Answer: The empirical formula is [tex]C_{17}H_{19}O_3N[/tex]
Explanation:
Mass of C= 17.900 g
Mass of H = 1.680 g
Mass of O = 4.225 g
Mass of N = 1.228 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{17.990g}{12g/mole}=1.5moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.680g}{1g/mole}=1.680moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{4.225g}{16g/mole}=0.264moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{1.228g}{14g/mole}=0.087moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{1.5}{0.087}=17[/tex]
For H = [tex]\frac{1.680}{0.087}=19[/tex]
For O =[tex]\frac{0.264}{0.087}=3[/tex]
For N = [tex]\frac{0.087}{0.087}=1[/tex]
The ratio of C : H: O: N = 17: 19: 3: 1
Hence the empirical formula is [tex]C_{17}H_{19}O_3N[/tex]
