Respuesta :
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
Let the formula of compound be [tex]C_{a}H_{b}O_{c}N_{d}[/tex]. The mass and molar mass of compound is 5.214 g and 129.1 g/mol respectively.
The combustion of compound gives 5.34 g of [tex]CO_{2}[/tex], 1.09 g of [tex]H_{2}O[/tex] and 1.70 g of [tex]N_{2}[/tex]. First number of moles of carbon, hydrogen, oxygen and nitrogen to compare the molar ratio.
Calculation for number of moles:
For number of moles of C, first calculate number of mole of [tex]CO_{2}[/tex]:
[tex]n=\frac{m}{M}[/tex]
Here, m is mass and M is molar mass.
Molar mass of [tex]CO_{2}[/tex] is 44.01 g/mol thus,
[tex]n=\frac{5.34 g}{44.01 g/mol}=0.1213 mol[/tex]
Since, 1 mol of [tex]CO_{2}[/tex] have 1 mole of C thus, number of C will be 0.1213 mol.
Or, [tex]n_{C}=0.1213 mol[/tex]
Convert this into mass as follows:
[tex]m=n\times M[/tex]
Molar mass of C is 12 g/mol thus,
[tex]m_{C}=0.1213 mol\times 12 g/mol=1.4556 g[/tex]
For number of moles of H, first calculate the number of moles of [tex]H_{2}O[/tex]:
[tex]n=\frac{m}{M}[/tex]
Molar mass of [tex]H_{2}O[/tex] is 18 g/mol thus,
[tex]n=\frac{1.09 g}{18 g/mol}=0.060 mol[/tex]
Since, 1 mol of [tex]H_{2}O[/tex] have 2 mole of H thus, number of H will be 2×0.060 mol=0.12 mol.
Or, [tex]n_{H}=0.12 mol[/tex]
Convert this into mass as follows:
[tex]m=n\times M[/tex]
Molar mass of H is 1 g/mol thus,
[tex]m_{H}=0.12 mol\times 1 g/mol=0.12 g[/tex]
For number of moles of N, first calculate the number of moles of [tex]N_{2}[/tex]:
[tex]n=\frac{m}{M}[/tex]
Molar mass of [tex]N_{2}[/tex] is 28 g/mol thus,
[tex]n=\frac{1.70 g}{28 g/mol}=0.060 mol[/tex]
Since, 1 mol of [tex]N_{2}[/tex] have 2 mole of N thus, number of N will be 2×0.060 mol=0.12 mol.
Or, [tex]n_{N}=0.12 mol[/tex]
Convert this into mass as follows:
[tex]m=n\times M[/tex]
Molar mass of N is 14 g/mol thus,
[tex]m_{N}=0.12 mol\times 14 g/mol=1.68 g[/tex]
Since, mass of compound is 5.214 g thus, mass of oxygen will be:
[tex]m_{O}=m_{compound}-m_{C}-m_{H}-m_{N}=(5.214-1.4556-0.12-1.68)g=1.9584 g[/tex]
Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:
[tex]n_{O}=\frac{1.9583 g}{16 g/mol}=0.1224 mol[/tex]
The ratio of number of moles of C, H,O and N will be:
[tex]C:H:O:N=0.1213:0.12:0.1224:0.12=1:1:1:1[/tex]
Thus, empirical formula of compound will be CHON.
According to above formula molar mass of compound will be 43 g/mol
Now, according to chemical formula, the molar mass is 129.1 g/mol taking the ratio:
[tex]x=\frac{129.1}{43}\approx 3[/tex]
Thus, chemical formula will be [tex]3\times CHON=C_{3}H_{3}O_{3}N_{3}[/tex]
Therefore, chemical formula of compound is [tex]C_{3}H_{3}O_{3}N_{3}[/tex].
