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Chapter 8. Chemical Composition. Students will learn…. Mole as a counting unit amu = atomic mass unit molar mass % composition of compounds Formulas of compounds empirical formulas molecular formulas. Mole (mol). A) Is a counting unit B) 1 mole = 6.02 x 10 23 particles
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Chapter 8 Chemical Composition
Students will learn… • Mole as a counting unit • amu = atomic mass unit • molar mass • % composition of compounds • Formulas of compounds • empirical formulas • molecular formulas
Mole (mol) A) Is a counting unit B) 1 mole = 6.02 x 1023 particles (ex) 1 mol of pencils = 6.02 x 1023 pencils 1 mol of iron = 6.02 x 1023 Fe atoms 1 mol of CO2 = 6.02 x 1023 CO2molecules 1 mol of Ca2+ ions = 6.02 x 1023 Ca2+ ions C) Also called Avogadro’s number
Know the Particle Types! • molecules: particles of covalent compounds (nonmetal + nonmetal) • formula units: particles of ionic compounds (metal + nonmetal) • ions: positively or negatively charged particles (cation & anion) • atoms: particles of element
Conversion Factors 1) always in fraction form and equal to 1 (Ex) 2) Two conversion factors are written from one equal relationship. (Ex) 3) Also called as unit or dimensional analysis
4) Always follow the same set-up. Given quantity Conversion Wanted quantity factor (Ex 1) How many oxygen atoms are in 4.5×108 formula units of CuSO4?
amu (atomic mass unit), u • a mass unit for a very small mass • defined by assigning 12.0 amu to the mass of a carbon-12 atom • 1 amu= mass of 1/12 of a C-12 atom =1.66053886 × 10-27kg • 1 proton ≈ 1 amu 1 neutron ≈ 1 amu 1 electron ≈ 0 amu
amu continues… • The amu of other element is determined by comparing its weight to the weight of a carbon-12 atom. • The decimal number in the periodic table is the average amu of the isotopes of each element • The word, average, will be left out. • Isotopes = atoms with the same atomic number but different atomic mass number
Molar Mass • The mass of 1 mole of a pure substance • The mass of 6.02x1023particles • Numerically the same as the average atomic mass unit (amu) (Ex) Mass of 1 C-12 atom = 12.0 amu Mass of 1 mol C-12 = 12.0 g
Example The (average) mass of one Cu atom =63.546 amu The(average) atomic mass of Cu = 63.546 g
Molar Mass of Compound • mass of H2O? 1 mol H2O = 2 mol H + 1 mol O • Formula weight of Ca(NO3)2? 1 mol Ca(NO3)2 = 1 mol Ca + 2 mol N + 6 mol O
Use the correct terms • For elements, molar mass = atomic mass • For compounds, molar mass = formula mass ** In chemistry, mass = weight
Practice Find the molar mass of: 1) C6H5CH3 (toluene) 2) CaHPO4 3) Ca3(PO4)2 4) CrSO4∙3H2O
Answers 1) 92.15 g 2) 136.1 g 3) 310.2 g 4) 202.1 g
Learned So Far mol Avogadro’s # Molar mass # of particles (atoms, molecules, formula units, ions) mass
Problem Solving Strategy • Know what are you solving for. • Know what is the given quantity to convert. • Know what is the conversion factor. • 1 mol = 6.02x1023 particles • 1 mol = mass of 1 mol of a substance • Use the same set-up format *Called the “box” or “railroad track”
Examples Mol grams 1.A certain sample of calcium phosphate contains 4.86 mol. What is the mass in grams of this sample?
Grams mol 2.The molar mass of sodium sulfate, Na2SO4. How many moles of sodium sulfate are in 300.0 g?
Grams # of particles 3.Isopentyl acetate, C7H14O2 (MW=130.2 g) the compound responsible for the scent of bananas, can be produced commercially. How many molecules of isopentyl acetate are in 1 µg?
% Composition of Compound • Mass percent of an element: • (Ex) For iron in iron(III) oxide, (Fe2O3):
Chemical Formulas • Empirical formula ≈ simplest formula • shows the lowest whole-number ratio of the atoms in a compound (Ex) a benzene molecule = C6H6 Empirical formula of benzene = CH • Molecular Formula • shows the actual composition of molecules (Ex) Molecular formula of benzene = C6H6
EF & MF Relationship • MF = n·EF • n = a whole number ≥ 1
From mass % to MF • Get the mass % of each element • Get the mass of each element in 100 g sample of compound • Convert the mass of each element to moles • Reduce the moles to the lowest whole numbers • Do not apply the “5 and up” rule.
Why “5 and up” can’t be applied? Compounds made of 2 elements, C and H: • 1 mol C: 0.5 mol H 1 mol C:1 mol H • 1 mol C: 1.2 mol H 1 mol C:1 mol H • 1 mol C: 1.33 mol H 1 mol C:1 mol H *Do you see a problem?
1 mol C: 0.5 mol H 10 mol C: 5 mol H 2 mol C: 1 mol H 1 mol C: 1.2 mol H 10 mol C:12 mol H 5 mol C: 6 mol H 1 mol C: 1.33 mol H 100 mol C:133 mol H 3 mol C: 4 mol H ** Handy to remember: 0.25 = ¼; 0.2 = 1/5; 0.33 = 1/3
How to round correctly? • Multiply the moles by a factor of 10 until all moles become whole numbers • Reduce the whole-number moles to the lowest whole numbers
Examples • mass % to EF The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). What is the empirical formula? C3H5O2
The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. The empirical formula is C3H5O2. What is the molecular formula? C6H10O4 • Mass % to MF