The elementary analysis of 22 g hydrocarbon sample results in 18 g C and 4 g H.
a) Choose the correct chemical formula of the hydrocarbon:
i) CH4
ii) C2H4
iii) C3H8
b) What is the amount of CO2 in grams formed by the total combustion of hydrocarbon? Assume complete
combustion, meaning that only CO2 and H2O is formed.
25g
To determine the correct chemical formula of the hydrocarbon, we need to use the given information about the masses of carbon and hydrogen in the sample.
a) Let's calculate the ratio of carbon to hydrogen in the hydrocarbon:
Mass of carbon (C) = 18 g
Mass of hydrogen (H) = 4 g
To find the ratio, we divide the mass of carbon by the atomic mass of carbon (12 g/mol) and the mass of hydrogen by the atomic mass of hydrogen (1 g/mol):
Ratio of carbon to hydrogen = (18 g / 12 g/mol) / (4 g / 1 g/mol)
Simplifying, we get: Ratio of carbon to hydrogen = 1.5
Now, let's compare this ratio to the ratios of carbon and hydrogen in the given chemical formulas:
i) CH4: Carbon to hydrogen ratio = 1:4
ii) C2H4: Carbon to hydrogen ratio = 2:4 (which simplifies to 1:2)
iii) C3H8: Carbon to hydrogen ratio = 3:8
Based on the given ratio of 1.5, none of the options have a matching ratio. Therefore, none of the given options (i, ii, iii) are correct.
b) To determine the amount of CO2 formed by the complete combustion of the hydrocarbon, we need to calculate the number of moles of carbon in the sample.
We can use the molar mass of carbon (12 g/mol) to find the number of moles of carbon:
Moles of carbon = Mass of carbon (C) / Molar mass of carbon
Moles of carbon = 18 g / 12 g/mol = 1.5 moles of carbon
Since complete combustion of one mole of carbon results in the formation of one mole of CO2, we can determine the amount of CO2 formed:
Amount of CO2 formed = Moles of carbon = 1.5 moles
To convert moles of CO2 to grams, we multiply by the molar mass of CO2 (44 g/mol):
Amount of CO2 formed (in grams) = Moles of CO2 x Molar mass of CO2
Amount of CO2 formed = 1.5 moles x 44 g/mol = 66 g
Therefore, the amount of CO2 formed by the total combustion of the hydrocarbon is 66 grams.