1. Use the Van der Waals equation to calculate the pressure exerted by 1.00 mol of Cl2; in 22.41 L at 0.0 degrees C. The constants for Cl are a = 6.49 L2 atm/mol2 and b = 0.0526 L/mol.
2. How much potassium chlorate is needed to produce 20.0 mL of oxygen at 670 mm Hg and 20 degrees C.
3. A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?
1. To calculate the pressure exerted by 1.00 mol of Cl2 using the Van der Waals equation, we can use the formula:
\[ \left( P + \frac{an^2}{V^2} \right) \left( V - nb \right) = nRT \]
where:
P = pressure
n = number of moles
V = volume
a = Van der Waals constant a (dependent on the gas)
b = Van der Waals constant b (dependent on the gas)
R = ideal gas constant (0.0821 L * atm /(mol * K))
T = temperature in Kelvin
Given:
n = 1.00 mol
V = 22.41 L
a = 6.49 L^2 atm/(mol^2)
b = 0.0526 L/mol
T = 0.0 degrees C = 273.15 K
Plugging in the values into the equation, we get:
\[ \left( P + \frac{6.49 \times (1.00)^2}{(22.41)^2} \right) \left( 22.41 - 1.00 \times 0.0526 \right) = 1.00 \times 0.0821 \times 273.15 \]
Solving this equation will give us the pressure, P.
2. To calculate the amount of potassium chlorate needed to produce 20.0 mL of oxygen, we need to use the ideal gas law equation.
\[ PV = nRT \]
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L * atm /(mol * K))
T = temperature in Kelvin
Given:
P = 670 mm Hg = 0.882 atm
V = 20.0 mL = 0.0200 L
T = 20 degrees C = 293.15 K
Rearranging the equation, we get:
\[ n = \frac{PV}{RT} \]
Plugging in the values, we can calculate the number of moles of oxygen produced.
3. To find the molecular formula of the compound given its empirical formula, we need to know the molar mass of the compound. We can calculate the molar mass using the empirical formula and the given vapor density.
The empirical formula CH indicates that the empirical formula mass is 12.01 g/mol (for carbon) + 1.008 g/mol (for hydrogen) = 13.018 g/mol.
Given the vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg, we can assume the compound exists as a gas under these conditions. The molar mass is simply the vapor density multiplied by the molar volume. Using the ideal gas law, we can calculate the molar volume.
\[ PV = nRT \]
where:
P = 760 mm Hg = 1 atm
V = molar volume
n = number of moles
R = ideal gas constant (0.0821 L * atm /(mol * K))
T = 100 degrees C = 373.15 K
Rearranging the equation, we get:
\[ n = \frac{PV}{RT} \]
Since we know the molar mass (molar mass = vapor density x molar volume), we can calculate the molar volume. Then, calculating the number of moles of the compound gives us the molecular formula.
1. To calculate the pressure exerted by 1.00 mol of Cl2 using the Van der Waals equation, we need to substitute the given values into the equation and solve for the pressure (P).
The Van der Waals equation is:
(P + (a * n^2 / V^2)) * (V - n * b) = nRT
Where:
P is the pressure
n is the number of moles
V is the volume
a and b are the constants for Cl2
R is the ideal gas constant (0.0821 L atm/(mol K))
T is the temperature
Given values:
n = 1.00 mol
V = 22.41 L
T = 0.0 degrees C = 273.15 K
a = 6.49 L^2 atm/(mol^2)
b = 0.0526 L/mol
Now we can substitute the values into the equation and solve for P:
(P + (6.49 L^2 atm/(mol^2) * (1.00 mol)^2 / (22.41 L)^2)) * (22.41 L - 1.00 mol * 0.0526 L/mol) = 1.00 mol * 0.0821 L atm/(mol K) * 273.15 K
Simplifying the equation will give us the value of P.
2. To calculate the amount of potassium chlorate needed to produce 20.0 mL of oxygen, we follow these steps:
Step 1: Convert the given volume to liters.
20.0 mL = 20.0 mL * (1 L / 1000 mL) = 0.0200 L
Step 2: Use the ideal gas law equation to calculate the number of moles of oxygen present.
PV = nRT
Where:
P is the pressure (670 mm Hg)
V is the volume (0.0200 L)
n is the number of moles (to be determined)
R is the ideal gas constant (0.0821 L atm/(mol K))
T is the temperature (20 degrees C = 293.15 K)
n=(P * V)/(R * T)
Step 3: Determine the molar ratio between oxygen and potassium chlorate.
From the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Thus, the molar ratio is 3 moles of O2 for every 2 moles of KClO3.
Step 4: Calculate the amount of potassium chlorate needed.
Since we know the molar ratio, we can set up the following proportion:
(3 moles of O2 / 2 moles of KClO3) = (0.0200 moles of O2 / x moles of KClO3)
Cross-multiplying and solving for x, we can find the number of moles of KClO3.
Step 5: Convert moles of KClO3 to grams.
To convert moles to grams, we need the molar mass of KClO3 (potassium chlorate). This can be found by adding up the atomic masses of the elements in the compound (K: 39.10 g/mol, Cl: 35.45 g/mol, O: 16.00 g/mol). Multiply the molar mass by the number of moles of KClO3 to get the mass in grams.
3. To determine the molecular formula of a compound based on its empirical formula and the given vapor density, we need to follow these steps:
Step 1: Calculate the molar mass of the empirical formula.
Calculate the molar mass of CH (carbon and hydrogen) by adding the atomic masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol). This gives a molar mass of 13.02 g/mol for CH.
Step 2: Calculate the molar mass of the molecular formula.
To do this, divide the given molecular mass (vapor density) by the calculated molar mass of the empirical formula. This will give you the ratio between the molar mass of the molecular formula and the empirical formula.
Step 3: Determine the molecular formula.
If the ratio obtained in Step 2 is a whole number, then the molecular formula is the same as the empirical formula. If it is not a whole number, multiply both elements in the empirical formula by the same factor to get the molecular formula.