Answer:
Empirical formula = [tex]\mathbf{C_{10}H_{12}N_{2}}[/tex]
Molecular formula = [tex]\mathbf{C_{20}H_{24}N_{4}}[/tex]
Explanation:
From the given information:
we need to estimate the mass of carbon C in 1.321 g of [tex]CO_2[/tex]
before that, the number of moles of C is:
[tex]C = 1.321 \ g \times \dfrac{1 \ mol \ CO_2}{44/010 \ g \ of \ CO_2}[/tex]
c = 0.03002 mol
we know that:
number of moles = mass/ molar mass
mass = number of moles × molar mass
mass = 0.03002 × 12.011g of C
mass of C = 0.3606 g
Similarly; for hydrogen
the number of moles of H = [tex]0.325 g \ of \ H_2O \times \dfrac{ 1\ mol \ H_2O}{18.02g \ of \ H_2O}\times \dfrac{2 \ mole \ of H }{1 \ mol \ H_2O }[/tex]
the number of moles of H = [tex]0.325 g \ of \ H_2O \times \dfrac{2 \ mole \ of H}{18.02g \ of \ H_2O}[/tex]
the number of moles of H = 0.0361 mol of H
mass of H = [tex]0.0361 \ mol \ of \ H \times \dfrac{1.008 g \ of \ H}{ 1 \ mol \ of \ H}[/tex]
mass of H = 0.0364 g
The mass of N will therefore be the difference the sample burnt with the mass of carbon and hydrogen.
i.e
mass of N = 0.487 g - 0.3606 of C - 0.0364 g of H
mass of N = 0.0900 g
however, the number of moles of nitrogen = mass/ molar mass
the number of moles = 0.0900 g /14.007 g
the number of moles of nitrogen = 0.00643 mol
Thus, the formula is: [tex]\mathsf{C_{0.03002}H_{0.0361}N_{0.00643}}[/tex]
If we divide by the smallest number (0.00643); we have:
[tex]\mathsf{C_{\dfrac{0.03002}{0.006432}}H_{\dfrac{0.0361}{0.00643}}N_{\dfrac{0.00643}{0.00643}}}[/tex]
= [tex]\mathsf{C_{4.7}H_{5.7}N}[/tex]
Thus, multiplying the subscript by 2.1, we have:
[tex]\mathsf{C_{4.7 \times 2.1}H_{5.7 \times 2.1}N_{1\times 2.1}}[/tex]
Thus, the empirical formula = [tex]\mathbf{C_{10}H_{12}N_{2}}[/tex]
The mass of the empirical formula is:
= (10 × 12.010 u) + (12 × 1.008 u) + ( 2 × 14.007 u)
= 160.21 u
Thus, because the molecular mass 324 g/mol is double the value of the empirical formula, the molecular mass is definitely double the empirical formula;
i.e
Molecular formula = [tex]\mathbf{C_{10\times 2}H_{12\times 2}N_{2\times 2}}[/tex]
Molecular formula = [tex]\mathbf{C_{20}H_{24}N_{4}}[/tex]
