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Chemical Quantities, the Mole, and Conversions

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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

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6.022 x 10 23 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

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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 = 6.022 x 10 23 molecules

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Number of particles to moles… RP x 1 mole =Moles 6.02 x 10 23 RPs How many moles are 2.80 x 10 24 atoms of silicon? 4.65 moles

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Moles to number of particles… moles x 6.02 x 10 23 RPs.= RPs 1 mole How many atoms are in 1.14 mol Co? 6.86 x 10 23 atoms Co

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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

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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!

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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 = 87.62 + 2(35.453) = 158.526 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)

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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(1.00797) + 15.9994 = 18.01534 amu

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Molar Mass is the mass in GRAMS of one mole (6.02 x 10 23 particles) numerically equal to the formula mass of a compound (units = g/mol)

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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 10 -3 mol C 20 H 42

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Step 1: Find the Molar Mass: 20 (12.01115) + 42 (1.00797) = 282.55774 g/mol Step 2: Convert: 4.52 x 10 -3 mol x 282.55774 g C 20 H 42 = 1 mol C 20 H 42 Answer: 1.28 g C 20 H 42

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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( 14.0067) + 3(15.9994) = 76.0116 g/mol Step 3: Convert: 75.0 g N 2 O 3 x 1 mole N 2 O 3 = 76.0116 g N 2 O 3 Answer: 0.987 mole N 2 O 3

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

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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 101.325 kPa (1 atm, 760 mmHg, 760 torr)

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Molar Volume = the 22.4 L of space occupied by 6.02 x 10 23 representative particles of any gas at STP Example – What is the volume of 3.20 x 10 -3 mol CO 2 at STP? 3.20 x 10 -3 mol CO 2 x 22.4 L = ? 1 mole Answer: 0.0717 L CO 2

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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

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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

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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

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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

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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

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

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What is the percent composition by mass of hydrogen in water? Mass %H = (1.00797g) x 2 x 100 = 18.01534g Answer: 11.19013 %H

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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

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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(22.98977) + 12.01115 + 3(15.9994) + 10(18.01534) = 286.14229g

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%Na = 2(22.98977) x 100 = 16.06877 % Na 286.14229 %C = 12.01115 x 100 = 4.197614 % C 286.14229 %O = 3(15.9994) x 100 = 16.77424 % O 286.14229 %H 2 O = 10(18.01534) x 100 = 62.95938 % H 2 O 286.14229

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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

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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

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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 40.08 g Ca 48.7 g F x 1 mole F = 2.56 mole F 18.998403 g F Step 3: Ratio mole F = 2.56 mole F = 2 - ratio is 1:2 mole Ca 1.28 mole Ca

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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

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C2H2C8H8C2H2C8H8

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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 )

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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

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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 = 18.01 g by analysis Formula Mass (Molar Mass) = 2(1.00797) + 15.9994 = 18.01534g Ratio of MM/FM = 18.01 18.0534 or 1:1 so this is also the molecular formula

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Hydrogen Peroxide – the simplest formula is HO Molecular Mass = 34.0g by analysis (given) Formula Mass (Molar Mass) = 1.00797 g + 15.9994 g = 17.00737 g Ratio of MM/FM = 34.0 17.00737 = 2:1 so the formula is doubled to H 2 O 2

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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 100.0 g = 40.1 g S 59.9% of 100.0 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 15.9994 g O

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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* = 32.06 g S + 3(15.9994 g)O = 80.0582 g SO 3 MM = 80.1g (given in problem) 1:1 so the molecular formula = the empirical

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