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Pre-­‐AP Chemistry Chemical Quan44es: The Mole Representa9ve par9cles can be atoms, molecules, formula units, or ions. For example… Atoms: Fe, Al, Mg, C, etc. Molecules: H20, CO, SO2, etc. Formula units: NaCl, KBr, FeSO4, etc. Ions: Na+, Cl-­‐, SO42-­‐, etc. A mole of substance represents 6.02 x 1023 representa9ve par9cles of a par9cular substance. The experimentally determined 6.02 x 1023 is called Avogadro s number. Mole – SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Diatomic Elements Certain elements are only stable in pairs or with other elements in a compound. These are known as the diatomic elements: Hydrogen (H2); Bromine (Br2); Oxygen: (O2); Nitrogen (N2); Chlorine (Cl2); Iodine (I2); and Fluorine (F2) 6.02X1023 602X1021 602,000,000,000, 000,000,000,000 1
Example: How many atoms are in 0.370 **Conversion factor between representa9ve par9cles and moles.** mol of aluminum? 0.370 mol Al 6.02x1023 Al atoms = 1 mol Al 2.23X1023 Al atoms 1 mol = 6.02 x 1023 representa9ve par9cles How many molecules are in 4.50 moles of water? 4.50 mol H2O 6.02x1023 H2O molec. = 1 mol H2O 2.71x1024 H2O molecules How many moles are 9.3 x 1015 atoms of lead? 9.3 x 1015 Pb atoms 1 mol Pb = 6.02x1023 Pb atoms 1.5x10-­‐8 Pb mol How many moles are in 8.5 x 1025 molecules of carbon dioxide? 25
8.5 x 10 molec. CO2 1 mol CO2 = 6.02x1023 CO2molec. 140 CO2 mole How many chloride ions are in 0.075 mol calcium chloride? 0.075 mol CaCl2 2 mol Cl-­‐ 6.02x1023 Cl-­‐ = 1 mol CaCl2 1 mol Cl-­‐ 9.0 x 1022 Cl-­‐ ions 2
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 Gram Molecular Mass (gmm) – the mass of 1 mol of a molecule. Sum of the atomic masses of each atom making up the molecule. Example: mass of ace9c acid, What is the gram molecular CH3COOH? 1 mol CH3COOH : 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 For example… Mg = 24.3 g = mass of 6.02 x 1023 atoms 24.3 g/mol is the gram atomic mass of Mg For example… 1 mol CO2: 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 CO2 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 60.0g CH3COOH 3
Example: Recap: What is the gram formula mass of ferric sulfate? Fe2(SO4)3 2 mol Fe 2 x 55.9 g = 111.8 g 3 mol S 3 x 32.1 g = 96.3 g 12 mol O 12 x 16.0 g =+192.0 g 400.1 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 CaCO3 = 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 = + 48.0g 100.1g CaCO3 Mole – Mass Conversions Conversion Factor! 1 mol = molar mass 4
Find the moles in 42.5 g of calcium. Example: Find the moles in 26.2 g of gold, Au. 26.2g Au = ? mol Au Conversion factor: 197.0 g Au= 1 mol Au Use Dimensional Analysis!!! 26.2 g Au 1 mol Au = 0.133 mol Au 197.0 g Au 42.5 g Ca 1 mol Ca = 40.1g Ca 1.06 mol Ca Find the mass in grams of 4.35 mol uranium, U. Find the mass in grams in 2.00 mol iron. 2.00 mol Fe 55.8g Fe = 1 mol Fe 112g Fe 4.35 mol U 238.0 g U = 1 mol U 1040g U Mul4-­‐Step Mole Problems: Calculate the number of atoms present in 1.50g silver. 1.50g Ag 1 mol Ag 6.02x1023 Ag atoms = 107.9g Ag 1 mol Ag 8.37x1021 Ag atoms Calculate the number of molecules present in 3.25g diphosphorous pentoxide. 3.25g P2O5 1 mol P2O5 6.02x1023 P2O5molec. 142.0g P2O5 1 mol P2O5 1.38x1022 P2O5 molecules 5
