Answer to Question #287434 in General Chemistry for Hardhe

Question

On combustion, 0.0065g of an organic compound containing C, H and N. Only yielded 0.0146g of CO2 and 0.0089g of H2o. What is the empirical formular of the compound?

Answer 1

Moles of CO2 = mass/Molecular mass

= 0.0146/44 = 0.000332

Mole ratio of CO2 : C = 1 : 1

Moles of CO2 = moles of Carbon

Moles of Carbon = 0.000332

Moles of H2O = mass/Molecular mass

0.0089/18 = 0.00494

Moles ratio of H2O : H = 1:2

Moles of Hydrogen = 2 ÷ 0.000494 =0.000247

Mass = moles × mass

Mass of Carbon = 0.000332 × 12 = 0.003984g

Mass of Hydrogen = 0.000247 g

Mass of Nitrogen = 0.0065 - ( 0.003984 + 0.000247) = 0.002269g

Moles of Nitrogen = 0.002269 ÷14 = 0.0001620714 moles

0.0001620714÷0.0001620714 = 1

0.000247 ÷ 0.0001620714 = 1.5

0.000332 ÷ 0.0001620714 = 2

Answer = C4H3N