Question

A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of the compound is 129.1 g/mol, what is the chemical formula of the compound?

Answer 1

Let the formula of compound be $C_{a}H_{b}O_{c}N_{d}$. 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 $CO_{2}$, 1.09 g of $H_{2}O$ and 1.70 g of $N_{2}$. 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 $CO_{2}$:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Molar mass of $CO_{2}$ is 44.01 g/mol thus,

$n=\frac{5.34 g}{44.01 g/mol}=0.1213 mol$

Since, 1 mol of $CO_{2}$ have 1 mole of C thus, number of C will be 0.1213 mol.

Or, $n_{C}=0.1213 mol$

Convert this into mass as follows:

$m=n\times M$

Molar mass of C is 12 g/mol thus,

$m_{C}=0.1213 mol\times 12 g/mol=1.4556 g$

For number of moles of H, first calculate the number of moles of $H_{2}O$:

$n=\frac{m}{M}$

Molar mass of $H_{2}O$ is 18 g/mol thus,

$n=\frac{1.09 g}{18 g/mol}=0.060 mol$

Since, 1 mol of $H_{2}O$ have 2 mole of H thus, number of H will be 2×0.060 mol=0.12 mol.

Or, $n_{H}=0.12 mol$

Convert this into mass as follows:

$m=n\times M$

Molar mass of H is 1 g/mol thus,

$m_{H}=0.12 mol\times 1 g/mol=0.12 g$

For number of moles of N, first calculate the number of moles of $N_{2}$:

$n=\frac{m}{M}$

Molar mass of $N_{2}$ is 28 g/mol thus,

$n=\frac{1.70 g}{28 g/mol}=0.060 mol$

Since, 1 mol of $N_{2}$ have 2 mole of N thus, number of N will be 2×0.060 mol=0.12 mol.

Or, $n_{N}=0.12 mol$

Convert this into mass as follows:

$m=n\times M$

Molar mass of N is 14 g/mol thus,

$m_{N}=0.12 mol\times 14 g/mol=1.68 g$

Since, mass of compound is 5.214 g thus, mass of oxygen will be:

$m_{O}=m_{compound}-m_{C}-m_{H}-m_{N}=(5.214-1.4556-0.12-1.68)g=1.9584 g$

Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:

$n_{O}=\frac{1.9583 g}{16 g/mol}=0.1224 mol$

The ratio of number of moles of C, H,O and N will be:

$C:H:O:N=0.1213:0.12:0.1224:0.12=1:1:1:1$

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:

$x=\frac{129.1}{43}\approx 3$

Thus, chemical formula will be $3\times CHON=C_{3}H_{3}O_{3}N_{3}$

Therefore, chemical formula of compound is $C_{3}H_{3}O_{3}N_{3}$.

Answer 2

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.