Complete combustion of 0.858 gram of compound X gives 2.63 gram of CO2...
Mass of carbon in 2.63 gm of CO2 = 2.63 x 12/44 = 0.717 gm
Mass of hydrogen in 1.28 gm of H2O = 1.28 x 2/18 =0.142 gm.
Percentage of carbon = 0.717 x 100/0.858 = 83.6
Percentage of hydrogen = 0.142 x 100/0.858= 16.6
The compound X can only be a hydrocarbon as the total percentages of both carbon and hydrogen is almost equal to 100.
C : H = 83.6/12 : 16.6/1 = 6.97 : 16.6 = 1 : 16.6/6.97 = 1 : 2.38.
On converting to whole number ratio, 1 : 2.38 = 5 : 12
So, the formula of the hydrocarbon X is C5H12.
Molecular mass of X = 12 x 5 + 1 x 12 = 72.
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Complete combustion of 0.858 gram of compound X gives 2.63 gram of CO2...
Given Information:
- Mass of compound X = 0.858 g
- Mass of CO2 produced = 2.63 g
- Mass of H2O produced = 1.28 g
Calculating Moles:
To determine the molecular mass of compound X, we need to calculate the number of moles of CO2 and H2O produced.
- Moles of CO2 = mass of CO2 / molar mass of CO2
= 2.63 g / 44.01 g/mol
= 0.0597 mol
- Moles of H2O = mass of H2O / molar mass of H2O
= 1.28 g / 18.02 g/mol
= 0.071 mol
Stoichiometry:
The balanced chemical equation for the combustion of compound X can be written as follows:
Compound X + O2 -> CO2 + H2O
From the balanced equation, we can see that the ratio of moles of CO2 to moles of X is 1:1, and the ratio of moles of H2O to moles of X is also 1:1.
Since the moles of CO2 and H2O produced are equal to the moles of compound X, the lowest molecular mass of X can be calculated using the formula:
Molecular mass (X) = mass of X / moles of X
Calculating Molecular Mass:
- Molecular mass (X) = 0.858 g / 0.0597 mol
= 14.4 g/mol
Explanation:
The lowest molecular mass of compound X is 14.4 g/mol. This means that X could be any compound with a molecular mass of 14.4 g/mol, or a multiple of it. Without further information, it is not possible to determine the exact identity of compound X.
The given data and calculations show that when 0.858 g of compound X undergoes complete combustion, it produces 2.63 g of carbon dioxide (CO2) and 1.28 g of water (H2O). The balanced chemical equation for the combustion reaction is not provided, but it can be inferred that the reaction involves the consumption of oxygen (O2) to produce CO2 and H2O.
The stoichiometry of the reaction allows us to determine the ratio of moles of CO2 and H2O to moles of X. By comparing the moles of CO2 and H2O produced in the reaction to the moles of X, we can calculate the molecular mass of X.
The molecular mass of X is determined by dividing the mass of X by the moles of X. In this case, the given mass of X is 0.858 g, and the calculated moles of X are 0.0597 mol. Dividing these values gives a molecular mass of 14.4 g/mol.
Therefore, the lowest molecular mass that compound X can have is 14.4 g/mol, but the exact identity of X cannot be determined without additional information.
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