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Chemical Quantities Calculating The Number of Atoms in a Specific Mass You have a 1.00 g sample of lead. How many atoms of lead are present? 1.00 g Pb 1 atom Pb = 207.2 amu 1.66 10-24 g = 1 amu 1.00 g Pb 1 amu 1.66 x 10-24 g 1 atom Pb 207.2 amu = 2.9 10 atoms Pb 21 Calculating Mass Example Calculate the mass (in amu) of 1.0 х 104 carbon atoms 4 1) Given: 1.0 10 atoms C 3) CF 1atom C = 12.01amu 2) Plan: Convert from atoms to amu 12.01 amu 1 C atom and 1 C atom 12.01 amu 4) Set Up Problem 12.01 amu 1.0 10 C atom = 1.2 10 5 amu 1 C atom 4 Formula Mass The sum of atomic masses of all atoms in its formula Important role in nearly all chemical calculations Can be calculated for compounds and diatomic elements Calculating Formula Mass Calculate the formula mass of calcium chloride Write the formula from the name given Ca2+ (from group II) and Cl- (from group VII) Formula is CaCl2 due to charge balance Formula mass: Sum of the atomic masses of atoms in the formula (1 Ca atom + 2 Cl atoms) 40.08 amu = 40.08 amu 1 Ca atom 35.45 amu 2 Cl atom = 70.90 amu 1 Cl atom 1 Ca atom 110.98 amu Formula mass of CaCl2 Counting Large Quantities Many chemical calculations require counting atoms and molecules It is difficult to do chemical calculations in terms of atoms or formula units Since atoms are so small, extremely large numbers are needed in calculations Need to use a special counting unit just as used for other items A ream of paper One dozen donuts A pair of shoes The Mole Mole: A unit that contains 6.022 х 1023 objects It is used due to the extremely small size of atoms, molecules, and ions 6.022x1023 particles in 1 mole Called Avogadro’s Number Periodic Table The average atomic mass in amu or g/mol (one atom) The weight of 1 mole of the element in grams Avogadro’s number provides the connecting relationship between molar masses and atomic masses Calculating the Number of Molecules in a Mole How many molecules of bromine are present in 0.045 mole of bromine gas? Given: 0.045 mol Br2 Equality: Avogadro’s number Need: molecules of Br2 1 mol Br2 = 6.022 10 23 molecules Br2 6.022 1023 molecules Br2 mol Br2 and Conversion factors: mol Br2 6.022 1023 molecules Br2 Set Up Problem: 0.045 mol Br2 6.022 10 23 molecules Br2 = mol Br2 2.7 10 22 molecules Br2 Subscripts State Moles of Elements The subscripts in a chemical formula indicate the number of atoms of each element present in a compound The subscripts in a chemical formula can also indicate the number of moles of atoms of each element present in one mole of a compound i.e. In one molecule of glucose (C6H12O6) there are 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen Calculating the Moles of an Element in a Compound How many moles of carbon atoms are present in 1.85 moles of glucose? Plan: moles of glucose subscript moles of C atoms Equality: (One) mol C6H12O6 = 6 mols C atoms Conversion Factors: Set Up Problem: mol C6H12O6 6 moles C atoms and mol C6H12O6 6 moles C atoms 1.85 mol C 6H12 O 6 6 mols C atoms = 11.1 mol C atoms mol C 6H12 O 6 Molar Mass The atomic mass of a carbon-12 atom is 12.00 amu The atomic mass of one mole of carbon-12 atoms 12.00 g One mole of any element is the amount of atoms (molecules or ions) that is equal to its atomic mass (in grams) This mass contains 6.022 х 1023 particles of that element Use the periodic table to obtain the molar mass of any element Molar Mass When the number of grams (weighed out) of a substance equals the formula mass of that substance, Avogadro’s number of molecules of that substance are present Molar Mass of a Compound Calculate the molar mass of iron (II) sulfate Formula is FeSO4 1) Calculate the molar mass of each element 2) Each element is multiplied by its respective subscript: (number of moles of each element) Moles of Compound Formula Subscript Moles of Element in Compound 3) The molar mass is calculated by the sum of the molar masses of each element Molar Mass of a Compound 1) Formula is FeSO4: The molar masses of iron, sulfur, and oxygen are 55.85 g Fe mol Fe 32.00 g S mol S 16.00 g O mol O 2) Multiply each molar mass by its subscript 1 mol Fe 55.85 g Fe = 55.85 g Fe mol Fe 4 mol O 16.00 g O = 64.00 g O mol O 1 mol S 32.00 g S = 32.00 g S mol S 3) Find the molar mass of the compound by adding the mass of each element 55.85 g 32.00 g 64.00 g = 151.85 g Calculations Using Molar Mass The three quantities most often calculated Number of particles Number of moles Number of grams Using molar mass as a conversion factor is one of the most useful in chemistry Can be used for g to mole and mole to g conversions Relationship between Moles, Molar Mass and Avogadro’s number Moles of substance Avogadro’s Number Particles of substance Molar Mass Moles of substance Grams of substance Converting Mass of a Compound to Moles International Foods Coffee contains 3 mg of sodium chloride per cup of coffee. How many moles of sodium chloride are in each cup of coffee? 