Chemical Quantities, the Mole, and Conversions

 Measuring Matter -The amount of something is usually determined one of three ways; by counting, by mass, or by volume. -Groups are made to make counting easier. (Remember the bean lab?) A bunch, a dozen, a ream, a can

 x of anything!  Known as Avogadro’s number  Representative Particles are the smallest division of a substance that maintains the characteristics of that substance -For molecular compounds R.Ps are molecules -For ionic compounds R.Ps are formula units -For elements R.Ps are atoms

 1 mole of any element is equal to the atomic mass on the periodic table  AKA molar mass  For example: 18g of H 2 O = x molecules

 Number of particles to moles… RP x 1 mole =Moles 6.02 x RPs  How many moles are 2.80 x atoms of silicon?  4.65 moles

 Moles to number of particles…  moles x 6.02 x RPs.= RPs 1 mole  How many atoms are in 1.14 mol Co?  6.86 x atoms Co

 The atomic mass is the mass of one atom of an element (in AMU)  Based on the mass of one atom of Carbon-12  AKA – relative mass since each element is compared to C-12  Not very practical when working with chemicals in real life

 Molar Mass is the atomic mass of an element expressed in grams or, the mass of one mole of atoms of a particular element expressed in grams  Numerically equal to the atomic mass in amu’s because the Mole is a constant number!

 Formula Mass is the sum of the average atomic masses of all the elements represented in the formula (in amu’s)  Example: Use the atomic masses in the periodic table to calculate the formula mass of SrCl 2 FM = (35.453) = amu  Importance – the formula mass of a compound gives us useful information about the quantities of reactants needed to make new products (must follow the Law of Conservation of Mass)

 Molecular Mass is the same as formula mass, but this term in used specifically for molecular compounds  Not used for ionic compounds  Example – water: H 2 O = 2( ) = amu

 Molar Mass is the mass in GRAMS of one mole (6.02 x particles) numerically equal to the formula mass of a compound (units = g/mol)

 Mole-Mass Relationship Use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance Moles to Mass (grams): Multiply by the Molar Mass Example – Find the mass, in grams, of 4.52 x mol C 20 H 42

 Step 1: Find the Molar Mass: 20 ( ) + 42 ( ) = g/mol  Step 2: Convert: 4.52 x mol x g C 20 H 42 = 1 mol C 20 H 42 Answer:  1.28 g C 20 H 42

 Grams to moles: Divide by the Molar Mass Example – Calculate the number of moles in 75.0 g of Dinitrogen trioxide. Step 1: Write the correct chemical formula: N 2 O 3 Step 2: Find the Molar Mass: 2( ) + 3( ) = g/mol Step 3: Convert: 75.0 g N 2 O 3 x 1 mole N 2 O 3 = g N 2 O 3 Answer: mole N 2 O 3

# of Moles x Avogadro’s # # of RPs x molar mass Mass (g)

 Avogadro’s Hypothesis states equal volumes of gases at the same temperature and pressure contain equal numbers of molecules  Standard Temperature and Pressure  (STP) = 0 ˚C (273 ˚K) and kPa (1 atm, 760 mmHg, 760 torr)

 Molar Volume = the 22.4 L of space occupied by 6.02 x representative particles of any gas at STP  Example – What is the volume of 3.20 x mol CO 2 at STP?  3.20 x mol CO 2 x 22.4 L = ? 1 mole Answer: L CO 2

 Example: – How many moles of CO 2 are in 57.0 L of CO 2 ?  57.0 L CO 2 x 1 mole = 22.4 L Answer: 2.54 mol CO 2

 Calculating Molar Mass from Density  Background – the density of gases is measured in g/L at a specific temperature, and, the density of a gas at STP is a constant value (characteristic property)  To find Molar Mass of a gas from density – multiply the density at STP by the molar volume at STP

 Example – A gaseous compound composed of sulfur and oxygen, which is linked to the formation of acid rain, has a density of 3.58 g/L at STP. What is the molar mass of this gas?  Molar Mass = 3.58 g of gas x 22.4 L = 1 L 1 mol Answer: 80.2 g/mol

