How many grams of oxygen are collected in a reaction where 247 mL of oxygen gas is collected over water at a temperature of 25°C and a total pressure of 697 torr? And would i just convert 697 torr into atm and use that for pressure? And how do i find the molar mass of the problem?

I think I worked through this earlier.

First you need to determine the gas pressure less the water vapor.
Ptotal = pO2 + pH2O
697 = pO2 -vapor pressure H2O @ 25C.
Then convert pO2 to atm (divide by 760).
Then use PV = nRT and solve for n = number of mols. Finally,
n = grams/molar mass. You know n and grams, solve for molar mass.

To find the number of grams of oxygen collected in the reaction, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin

First, let's find the number of moles of oxygen gas by rearranging the ideal gas law equation:

n = PV / RT

In this case, the pressure (P) is given as 697 torr, but it needs to be in atm for the ideal gas law. So, we convert 697 torr to atm:

697 torr * (1 atm / 760 torr) = 0.917 atm

Now, we can calculate the number of moles of oxygen:

n = (0.917 atm * 0.247 L) / (0.0821 L·atm/mol·K * 298 K)
n ≈ 0.0100 moles

To find the molar mass of oxygen, we can use the periodic table. And the molar mass of oxygen is approximately 32 g/mol.

Finally, we can find the mass of oxygen gas in grams:

mass = n * molar mass
mass = 0.0100 mol * 32 g/mol
mass = 0.32 grams

Therefore, approximately 0.32 grams of oxygen gas are collected in the reaction.

To summarize:
1) Convert the pressure to atm (if needed).
2) Use the ideal gas law equation to find the number of moles of the gas.
3) Find the molar mass of the gas using the periodic table.
4) Multiply the number of moles by the molar mass to find the mass of the gas in grams.