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MODULE 5 UNIT 1Assignment Part D1.A compound whose empirical formula is CH3has a molar mass of 30 g/mol. What is its molecular formula? (3 points)C: 12.01gH:1.01g The empirical formula is C (12.01g) + H3(3.03g) = 15.04gMolar mass/empirical mass = molecular formula30g./15.04g = 1.99 = 2 x empirical formula of CH3 = C2H6The molecular formula is C2H62.Find the molecular formula for each of the following compounds: (6 points)a)30.5% nitrogen; 69.5% oxygen; molar mass of 46 g/molN: 14.0g/mol. 30.5g/14g/mol. = 2.2 molesO: 16.0g/mol. 69.5g/16g/mol. = 4.3 moles4.3moles /2.2moles = 1.9 = 2 to 1 ratioNO2is the empirical formula N (14g) + O x2 (30g) = 45g/mol.The molecular formula is NO2b)2.67 g of carbon; 0.22 g of hydrogen; 7.11 g of oxygen; molar mass of 90 g/molC: 12.01g/mole 2.67g/12.01g/mol. = 0.22 moleH: 1.01g/mole 0.22g/1.01g/mol. = 0.22 moleO: 16g/mole 7.11g/16g/mol. = 0.44 mole0.44mol./0.22mol. = 2 = 2 to 1 ratio CHO2is the empirical formula with a mass of 45g/mol The molecular formula is C2H2O4 with a molar mass of 90gHint:Start by calculating the empirical formula.