The molecular mass of nicotine is $162 g / m o l$ . It contains $74 %$ carbon, $8.7 %$ hydrogen and $17.3 %$ nitrogen. Determine nicotine's empirical and molecular formula.

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Solution

$C$ $H$ $N$
$74$ $8.7$ $17.3$ Percentage
$12$ $1$ $14$ Atomic Mass
$6.17$ $8.70$ $1.23$ Number of Moles
$5.01$ $7.07$ $1$ Simple molar ratio
( $\div$ by $1.23$ )
Hence, $C : H : N = 5 : 7 : 1$
Empirical formula of nicotine is $C_{5} H_{7} N$ .
Also, molecular formula is same as $C_{5} H_{7} N$ .

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$C$	$H$	$N$
$74$	$8.7$	$17.3$	Percentage
$12$	$1$	$14$	Atomic Mass
$6.17$	$8.70$	$1.23$	Number of Moles
$5.01$	$7.07$	$1$	Simple molar ratio ( $\div$ by $1.23$ )

The molecular mass of nicotine is 162g/mol. It contains 74% carbon, 8.7% hydrogen and 17.3% nitrogen. Determine nicotine's empirical and molecular formula.

C H N 748.7 17.3 Percentage 121 14 Atomic Mass 6.178.70 1.23 Number of Moles 5.017.071 Simple molar ratio( ÷ by 1.23)Hence, C:H:N=5:7:1Empirical formula of nicotine is C5H7N.Also, molecular formula is same as C5H7N.

The molecular mass of nicotine is $162 g / m o l$ . It contains $74 %$ carbon, $8.7 %$ hydrogen and $17.3 %$ nitrogen. Determine nicotine's empirical and molecular formula.

$C$ $H$ $N$
$74$ $8.7$ $17.3$ Percentage
$12$ $1$ $14$ Atomic Mass
$6.17$ $8.70$ $1.23$ Number of Moles
$5.01$ $7.07$ $1$ Simple molar ratio
( $\div$ by $1.23$ )
Hence, $C : H : N = 5 : 7 : 1$
Empirical formula of nicotine is $C_{5} H_{7} N$ .
Also, molecular formula is same as $C_{5} H_{7} N$ .