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Chapter 3 Chemical Reactions and Reaction Stoichiometry Atomic mass calculation: Ex: Carbon occur as a mixtures of three isotopes in nature. 98.89% 12C, 1.11% 13C, 0.01% 14C . Calculate its atomic mass. Atomic Mass of Carbon =(.9889x12)+(.0111X13)+(.0001x14) ( =12.01 amu 2 Average Atomic Mass Practice An element occurs in nature as a mixture of two isotopes such as 7.50% with an atomic mass of 6.01521 grams and 92.50% with a mass of 7.016003 grams. Calculate the average atomic mass of the element. • The mole is a unit that allows us to count the number of atoms or molecules in a substance • The mole is also related to mass Avogadro’s Number( Mole) Avogadro’s number is the number of atoms in 12.01 grams of carbon.12.01 g sample of carbon contains 6.02 x 1023 carbon atoms. Mole (mol) is a unit of measure for an amount of a chemical substance. Mole is Avogadro’s number of particles. 1 mol = Avogadro’s number = 6.02 x 1023 particles Mole Calculation Steps 1 mol = 6.02 x 1023 particles(molecules) 1 mol = GFM( gram formula mass) First we need to find the conversion factor that allows us to convert between mass and moles Molecules Moles (mol) Mass (g) The conversion factor is called: gram formula mass (GFM) or molecular mass grams The unit for gram formula mass is: mol Samples of One Mole Quantities = 6.02 x 1023 C atoms 1 mole Al = 6.02 x 1023 Al atoms 1 mole S 6.02 x 1023 S atoms 1 mole C = 1 mole H2O = 6.02 x 1023 H2O molecules 1 mole CCl4 = 6.02 x 1023 CCl4 molecules Calculating Molar Mass/ molecular mass/ gram formula mass(GFM) The molar mass of a substance is the sum of the molar masses of each element. Ex: What is the gfm of iron(II) nitrate? The sum of the atomic masses is as follows: 55.85+ 2(14.01) + 6(16.00 )= 179.87 amu The molar mass for Fe(NO3)2 = 179.87 g/mol 1 mol Fe(NO3)2 = 179.87g Mole calculation Steps 1. Write down given and find 2. Choose correct conversion factor: – “Avogadro’s #” or “GFM”( gram formula mass) 3. Multiply given by correct conversion factor Mole Calculations • How many potassium atoms are in 3.6740 mol of K? 3.6740 mol K x 6.02 x 1023 atoms 1 mol K = 2.2117 x 1024 atoms Mole Calculations Avogadro’s number is used to convert particles of a substance to moles. Ex:-How many moles of CO2 are in 2.50 x 1024 molecules CO2? 2.50 x 1024 molecules CO2 x 1 mole CO2 6.02 x 1023 molecules CO2 = 4.15 mole CO2 Using Avogadro’s Number Avogadro’s number is used to convert moles of a substance to particles. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 1023 Cu atoms 1 mole Cu = 3.0 x 1023 Cu atoms Moles Practice How many atoms are in 5.00 moles of aluminum nitrate? Moles Molecules Atoms Moles Practice How many moles of oxygen gas (O2) are in 7.5 x 1024 molecules? Percent Composition Mass percent of an element: mass of element in compound mass % = × 100% mass of compound Ex: What is the % of iron in iron(II) nitrate? % Fe = 55.85x100/179.87 = 31.05 % 16 Mass Percent Practice What is the percentage of sodium in sodium sulfate? Empirical Formula(EF) • The empirical formula of a compound is the simplest whole number ratio of atoms of each element in a molecule. • The molecular formula of benzene is C6H6. EF= CH. • The molecular formula of octane is C8H18. EF= C4H9. Ex: 1) Which of these is NOT empirical formula: • • • • NH CH3O2 N6H9O3 CH8O12 2) Which of these is an empirical formula? • N2H5O7 • N2H4O8 • C2H4 • C2O4 Empirical Formula Calculations What is the empirical formula for a compound that is 60.3% Mg and 39.7% O? Element Mass (g) or Mass % g or %_____ atomic mass mol Simple Ratio (divide by smallest #) Mg 60.3 60.3/24.31 2.480 1 O 39.7 39.7/16 2.481 1 Answer: EF = MgO Empirical Formula Practice Aspirin contains 60.0% carbon, 4.48% hydrogen, and 35.5% oxygen. What is its empirical formula? Molecular Formula (MF) The actual molecular formula is some multiple of the empirical formula, ( EF)n. Molecular Formulas Remember, these give the actual number of atoms, not just the ratio. 1. Determine the empirical formula. 2. Find the empirical formula mass 3. Divide the molecular mass by empirical formula mass & multiply your empirical formula by this number Molecular Formula Calculations Ex: The empirical formula is SO2 and the molecular mass of the compound was experimentally determined to be 128.128g/mol. What is the molecular formula? EF mass = S x 1 = 32.066 x 1 = 32.066 O x 2 = 15.999 x 2 = 31.998 EF mass = 64.064 n= MF mass/ EF mass = 128.128/64.064 = 2 Molecular Formula =( SO2 )2 = S2O4 Molecular Formula Practice A compound has 49.3% carbon, 6.90% hydrogen, and 43.8% oxygen. The molar mass of the compound is 146 grams/mol. Determine the empirical and molecular formulas. Balancing chemical equations Potassium iodide reacts with lead nitrate to yield lead iodide and potassium nitrate. 2 KI + Pb(NO3)2 PbI2 +2 K(NO3) Balancing Practice 1) Methane (CH4) reacts with oxygen to form carbon dioxide and water. 2) Nitrogen gas reacts with hydrogen gas to form nitrogen trihydride. Stochiometry 2 NO(g) + O2(g) → 2 NO2(g) In a balanced chemical equation “2 moles of NO gas react with 1 mole of O2 gas to produce 2 moles of NO2 gas.”The coefficients indicate the mole ratio, of reactants and products. Stochiometry Practice 15.0 grams of copper (II) chloride react with 20.0 grams of sodium nitrate to form copper(II) nitrate and sodium chloride. a) How much sodium chloride can be formed? b) What is the limiting reactant for the reaction? c) If 11.3 g of sodium chloride are formed in the reaction, what is the percentage yield of the reaction? Stochiometry Practice Sodium reacts with water to form sodium hydroxide and hydrogen gas. How many moles of hydrogen gas are produced from 5.4 grams of sodium? Stochiometry Practice Copper (II) chloride reacts with aluminum to form aluminum chloride and copper. How many grams of aluminum would it react to form 145 grams of copper? Percent Yield When you perform a laboratory experiment, the amount of product collected is the actual yield. The amount of product calculated from a limiting reactant problem is the theoretical yield. The percent yield is the amount of the actual yield compared to the theoretical yield. actual yield x 100 % = percent yield theoretical yield Percent Yield & Limiting Reactant Practice Ex: 54.0 grams of silver chloride combine with 32.0 grams of magnesium nitrate to produce silver nitrate. a) How many grams of magnesium chloride will be produced? b) Which reactant is the limiting reactant? c) After the reaction, you only obtained 14.35 g of magnesium chloride. What is percent yield?