a. a compound weighing 5.714 mg produced 14.414 mgCO 2 (44.010 g/mol) & 2.529 mg

Question

a. a compound weighing 5.714 mg produced 14.414 mgCO2 (44.010 g/mol) & 2.529 mg of H20(18.015 g/mol) on combustion. Find the weight % of C (12.0107 g/moland H (1.00704 g/mol) in the sample. b. a 10 ml solution containing Cl- was treated withexcess AgNO3 to precipitate 0.436 g of AgCl (143.321g/mol). What was the molarity of Cl- in the unkownsolution? a. a compound weighing 5.714 mg produced 14.414 mgCO2 (44.010 g/mol) & 2.529 mg of H20(18.015 g/mol) on combustion. Find the weight % of C (12.0107 g/moland H (1.00704 g/mol) in the sample. b. a 10 ml solution containing Cl- was treated withexcess AgNO3 to precipitate 0.436 g of AgCl (143.321g/mol). What was the molarity of Cl- in the unkownsolution?

Explanation / Answer

a. uncompound .........> CO2 + H2O Hence the number of moels of C present in the unknown compoundequals to number of moles of C present in the 14.414mg of CO2
Weight of un known compound = 5.714mg     we know that moles = mass / molaramss so moles of CO2 = (5.714mg (1g/1000mg) ) / (44.010g/mol)                           = 1.30*10^-4 mol C Mass of 1.30*10^-4 mol C = 1.30*10^-4 mol C *12g/mol                                            = 0.0016g so mass percent of C = ({0.0016g (1000mg/1g) ) /(5.714mg)}*100                                 = 28.001% b. 0.436g of AgCl = 0.436g / (143.321g/mol)                             =0.00304mol so moles of Cl- = 0.00304 mol concentration = moles of solute / volume of solution so [Cl-] = 0.00304mol / 10mL(1L/1000mL)               = 0.00304mol / 0.01L               = (0.00304/0.01)mol/L               = 0.304M

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