A2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen to yield 4.23 g of co2 and 1.01 g of h2o. another sample of the same compound, of mass 4.14 g, yielded 2.11 g of so3. a third sample, of mass 5.66 g, was burned under different conditions to yield 2.27 g of hno3 as the only nitrogen containing product. determine the empirical formula of the compound.

Empirical formula of the compund is C₆H₇SNO₂

Explanation:

The moles produced of CO₂ are the same than moles of C and the moles of H₂O are half moles of H.

4,23 g CO₂ × = 0,096 moles of CO₂ ≡ moles of C×12,01 g/mol = 1,15296 g of C/ 2,52*100 = 45,80 wt%

1,01 g H₂O × = 0,056 moles of H₂O = 0,112 moles of H×1,01 g/mol= 0,1132 g of H/2,52*100 = 4,49 wt%

Moles of SO₃ are the same than moles of S and moles of HNO₃ are the same than moles of N.

2,11 g SO₃ × = 0,026 moles of SO₃ = 0,026 moles of S×32,065 g/mol= 0,8457 g of S/4,14*100 = 20,43 wt%

2,27 g HNO₃ × = 0,036 moles of H₂O = 0,036 moles of N×14 g/mol= 0,504g of H/5,56*100 = 9,07 wt%

wt% of Oxygen is 100-9,07-20,43-4,49-45,80 = 20,21 wt%

With wt% it is possible to obtain relative moles dividing in molar mass of each atom thus:

moles of C: 45,80/12,01 = 3,81

moles of H: 4,49/1,01 = 4,44

moles of S: 20,43/32,05 = 0,64

moles of N: 9,07/14 = 0,65

moles of O: 20,21/16 = 1,26.

Dividing in the least number:

C: 3,81/0,64 = 6

H: 4,44/0,64 = 7

S: 0,64/0,64 = 1

N: 0,65/0,64 = 1

O: 1,26/0,64 = 2

Thus, empirical formula of the compound is:

C₆H₇SNO₂

I hope it helps!

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.