Respuesta :
Answer:
Molecular formula of the compound is C₃H₆O
Explanation:
Firstly let's determine the moles of the vapor (gas) with the Ideal Gases Law, so it can give us the molar mass with the mass and afterwards we can work with the percent composition.
Pressure . Volume = moles . Ideal Constant Gases . Temperature in K
Temperature in K = T°C + 273 → 150°C + 273 = 423K
P . V = n . R . T
n = (P .V) / (R. T)
n = 1 atm . 0.5L / (0.082 . 423K)
n = 0.0144 moles
These are the moles for 0.8365 g, so let's determine the molar mass
Molar mass (g/mol) = 0.8365 g / 0.0144 mol → 58.02 g/mol
Percent composition means:
100 g of compound have 62.04 g of C
100 g of compound have 10.41 g of H
100 g of compound have 27.54 g of O
Let's make the rule of three:
100 g of compound have __ 62.04 g of C __ 10.41 g of H __ 27.54 g of O
The 58.02 g of compound must have:
(58.02 g . 62.04 g) / 100 g = 36 g of C
(58.02 g . 10.41 g) / 100 g = 6 g of H
(58.02 g . 27.54 g) / 100 g = 16 g of O
Let's find out the moles of each
Mass / Molar mass
36 g / 12 g/mol = 3 C
6 g / 1 g/mol = 6 H
16g / 16 g/mol = 1 O
The molecular formula of the given compound is C₃H₆O. The molecular formula can be determined by finding the ratio of each element in the compound.
How to determine the Molecular formula of a compound?
It can be determined by finding the ratio of each element in the compound.
First, calculate the moles of the compound from the ideal gas formula,
[tex]n = \dfrac {1 {\rm\ atm \times 0.5L} }{(0.082 \times 423{\rm \ K})}\\\\n = 0.0144 \rm \ moles[/tex]
Then calculate the molar mass of the compound,
[tex]m = {\rm \dfrac {0.8365 \ g }{0.0144 \ mol}} \\\\m = 58.02 \rm \ g/mol[/tex]
Then calculate the mass of individual elements in the 58.02 g of compound:
[tex]\text{ Mass of Carbon} = \dfrac {58.02 {\rm \ g} \times 62.04 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Carbon} = 36 \rm \ g[/tex]
[tex]\text{ Mass of Hydrogen } = \dfrac {58.02 {\rm \ g} \times 10.14 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Hydrogen } = 6 \rm \ g[/tex]
[tex]\text{ Mass of Oxygen } = \dfrac {58.02 {\rm \ g} \times 27.54 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Oxygen } = 16 \rm \ g[/tex]
Now find the moles of each element, we get
3 moles of Carbon
6 moles of Hydrogen
1 mole of Oxygen
Therefore, the molecular formula of the given compound is C₃H₆O.
Learn more about the molecular formula,
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