Calculate the mass in grams of 3.00 x 1020 nitrogen molecules. 3.00x1020 N2 molec 1 mol N2 28.0g N2 = 6.02x1023 N2 molec. 1 mol N2 0.0140g N2 The Volume of a Mole of Gas Standard Temperature and Pressure (STP) : 0oC and 1 atm The volume of a gas is usually measured at STP. Since 1 mol of any substance contains Avogadro s number of par9cles, 22.4 L of any gas at STP contains 6.02 x 1023 par9cles of gas. How many atoms of nitrogen are in 1.2 g of aspartame, C14H18N2O5? 1.2g C H N O 1 mol C H N O 14 18 2
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14 18 2
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294g C14H18N2O5 2 mol N 6.02x1023 N atoms = 1mol C14H18N2O5 1 mol N 4.9x1021 N atoms 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 1023 par9cles of the gas. n 22.4 L of a gas has a mass equal to the gfm of the gas. -­‐-­‐ Conversion factor-­‐-­‐ 1 mol of any gas at STP = 22.4 L 6
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 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). 22.4 L/mol x density (g/L) = gfm Calculate the density of methane gas, CH4, at STP. CH4 = (1x12.0g) + (4x1.0g) =16.0 g/mol 16.0 g CH4 1 mol CH4 = 1 mol CH4 22.4L CH4 Determine the number of moles in 625 L H2 gas at STP? 625 L H2 1 mol H2 = 22.4L H2 27.9 mol H2 Example: The density of a gaseous compound is 1.623 g/L at STP. Determine the gfm of this compound. 22.4 L 1.623 g = 36.36 g/mol 1 mol 1L Percent Composi4on The percent by mass of each element in a compound is known as percent composi9on. The percent must total 100% 0.714 g/L CH4 7
Example: Find the % composi4on for a compound that is formed from 222.6 g of Na and 77.4 g O 222.6 g Na + 77.4g O = 300.0 g total 222.6 g Na x 100 = 74.2% Na 300.0g total 77.4g O x 100 = 25.8% O 300.0 g total What is the percent composi4on of Mg(NO3)2. Molar Mass of Mg(NO3)2 = 148.3 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 OR solve for percent composi9on from the chemical formula: % mass = grams of element in 1 mol of compound x 100 molar mass of compound 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: CO2 is empirical and molecular N2H4 is a molecular formula but its empirical formula would be NH2 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
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 : 4.54 2.28 : 2.28 1 : 2 CO2 carbon dioxide 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 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. 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. A compound has the empirical formula C2H3O and a gram formula mass of 172g. What is its molecular formula? efm = 1 mol C2H3O = 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) C8H12O4 71.65g Cl 1 mol Cl = 2.018 mol Cl 35.5g Cl 24.27g C 1 mol C = 2.023 mol C 12.0g C 4.07g H 1 mol H = 4.07 mol H 1.0g H Cl : C : H 2.018 : 2.023 : 4.07 2.018 2.018 2.018 1: 1: 2 ClCH2 empirical formula efm = (1x35.5g) + (1x12.0g) + (2x1.0g) = 49.5g n= gfm = 98.96g = 2 efm 49.5g Cl2C2H4 9
Combus4on Analysis Complete combus9on (reac9on with O2) of a sample of propane gas produced 2.641g of CO2 and 1.442g of water as the only products. Find the empirical formula for propane. CxHy + O2 à CO2 + H2O 2.641g 1.442g 2.641g CO2 1 mol CO2 1 mol C = 0.0600 mol C 44.0g CO2 1 mol CO2 1.442g H2O 1 mol H2O 2 mol H = 0.160 mol H 18.0g H2O 1 mol H2O C : H 0.0600 : 0.160 0.0600 0.0600 1 : 2.667 mul9ply by 3 3 : 8 C3H8 10