Here in this example question, it says a sample of oxygen gas has a measured volume of 325 MLS at STP. How many grams are present? All right, so here they're giving us volume in the forms of milliliters. An STP is giving us temperature plus pressure. We know from the ideal gas law with these values given to us or these variables given to us. The only thing that's missing is our moles, because we have a volume already, we have pressure and temperature, and we always know what R is.
So we're going to divide OPT RT and when we do that we're going to get our moles. Moles here will equal pressure times volume over R * T. At STP our pressure is 1 atmosphere. Our volume we just change milliliters to liters, so 0.325 liters. We have our R constant which is .08206 liters times atmospheres over moles times K. Then remember we're dealing with temperature at T we use the units of Kelvin. So that's 273.15 Kelvin.
So here, Kelvin's cancel out, liters cancel out, atmospheres cancel out, and we'll be left with the moles of our oxygen gas. When we plug that in, we get 0.01450 moles of O2. But here the question is not asking us to determine the moles of oxygen gas, but instead the grams of oxygen gas. So we just need to do a simple conversion. We say that for every one mole of O2, it weighs 32g because there's two oxygens. Those cancel out and what we get at the end is 0.464 grams of O2.
Here our answer has three significant figures because the value of 325 has three significant figures as well. So just remember when we're faced a question like this, look and see what variables are given. Since we're dealing with the ideal gas law, we see that the value that's missing is moles. Then go for moles to grams to get your final answer.