Respuesta :
The empirical formula is [tex]C_2H_6N[/tex].
Explanation:
Putrescine has the elements like Carbon, Nitrogen and Hydrogen present in them. So in order to determine the empirical formula, we first have to find the number of moles present in the putrescine. As the percentage of C, H and N present in the chemical is given as 54.50%, 13.73% and 31.77%, we assume that 100 g of Putrescine is taken as sample.
Then the mass of C, H and N present in Putrescine will be 54.50 g, 13.73 g and 31.77 g. We know that the molar mass of C is 12 g/mol, H is 1 g/mol and N is 14 g/mol. So divide the mass with the molar mass of the respective elements to determine the number of moles of these elements present in the sample.
[tex]No.\ of\ moles\ of\ C=\frac{\text { Mass of } C}{\text { Molar mass of } C}=\frac{54.50 \mathrm{g}}{12 \mathrm{g} / \mathrm{mol}}=4.54\ moles[/tex]
Similarly, the number of moles of H and N present is determined.
[tex]\text { No. of moles of } H=\frac{\text { Mass of } H}{\text { Molar mass of } H}=\frac{13.73 \mathrm{g}}{1 \mathrm{g} / \mathrm{mol}}=13.73 \text { moles }[/tex]
[tex]No.\ of\ moles\ of\ N=\frac{\text { Mass of } N}{\text { Molar mass of } N}=\frac{31.77 \mathrm{g}}{14 \mathrm{g} / \mathrm{mol}}=2.27\ moles[/tex]
Then the empirical formula can be determined by dividing the number of moles of all elements with the least number of moles that is 2.27.
[tex]\begin{aligned}&\text { No. of atoms of } C=\frac{4.54}{2.27}=2\\&\text { No. of atoms of } H=\frac{13.73}{2.27}=6\\&\text { No. of atoms of } N=\frac{2.27}{2.27}=1\end{aligned}[/tex]
So, the empirical formula is [tex]C_2H_6N[/tex].
