Question

4.24mg of a sample of organic compounds completely burnt in oxygen, it gives 8.45 mg of CO2 & 3.46mg . determine the empirical & molecular formula of this compound, if molecular mass is 88u​

Answer 1

Answer:

number of moles = mass / molar mass

molar mass of CO2 = 44 g/ mole

0.00845 g of CO2 has 0.008 /44 = 0.00019 moles of CO2

there is 1 mole of C in CO2 and all the C from the compound becomes CO2

so moles of C in the compound = 0.00019 moles

mass of C = 0.00019 x 12 = 0.00231 g

molar mass of H2O = 18 g/ mole

0.00346 g of H2O has 0.00346 / 18 = 0.00019 moles of H2O

there are 2 moles of H in H2O so moles of H in the compound = 0.00038 moles

mass of H = 0.00038 x 1.0079 = 0.0004 g

mass of H + C = 0.00269 g

mass of sample = 0.00424 g

mass of O by difference = 0.00155 g

moles of O = 0.00155 /16 = 0.00010 moles

molar ratio of C : H : O = 0.00019 : 0.00038 : 0.00010

smallest number 0.00010

divide the ratio by the smallest number we get

molar ratio of C : H : O = 2.0 : 4.0 : 1.0

empirical formula is C2H4O

Answer 2

$Step I : Calculation of mass percentage of different elements Percentage of carbon. Percentage of carbon can be calculated as follows : </p><p></p><p> 44 parts 12 parts or 44 mg 12 mg 44 mg of</p><p></p><p>contain C = 12 mg </p><p></p><p>g of</p><p></p><p>contain</p><p></p><p>mg Percentage of</p><p></p><p> Percentage of hydrogen. Percentage of hydrogen can be calculated as : </p><p></p><p> 18 parts 2 parts 18 mg 2 mg 18 mg of</p><p></p><p>contain H = 2mg </p><p></p><p>of</p><p></p><p>contain</p><p></p><p>mg Percentage of</p><p></p><p> The sum of the percentage of C and H</p><p></p><p> As this comes out to be less than 100, the balance must be oxygen. </p><p></p><p>Percentage of</p><p></p><p></p><p>$