A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O.
1 answer:
Answer:
Explanation:
Mass of water obtained = 0.0609 g
Molar mass of water = 18 g/mol
Moles of = 0.0609 g /18 g/mol = 0.00338 moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 0.00338 = 0.00676 moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 0.00676 x 1.008 = 0.00681 g
Mass of carbon dioxide obtained = 0.1486 g
Molar mass of carbon dioxide = 44.01 g/mol
Moles of = 0.1486 g /44.01 g/mol = 0.00337 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.00337 moles
Molar mass of C atom = 12.0107 g/mol
Mass of C in molecule = 0.00337 x 12.0107 = 0.04047 g
Given that the compound only contains hydrogen, oxygen and carbon. So,
Mass of O in the sample = Total mass - Mass of C - Mass of H
Mass of the sample = 0.1014 g
Mass of O in sample = 0.1014 - 0.04047 - 0.00681 = 0.05412 g
Molar mass of O = 15.999 g/mol
Moles of O = 0.05412 / 15.999 = 0.00338 moles
Taking the simplest ratio for H, O and C as:
0.00676 : 0.00338 : 0.00337
= 2 : 1 : 1
The empirical formula is =
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12 + 2×1 + 16= 30 g/mol
Molar mass = 180 g/mol
So,
Molecular mass = n × Empirical mass
180 = n × 30
⇒ n = 6
The formula of compound =
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Answer:
Complete Question:
Equimolar samples of CH3OH(l) and C2H5OH(l) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is attached to a pressure gauge, and the temperatures are kept at 300 K. In both vessels, liquid is observed to remain present at the bottom of the container at all times. The change in pressure inside the vessel containing CH3OH(l) is shown below.
Compared to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is
ANSWER : lower, because London dispersion forces among C2H5OH molecules are greater than those among CH3OH molecules
Explanation:
In Order to get a good understanding the answer above let first define some terms
The London dispersion force is the weakest inter-molecular force. The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.
The vapor pressure of a liquid is the equilibrium pressure of a vapor above its liquid (or solid); that is, the pressure of the vapor resulting from evaporation of a liquid (or solid) above a sample of the liquid (or solid) in a closed container
The pressure will be Lower, because London dispersion forces among C2H5OH molecules are greater than those among CH3OH molecules ,What this means is that the stronger the inter-molecular forces, the stronger the interactions that hold the substance together, hence the lower the vapor pressure of the liquid at a given temperature, and the harder it is to vaporize the substance.
Note: London dispersion force among C2H5OH molecules is higher than that among CH3OH molecules because the number of electrons in C2H5OH is greater than that in CH3OH.
