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?

Use PV = nRT and solve for n.

P = (697-vapor pressure H2O @ 25C/760)
V = 0.247 L
T = 298 K
n = number of mols. Then n = grams/molar mass. You know molar mass and n, solve for grams.

So would i just convert 697 torr into atm and use that for pressure? or what do i do with the 25C/760?

It isn't 25C/760.

See the later post where I broke this down. If you still have trouble, repost and make a note about what you don't understand.
Ptotal = pgas + pH2O
697 = pgas + vapor pressure H2O @ 25C.
Solve for pgas. Then divide that number by 760 to convert to atm.

To find the number of grams of oxygen collected in the reaction, we first need to use the ideal gas law to determine the number of moles of oxygen. Then, we can convert moles to grams using the molar mass of oxygen.

1. Convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15 = 298.15 K

2. Convert the pressure from torr to atmospheres:
P(atm) = P(torr) / 760
P(atm) = 697 / 760 = 0.916 atm

3. Calculate the partial pressure of oxygen:
P(oxygen) = P(total) - P(water vapor)
The vapor pressure of water at 25°C is approximately 23.76 torr.
P(oxygen) = 0.916 atm - (23.76 torr / 760) = 0.887 atm

4. Apply the ideal gas law to calculate the number of moles of oxygen:
PV = nRT
n = PV / RT
P = pressure of oxygen (in atm)
V = volume of oxygen (in L)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in K)

n = (0.887 atm) * (0.247 L) / (0.0821 L·atm/(mol·K)) * (298.15 K)
n ≈ 0.0294 mol

5. Calculate the mass of oxygen:
To calculate the mass of oxygen, we need to use the molar mass of oxygen, which is approximately 32 g/mol.

mass = moles * molar mass
mass = 0.0294 mol * 32 g/mol ≈ 0.941 g

Therefore, approximately 0.941 grams of oxygen are collected in the reaction.