Question

# Combustion analysis of a compound yielded 269.01 g CO2, 55.09 g H2O, 70.30 g NO2, and...

Combustion analysis of a compound yielded 269.01 g CO2, 55.09 g H2O, 70.30 g NO2, and 97.90 g SO2.

(a) What is the empirical formula of the compound? (Assume it contains no oxygen.)

(b) If the molar mass of the compound is 196.30 g/mol, what is the molecular formula of the compound?

Mass if CO2 obtained is 269.01

Molar mass of CO2 is 44 g/mole

Number of moles of CO2 equals 6.11

Similarly, number of moles of water equal s 55.09/18= 3.06

Number of moles of NO2 = 70.3/46=1.528

Number of moles of SO2 = 97.9/64 = 1.528

Divide all number of moles by minimum ie 1.528

Then CO2 will come 4

Water will come 2

SO2 will come 1

NO 2 will come 1

Kindly note that one mole carbon in the molecule produce one mole CO2

2 mole hydrogen in mole produce one mole water

1mole N produced one mole NO2

One mole S produces one mole SO2

Empirical formula is thus C4H4SN

Empirical formula mass =98 (note molar mass is double of empirical mass)

Thus molecular formula is C8H8N2S2

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