Question

An organic compound contains C, H, N, and O. Combustion of 0.3069 g of the compound produces 0.8301 g CO2 and 0.2975 g H2O. A sample of 0.5322 g of the compound was analyzed for nitrogen. At STP, 30.54 mL of dry N2 (g) was obtained. In a third experiment, the density of the compound as a gas was found to be 3.68 g/L at 143 deg C and 244 torr. Calculate the empirical formula and the molecular formula of the compound.

Answer 1

Calculate the number of moles of CO2 and H2O generated

moles of CO2 = mass in grams / molar mass = 0.8301 / 44 = 0.01886moles

moles of H2O = 0.2975 / 18 = 0.0165 moles

One molecule of CO2 is produced by one atom of carbon and in organic compound. Similarly one molecule of H2O is produced by 2 atoms of Hydrogen.

Therefore moles of C in the compound = 0.01886

moles of H in the compound = 2 X moles of water = 2 X 0.0165 = 0.033 moles

Moles of N2 = Volume in mL / 22400 mL

Because 1 mole of a gas at STP = 22400mL

Moles of N2 = 30.54 / 22400 = 0.00136 moles

One N2 molecule contains two atoms,

Therefore moles of N atoms = 2 X 0.00136 = 0.0027 moles

Convert the No. of moles to % mass

% of C by mass = (Mass of carbon / mass of the sample) X 100

% of C by mass = (No. of moles X atomic mass of C / mass of the sample) X 100

Because mass of the element = No. of moles X Atomic mass

% of C by mass = (0.01886 X 12 / 0.3069) X 100 = 73.74%

% of H by mass = (0.033 X 1.008 / 0.3069) X 100 = 10.83%

% of N by mass =(0.0027 X 14 / 0.5322) X 100 = 7.10%

Sum of percentages = 73.74 + 10.83 + 7.10 = 91.67

Therefore % of O = 100 - 91.67 = 8.33%

Mass of O = % mass X mass of the sample = 0.0833 X 0.3069 = 0.0255

No. of moles of O = 0.0255 / 16 = 0.0016

Atoms C H N O

Moles 0.01886 0.033 0.0027 0.0016

Simple ratio 11.78 20.62 1.68 1

12 21 2 1

Emperical formulla = C₁₂H₂₁N₂O

To calculate the chemical formulla molecular mass must be given