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The concepts of empirical and molecular formulas, their relationship, and how to find the empirical formula given the molecular weight or the percentage composition of a compound. Several examples are provided to illustrate the concepts.
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that can be reduced (ex: formaldeyhyde,CH 2 O). They are different in any case where the molecular formula may be reduced to a smaller whole-number ratio of elements (ex: acetic acid, C 2 H 4 O 2 has the same empirical formula as formaldehyde, CH 2 O).
NH =15 g/ (^) mol. n = 30.0 g/ (^) mol 15.0 g/ (^) mol = 2 2 x NH = N 2 H 2
CH 3 =15 g/ (^) mol. n = 30.0 g/ (^) mol 15.0 g/ (^) mol = 2 2 x CH 3 = C 2 H 6
Assume 100g of the compound to simplify the problem. 81.7g C x (1 mol C / 12.01g C) = 6.803 mol C … divide both by 6.803 to reduce the ratio = 1 mol C 18.3g H x (1mol H / 1.01g H) = 18.12 mol H … to 2.663 mol H Multiplying by 3 will give the smallest whole number ratio: 3 (C 1 H2.663) = C 3 H 8
g/mol, so the empirical and molecular formulas are the same.
54.5g C x (1 mol C / 12.01g C) = 4.538 mol C … divide all by 2.275 to reduce the ratio = 1.995 mol C 9.09g H x (1mol H / 1.01g H) = 9.000 mol H … to 3.956 mol H 36.4g O x (1mol H / 16.00g H) = 2.275 mol O … to 1 mol O This gives the smallest whole number ratio: C 2 H 4 O
Empirical molar mass = 44. g/mol n = 88.0 g/ (^) mol 44.0 g/ (^) mol = 2 2 x C 2 H 4 O = C 4 H 8 O (^2)
CxHyOz + O 2 CO 2 + H 2 O 11.63g 25.5g 14.0g Convert to moles:??? 0.5795mol 0.7778mol
You can assume all the moles of C and all the moles of H came from the isopropanol. 0.5795 mol C x (12.01g C / 1 mol C) = 6.96g carbon 0.7778 mol H 2 O x (2 mol H / 1 mol H 2 O) = 1.5556mol H …x (1.01 g H / 1 mol H) = 1.57 g hydrogen 11.63g total – (6.96g C + 1.57g H) = 3.10g oxygen … (1 mol O / 16.00g O) = 0.1937 mol O Dividing all moles by 0.1937 to reduce the ratio gives: C 3 H 8 O
The molar mass of the empirical formula C 3 H 8 O is 60.0 g/mol, so the molecular formula is the same.