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Nicotine, a component of tobacco, is composed of C, H, and N. A 3.150-mg sample of nicotine was combusted, producing 8.545 mg of co and 2.450 mg of H20. what is the empirical formula for nicotine?

Md Ariful Islam

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Md Ariful Islam

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And so we got $C_xH_yN_n$

We combust a $3.159*mg$ mass of the stuff to get....

$(8.545*mg)/(44.01*g*mol^-1)=1.94xx10^-4*mol$ $CO_2$...i.e. a mass with respect to carbon of $2.33*mg$....($CO$ would not result from the combustion.)

$(2.450*mg)/(18.01*g*mol^-1)=1.36xx10^-4*mol$ $H_2O$...i.e. a mass with respect to hydrogen of $0.274*mg$, $0.272*mg$..

The balance of the mass was due to NITROGEN....${3.159-2.33-0.272}*mg=0.557*mg-=3.98xx10^-5*mol$

We divide the molar quantities thru by the LEAST molar quantity to get an empirical formula of....

$C_((1.94xx10^-4*mol)/(3.98xx10^-5*mol))H_((2.74xx10^-4*mol)/(3.98xx10^-5*mol))N_((3.98xx10^-5*mol)/(3.98xx10^-5*mol))-=C_(4.87)H_(6.88)N~=C_5H_7N$

$"Whew, arithmetic.....!!"$ I take it this is first year....?

Ordinarily combustion analysis gives you $%N$...here we had to interpolate the nitrogen mass by subtracting the calculated masses of carbon, and hydrogen, from the starting mass....

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