Question

3 grams of an organic compound on combustion give 8.8 grams of co2 and 5.4 grams of water the empirical formula of the compound is​

Answer 1

Answer:

ch3

Explanation:

mass of CO2 = 8.8 gm

or since the molar mass of CO2 is 44gm / mole we have 0.2 moles of CO2

since there is 1 mole so C in CO2we have 0.2 moles of C or 2.4 gm of C

mass of H2O = 5.4 gm or since the molar mass of water is 18 gm / mol we have 0.3 moles of H2O

since there are 2 moles of H in 1 moles of H2O =0.6 moles of H or 0.6gm of H

molar ratio of C : H = 0.2 : 0.6

mass of C+H = 3.0 gmor after dividing by the smallest 0.2molar ratio

C : H = 1 : 3

Hence empirical formula of the compound is CH3.

{ *this is based on combustion of alkane.}

your welcome  :) :) :)

Answer 2

Answer:

CH₃

Explanation:

• Mass of CO₂ =8.8g
• Molar mass of CO₂=44g
• No. Of moles of CO₂=$\frac{given mass}{molar mass}$

                                          =$\frac{8.8}{44}$

                                    =0.2 moles of CO₂

• Similarly, Mass of H₂O=5.4g
• molar mass of H₂O = 18 g
• No. Of moles of CO₂=$\frac{given mass}{molar mass}$

                                           =$\frac{5.4}{18}$

• No. Of moles of H₂O= 0.3 moles
• No. of moles of C=0.2 (one mole of carbon in one mole of CO₂
• No. of moles of H=0.6( 1 mole water has 2 moles of hydrogen So, 0.3 moles of water has 0.6 moles of hydrogen)

• Given C:H=0.2:0.6 =1:3

• Emperical formulae of compound is = CH₃