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Starter S-93 1. What is the molar mass of H2S? 2. How many moles of H2S would be found in 100 g? 3. How many formula units (particles) of H2S would be in 100g? Chapter 10 Chemical Quantities Chapter 10 10.1 The Mole: A Measurement of Matter 10.1 The Mole Quantitative – yield a number value Most common methods count mass volume 10.1 The Mole Words often mean a quantity pair – 2 dozen – 12 mole? 10.1 The The Mole 10.1 Mole Mole – means a specific number of particles 6.022 x10 23 This is called Avogadro’s Number The unit is used for atoms molecules (covalent compounds) formula units (ionic compounds) 10.1 The The Mole 10.1 Mole To convert numbers of particles to moles we need an equality 1mol 6.022 x10 23 atoms This could also be 1mol 6.022 x10 23 molecules Or 1mol 6.022 x1023 formula.units Depending on what type of particle you are trying to convert 10.1 The The Mole 10.1 Mole The rest is just the conversion we have always done remember – the unit you have goes on the bottom of the fraction -the unit you are converting to goes on the top So to convert 2500 atom of C to moles 1molC 2500atomC 23 6.022 x10 atomsC 4.2 x10 21 molC 10.1 The The Mole 10.1 Mole Converting from moles to partciles follows the same process How many molecules of Carbon Dioxide are in 1.55 mol? 6.022 x1023 moleculesC O2 1.55molCO2 1molCO2 9.33x1023 moleculesCO2 10.1 The The Mole 10.1 Mole Convert the following quantities to moles 1. 9,487,212 formula units of AgNO3 1.575 x10 17 molAgNO3 2. 5.78x1023 atoms of Nitrogen 0.960molN 3. 4.1x1024 molecules of Carbon Monoxide 6.8molCO 10.1 The The Mole 10.1 Mole Convert the following to the correct type of particle 1. 95 moles of CCl4 25 5.7 x10 moleculesCCl4 2. 7.211x10-3 moles of CuCO3 21 4.342 x10 formula.unitsCuCO3 3. 0.08 moles of Helium 22 5 x10 atomsHe 10.1 The The Mole 10.1 Mole By definition, the atomic mass of an element in grams is the mass of one mole of the element. This is called the molar mass 10.1 The The Mole 10.1 Mole For a compound we must calculate the molar mass 1. Write down what type of atoms, and how many of each are present CH4 C 1 H 4 2. Multiply by the molar mass of each element 10.1 The The Mole 10.1 Mole For a compound we must calculate the molar mass 1. Write down what type of atoms, and how many of each are present CH4 C 1 x 12.0107g = H 4 x 1.00794g = 2. Multiply by the molar mass of each element 10.1 The The Mole 10.1 Mole For a compound we must calculate the molar mass 1. Write down what type of atoms, and how many of each are present CH4 C 1 x 12.0107g = 12.0107g H 4 x 1.00794g = 4.03176g 2. Multiply by the molar mass of each element 3. Add to get a total 10.1 The The Mole 10.1 Mole For a compound we must calculate the molar mass 1. Write down what type of atoms, and how many of each are present CH4 C 1 x 12.0107g = 12.0107g unit H 4 x 1.00794g = 4.03176g 2. Multiply by the molar mass of each element 3. Add to get a total 16.0425g CH4 Starter S-94 1. What is the molar mass of CO? 2. How many moles of CO would be found in 0.56 g? 3. How many molecules of CO would be found in 3.51 moles? Chapter 10 10.2 Mole-Mass and Mole-Volume 10.2 Mole-Mass andMole Mole-Volume 10.1 The We don’t convert from particles to moles nearly as often as we do from moles to grams. The reason is that we usually measure the amount of a substance on the balance We need to know numbers of particles so that we can compare ratios of atom or compounds 10.2 Mole-Mass andMole Mole-Volume 10.1 The This is just another conversion problem The equality is Value Molar Mass = 1 mole Again, the quantity you have goes on the bottom, the quantity you want goes on top So if you have 9.5g of Carbon 1molC 0.79molC 9.5 gC 12.0107 gC 10.2 Mole-Mass andMole Mole-Volume 10.1 The If you have 2.2 moles of Silver Nitrate convert to mass First we need to know the formula of Silver Nitrate AgNO3 Then the molar mass 169.8731g And finally we can convert 169.8731gAgNO3 370 gAgNO3 2.2molAgNO3 1molAgNO3 10.2 Mole-Mass andMole Mole-Volume 10.1 The Try the following example: How many moles is 8.2g of Copper (II) Chloride Formula CuCl2 Molar mass 134.451g Moles 1molCuCl2 0.061molCuCl2 8.2 gCuCl2 134.451gCuCl2 10.2 Mole-Mass andMole Mole-Volume 10.1 The And another one: How many grams is 2.4 mol of Iron (III) Sulfate Formula Fe2 SO4 3 Molar mass 399.881g Moles 399.881gFe2 SO4 3 960 gFe2 SO4 3 