In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

Explanation

Here, mass of the substance taken = 0.30g
Volume of nitrogen collected = 50 mL, Atmospheric pressure = 715mm Hg
Room temperature = 300K
Aqueous tension at 300 K = 15mm `:.` Actual pressure of the gas (dry gas) = `715 - 15 = 700mm Hg`
Step 1. To convert the volume at experimental conditions to volume at STP.
`{:("Experimental values",,"At STP"),(P_(1)=700 mm,,P_(2) = 760 mm),(V_(1) = 50 mL,,V_(2) = ?),(T_(1) = 300 K,,T_(2) = 273 K):}`
Substituting these values in the gas equation.
`(P_(2)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`, we get `(700 mm xx 50 mL)/(300 K) = (760 mm xx V_(2) mL)/(273 K)` or `V_(2) = (273 xx 700 xx 50)/(300 xx 760) = 41.9 mL`.
Step 2.To convert the volume at STP into mass.
According to the definition of GMV, 22400 mL of nitrogen at STP weigh = 28 g
`:. 41.9 mL` of nitrogen at STP will weigh `= (28 xx 41.9)/(22400) g`
Step 3. To calculate the percentage of nitrogen.
Percentage of nitrogen `= ("Mass of " N_(2) " at STP")/("Mass of the substance taken") xx 100 = (28 xx 41.9 xx 100)/(22400 xx 0.3) = 17.46`