Question

A compound contains 48.64% C, 8.16% H and 43.20% O. The molar mass is 148g. What is the molecular formula? C2H302 CH,02 C6H1204 C6H1206 * Previous

Answer 1

solution Since the scale for percentage is 100, it is the most Convenient to caleulate the massot elements present in a sample g. the calculation is most convenients because, per the definition for percent composition, the mass of the givenelement in grams is rumerically equivalent to the elements mass percentage The mass percentage provided may be more conveniently expressed as fractions 48.64% C = 48.69 g - 0.4864 100 g compound 43.20% OF 43.20 go 0.4320 100 g compound 8.16 % H = 2.16gH = 0.0816 100g compound The molar amounts of carbon , oxygen and hydrogen in 100 g sample are caleulated by dividing each element's mass by its molar mass 48.64 % (mole) = 4.0499 mole 12.01g . - 43,20% 0 (molo = 2.7 molo 16100g 8.16 % H (molt ) = 8.095 mol H. Coefficients of the tentative empirical formula are devided by 2 , then we get

moles of @ = moles of o= moles of H = 4.0 499 mot 2 moles 2.7mden 1.35 moles 2. 8.095 A moles empirical formula = (C2Hq 01.35 de. molar mass of the compound - 1989, Caleulating mass from empirical formula = [(12x2]n+ an+ (76X1:35)n] g. = (24+4 +21.6 Dng = 49.60 g. u. From the questions we get 49.62 g = 1989 → 49.60 =198 3 = 2.983 ~ 3. 2. R = 3, so molecular formula = (C2H2 0 1.35) Here (1.35 x 3) = 4.05 ~ 4. = CH1 24. Hence molecular formula is C6H12O4 Final any way O (2 H3 L ос, Н, О, C6H12O4 O C6H12O6 na 198 - 4916 - 203 – . ООО