2.4molFe2 SO4 3 1 molFe SO 2 4 3 10.2 Mole-Mass andMole Mole-Volume 10.1 The In 1811, Amedeo Avogadro proposed Avagador’s Hypothesis – equal volumes of gases at the same temperature and pressure contain equal numbers of particles 10.2 Mole-Mass andMole Mole-Volume 10.1 The Standard Temperature and Pressure (STP) T – 0oC or 273K P – 101.3 kPa, or 1 atm At STP the volume of one mole is 22.4L So the equality for conversion is 1mol 22.4L 10.2 Mole-Mass andMole Mole-Volume 10.1 The To do these problems, the identity of the gas doesn’t really matter. If we have 15 L of Chlorine gas Cl2 The number of moles would be 1molCl2 0.67molCl2 15LCl2 22.4 LCl2 10.2 Mole-Mass andMole Mole-Volume 10.1 The If instead we calculate for a more complicated gas such as propane And we also have 15 L of propane gas C3 H 8 The number of moles would be 1molC3 H 8 0.67molC3 H 8 15LC3 H 8 22.4LC3 H 8 Starter S-95 1. What is the molar mass of Pb(SO4)2? 2. How many moles of Pb(SO4)2 would be found in 250 g? 3. How many moles of H2 gas are found in 250 L? Chapter 10 10.3 Percent Composition and Chemical Formulas 10.3 10.1 Percent TheComposition Mole The relative amounts of the elements in a compound is called the percent composition The percent by mass of an element is the number of grams of the element divided by the mass in grams of the compound multiplied by 100% mass.element %mass x100% mass.compound 10.3 10.1 Percent TheComposition Mole Calculating the mass percent from a formula CH 4 1. Formula 2. Calculate the total C 1x12.0107 g 12.0107 g mass of each element H 4 x1.00794 4.03176 g 3. Calculate the total CH 4 12.0107 4.03176 16.0425g mass of the compound 12.0107 C 4. Calculate the x100% 74.8680% 16.0425 percent by 4.03176 H mass for x100% 25.1317% 16.0425 each element 10.3 10.1 Percent TheComposition Mole Calculate the mass percent C 1 x 12.0107 g 12.0107 g CH3OH 1. Formula 2. Mass of each element H 4 x1.00794 4.03176 g O 1x15.9994 15.9994 g 3. Total mass CH3OH 12.0107 4.03176 15.9994 32.0419 g 4. Mass Percent 12.0107 C x100% 37.4844% 32.0419 4.03176 H x100% 12.5828% 32.0419 15.9994 O x100% 49.9327% 32.0419 Starter S-98 What is the percent by mass of the all the elements in Cu(NO3)2. 10.3 10.1 Percent TheComposition Mole Empirical Formula – smallest whole number ratio of the elements in a Compound Empirical compound Formula The empirical formula H2O H2O can be calculated CH3COOH CH2O from the percent composition CH2O CH2O C6H12O6 CH2O S8 S 10.3 10.1 Percent TheComposition Mole To calculate the empirical formula N 25.9% 1. List the elements and their O 74.1% percent compositions 1molN N 25.9 gN 1.85molN 2. Convert the percent 14.0 gN 1molO compositions to moles O 74.1gO 4.63molO 16.0 gO 3. Calculate the mole ratio (divide N 1.85mol 1.00 N 2mol 1.85 by the smallest number of moles) 4.63 mol O 5 O 2.50 4. Smallest Whole Number ratio 1.85mol 5. Write the Formula 2 5 NO 10.3 10.1 Percent TheComposition Mole To calculate the empirical formula 1. Elements 2. Convert to Moles Hg 67.6% S 10.8% O 21.6% 1molHg Hg 67.6 gHg 0.337 molHg 200.59 gNg 1molS S 10.8 gS 0.337 molS 32.06 gS 1molO O 21.6 gO 1.35molO 16.0 gO 3. Mole ratio 4. Whole Number ratio 5. Write the Formula HgSO4 0.337 mol Hg 1.00 Hg 0.337 1 mol 0.337mol S S 1 1.00 0.337mol O 1.35 4 mol 4.01 O 0.337mol Starter S-99 What is the empirical formula if Lead is 59.7% Hydrogen is 2.9% Arsenic is 21.6% Oxygen is 18.4% 10.3 10.1 Percent TheComposition Mole The molecular formula can be calculated from the empirical formula and the molar mass Comparison of Empirical and Molecular Formulas Formula Classification Molar Mass CH Empirical 13 C2H2 Molecular 26 (2x13) C6H6 Molecular 78 (6x13) CH2O Empirical 30 C2H4O2 Molecular 60 (2x30) C6H12O6 Molecular 180 (6x30) 10.3 10.1 Percent TheComposition Mole Steps in calculations (if mass=60.0g) 4 1. Determine the empirical formula C 1x12.0 12.0 gC 2. Calculate the mass H 4 x1.01 4.04 gH of the empirical formula 4 CH N CH N 30.0 g N 1x14.0 14.0 gN 3. Divide the actual molar mass by this number 4. Multiply the empirical formula 60.0 g 2.00 30.0 g C2 H 8 N 2 Starter S-102 What is the empirical formula if Silver – 63.5% Nitrogen – 8.2 % Oxygen – 28.2% If the formula mass is 170g, what is the formula of this compound? Starter S-103 What is the empirical formula if Carbon – 49.5% Hydrogen – 5.2% Nitrogen – 28.9% Oxygen – 16.5% What if the molecular formula, if the molar mass is 194g?