# 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?

May 31, 2018

I make it ${C}_{5} {H}_{7} N$...

#### Explanation:

And so we got ${C}_{x} {H}_{y} {N}_{n}$

We combust a $3.159 \cdot m g$ mass of the stuff to get....

$\frac{8.545 \cdot m g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 1.94 \times {10}^{-} 4 \cdot m o l$ $C {O}_{2}$...i.e. a mass with respect to carbon of $2.33 \cdot m g$....($C O$ would not result from the combustion.)

$\frac{2.450 \cdot m g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 1.36 \times {10}^{-} 4 \cdot m o l$ ${H}_{2} O$...i.e. a mass with respect to hydrogen of $0.274 \cdot m g$, $0.272 \cdot m g$..

The balance of the mass was due to NITROGEN....$\left\{3.159 - 2.33 - 0.272\right\} \cdot m g = 0.557 \cdot m g \equiv 3.98 \times {10}^{-} 5 \cdot m o l$

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

${C}_{\frac{1.94 \times {10}^{-} 4 \cdot m o l}{3.98 \times {10}^{-} 5 \cdot m o l}} {H}_{\frac{2.74 \times {10}^{-} 4 \cdot m o l}{3.98 \times {10}^{-} 5 \cdot m o l}} {N}_{\frac{3.98 \times {10}^{-} 5 \cdot m o l}{3.98 \times {10}^{-} 5 \cdot m o l}} \equiv {C}_{4.87} {H}_{6.88} N \cong {C}_{5} {H}_{7} N$

$\text{Whew, arithmetic.....!!}$ 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....