Chap apter ter 10 Chemi mical cal Quan antities tities Se - PowerPoint PPT Presentation

Chap apter ter 10 Chemi mical cal Quan antities tities Se Sectio ction n 10.1 The Mol ole: : A m A meas asuremen ement t of of Mat atter ter

What at is is a a Mole le? A mole (mol) of a substance is 6.02 X 10 23 representative particles of that substance. Usually atoms or molecules. 6.02 X 10 23 is called Avogadro’s number

Converting Number of Particles to Moles Number of Particles (Atoms or Molecules) = Number of Moles Avogadro' s Number Molecules or atoms Particles Av. Moles number e.g. How many moles of magnesium in 1.25 X 10 23 atoms of magnesium 23 1.25 x 10 = Number of Moles = 0.208 mol Mg 23 6.02 x 10 Page e 291: #3 & 4 4 Moles - Particles

Converting Moles to Number of Particles = 23 Number of Particles (Atoms or Molecules) Number of Moles x 6.02 x 10 e.g. . How many atoms ms are in 2.12 mol of propane ane (C 3 H 8 )? )? Number of atoms = 2.12 x 6.02 x 10 23 x 11 = 1.4 x 10 25 atoms Page: e: 292 #5 & 6 6

The mass of a Mole of an Element The mass of a mole of an element is its mo molar ar ma mass. How can you calculate the molar mass of a compound? To calculate the molar mass of a compound: 1. Find the number of grams of each element. 2. Add the masses of the elements in the compound. e.g. . what is the molar ar mass s of calcium lcium oxide? de? Chemical formula = CaO, Ca = 40, O = 16 Molar mass = 40 + 16 = 56g Page: e: 296, #7 & 8

Moles – Atoms If atoms Atoms or Molecules or Multiplies by number of atoms Particles Number of Avogadro’s Moles Number

Se Sect ction ion 10.2 .2 le – Ma Mo Mole Mass ss & le – Vol Mo Mole olume me Re Relation lationships ships

Converting Mass ss to Moles les Mass of the substance = Number of Moles Molar mass e.g. . How many moles es of iron(II (III) ) oxide de are contain tained ed in n 92 92.2g g of pur ure e Fe Fe 2 O 3 ? Mass = 92.2g Molar mass = 2x55.8 + 3x16 = 159.6g/mol Number of moles = 92.2/159.6 = 0.578 mol Fe 2 O 3 Page: e: 299, # 18 & 19 Moles - Mass

Converting Moles les to Mass ss Mass = number of Moles x molar mass e.g. . Wha hat t is th the ma mass s of 9. 9.45 45 mo mol of alumi uminu num m oxide? de? Chemical Formula = Al 2 O 3 Number of moles = 9.45 mol Molar mass = 2x27 + 3x16 = 102g Al 2 O 3 Mass = 9.45 x 102 =964g Al 2 O 3 Page: e: 298, # 16 & 17

The volume of a Mole of an Element The volume of a mole of an element is called molar volume The molar volume of a gas at STP is 22.4 L Standard Temperature and Pressure To calculate the volume of a gas at STP: Volume of gas = Number of Moles x 22.4 e.g. . Deter termine mine th the volum ume e of 0. 0.6 6 mo mol sul ulfur ur dioxide ide gas at t STP. Volume of SO 2 = 0.6 x 22.4 = 13.4 L SO 2 Page: e: 301, #20 & 21 Moles - Volume

Calculating Molar Mass from Density Molar mass = density at STP x molar volume at STP e.g. . The he densi nsity ty of a gaseous eous compound mpound carbon bon and nd oxygen en is foun und d to be 1.964g/L g/L at STP. . What is the molar ar mass s of the compound? pound? Molar mass = 1.964 x 22.4 = 44g/mol Page: e: 302, #22 & 23 Moles - Density

Moles – Mass Mass Number of Molar Moles Mass

Moles – Volume Volume Number of 22.4 Moles

Moles – Density Molar mass Density 22.4

Se Sect ction ion 10.3 .3 Per ercent cent Co Composit position ion and d Ch Chemical emical Fo Formulas mulas

The e percent cent comp mpos ositio ition is the relative amounts of the elements in a compound. The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100 Calculating Percent Composition from Mass Data mass of element =  % mass of element 100 mass of compound e.g. When a 13.6 g sample of a compound containing only magnesium and oxygen. 5.4 g of oxygen is obtained. What is the percent composition of this compound? Mass of compound = 13.6 g Mass of oxygen = 5.4 g Mass of magnesium = 13.6 – 5.4 = 8.2 g mass of Mg 8 . 2 =  =  = % Mg 100 100 60 . 3 % mass of compound 13 . 6 mass of O 5 . 4 =  =  = % O 100 100 39 . 7 % mass of compound 13 . 6 Page e 30 306, 6, # 32 # 32 & 33 33

Calculating Percent Composition from the Chemical Formula mass of element in 1 mole compound =  % mass of element 100 molar mass of compound e.g. Propane (C 3 H 8 ) is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Mass of C in C 3 H 8 = 36.0 g Mass of H in C 3 H 8 = 8.08 g Molar mass of C 3 H 8 = 44.08 g/mol mass of C 36 . 0 =  =  = % C 100 100 81 . 7 % mass of propane 44 . 08 mass of H 8 . 08 =  =  = % H 100 100 18 . 3 % mass of propane 44 . 08 Page: e: 307, # 34 & 35

Empirical Formula The empirical formula of a compound shows the smallest whole - number ratio of the atoms in the compound. It shows the kinds and lowest relative count of atoms or moles of atoms in molecules or a compound. An empirical formula may or may not be the same as a molecular formula. For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1 So the empirical formula of hydrogen peroxide is HO The actual molecular formula of hydrogen peroxide has twice the number of atoms as the empirical formula. The molecular formula is (HO) x 2 or H 2 O 2 Notice ice that the ratio of hydrogen to oxygen is still the same. The molecular formula tells the actual number of each kind of atom present in a molecule of the compound. For carbon dioxide, the empirical and molecular formulas are the same – CO2

e.g. A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? 25.9% N 74.1% O Step 1. Assume that you have 100 g of the compound, 25.9 g 74.1 g Step 2. Divide each by its atomic mass. 25.9 74.1 14 16 1.85 mol 4.63 mol Step 3. Divide each by the smallest number of moles. 4.63 1.85 1.85 1.85 1 2.5 • Multiply all by 2 The empirical formula is N 2 O 5 Page: e: 310, # 36 & 3 37

Molecular ecular Fo Formu mula la is the same as the empirical formula or it is a simple whole - number multiple of its empirical Once you determined the empirical formula of a compound, you can determine its molecular formula, but you must know the compound ’ s molar mass and empirical formula mass. e.g. Calculate the molecular formula of a compound whose molar mass is 60 g/mol and empirical formula is CH 4 N Empirical formula = CH 4 N Molar mass = 60 g/mol Molecular formula = C ? H ? N ? Ste tep 1. Calculate the empirical formula mass Empirical formula mass = 1x12 + 4x1.01 + 1x14 = 30.04 g/mol Step 2. Divide the molar mass by the empirical formula mass 60 = 2 30.04 Step 3. Multiply the formula subscripts by this value (factor) Molecular formula = 2 x CH 4 N = C 2 H 8 N 2 Page: e: 312, # 38 & 3 39