Answer:
A. [tex]C_5H_7N[/tex]
B. [tex]C_4H_5N_2O[/tex]
Explanation:
Hello,
a. In this case, based on the given percentages, one computes the moles of C, H and N as follows:
[tex]n_C=0.7403gC*\frac{1molC}{12gC}=0.062molC\\n_H=0.087gH*\frac{1molH}{1gH}=0.087molH\\n_N=0.1727gN*\frac{1molN}{14gN}=0.0123molN\\[/tex]
Now, we divide each element's moles by the smallest amount of moles, that is nitrogen's moles:
[tex]C=\frac{0.062}{0.0123}=5;H=\frac{0.087}{0.0123}=7;N=\frac{0.0123}{0.0123}=1[/tex]
So the empirical formula is:
[tex]C_5H_7N[/tex]
b. In this case, based on the given percentages, one computes the moles of C, H, N and O as follows:
[tex]n_C=0.4948gC*\frac{1molC}{12gC}=0.0396molC\\n_H=0.0519gH*\frac{1molH}{1gH}=0.0519molH\\n_N=0.2885gN*\frac{1molN}{14gN}=0.021molN\\n_O=0.1648gO*\frac{1molO}{16gO}=0.0103molO\\[/tex]
Now, we divide each element's moles by the smallest amount of moles, that is oxugen's moles:
[tex]C=\frac{0.0396}{0.0103}=4;H=\frac{0.0519}{0.0103}=5;N=\frac{0.021}{0.0103}=2;O=\frac{0.0103}{0.0103}=1[/tex]
So the empirical formula is:
[tex]C_4H_5N_2O[/tex]
Best regards.
