Chemistry, 03.08.2019 05:30 krystlemiller4307

Acombustion 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?

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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 from: Quest

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

Answer from: Quest

0.50 hope this

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Acombustion 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 ....

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