ShmoopTube

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Here's your shmoop du jour, brought to you by balloons. Aside from being kicked in a

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sensitive area, they're still the number one way to talk in a high-pitched voice.

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A party is thrown with balloons. The balloon salesman tells the party planners that he

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is out of helium cylinders, but he has a stock of other gases: hydrogen, argon, and chlorine.

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The party planners see an unlabeled gas cylinder in the back of the store. They are curious

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as to which gas it is. They fill a balloon to 1 liter and weigh it.

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NOTE: We do NOT recommend filling balloons with random, unlabeled gas. These guys are

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professional idiots...do not try this at home... Anyway, after subtracting the weight of the

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balloon, they find that the weight of the gas in the balloon is 0.819 grams

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It is 27 degrees Celsius outside with standard atmospheric pressure. Assume ideal behavior

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for all gases. Air has a density of 1.161 grams per Liter under these conditions.

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So for Part A, we need to perform the following calculations:

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1 - Calculate the density in grams per Liter of hydrogen.

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2 - Calculate the density in grams per Liter of argon.

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3 - Calculate the density in grams per Liter of chlorine.

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First things first: Calculate the density in grams per liter of hydrogen.

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Okay, so this problem mentions that we're working with gases, and that we can assume

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ideal behavior for all gases. So...we can assume that none of them is going

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to be resting their elbows on the dinner table. We should automatically be thinking about

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the ideal gas law... PV equals nRT. This law allows us to find certain properties

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about an ideal gas...where p equals pressure, v equals volume in liters, n equals number

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of moles, R is the ideal gas constant, and T is temperature in Kelvin.

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What's an ideal gas, you might ask? It's a hypothetical gas in which atoms are perfectly

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elastic and there are no intermolecular attractive forces between molecules. They're all...VERY

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attracted to each other. But back to solving the problem.

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Ok, so we're given standard atmospheric pressure, which means that P equals 1 atmosphere, the

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gas constant R equals .0821, and the temperature equals 27 degrees Celsius.

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But we have to convert temperature to Kelvin...

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and to do so, we can add 273 to 27...to get 300 Kelvin.

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Plugging all these numbers into the equation,

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we find that V equals n times .0821 times 300. But that doesn't look anything like density,

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which is in units: grams per liter. So what else do we know about hydrogen gas

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that can help us? It's molecular weight -- bingo.

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We can calculate the molecular weight of hydrogen gas, which is 2 times the weight of two hydrogen

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atoms, using the periodic table. This equals 2.106 grams per mole.

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Using dimensional analysis, to convert grams per mole into grams per liter, we know that

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we have to have moles on the top, and liters on the bottom, which is equivalent to n over V.

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Now we know what to do with our ideal gas

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law equation. N over V equals 1 over .0821 times 300, or .0406 moles per Liter.

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Multiplying this by the molecular weight of hydrogen gas, we can convert our value into density.

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So we have .0406 moles per liter times 2.106 grams per

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mole equals .0855 grams per liter. But we can't forget about significant figures

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either! All the numbers we're given in the problem have three sig figs, so our answer

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should have three as well. So great.

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Now we just need to do the same thing

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for the other 2 calculations... Let's find ourselves the density of argon.

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Because we are assuming ideal gas behavior, this calculation will be the same as the calculation

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in part 1 to obtain the value for n over v... ... .0406 moles per liter.

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For ideal gases, this value is independent of the identity of the gas.

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We use the molecular weight of argon -- 39.948 grams per mole...which we can find from the

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periodic table again....to convert this value to density.

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0.0406 moles per liter times 39.948 grams per mole equal 1.62 grams per liter.

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Okay, and now for the density of chlorine. Same deal, but this time we use the molecular

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weight of chlorine - 70.905 grams per mole. We can multiply .0406 times 70.905 to get

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2.88 grams per liter.

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OK...Now for part B of this question...

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Assuming the party planners want their balloons to float, which of the three listed gases

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should they purchase? Justify your answer.

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Well, if we ever want something to float in

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another, we need the thing we want to float to have a LOWER density than the other.

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Since air has a density of 1.161 grams per Liter, as given in the problem...we want the

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gas that has a density less than that....

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Hydrogen has a density of 0.0855 grams per Liter, so that's our answer!

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Time for Part C...

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Let's take question 1 first... what is the molecular weight and the identity of the gas

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in the unlabeled cylinder, assuming it is monatomic and acts as an ideal gas?

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Ok...so this question is asking us to figure out the molecular weight and identity of the

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unknown gas. We're given that the weight of the gas in the balloon is 0.819 grams.

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But molecular weight is in the unit: grams per mole...so all we have to do is figure

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out the number of moles, using the ideal gas law again, PV equals nRT.

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Rearranging the equation to find moles, n equals PV divided by RT.

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We have all the information we need to solve: standard pressure -- P equals 1 atmosphere...

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volume -- V equals 1 Liter, gas constant -- (R = 0.0821 liters atmospheres over moles times

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Kelvin), and temperature -- T equals 300 Kelvin.

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We plug in the numbers and solve for n:

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1 times 1 over .0821 times 300 equals .0406 moles.

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So the molecular weight is grams per mole.... .819 grams divided by .0406 moles equals 20.2

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grams per mole. Looking at the periodic table, the element

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with that molecular weight is Neon.

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Now the question wants to know...will it float?

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Well, like part B of this question, the key to figuring out if a gas will float is if

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it is less than the density of air. The density of neon is in grams per liter...so

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.819 grams divided by 1 liter is .819 grams per liter.

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0.819 grams per liter is less than 1.161 grams per liter...so yeah, it totally floats!

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Finally... Part D... Question 1 of part D... Write the balanced,

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net ionic equation for the reaction that occurs when chlorine gas is bubbled into a solution

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of sodium bromide.  Alright, to be honest, we know this question

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is kinda unrelated to the rest of the problem. But sometimes the test makers are like that,

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and we just have to... deal with it. So to find the net ionic equation, we first have to start

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with the balanced molecular equation. Chlorine gas into sodium bromide...chlorine

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gas is a diatomic molecule, so we have Cl2. The formula for sodium bromide is NaBr.

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This is an example of a single displacement chemical equation...so the chlorine replaces

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the bromide to make sodium chloride gas and bromine gas...which is a diatomic molecule.

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So we have two bromine atoms, not just one. To balance this equation, we can add a 2 in

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front of the sodium bromide and 2 in front of the NaCl.

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To find the net ionic equation, we have to break all the soluble electrolytes into their

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ions... ...which means the sodium ions on both sides

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need to be taken out of their salts.

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Just like a normal math equation, we can cancel

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out the two sodium ions on the left and right...and we're left with chlorine gas plus 2 bromide

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ions, which makes bromine gas and 2 chlorine ions.

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Number 2 in part D. Last question...asks us which species is oxidized in the equation.

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Well, oxidation occurs when an element becomes positive because they are LOSING electrons.

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Good for them. We don't become positive when we lose ANYTHING.

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If we look at our net ionic equation, the bromide ions are being oxidized because, on

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the left, they have a negative charge, but then after the ions react with the chlorine

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gas, they have no charge. They are becoming more positive, or being oxidized!

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Phew. So thanks for hanging in there with us for that one. It was a beast.

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Hello? Oh...don't tell us we have to say all that all over again...

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