Using the answer from problem 8.61, how many grams of nitrogen are in Whitney's lungs at STP if air contains 78% nitrogen?

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Welcome back everyone. A diving cylinder has an air capacity of 3.50 liters at S T P. We need to calculate the mass of oxygen in grams present in the cylinder at T P. If it contains 18% oxygen. Let's recall that since our conditions are S T P, we have a pressure equal to one point oh ATM. And our temperature at S T P, we recall is equal to 2 73.15. Kelvin. Now, let's begin by utilizing our ideal gas equation in which we should recall that our pressure times volume is equal to our moles of our gas times our gas constant R times our temperature. We're going to need to solve for total moles of air in our cylinder. So we would have to isolate for N, we're going to divide both sides by R times T. So that those terms will cancel out on the right. And then we'll find that our moles, we say total moles of air would equal our pressure times volume divided by our gas constant R times temperature. So plugging in our given variables for pressure because we are at S T P. We have a pressure of 1.0 ATM multiplied by our given volume for the air capacity in the diving cylinder. From the prompt being 3.50 liters. This is divided by our gas constant R which from our textbooks we should recall is the value 0.82 oh six liters times ATM S divided by moles times Kelvin. And then we would multiply by our temperature at S T P, which as we noted above is 2 73.15, Kelvin. So notice that we can cancel our units of Kelvin as well as ATM and as well as liters leaving us with moles, which is what we need for total moles of air. So in our next line, this will simplify so that we have our total moles of air equal to 0.156148. So we'll shorten that to 0.15615. And now going back to our prompt notice that we're told that the cylinder contains 18% oxygen. So this 18% of oxygen in our cylinder refers to our percent volume in the cylinder. So we're going to need to recall Avogadro's law, which relates our percent volume expressed by our initial volume divided by initial moles, which is equivalent to our final volume divided by final moles. And this is true at constant temperature and pressure which we would have at S T P. Now we can set up a more proportional setup so that we have V one over V two equivalent to N one over N two. So this is just a rearranged version of our Avogadro's law. And what we can see from this proportion or from these two proportions is that we have an equivalence between our percent volume. V one over V two to our percent mole and one over N two. So noting that down percent volume is equivalent to percent mole. And so we can interpret our 18% volume of oxygen as 18% moles of oxygen. So we'll say this is also equal to our percent mole of oxygen. So now after step one, finding total moles of air, we can move on to step two, which is to find our moles of oxygen 02. And so we would take our percent moles of oxygen from our percent volume, which is equivalent to one another. And we would find that interpreting this percent, we would take a proportion of 18, over 100 to express our percent more. And we can multiply this by our total moles of air which we determined above to be 0.15615. And this is moles as our final unit as we determined above. And sorry ATM was canceled out in our first step. So going back to our equation for moles of oxygen, when we simplify, we would find that from our product in the numerator, we have a value of zero point or sorry, two 0. moles which is divided by 100. And in our next line, we would simplify to take the value of the equation where we would get a result of 0.2811. And this is units of mose of our oxygen. Next, we want to note our molar mass of oxygen from the periodic table and sorry, this is O sub two where we would recall that for one mole of oxygen found in group six A, we have a molar mass of 16.0 g per mole. However, in our chemical formula of oxygen, we have a subscript of two. So we're going to multiply the molar mass by two. And this would give us a molar mass of 32 point oh grams per mole. So now with our molar mass of oxygen, we can use this as a conversion factor to get to our final unit being grams. Meaning taking our moles of oxygen moving on to step three, we have 0.2811 moles of oxygen in which we can multiply to go from our molar mass of 32. And actually, we would want to place 32 in the numerator. So we would have 32 point oh oh grams and this is per mole of oxygen in our denominator. So canceling out our units of moles of oxygen, we're left with grams of oxygen as our final unit. And as a result in our next line, we would find that our mass of oxygen is equal to our final answer of 0.900 g. So this would be our final answer as our mass of oxygen in our diving cylinder, given our standard temperature and pressure conditions. I hope that this made sense and let us know if you have any questions.

Using the answer from problem 8.61, how many grams of nitrogen are in Whitney's lungs at STP if air contains 78% nitrogen?

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Key Concepts

Standard Temperature and Pressure (STP)

Composition of Air

Molar Mass of Nitrogen