Respuesta :
The general equation for a combustion reaction would be expressed as:
CxHy + (x+y)O2 = xCO2 + y/2H2O
To determine the empirical formula of the substance, toluene, we calculate the moles of C and H in the product. The only source of carbon would be from carbon dioxide and for hydrogen, from water. We do as follows:
1.37 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H2O ) = 0.1521 mol H
5.86 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.1332 mol C
So we have,
C = 0.1332 C / 0.1332 = 1
H = 0.1521 H / 0.1332 = 1
Thus, the empirical formula woud be CH.
CxHy + (x+y)O2 = xCO2 + y/2H2O
To determine the empirical formula of the substance, toluene, we calculate the moles of C and H in the product. The only source of carbon would be from carbon dioxide and for hydrogen, from water. We do as follows:
1.37 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H2O ) = 0.1521 mol H
5.86 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.1332 mol C
So we have,
C = 0.1332 C / 0.1332 = 1
H = 0.1521 H / 0.1332 = 1
Thus, the empirical formula woud be CH.
Answer: The empirical formula for the given organic compound is [tex]CH[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:
[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.
We are given:
Mass of [tex]CO_2=5.86mg=5.86\times 10^{-3}g[/tex] (Conversion factor: 1 g = 1000 mg)
Mass of [tex]H_2O=1.37mg=1.37\times 10^{-6}g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in [tex]5.86\times 10^{-3}g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g[/tex] of carbon will be contained.
- For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in [tex]1.37\times 10^{-3}g[/tex] of water, [tex]\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g[/tex] of hydrogen will be contained.
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles[/tex]
- Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is [tex]0.133\times 10^{-3}moles[/tex]
For Carbon = [tex]\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1[/tex]
For Hydrogen = [tex]\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 1
Hence, the empirical formula for the given compound is [tex]C_1H_{1}=CH[/tex]
