Presentation on theme: "Chapter 11. Mole SI base unit for measuring the amount of substance The number of representative particles in exactly 12 grams of pure carbon-12 1 mole."— Presentation transcript:
Mole SI base unit for measuring the amount of substance The number of representative particles in exactly 12 grams of pure carbon-12 1 mole = 6.02 x 10 23 representative particles
Avogadro’s Number 6.02 x 10 23 is called Avogadro’s number It is the number of representative particles equal to 1 mole A representative particle is any kind of particle and depends on the substance – Water = molecule – Copper = atom – NaCl (salt) = formula unit
Molar/Formula Mass Molar/Formula mass is the mass in grams of 1 mole of any pure substance. – units are g/mol Molar mass is equal to the atomic mass on the periodic table Gram atomic mass: (found on periodic table) – Ex: C = 12.011 amu This means 1 mole of Carbon = 12.011 g
Let’s practice: Find the # of grams 1 mole of the following is equal to – Al – H – O – P
Formula Mass/Molar Mass Formula or Molar mass is the mass of 1 mole of a pure substance H 2 O – covalent compound (metal/nonmental) H=2x1 g= 2 g O=1x16 g= 16 g total = moles x avg. atomic mass = 18 g
Let’s practice: Find the # of grams 1 mole of the following is equal to – H 2 O 2 – NaCl – Ca(NO 3 ) 2 – (NH 4 ) 3 PO 4 – N 2 O 4
Percent Composition You can calculate the percentage of a certain element in a compound Use the % Formula: (Part/Whole) x 100 = 1 st - we need the formula mass or the whole 2 nd – we need the part found in the compound 3 rd – divide and multiple by 100
Example: What is the % of Hydrogen in H 2 O? 1 st : Gram atomic Mass – H 2 O – covalent compound (metal/nonmental) H=2x1= 2 O=1x16= 16 total = moles x avg. atomic mass = 18 g 2 nd : Hydrogen has a mass of 2g 3 rd : %H = (2g/18g) x 100 = 11.11%
What is the percent of oxygen in in H 2 O? The percent of each element in a compound should add and equal 100.
Lets practice: Find the percent composition of each of the elements in the following compounds. – Fe 2 O 3 – Cu 2 S
Molar Mass can be used as a Conversion Factor What is the mass in grams of 2.5 mol of NaCl? Steps – 1 st : get the formula/molar mass of the compound – 2 nd : convert Na= 1 x 23 = 23 Cl = 1 x 35.5 = 35.5 = 58.5g ?g = 2.5 mol NaCl58.5 g NaCl=146 g NaCl 1 mol NaCl
Avogadro’s number as a conversion factor 1 mole = 6.02 x 10 23 representative particles Representative particles can be atoms, molecules, particles, formula units, pieces, etc How many molecules are in 3.5 moles of CO 2 ? ?molecules = 3.5 mol CO 2 6.02 x 10 23 molecules = 1 mol NaCl 2.1 x 10 24 molecules CO 2
Volume Conversions At STP 1 mol = 22.4 liters of a gas What volume will 54.6 grams of CO 2 occupy at STP? ?liters = 54.6 g CO 2 1 mol CO 2 22.4 liters CO 2 = 44 g CO 2 1 mole CO 2 27.8 liters of CO 2
Let’s practice: 1.How many grams are in 0.817 moles of C 2 H 2 O 6 ? 2.How many moles are in 60 grams of CaC0 3 ? 3.How many formula units of CaCl 2 are in 45 grams?
Empirical Formulas The simplest whole number formula/ratio of a compound. This means the equation cannot be reduced anymore. Smallest whole number ratio Example: An unknown sample contains 25% H and 75% C. What is the empirical formula
Steps: 1 st Convert % to grams (Assume you have 100 grams total) 2 nd Convert grams to moles by using the atomic mass 25 % = 25 grams H 75 % = 75 grams C 25 % = 25 g H x (1mol H/1g H) = 25 mol H 75 % = 75 g C x (1mol C/12 g C) = 6.25 mol O
3 rd Divide by the smallest # of moles to get the ratio 4 th Put numbers as subscripts of elements in formula 5 th if a whole number is not found after dividing, multiply by a factor to make it a whole # (must multiply all numbers by the same factor) H = 25 mol/ 6.25 mol = 4 C = 6.25 mol / 6.25 mol = 1 CH 4
Molecular Formula Indicates the actual amount of atoms present Must have molecular weight to determine molecular formula
Steps: 1.Determine the empirical formula 2.Determine the empirical weight (molar mass of the empirical formula) 3.Divide the molecular weight given by the empirical weight you calculated – If one, then the empirical formula is the same as the molecular formula
4.If the answers from step 3 is a whole number, then take the answer and multiply all coefficients in the equation by the number. 5.If it is not a whole number you must multiple by a factor to make it a whole number then multiply each of the subscripts by it. This will give the molecular formula.
Molecular Formula Example What is the molecular formula of a compound with a molecular mass of 64 g/mol and an empirical formula of CH 4 ? CH 4 empirical mass = (12) + (4 x 1) = 16g/mol Molecular mass = 64 g/mol 64 / 16 = 4 Multiply subscripts by 4 New formula is C 4 H 16
Hydrates A compound that contains a certain amount of water molecules Example:CuSO 4. 5H 2 O Copper sulfate pentahydrate Hydrates can be dehydrated by heating The dehydrated form can be used to absorb moisture from the air – like those packets you find in your shoe boxes – deliquescent