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Chapter 3: Problem 96

An organic compound was found to contain only \(\mathrm{C}, \mathrm{H},\) and Cl. When a \(1.50-\mathrm{g}\) sample of the compound was completely combusted in air, 3.52 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment, the chlorine in a \(1.00-\mathrm{g}\) sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the organic compound containing only carbon, hydrogen, and chlorine is C8H50Cl.

Step by step solution

01

Calculate the moles of carbon in CO2

Given that 3.52 g of CO2 is produced from combustion of the 1.50 g sample, we can calculate the moles of carbon in CO2 using the molar mass of carbon (12.01 g/mol) and CO2 (44.01 g/mol): Moles of carbon = (3.52 g CO2) x (1 mol CO2 / 44.01 g CO2) x (1 mol C / 1 mol CO2) = 0.0800 mol C
02

Calculate the moles of chlorine in AgCl

We are given that 1.27 g of AgCl is produced from the 1.00 g sample of the compound. We can calculate the moles of chlorine in AgCl using the molar mass of chlorine (35.45 g/mol) and AgCl (143.32 g/mol): Moles of chlorine = (1.27 g AgCl) x (1 mol AgCl / 143.32 g AgCl) x (1 mol Cl / 1 mol AgCl) = 0.00999 mol Cl
03

Calculate the moles of hydrogen

We have the mass of the original compound (1.50 g) and the moles of carbon and chlorine calculated in Steps 1 and 2. Now, we can find the mass of hydrogen and calculate the moles of hydrogen using the molar mass of hydrogen (1.01 g/mol): Mass of hydrogen = 1.50 g (total mass) - (0.0800 mol C x 12.01 g/mol C) - (0.00999 mol Cl x 35.45 g/mol Cl) = 0.506 g H Moles of hydrogen = (0.506 g H) / (1.01 g/mol H) = 0.501 mol H
04

Find the empirical formula

We have the moles of every element: carbon (0.0800 mol), hydrogen (0.501 mol), and chlorine (0.00999 mol). To find the empirical formula, we'll divide all the calculated moles by their lowest common multiple: C: (0.0800 mol) / (0.00999 mol) ≈ 8 H: (0.501 mol) / (0.00999 mol) ≈ 50 Cl: (0.00999 mol) / (0.00999 mol) = 1 Therefore, the empirical formula of the compound is C8H50Cl.

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Most popular questions from this chapter

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