Question

1. A gaseous compound weighing 1.080 g and occupying 213 mL of volume at 131 ̊C and 0.985 atm of pressure is found by elemental analysis to be comprised of 0.6781 g C, 0.1365 g H, and 0.2651 g N

a) what are its empirical and molecular formulas?

b) What is the density of this gas at STP?

c) The % mass composition of C in the compound?

Answer 1

We have

C = 0.6781 g

H = 0.1365 g

N = 0.2651 g

Dividing by atomic numbers of each

C = 0.6781 / 12 = 0.05650

H = 0.1365 /1 = 0.1365

N = 0.2651 / 14 = 0.01894

dIvding by smallest numer to all

We get

C = 0.05650 /0.01894 = 2.98 approx = 3

H = 0.1365 /0.01894 = 7.20 approx = 7

N = 0.01894 / 0.01894 = 1

So emperical formula is

C₃H₇N

b)

Now density = mass/volume

So

Density = 1.080 g / 213 mL

= 0.005 g/mL

c)

% mass composition of C in compound = mass of C / mass of compound x 100

= (0.6781 g / 1.080 g) x 100

= 62.79 %