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A hydrocarbon that contains 79.85% c and 20.15% h. The molar mass of thernhydrocarbon is 30.08 g/mol.rna. Determine the molecular formula of the hydrocarbon.rnb. Write the chemical equation for the combustion of this hydrocarbon.rnc. What is the theoretical yield of CO2 from this combustion reaction if 2.5rngrams of the hydrocarbon reacts with 2.5 grams of O2?
Accepted Answer
a. Determine the molecular formula of the hydrocarbon.
Assuming the empirical formula is CHx, we can calculate the empirical formula mass as:
Empirical formula mass = (12.01 g/mol) + (x * 1.01 g/mol)
To find the value of x, we can use the given percentage composition:
%C = (12.01 g/mol) / (Empirical formula mass) * 100
%H = (x * 1.01 g/mol) / (Empirical formula mass) * 100
Substituting the given percentages, we get:
79.85% = (12.01 g/mol) / (Empirical formula mass) * 100
20.15% = (x * 1.01 g/mol) / (Empirical formula mass) * 100
Solving for x, we find that x = 4.
Therefore, the empirical formula is CH4.
To find the molecular formula, we need to divide the molar mass of the hydrocarbon by the empirical formula mass:
Molecular formula mass / Empirical formula mass = n
Substituting the given values, we get:
30.08 g/mol / (12.01 g/mol + 4 * 1.01 g/mol) = n
n = 2
Therefore, the molecular formula of the hydrocarbon is C2H8.
b. Write the chemical equation for the combustion of this hydrocarbon.
C2H8 + 5O2 → 2CO2 + 4H2O
c. What is the theoretical yield of CO2 from this combustion reaction if 2.5 grams of the hydrocarbon reacts with 2.5 grams of O2?
First, we need to calculate the moles of the hydrocarbon and oxygen:
Moles of C2H8 = 2.5 g / 30.08 g/mol = 0.083 mol
Moles of O2 = 2.5 g / 32.00 g/mol = 0.078 mol
According to the balanced chemical equation, 1 mole of C2H8 reacts with 5 moles of O2. Therefore, the limiting reactant is O2.
The theoretical yield of CO2 is:
Moles of CO2 = 0.078 mol O2 × (2 mol CO2 / 5 mol O2) = 0.0312 mol
Mass of CO2 = 0.0312 mol × 44.01 g/mol = 1.37 g
Therefore, the theoretical yield of CO2 is 1.37 g.
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