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Answer: The empirical formula of the compound becomes [tex]C_5H_{10}O_2[/tex]
Explanation:
The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.
Let the mass of the compound be 100 g
Given values:
% of C = 58.8%
% of H = 9.87%
% of O = [100 - 58.8 - 9.87] = 31.33%
Mass of C = 58.8 g
Mass of H = 9.87 g
Mass of O = 31.33 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol[/tex]
[tex]\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol[/tex]
[tex]\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol[/tex]
- Step 2: Calculating the mole ratio of the given elements.
Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles
[tex]\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5[/tex]
[tex]\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5[/tex]
[tex]\text{Mole fraction of O}=\frac{1.96}{1.96}=1[/tex]
Converting the mole fraction into whole numbers by multiplying them with 2.
[tex]\text{Mole fraction of C}=2.5\times 2=5[/tex]
[tex]\text{Mole fraction of H}=5\times 2=10[/tex]
[tex]\text{Mole fraction of O}=1\times 2=2[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 5 : 10 : 2
Hence, the empirical formula of the compound becomes [tex]C_5H_{10}O_2[/tex]
The empirical formula of the substance is[tex]C_5H_{10}O_2[/tex].
Explanation:
Given:
A compound made up of 58.8% of carbon, 9.87% hydrogen, and oxygen.
To find:
The empirical formula of this substance.
Solution
The percentage of carbon in a given substance by mass = 58.8%
The percentage of hydrogen in a given substance by mass = 9.87%
The percentage of oxygen in a given substance by mass :
[tex]= 100\%-58.8\%- 9.87\%=31.33\%[/tex]
Consider 100 grams of a substance.
In 100 grams of substance:
The mass of carbon =58.8% of 100 g = 58.8 g
The mass of hydrogen = 9.87% of 100 g = 9.87 g
The mass of oxygen = 31.33% of 100 g= 31.33 g
The moles of carbon [tex]=\frac{58.8 g}{12.0107 g/mol}=4.90 mol[/tex]
The moles of hydrogen[tex]=\frac{9.87g}{1.00784g/mol}=9.79 mol[/tex]
The moles of oxygen[tex]=\frac{31.33g}{15.999g/mol}=1.96 mol[/tex]
let the empirical formula of the substance = [tex]C_xH_yO_z[/tex]
The value of subscript x:
[tex]=\frac{4.90 mol}{1.96 mol}=2.5[/tex]
The value of subscript y:
[tex]=\frac{9.79mol}{1.96 mol}=5[/tex]
The value of subscript z:
[tex]=\frac{1.96mol}{1.96 mol}=1[/tex]
The empirical formula of a substance :
[tex]C_{2.5}H_5O_1=C_{\frac{25}{10}}H_5O_1=C_{25}H_{50}O_{10}[/tex]
Reducing to the lowest whole numbers:
[tex]C_{25}H_{50}O_{10}=C_5H_{10}O_2[/tex]
The empirical formula of the substance is [tex]C_5H_{10}O_2[/tex].
Learn more about the empirical formula here:
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