Chem 1 Lecture Notes on Molecular Weight, Molar Mass, and Moles PDF

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Lecture notes on molecular weight, percent composition, Avogadro's number, and empirical formula from Dr. Ariyadasa's course...

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Calculate the formula weight of Al2O3: ● (Atomic weight of Al x 2) + (atomic weight of O x 3) ● (26.9815 amu x 2) + (15.994 amu x 3) = 101.9612 amu Molecular weight- the sum of the atomic weights of the atoms in a molecule (also called formula weight) Calculate the molecular weight of benzene: ● C6H6 ● (AW of C x 6) + (AW of H x 6) ● (12.0107 amu x 6) + (1.00794 amu x 6) = 78.11184 amu Percent composition: % element = ((# of atoms x atomic weight) / MW of the compound) x 100 Calculate the percent composition of carbon in benzene: ● ((6 x 12.0107 amu) / 78.1118 amu) x 100 = 92.2579% Avogadro’s number: One mole (abbreviated as mol) is the number of particles found in exactly 12g of 12C 6.022 x 10^23 atoms or molecules is the number of particles in one mole Molar mass: The mass of 1 mol of substance The molar mass of an element is the atomic weight for the element from the periodic table The formula weight will be the same number as the molar mass Calculate the molar mass of water: ● H2O ● 2 moles of hydrogen to 1 mole of oxygen ● (1.00794 g/mol x 2) + (15.9994 g/mol) = 18.01528 g/mol ● 1 mol H2O = 18.01528 g Find the mass in grams of 2.35 x 10^-2 mol CaCO3: ● Molar mass of CaCO3 = 40.078 g/mol + 12.0107 g/mol + (15.9994 g/mol x 3) = 100.087 g/mol ● 2.35 x 10^-2 mol CaCO3 x (100.087 g CaCO3 / 1 mol CaCO3) = 2.35 g CaCO3 Find the number of moles in 10.5 g of Al(NO3)3: ● (MW of Al) + (MW of N x 3) + (MW of O x 9) ● (26.981538 g/mol) + (14.0067 g/mol x 3) + (15.9994 g/mol x 9) = 212.9963 g Al(NO3)3 ● 10.5 g Al(NO3)3 x (1 mol Al(NO3)3 / 212.9963 g Al(NO3)3) =0.0493 mol Al(NO3)3 Calculate the number of molecules in 0.456 mol of benzene: ● C6H6 ● 1 mol C6H6 = 6.022 x 10^23 molecules of C6H6 ● 0.456 mol C6H6 x (6.022 x 10^23 molecules of C6H6 / 1 mol C6H6) = 2.75 x 10^23 molecules C6H6 How many atoms in 3 g of copper: ● 1 mol Cu = 63.546 g Cu ● 1 mol Cu = 6.022 x 10^23 Cu atoms ● 3 g Cu x (1 mol Cu / 63.546 g Cu) x (6.022 x 10^23 Cu atoms x 1 mol Cu) = 3 x 10^22 atoms Cu

Empirical Formula- the lowest whole number ratio possible which gives the relative number of atoms of each element one molecule contains The empirical formula of H2O2 would be HO because the ratio of the molecular formula is 2:2 and that can be reduced to 1:1 The empirical formula for C6H12O6 would be CH2O because the ratio of the molecular formula is 6:12:6 and that can be reduced to 1:2:1 A sample was found to contain 0.423 g of C, 2.50 g of Cl, and 1.34 g of F. What is the empirical formula of the compound?...