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=(m)/(M)`

Here, m is mass and M is molar mass.

Molar mass of
`CO_(2)` is 44.01 g/mol thus,

`n=(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* M`

Molar mass of C is 12 g/mol thus,

`m_(C)=0.1213 mol* 12 g/mol=1.4556 g`

For number of moles of H, first calculate the number of moles of
`H_(2)O`:

`n=(m)/(M)`

Molar mass of
`H_(2)O` is 18 g/mol thus,

`n=(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* M`

Molar mass of H is 1 g/mol thus,

`m_(H)=0.12 mol* 1 g/mol=0.12 g`

For number of moles of N, first calculate the number of moles of
`N_(2)`:

`n=(m)/(M)`

Molar mass of
`N_(2)` is 28 g/mol thus,

`n=(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* M`

Molar mass of N is 14 g/mol thus,

`m_(N)=0.12 mol* 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)=(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=(129.1)/(43)\approx 3`

Thus, chemical formula will be
`3* 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.