Answer :
Answer: C₃H₈
Explanation:
From PV = nRT
Moles of CO2 = PV / RT =(1 x 1.037)/(0.08206 x ( 273+98.3)) = 0.0340 moles of CO2
CxHy + (x + (y/2))O₂ ----> xCO₂ + yH₂O
there is 1 mole of C in each mole of CO2 so moles of C in CO₂ = 0.0340 moles
mass of C in CO₂ = 0.0340 x 12 = 0.41 g
This means mass of C in the hydrocarbon = 0.41g
so mass of H in the hydrocarbon = 0.50 - 0.41= 0.09 g
moles of H in the hydrocarbon = mass/molar mass = 0.09/1 = 0.09 moles
molar ratio of C:H = 0.034 : 0.09 = 1 :2.67
or 3 :8
empirical formula is C3H8
Final answer:
To find the empirical formula of the hydrocarbon, calculate the moles of carbon dioxide produced using the ideal gas law and the moles of carbon and hydrogen in the hydrocarbon. The empirical formula is C12H1.
Explanation:
In order to determine the empirical formula of the hydrocarbon, we need to find the mole ratio of carbon to hydrogen in the compound. To do this, we first calculate the moles of carbon dioxide (CO2) produced using the ideal gas law:
n = (PV)/(RT) = (1.000 atm)(1.037 L)/(0.08206 L.atm/mol.K)(98.3 + 273.15 K) = 0.0443 mol
Since the combustion of 1 mole of hydrocarbon produces 1 mole of CO2, we can conclude that the moles of hydrocarbon consumed is also 0.0443 mol. Next, we calculate the moles of carbon and hydrogen in the hydrocarbon:
moles of carbon = (mass of carbon in hydrocarbon) / (molar mass of carbon) = (12.01 g/mol)(0.5000 g) / (1 g) = 6.003 mol
moles of hydrogen = (mass of hydrogen in hydrocarbon) / (molar mass of hydrogen) = (1.008 g/mol)(0.5000 g) / (1 g) = 0.5008 mol
The empirical formula of the hydrocarbon is then C6H0.5008. We can simplify this ratio by multiplying all the subscripts by 2 (the smallest whole number ratio) to get the empirical formula: C12H1.
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