Chemistry, 10.07.2019 10:40 taysomoneyyy

Combustion of 25.0 g of a hydrocarbon produces 86.5 g of co2. what is the empirical formula of the compound?

Answers: 2

Show answers

Answers

Answer from: itzyagirlshy

The empirical formula of the hydrocarbon is $\boxed{{{\text{C}}_7}{{\text{H}}_5}}$ .

Further explanation:

Empirical formula:

It is atom’s simplest positive integer ratio in the compound. It may or may not be same as that of molecular formula. For example, empirical formula of sulfur dioxide is SO.

Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.

Example of combustion reactions are as follows:

(a) ${\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}$

(b) ${{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+ 4{{\text{H}}_2}{\text{O}}$

${\text{C}}{{\text{O}}_2}$ is formed as a product during combustion reactions.

Step 1: ${\text{C}}{{\text{O}}_2}$ is formed as a product during combustion reactions. Initially, we have to calculate the moles of ${\text{C}}{{\text{O}}_2}$ . The formula to calculate moles of ${\text{C}}{{\text{O}}_2}$ is as follows:

${\text{Moles of C}}{{\text{O}}_2}=\dfrac{{{\text{Given mass of C}}{{\text{O}}_2}}}{{{\text{Molar mass of C}}{{\text{O}}_2}}}$ ...... (1)

The given mass of ${\text{C}}{{\text{O}}_2}$ is 86.5 g.

The molar mass of ${\text{C}}{{\text{O}}_2}$ is 44 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Moles of C}}{{\text{O}}_2}&=\left( {{\text{86}}{\text{.5 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{44 g}}}}} \right)\\&={\text{1}}{\text{.9659 mol}}\\ &\approx{\text{1}}{\text{.966 mol}}\\\end{aligned}$

Step 2: During combustion, one mole of carbon reacts to form one mole of ${\text{C}}{{\text{O}}_2}$ .So the mass of C in the hydrocarbon is calculated as follows:

${\text{Mass of C}}=\left( {{\text{Moles of C}}{{\text{O}}_{\text{2}}}}\right)\left( {\dfrac{{{\text{Moles of C}}}}{{{\text{Moles of C}{{\text{O}}_{\text{2}}}}}}\right)\left( {{\text{Molar mass of C}}}\right)$ ...... (2)

The moles of ${\text{C}}{{\text{O}}_2}$ is 1.966 mol.

The molar mass of C is 12 g/mol.

The mole of C is 1 mol.

The moles of ${\text{C}}{{\text{O}}_2}$ is 1 mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{Mass of C}}&=\left( {{\text{1}}{\text{.966 mol}}} \right)\left( {\frac{{{\text{1 mol of C}}}}{{{\text{1 mol of C}}{{\text{O}}_{\text{2}}}}}} \right)\left( {\frac{{{\text{12 g}}}}{{{\text{1 mol}}}}} \right)\\&= {\text{23}}{\text{.592 g}}\\&\approx{\text{23}}{\text{.59 g}}\\\end{aligned}\\$

Step 3: Since the hydrocarbon consists of only carbon and hydrogen. The mass of hydrogen is calculated as follows:

${\text{Mass of H}}={\text{Mass of hydrocarbon}}-{\text{Mass of C}}$ ...... (3)

The mass of hydrocarbon is 25 g.

The mass of carbon is 23.59 g.

Substitute these values in equation (3).

$\begin{aligned}{\text{Mass of H}}&={\text{25 g}}-{\text{23}}{\text{.59 g}}\\&= {\text{1}}{\text{.41 g}}\\\end{aligned}$

The formula to calculate moles of H is as follows: ${\text{Moles of H}}=\dfrac{{{\text{Given mass of H}}}}{{{\text{Molar mass of H}}}}$ ...... (4)

The given mass of H is 1.41 g.

The molar mass of H is 1.01 g/mol.

Substitute these values in equation (4).

$\begin{aligned}{\text{Moles of H}}&=\left({{\text{1}}{\text{.41 g}}}\right)\left( {\frac{{{\text{1 mol}}}}{{{\text{1}}{\text{.01 g}}}}}\right)\\&={\text{1}}{\text{.396 mol}}\\ \end{aligned}$

The moles of carbon and hydrogen present in hydrocarbon are to be written with their corresponding subscripts. So the preliminary formula becomes,

${\text{Preliminary formula of hydrocarbon}}={{\text{C}}_{1.966}}{{\text{H}}_{1.396}}$

Step 4: Each of the subscripts is divided by the smallest subscript to get the empirical formula. In this case, the smallest one is 1.39. So the empirical formula of hydrocarbon is written as follows:

$\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{\dfrac{{1.966}}{{1.396}}}}{{\text{H}}_{\dfrac{{1.396}}{{1.396}}}}\\&= {{\text{C}}_{1.408}}{{\text{H}}_1}\\\end{aligned}$

Step 5: Multiply each subscript of the empirical formula by 5, we get the final empirical formula as follows:

$\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{5\left( {1.408} \right)}}{{\text{H}}_{5\left( 1 \right)}}\\&={{\text{C}}_7}{{\text{H}}_5}\\\end{aligned}$

Therefore, the empirical formula of hydrocarbon is ${{\mathbf{C}}_{\mathbf{7}}}{{\mathbf{H}}_{\mathbf{5}}}$ .

Learn more:

1. Calculate the moles of ions in the solution:

2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stoichiometry of formulas and equations

Keywords: empirical formula, C, H, C7H5, moles of CO2, C, H, 5, preliminary formula, whole number.

Answer from: Quest

What is the gold assay

Answer from: Quest

Hi; /c

Another question on Chemistry

Chemistry, 22.06.2019 08:00

Define dew point. i am writing this part to be able to ask the question

Answers: 1

Answer

Chemistry, 22.06.2019 09:00

Chen drew a diagram to compare the ways in which different organisms obtain nitrogen. which label belongs to the area marked z?

Answers: 3

Answer

Chemistry, 22.06.2019 10:20

Gwhich r group would most likely be found in a hydrophobic area of the tertiary structure of a globular protein? which r group would most likely be found in a hydrophobic area of the tertiary structure of a globular protein? −ch2−oh −ch2−o||c−nh2 −ch2−coo− −ch2−ch2−ch2−ch2−n+h3

Answers: 3

Answer

Chemistry, 22.06.2019 10:30

What determines the average kinetic energy of the particles in a gas? a. the number of collisions b. the number of particles c. the size of the particles d. the temperature

Answers: 1

Answer

You know the right answer?

Combustion of 25.0 g of a hydrocarbon produces 86.5 g of co2. what is the empirical formula of the c...

Questions