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

1 Answer
Sep 29, 2016

C_10H_12NOC10H12NO

Explanation:

As with all problems of this type, we assume a 100*g100g mass of compound, and we determine the molar composition with respect to each element:

And thus:

"Moles of carbon"Moles of carbon == (74.0*g)/(12.011*g*mol^-1)74.0g12.011gmol1 == 6.16*mol6.16mol

"Moles of hydrogen"Moles of hydrogen == (7.4*g)/(1.00794*g*mol^-1)7.4g1.00794gmol1 == 7.34*mol7.34mol

"Moles of nitrogen"Moles of nitrogen == (8.6*g)/(14.01*g*mol^-1)8.6g14.01gmol1 == 0.614*mol0.614mol

"Moles of oxygen"Moles of oxygen == (10.0*g)/(16.00*g*mol^-1)10.0g16.00gmol1 == 0.625*mol0.625mol

So we have the molar composition. We simply divide thru by the smallest molar quantity to gets our empirical formula:

Dividing thru by 0.614*mol0.614mol we get C_10H_12NOC10H12NO (clearly, I have done some rounding off!), as the empirical formula.