Question

A 3.09 g

3.09 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 3.34 g

3.34 g 

CO2 and 1.82 g

1.82 g H

2

O

H2O. A second sample of this compound with a mass of 4.50 g

4.50 g produced 2.95 g

2.95 g 

SO3. A third sample of this compound with a mass of 8.33 g

8.33 g produced 4.30 g

4.30 g HNO

3

HNO3. Determine the empirical formula of the compound. Enter the correct subscripts on the given chemical formula.

Answer 1

Solution:

$\dfrac{3.34g}{44\;(\frac{g}{mole})}*(\dfrac{1\;mole\;C}{1\;mole\;CO_2})=0.0758925\;mole\;of\;C\\(0.0758925\;mole\;C)*(12.01078\;(\frac{g}{mole}))=0.911528\;g\;of\;C$

$\dfrac{1.82g}{18.01532\;(\frac{g}{mole})}*(\dfrac{2\;mole\;H}{1\;mole\;H_2O})=0.202050\;mole\;of\;H\\(0.202050\;mole\;H)*(1.007947\;(\frac{g}{mole}))=0.203656\;g\;of\;H$

$\dfrac{2.95g}{80.0632\;(\frac{g}{mole})}*(\dfrac{1\;mole\;S}{1\;mole\;SO_3})=0.0253008\;mole\;of\;S\\(0.0253008\;mole\;S)*(32.0655\;(\frac{g}{mole}))=0.811283\;g\;of\;S$

$\dfrac{4.3g}{63.01296\;(\frac{g}{mole})}*(\dfrac{1\;mole\;N}{1\;mole\;HNO_3})*(\dfrac{3.09g}{8.33g})=0.0253135\;mole\;of\;N\\(0.0253135\;mole\;N)*(14.00672\;(\frac{g}{mole}))=0.354559\;g\;of\;N$

(3.09 g total) - (0.911528 g C) - (0.203656 g H) - (0.811283 g S) - (0.354559 g N) =

0.808974 g O

$\dfrac{0.808974\;g\;O}{15.99943\;(\frac{g}{mole})}=0.0505627\;mole\;of\;O$

Divide by the smallest number of moles:

$\dfrac{0.0758925\;mole\;C}{0.0253008\;mole}=2.9996$

$\dfrac{0.202050\;mole\;H}{0.0253008\;mole}=7.9859$

$\dfrac{0.0253008\;mole\;S}{0.0253008\;mole}=1$

$\dfrac{0.0253135\;mole\;N}{0.0253008\;mole}=1.0005$

$\dfrac{0.0505627\;mole\;O}{0.0253008\;mole}=1.9985$

Round to the nearest whole numbers to find the empirical formula:C₃H₈NO₂S