Expt #2-Electrophilic nitration of a mono-substituted benzene Combustion analyti
ID: 570264 • Letter: E
Question
Expt #2-Electrophilic nitration of a mono-substituted benzene Combustion analytical data of the mono-substituted benzene sample Series 100 contains carbon, hydrogen and oxygen, and has a molecular weight of 136 g/mol. In the stion of a 5.000 g sample, 12.943 g of CO, and 2.655 g H,O were produced. Series 200 contains carbon, hydrogen, oxygen and nitrogen, and has a molecular weight of 121 g/mol. In was found to contain 0.580 g of nitrogen stion of a 5.000 g sample, 12.723 g of CO and 2.610 g H,0 were produced. This sample also 140-win 300 contains carbon, hydrogen and nitrogen, and has a molecular weight of 103 g/mol. In the f a 5.000 g sample, 14.945 g of CO2 and 2.204 g H0 were produced. This sample also was found to contain 0.0.679 g of nitrogen. From the a bove data, calculate the empirical and molecular formulas of your unknown sample. Use the 4.me 14-17 following atomic weights: C-12.01 H- 1.01 0-16.00 N-14.01Explanation / Answer
100.
% C by mass = 12 *mass of CO2*100 / (44*mass of sample) = 12 * 12.943 * 100 / ( 44 * 5.000) = 70.6 %
% H by mass = 2 * mass of H2O*100/ (18*mass of sample) = 2 * 2.655 * 100 / (18 * 5.000) = 5.90 %
% by mass of O = 100 - 70.6 - 5.90 = 23.5 %
Element Mass Number of moles Division by small number Nearest whole number
C 70.6 70.6 / 12 = 5.88 5.88 / 1.47 = 4.00 4
H 5.90 5.90 / 1 = 5.90 5.90 / 1.47 = 4.02 4
O 23.5 23.5 / 16 = 1.47 1.47 / 1.47 = 1.00 1
The emperical formula = C4H4O
Emperical formula mass = 4 ( 12 ) + 4 ( 1 ) + 1 ( 16 ) = 68 g/mol
Molecular mass = 136 g/mol
n = molecular mass / emperical formula mass
n = 136 / 68
n = 2
Therefore,
Molecular formula = [Emperical formula]n
Molecular formula = [C4H4O]2 = C8H8O2
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