A 36.1 mL sample of oxygen gas contains 0.529 mol of oxygen. If enough oxygen is added to the sample to obtain a volume of 50.5 mL, what must the final mass of the sample be? (The pressure and temperature is constant.)

Well, it seems like oxygen is feeling a bit weight-conscious. Let's calculate its final mass and help it embrace its size!

First, let's find the initial density of the oxygen sample:
Density = Mass / Volume

Given that the initial volume is 36.1 mL and the number of moles is 0.529 mol, we can use the ideal gas equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the pressure and temperature are constant, we can write the equation as:

V(initial) / n = V(final) / n(final)

Plugging in the known values:
36.1 mL / 0.529 mol = 50.5 mL / n(final)

Solving for n(final), we find:
n(final) = (0.529 mol * 50.5 mL) / 36.1 mL

Now that we know the final number of moles, we can determine the final mass. Since the molar mass of oxygen (O2) is about 32 g/mol, we can calculate the final mass using the formula:

Mass = n(final) * Molar Mass

Mass = (0.529 mol * 50.5 mL / 36.1 mL) * 32 g/mol

Calculating the final mass gives us the answer oxygen is dying to know!

To solve this problem, we need to use the ideal gas law equation, which is given by:

PV = nRT

Where:
P is the pressure (which is constant in this case)
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature (which is constant in this case)

First, let's calculate the number of moles of oxygen in the original sample:
n1 = 0.529 mol

Next, let's calculate the volume of the oxygen after it has been added to the sample:
V2 = 50.5 mL

Now, let's calculate the number of moles of oxygen in the final sample using the ideal gas law equation:

n2 = (n1 * V1) / V2

where V1 is the initial volume of the sample, which is 36.1 mL.

n2 = (0.529 mol * 36.1 mL) / 50.5 mL
n2 = 0.378 mol

Now, let's calculate the final mass of the sample. We can use the molar mass of oxygen, which is approximately 32 g/mol.

m2 = n2 * M

where M is the molar mass of oxygen.

m2 = 0.378 mol * 32 g/mol
m2 = 12.096 g

Therefore, the final mass of the sample must be 12.096 g.

To find the final mass of the sample, we first need to find the molar mass of oxygen gas.

The molar mass of oxygen (O2) is approximately 32 g/mol. This means that one mole of oxygen gas has a mass of 32 grams.

Now, let's find the mass of the initial sample:

Mass of initial sample = Number of moles x Molar mass
Mass of initial sample = 0.529 mol x 32 g/mol
Mass of initial sample = 16.928 g

Next, we need to calculate the final number of moles of oxygen gas in the sample. To do this, we'll use the ideal gas law, which states:

PV = nRT

Where:
P = pressure (constant)
V = volume (changing)
n = number of moles (changing)
R = ideal gas constant
T = temperature (constant)

Since the pressure and temperature are constant, we can rewrite the ideal gas law as:

(V1/n1) = (V2/n2)

Where:
V1 = initial volume
n1 = initial number of moles
V2 = final volume
n2 = final number of moles

Plugging in the values we have, we get:

(36.1 mL / 0.529 mol) = (50.5 mL / n2)

Cross-multiplying and solving for n2, we find:

n2 = (50.5 mL x 0.529 mol) / 36.1 mL
n2 = 0.737 mol

Finally, let's calculate the final mass of the sample:

Mass of final sample = Number of moles x Molar mass
Mass of final sample= 0.737 mol x 32 g/mol
Mass of final sample = 23.6 g

Therefore, the final mass of the sample must be 23.6 grams.