: What is the molar mass of calcium carbonate? 1 mol CaCO 3 = 1 mol Ca = 40.1 g Ca/mol = 40.1g 1 mol C = 12.0 g C/mol = 12.0g 3 mol O = 16.0 g O/mol = g 100.1g CaCO 3 Mole Mass Conversions Conversion Factor! 1 mol = molar mass 4
5 Example: Find the moles in 26.2 g of gold, Au. 26.2g Au =? mol Au Conversion factor: g Au= 1 mol Au Use Dimensional Analysis!!! 26.2 g Au 1 mol Au = mol Au g Au Find the moles in 42.5 g of calcium g Ca 1 mol Ca = 40.1g Ca 1.06 mol Ca Find the mass in grams of 4.35 mol uranium, U mol U g U = 1040g U 1 mol U Find the mass in grams in 2.00 mol iron mol Fe 55.8g Fe = 1 mol Fe 112g Fe Mul4- Step Mole Problems: Calculate the number of atoms present in 1.50g silver. 1.50g Ag 1 mol Ag 6.02x10 23 Ag atoms = 107.9g Ag 1 mol Ag Calculate the number of molecules present in 3.25g diphosphorous pentoxide. 3.25g P 2 O 5 1 mol P 2 O x10 23 P 2 O 5 molec g P 2 O 5 1 mol P 2 O x10 22 P 2 O 5 molecules 8.37x10 21 Ag atoms 5
6 Calculate the mass in grams of 3.00 x nitrogen molecules. 3.00x10 20 molec 1 mol 28.0g = 6.02x10 23 molec. 1 mol g How many atoms of nitrogen are in 1.2 g of aspartame, C 14 H 18 O 5? 1.2g C 14 H 18 O 5 1 mol C 14 H 18 O 5 294g C 14 H 18 O 5 2 mol N 6.02x10 23 N atoms = 1mol C 14 H 18 O 5 1 mol N 4.9x10 21 N atoms The Volume of a Mole of Gas Standard Temperature and Pressure (STP) : 0 o C and 1 atm The volume of a gas is usually measured at STP. At STP, 1 mol of any gas occupies a volume of 22.4 L n 22.4 L is called the molar volume of a gas. n 22.4 L of a gas at STP contains 6.02 x par9cles of the gas. n 22.4 L of a gas has a mass equal to the gfm of the gas. Since 1 mol of any substance contains Avogadro s number of par9cles, 22.4 L of any gas at STP contains 6.02 x par9cles of gas. - - Conversion factor- - 1 mol of any gas at STP = 22.4 L 6
7 Example: What is the volume (in liters at STP) of 2.50 mol of carbon monoxide? 2.50 mol =? L 2.50 mol CO 22.4 L CO = 56.0 L 1 mol CO Determine the number of moles in 625 L H 2 gas at STP? 625 L H 2 1 mol H 2 = 22.4L H mol H 2 Density of a Gas - - Units : g/l (gram per liter) - - We can use the experimentally determined density to calculate the molar mass of a gas at STP (or vice versa) L/mol x density (g/l) = gfm Example: The density of a gaseous compound is g/l at STP. Determine the gfm of this compound L g = g/mol 1 mol 1L Calculate the density of methane gas, CH 4, at STP. CH 4 = (1x12.0g) + (4x1.0g) =16.0 g/mol 16.0 g CH 4 1 mol CH 4 = 1 mol CH L CH 4 Percent Composi4on The percent by mass of each element in a compound is known as percent composi9on. The percent must total 100% g/l CH 4 7
8 Example: Find the % composi4on for a compound that is formed from g of Na and 77.4 g O g Na g O = g total OR solve for percent composi9on from the chemical formula: g Na x 100 = 74.2% Na 300.0g total % mass = grams of element in 1 mol of compound x 100 molar mass of compound 77.4g O x 100 = 25.8% O g total What is the percent composi4on of Mg(NO 3 ) 2. Molar Mass of Mg(NO 3 ) 2 = g/mol % Mg = 24.3g x 100 = 16.4% Mg 148.3g %N = 28.0g x 100 = 18.9% N 148.3g %O = 96.0g x 100 = 64.7% O 148.3g Calcula4ng Empirical Formulas Empirical formula gives the lowest whole number ra4o of the elements in a compound. **May or may not be the same as molecular formula Example: CO 2 is empirical and molecular H 4 is a molecular formula but its empirical formula would be NH 2 Steps to calculate empirical formulas: 1. Find moles of each element 2. Set up mole ra9o 3. Simplify mole ra9o (divide by smallest). Answers must be whole numbers. If they are not, mul9ply by 2, 3, 4, or 5 to get whole numbers. 4. Use mole ra9o as subscripts in the formula If given % composi4on, assume 100g of compound 8
9 What is the empirical formula of a compound that is 27.3% C and 72.7% O? 27.3g C 1 mol C = 2.28 mol C 12.0g C 72.7g O 1 mol O = 4.54 mol O 16.0g O C:O 2.28 : : : 2 CO 2 carbon dioxide Calcula4ng Molecular Formulas Molecular formula actual formulas. They may be the same as its empirical formula or a mul9ple of it. Molecular formulas can be determined from its empirical formula and molar mass. Steps to calculate molecular formula: 1. Calculate empirical formula mass, efm. This is simply the molar mass of the empirical formula. 2. Divide gfm, gram formula mass (given), by the efm to find the mul9ple (n). n = gfm efm 3. Mul9ply each atom in the empirical formula by n to get the molecular formula. A compound has the empirical formula C 2 H 3 O and a gram formula mass of 172g. What is its molecular formula? efm = 1 mol C 2 H 3 O = 2 mol C = 2 x 12.0g = 24.0g 3 mol H = 3 x 1.0g = 3.0g 1 mol O = 1 x 16.0g =+16.0g 43.0g n = gfm = 172 g = 4 efm 43.0g C (2x4) H (3x4) O (1x4) C 8 H 12 O 4 A compound used as an addi9ve for gasoline to help prevent engine knock shows the following percentage composi9on: 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96g. Determine the empirical formula and the molecular formula for this compound g Cl 1 mol Cl = mol Cl 35.5g Cl 24.27g C 1 mol C = mol C 12.0g C 4.07g H 1 mol H = 4.07 mol H 1.0g H Cl : C : H : : : 1: 2 ClCH 2 empirical formula efm = (1x35.5g) + (1x12.0g) + (2x1.0g) = 49.5g n= gfm = 98.96g = 2 efm 49.5g Cl 2 C 2 H 4 9
10 Combus4on Analysis Complete combus9on (reac9on with O 2 ) of a sample of propane gas produced 2.641g of CO 2 and 1.442g of water as the only products. Find the empirical formula for propane. C x H y + O 2 à CO 2 + H 2 O 2.641g 1.442g 2.641g CO 2 1 mol CO 2 1 mol C = mol C 44.0g CO 2 1 mol CO g H 2 O 1 mol H 2 O 2 mol H = mol H 18.0g H 2 O 1 mol H 2 O C : H : : mul9ply by 3 3 : 8 C 3 H 8 10
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