 Definition – relative amounts of substances in a compound expressed as percents by mass  Percentage Composition from Mass Data  Equation: Mass of Element x 100 = % Mass Compound

 It has been measured in lab that 1.000g of water contains 0.112g Hydrogen. Find the mass percentage of hydrogen  Mass %H = 0.112g x 100 = 11.2 %H 1.000g

 Percentage Composition from the Chemical Formula (Atomic mass of elem.) x (# atoms of elem.) x 100 Molar mass of the compound

 What is the percent composition by mass of hydrogen in water?  Mass %H = ( g) x 2 x 100 = g  Answer: %H

 Percentage Composition as a Conversion Factor What is the mass of hydrogen in 30.0g of water? Mass H = (Mass %  100) x mass sample = (11.2%  100) x 30.0g = Answer: 3.36gH

 Hydrates (compounds containing water): You must include the water of hydration in the formula mass  What are the mass percentages of the elements in Na 2 CO 3  10 H 2 O?  MM = 2( ) ( ) + 10( ) = g

 %Na = 2( ) x 100 = % Na  %C = x 100 = % C  %O = 3( ) x 100 = % O  %H 2 O = 10( ) x 100 = % H 2 O

 A chemical formula showing the smallest (simplest) whole-number ratio of atoms in a compound  Steps for determining Empirical Formula from % composition Step 1: Determine the masses of the elements in a 100g (since its based on a percent by mass) sample Step 2: Convert these masses in grams to moles by dividing by the atomic mass

Step 3: Determine the simplest mole ratio between the elements by dividing each by the smallest number of moles Step 4: Use these mole ratios to write the subscripts in the formula

 Calcium Fluoride occurs as the mineral Fluorite. It contains 51.3% by mass of Ca and 48.7% by mass of F. Determine the Empirical formula.  Step 1: Start with a 100g sample of calcium fluoride 51.3% of 100g = 51.3 g Ca 48.7% of 100g = 48.7 g F  Step 2: Convert to Moles 51.3 g Ca x 1 mole Ca = 1.28 mole Ca g Ca 48.7 g F x 1 mole F = 2.56 mole F g F  Step 3: Ratio mole F = 2.56 mole F = 2 - ratio is 1:2 mole Ca 1.28 mole Ca

 CaF 2 is the empirical formula  Another Example! Substance X and Y. Suppose you found the number of moles of X and Y in 100.0g of a compound to be 1.96 mole X and 2.94 mole Y. What is the empirical formula? Answer: X2Y3X2Y3

C2H2C8H8C2H2C8H8

 The actual formula of a molecular compound which is often a multiple of the empirical formula (E.g. C 2 H 6 instead of CH 3 )

 The Molecular Mass must be determined by experimental analysis or be given in the problem  Divide the Molecular Mass by the Formula Mass (Molar Mass) and adjust the subscripts in the Empirical Formula by multiplying by this resulting number

Water – the simplest formula is H 2 O, but could the molecular formula be H 4 O 2 or H 8 O 4 ?  Molecular Mass = g by analysis  Formula Mass (Molar Mass) = 2( ) = g  Ratio of MM/FM =  or 1:1 so this is also the molecular formula

Hydrogen Peroxide – the simplest formula is HO  Molecular Mass = 34.0g by analysis (given)  Formula Mass (Molar Mass) = g g = g  Ratio of MM/FM = 34.0  = 2:1 so the formula is doubled to H 2 O 2

A compound is found by analysis to consist of 40.1% S and 59.9% O. The Molecular Mass is 80.1g. Determine the Empirical Formula, the Molecular Formula, and name the compound  Step 1: in a 100.0g sample 40.1% of g = 40.1 g S 59.9% of g = 59.0 g O  Step 2: convert to moles 40.1 g S x 1 mole S = 1.25 mole S 32.06g S 59.9 g O x 1 mole O = 3.74 mole O g O

 Step 3: ratio 3.74 mole O  1.25 mole S = 2.992:1 or 3:1  Step 4: empirical formula SO 3  Step 5: mass ratios FM* = g S + 3( g)O = g SO 3 MM = 80.1g (given in problem) 1:1 so the molecular formula = the empirical