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Combustion data of an organic compound yielded #74.0%, C;# #7.4%, H;# #8.6%, N;# and the balance oxygen. What is the empirical formula?

Answer 1

#C_10H_12NO#

As with all problems of this type, we assume a #100*g# mass of compound, and we determine the molar composition with respect to each element:

Consequently:

#"Moles of carbon"# #=# #(74.0*g)/(12.011*g*mol^-1)# #=# #6.16*mol#

#"Moles of hydrogen"# #=# #(7.4*g)/(1.00794*g*mol^-1)# #=# #7.34*mol#

#"Moles of nitrogen"# #=# #(8.6*g)/(14.01*g*mol^-1)# #=# #0.614*mol#

#"Moles of oxygen"# #=# #(10.0*g)/(16.00*g*mol^-1)# #=# #0.625*mol#

After obtaining the molar composition, we can obtain our empirical formula by simply dividing by the smallest molar quantity.

Dividing thru by #0.614*mol# we get #C_10H_12NO# (clearly, I have done some rounding off!), as the empirical formula.

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Answer 2

Cooper Adams

The empirical formula of the organic compound is CH2NO2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.