2.0 g of a bromo hydrocarbon (X) was subjected to Carius analysis, gave 3.36 g of AgBr. The percentage of carbon in the compound (X) is 26.7\%. Total number of carbon atoms in the empirical formula for compound (X) is _____. (Given molar mass in g mol^ -1 H:1, C:12, Br:80, Ag:108)
A. A
Mass of AgBr formed = 3.36 g Molar mass of AgBr = 108 + 80 = 188 g mol^ -1 Moles of AgBr formed = 3.36 188 0.01787 mol Since 1 mole of AgBr contains 1 mole of Br, moles of Br in the compound = 0.01787 mol. Mass of Br = 0.01787 80 = 1.43 g Percentage of Br in compound (X) = 1.43 2.0 100 = 71.5\% Given percentage of C = 26.7\% Percentage of H = 100\% - (71.5\% + 26.7\%) = 1.8\% Now, we find the molar ratio of the elements in the compound: Moles of C = 26.7 12 = 2.225 Moles of H = 1.8 1 = 1.8 Moles of Br = 71.5 80 = 0.893 Dividing by the smallest value (0.893) to get the simplest ratio: C : H : Br = 2.225 0.893 : 1.8 0.893 : 0.893 0.893 C : H : Br 2.5 : 2 : 1 Multiplying by 2 to obtain whole numbers, we get the ratio 5 : 4 : 2. The empirical formula of the compound is C_5H_4Br_2. The total number of carbon atoms in the empirical formula is 5. Answer: 5
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