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Combustion is a reaction between a combustible substance and oxygen, to ultimately produce carbon dioxide and water. Reaction between carbon and oxygen would give,
C + O2 ------> CO2
Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2.
86.5 g CO = 1.966 moles CO2
44 g CO2/ mole
Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel.
1.966 moles CO2 x 1 mole C x 12 g C = 23.59 g C
1 mole CO2 1 mole C
We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams.
25.0 g - 23.59 g = 1.41 grams Hydrogen
To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen.
To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel,
Carbon: 1.966 mol = 1.408 x 5 (factor) = 7
1.396 mol
Hydrogen: 1.396 mol = 1.00 x 5 (factor) = 5
1.396 mol
Thus, the empirical formula is C7H5
C + O2 ------> CO2
Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2.
86.5 g CO = 1.966 moles CO2
44 g CO2/ mole
Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel.
1.966 moles CO2 x 1 mole C x 12 g C = 23.59 g C
1 mole CO2 1 mole C
We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams.
25.0 g - 23.59 g = 1.41 grams Hydrogen
To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen.
To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel,
Carbon: 1.966 mol = 1.408 x 5 (factor) = 7
1.396 mol
Hydrogen: 1.396 mol = 1.00 x 5 (factor) = 5
1.396 mol
Thus, the empirical formula is C7H5
The empirical formula of the hydrocarbon is [tex]\boxed{{{\text{C}}_7}{{\text{H}}_5}}[/tex].
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) [tex]{\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}[/tex]
(b) [tex]{{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+ 4{{\text{H}}_2}{\text{O}}[/tex]
[tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions.
Step 1: [tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions. Initially, we have to calculate the moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] . The formula to calculate moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is as follows:
[tex]{\text{Moles of C}}{{\text{O}}_2}=\dfrac{{{\text{Given mass of C}}{{\text{O}}_2}}}{{{\text{Molar mass of C}}{{\text{O}}_2}}}[/tex] ...... (1)
The given mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 86.5 g.
The molar mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 44 g/mol.
Substitute these values in equation (1).
[tex]\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}[/tex]
Step 2: During combustion, one mole of carbon reacts to form one mole of [tex]{\text{C}}{{\text{O}}_2}[/tex] .So the mass of C in the hydrocarbon is calculated as follows:
[tex]{\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)[/tex] ...... (2)
The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1.966 mol.
The molar mass of C is 12 g/mol.
The mole of C is 1 mol.
The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1 mol.
Substitute these values in equation (2).
[tex]\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}\\[/tex]
Step 3: Since the hydrocarbon consists of only carbon and hydrogen. The mass of hydrogen is calculated as follows:
[tex]{\text{Mass of H}}={\text{Mass of hydrocarbon}}-{\text{Mass of C}}[/tex] ...... (3)
The mass of hydrocarbon is 25 g.
The mass of carbon is 23.59 g.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Mass of H}}&={\text{25 g}}-{\text{23}}{\text{.59 g}}\\&= {\text{1}}{\text{.41 g}}\\\end{aligned}[/tex]
The formula to calculate moles of H is as follows:[tex]{\text{Moles of H}}=\dfrac{{{\text{Given mass of H}}}}{{{\text{Molar mass of H}}}}[/tex] ...... (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).
[tex]\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}[/tex]
The moles of carbon and hydrogen present in hydrocarbon are to be written with their corresponding subscripts. So the preliminary formula becomes,
[tex]{\text{Preliminary formula of hydrocarbon}}={{\text{C}}_{1.966}}{{\text{H}}_{1.396}}[/tex]
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:
[tex]\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}[/tex]
Step 5: Multiply each subscript of the empirical formula by 5, we get the final empirical formula as follows:
[tex]\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}[/tex]
Therefore, the empirical formula of hydrocarbon is [tex]{{\mathbf{C}}_{\mathbf{7}}}{{\mathbf{H}}_{\mathbf{5}}}[/tex] .
Learn more:
1. Calculate the moles of ions in the solution: https://brainly.com/question/5950133
2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
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.