3 mg NaCl 3 mg NaCl moles of NaCl 1g NaCl = 0.003 g NaCl 1000 mg NaCl MM NaCl = 1(Na) 1(Cl) = 1(22.99) 1(35.45) = 58.44 Equality: 1 mol NaCl = 58.44 g g NaCl mol 58.44 g NaCl mol NaCl and mol NaCl 58.44 g NaCl 0.003 g NaCl 1mol NaCl = 5.13 10 -5 mol NaCl 58.44 g NaCl Converting Grams to Particles Ethylene glycol (antifreeze) has the formula C2H6O2. How many molecules are present in a 3.86 × 10-20 g sample? Plan: convert g Molar mass moles Avog Number Equality 1:1mol C2H6O2 =62.05 g C2H6O2 molecules of ethylene glycol Conversion 62.05 g C2H6O2 mol C2H6O2 and mol C2H6O 2 62.05 g C2H6O 2 Factor 1 Equality 2: 1mol C2H6O2 =6.022 1023 molecules Conversion 6.022 10 23 molecules 1mol C 2H6 O 2 or Factor 2 1mol C 2H6 O 2 6.022 10 23 molecules 3.86 10 -20 mol C 2H6 O 2 6.022 10 23 molecules g C 2H 6 O 2 = 375 molecules 62.05 g C 2H6 O 2 mol C 2H6 O 2 Percent Composition Percent Composition Sometimes it’s useful to know the composition of a compound in terms of what percentage of the total is each element Percent “Parts per 100” The number of specific items per a group of 100 items 50% of $100 is $50 (50 items/100 total items) Percent Example You have 4 oranges and 5 apples. What percent of the total is oranges? 4 oranges 100% = 44% oranges 9 total In “parts per 100” 44 oranges 100% = 44% oranges 100 total Percent Composition It is the percent by mass of each element in a compound Can be determined By its chemical formula Molar masses of the elements that compose the compound The percent of each element contributes to the mass of the compound mass of each element mass percent of each = 100% element in a compound molar mass of the compound Calculating Percent Composition Example What is the percent composition of each element in NH4OH? Determine the contribution of each element 14.01g N: 100% = 39.97% N 35.05 g H: 5.04 g 100% = 14.38% H 35.05 g 16.00 g O: 100% = 45.65% O 35.05 g N : 1 14.01g = 14.01g H : 5 1.0078 g = 5.04 g O : 1 16.00 g = 16.00 g Molar mass = 35.05 g Empirical Formulas The simplest ratio of elements in a compound It uses the smallest possible whole number ratio of atoms present in a formula unit of a compound If the percent composition is known, an empirical formula can be calculated Empirical Formulas To Determine the empirical formula: 1) Calculate the moles of each element Use molar mass (atomic mass) 2) Calculate the ratios of the elements to each other 3) Find the lowest whole number ratio Divide each number of moles by the smallest number of moles present Empirical Formula: Converting Decimal Numbers to Whole Numbers The subscripts in a formula are expressed as whole numbers, not as decimals The resulting numbers from a calculation represent each element’s subscript If the number(s) are NOT whole numbers, multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is obtained Relating Empirical and Molecular Formulas n represents a whole number multiplier from 1 to as large as necessary molar mass ( g / mol ) n= empirical formula mass ( g / mol ) Calculate the empirical formula and the mass of the empirical formula Divide the given molecular mass by the calculated empirical mass Answer is a whole number multiplier Relating Empirical and Molecular Formulas Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula Calculate Empirical Formula from Percent Composition Lactic acid has a molar mass of 90.08 g and has this percent composition: 40.0% C, 6.71% H, 53.3% O What is the empirical and molecular formula of lactic acid? Assume a 100.0 g sample size Convert percent numbers to grams Calculate Empirical Formula from Percent Composition Convert mass of each element to moles Divide each mole quantity by the smallest number of moles 40.0 g C 6.71 g H mol C = 3.33 mol C 12.0 g C mol H = 6.66 mol H 1.008 g H 53.3 g O mol O = 3.33 mol O 16.00 g O The ratio of C to H to O is 1 to 2 to 1 3.33 = 1.00 3.33 6.66 H: = 2.00 3.33 C: O: 3.33 = 1.00 3.33 Empirical formula is CH2O Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol Determination of the Molecular Formula Obtain the value of n (whole number multiplier) Multiply the empirical formula by the multiplier molar mass ( g / mol ) n= empirical formula mass ( g / mol ) 90.08 g / mol =3 30.03 g / mol Molecular formula = n х empirical formula Molecular formula = 3 (CH2O) C3H6O3 Formulas for Compounds Empirical Formula Smallest possible set of subscript numbers Smallest whole number ratio All ionic compounds are given as empirical formulas Molecular Formulas The actual formulas of molecules It shows all of the atoms present in a molecule It may be the same as the EF or a wholenumber multiple of its EF Molecular formula = n х Empirical formula