Answer:
mass of sample, M = 1.2359g
mass of CO2, mCO2 = 2.241g
mass of H, mH = 0.0648g
Since the compound is an organic compound,
when organic cpd is combusted it releases CO2 and H2O and NO2.
i) Mass of C, mC = unknown
CO2 molar mass, MCO2 = 44g/mol
C molar mass, MC = 12g/mol
mC = mCO2 × MC
MCO2
mC = 2.241g × 12g of C
44g of CO2
mC = 0.61g of C
C% = 49.36%
ii) mass percent of N = 28.84% of N
mass of N, mN = unknown
NO2 molar mass, MNO2 = 46g/mol
N molar mass, MN = 14g/mol
mN% = mN
M
where, M is mass of the sample.
mN = mN% × M
mN = 28.84% × 1.2359g
mN = 0.356g
iii) mass of O, mO = unknown
mass of sample = mC + mN + mO + mH
1.2359g = 0.61g + 0.356g + mO + 0.0648g
mO = 1.2359g - 1.0308g
mO = 0.2051g
O% = 16.56%
H% = 5.24%
iv) Empirical formula,
Element | C | H | N | O |
% |49.36%|5.24%|28.84%|16.56% |
A. M| 12 | 2 | 14 | 16 |
|49.36/12| 5.24/2 | 28.84/14 | 16.56/1|
n= |C= 4.1 | H= 2.6 |N= 2.1 |O= 1|
A.R |C=4.1|H=2.6|N=2.1|O=1|
Simplest | 4 | 3 | 2 | 1|
ratio
C4H3N20 is the empirical formula.
n = number of moles and at A. R is atomic ratio and I divided by smallest number of moles.
A. M = atomic mass.
