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Unit 8: Chemical Formulas Chapter 11 Chemistry 1K Cypress Creek High School The Mole • In the same way a dozen is worth 12, a mole is worth 6.02 x 1023 • The mole is a large number because particles are so small, it takes many of them make up an amount we can see and understand • We can use it to count anything! Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 1023 cookies 1 gross cars = 144 cars 1 mole of cars = 6.02 X 1023 cars 1 ream Al particles = 500 Al particles 1 mole of Al particles = 6.02 X 1023 particles Mole Conversions 23 22.4 L 6.02 X 10 atoms formula units, 1 MOLE molecules Molecular mass in grams The Mass of a Mole • How do we “count” a mole? • By measuring mass (just like how they count aluminum cans for recycling) • You can find atomic masses on the periodic table. • 1 mole of Mg atoms = 24.3 g • 1 mole of Cu atoms = 63.5 g Moles to Mass • Calculate the mass of 0.625 moles of calcium. • According to the periodic table, the atomic mass of calcium is 40.078 amu, so the molar mass (mass of 1 mole) of calcium is 40.078 g. X g calcium 40.078 g calcium = 0.625 mol calcium 1 mol calcium = 25.0 g calcium Mass to Particles • Calculate the number of atoms in 4.77 g lead. • First determine how many moles are in 4.77 g lead. X moles Pb 1 mole Pb = 4.77 g Pb 2.7.2 g Pb = 0.0230 mol Pb • Now use a second conversion factor to convert moles to number of particles. X atoms Pb 6.02 x 1023 atoms Pb = 0.0230 mol Pb 1 mol Pb = 1.38 x 1022 atom Pb Chemical Formula • A Chemical Formula is the abbreviated way of naming a compound using numbers and element symbols. • Subscripts denote number of each element in a compound • Ex: H2O has 2 hydrogens and 1 oxygen • Ex: Al2(SO4)3 has 2 aluminums, 3 sulfurs, and 12 oxygens • Coefficients act as a scalar (also called a multiplier) • Ex: 6H2O has 12 hydrogens and 6 oxygens • Ex: 3Al2(SO4)3 has 6 aluminums, 9 sulfurs, and 36 oxygens Moles of Compounds • If the substance is a molecular compound, such as ammonia (NH3), a mole is 6.02 x 1023 molecules of ammonia. • A mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound. • 1 mole of NH3 •1 mole of N • 3 moles of H Molar Mass of a Compound • The molar mass of a compound is the mass of a mole of the representative particles of the compound. • In the case of NH3. Molar mass of NH3 = molar mass of N + 3(molar mass of H) Molar mass of NH3 = 14.007 g + 3(1.008 g) = 17.031 g/mol • You can use the molar mass of a compound to convert between mass and moles. Molar Mass Practice • Find the molar mass for: • SO3 • Na2SO4 • Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Percent Composition • Percent Composition is the percentage by mass of each element in a compound • The percent composition is found by using the following formula: Total mass of element % Mass 100 molar mass of compound Calculating Percent Composition • What is the percent of each element in sodium sulfite, Na2SO3? • 2 mole Na x 22.990 g/mol Na = 45.980 g Na • 1 mole S x 32.066 g/mol S = 32.066 g S • 3 moles O x 15.999 g/mol O = + 47.997 g O Molar Mass of Na2SO3 = 126.043 g • Sodium • % = (45.980 g / 126.043 g) x 100 = 36.486 % Na • Sulfur • % = (32.066 g / 126.043 g) x 100 = 25.441 % S • Oxygen • % = (47.997 g / 126.043 g) x 100 = 38.080 % O % Composition Practice • What is the percent of C & H in C2H6? • What is the percent of each element in Na2SO3? Law of Definite Proportions • Joseph L. Proust was born in 1754 in France and worked as an apothecary. • Proust made artificial copper carbonate and compared it to natural copper carbonate. • He showed that each had the same proportion of weights between the three elements involved. • This was called the Law of Definite Proporitons Types of Formulas • There are 2 types of formulas that can be used to describe compounds. • Empirical Formula • Molecular Formula • An Empirical Formula is the LOWEST whole number ratio of the elements in a compound. • The empirical formula for C6H6 is CH • A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound. • The actual formula Examples of Empirical & Molecular Formulas • Example 1: Ethene • Empirical: CH2 • Molecular: C2H4 • Example 2: Caffeine • Empirical: C4N2OH5 • Molecular: C8N4O2H10 Back = Carbon Blue = Nitrogen Red = Oxygen Grey = Hydrogen Empirical Formulas • An Empirical Formula is the LOWEST whole number ratio of the elements in a compound. • The empirical formula for C6H6 is CH • C6H6 is a completely different chemical than CH! • The subscript cannot be a fraction • Ex: There is no CO2H1.5 - it must be C2O4H3 • What is the empirical formula for each? • C2H6 • C6H12O6 Empirical Formula Calculations • There are 3 types of empirical calculations: • Mole Ratio • Grams • Percent Empirical Formula from Mole Ratios • To calculate the empirical formula from mole ratios, follow these steps: 1) Divide each mole quantity by the smallest mole quantity 2) Assign ratios as subscripts 3) If any of the ratios are not even (ie: .5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts. Calculating Empirical Formula from Mole Ratio • Find the empirical formula for a compound containing 2 mole carbon and 6 mole hydrogen. LOWEST •C= 2/2 = 1 •H = 6/2 = 3 The formula is = CH3 Empirical Formula from Grams • To calculate the empirical formula from grams, follow these steps: 1) Convert the grams to moles (divide grams by molar mass). 2) Divide each mole quantity by the smallest mole quantity. 3) Assign ratios as subscripts. 4) If any of the ratios are not even (ie: .5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts. Calculating Empirical Formula from Grams •Calculatethe empirical formula of a compound that contains 13.5 grams of calcium,4.05 grams of carbon, and 16.2 grams of oxygen. calcium carbon oxygen 13.5 g = 0.337 4.05 g = 0.337 16.2 g = 1.01 40.08 g/mol mol 12.011 g/mol mol 15.999 g/mol mol LOWEST 0.337 = 1 0.337 1.01 = 3 =1 0.337 0.337 0.337 empirical formula = CaCO3 Empirical Formula from Percent Composition • To calculate the empirical formula from percent composition, follow these steps: 1) Change % sign to grams. You assume 100 g of the compound. 2) Convert the grams to moles (Divide grams by molar mass). 3) Divide each mole quantity by the smallest mole quantity. 4) Assign ratios as subscripts. 5) If any of the ratios are not even (ie: .5, 1.25, 1.3), multiply them all by the lowest common multiple to achieve whole numbers then place these numbers as the subscripts. Calculating Empirical Formula from % Composition • Calculate the empirical formula of a compound containing 18.8% nickel and 81.2% iodine. nickel iodine = 0.320 = 0.640 18.8 g 81.2 g mol mol 58.69 g/mol LOWEST 126.905 g/mol 0.320 = 1 0.640 = 2 0.320 0.320 empirical formula = NiI2 Practice Empirical Formula • Find the empirical formula given 65.2% Sc and 34.8 % O Formula for a Hydrate • Many compounds have a certain number of water molecules that are associated with them. These are called hydrates. • In the formula for a hydrate the number of water molecules associated with 1 unit of the compound is written following a dot. • CaCl2 3H2O Determining the Formula of a Hydrate 1) Start with a known mass of a hydrate. 2) Heat the hydrate to remove the water. 3) Subtract the mass of the compound without water from the original mass. 1) This is the mass of water. 4) Covert the masses to moles by dividing by the molar mass of each compound. 5) Calculate the mole ratio by dividing by the smallest number of moles. Calculating Molecular Formula • A hydrate of BaCl2 XH2O has a mass of 5.00g. After heating until all the water is lost, the compound has a mass of 4.26g. What is the formula of the hydrate? • Mass before heating 5.00g • Mass after heating 4.26g • Mass of water = 0.74g BaCl2 H2O 4.26 g 0.74 g 208.23 g/mol 18.02 g/mol 0.0205 mol = 1 0.041 mol = 2 0.0205 0.0205 • Formula of Hydrate = BaCl2 2H2O Practice Molecular Formula • A mass of 2.50g of blue, hydrated CuSO4 XH2O is placed in a crucible and heated. After heating, 1.59g of white anhydrous CuSO4 remains. What is the formula for the hydrate? • CuSO4 5H2O Molecular Formulas • For many compounds, the empirical formula is not the true formula. • A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound. • The actual formula • The molecular formula for a compound is always a whole-number multiple of the empirical formula. Determining Molecular Formulas 1) Determine the empirical formula 2) Find the empirical formula mass (you are given the molecular formula mass) 3) Divide the molecular formula mass by the empirical formula mass. This is the scalar (multiplier). 4) Multiply the subscripts in the empirical formula by the multiplier. Calculating Molecular Formula • A compound has an empirical formula of Sc2O3. The molecular mass is 414 g/mol. What is the molecular formula? • Empirical Formula = Sc2O3 • Empirical Formula Mass = •2 mol Sc x 44.956 g/mol Sc = 89.912 g Sc • 3 mol O x 15.999 g/mol O = + 47.997 g O molar mass = 137.909 g Sc2O3 • Find the multiplier… • 414 g / 137.909 g = 3 (the multiplier is 3) • Molecular Formula = Sc6O9 Practice Molecular Formula • A molecule has 85.6% carbon and 14.5% hydrogen. If the molecule has a mass of 42.1 grams, what is the molecular formula? End of Unit 8 Be Prepared for Unit 8 Test.

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