What volume of carbon dioxide gas would be released from a one cc. cube of dry ice at room temperature? --Adam Smith (pinecone@sowega.net)

Adam, Ask yourself the following questions:
  1. How many grams of CO2 are in one cc of dry ice? (You must look up the density of dry ice.)
  2. How many moles of CO2 are in that number of grams?
  3. How many liters of CO2 is that, if I assume the gas is at 1 atm and 298 K and that it behaves ideally?

If one gram of sodium bicarbonate reacts with excess vinegar solution that is 5% acetic acid, how much gas will be released in cc's? & How many cc's of vinegar are required? -Adam Smith (pinecone@sowega.net)

These are reaction stoichiometry problems (you're relating the amounts of different substances involved in a chemical reaction). To solve the first problem, follow these steps:

  1. Pick out the target. You want cc CO2.
  2. List the given information. You have 1 g of NaHCO3. You have an excess of 5% acetic acid solution (which means every 100 g vinegar contains 5 g of acetic acid).
  3. Connect the given information with the target. You're trying to convert g NaHCO3 into cc CO2. Whenever the problem involves a connection between two different substances, you must have a mole-to-mole relationship between the two to solve the problem. Write and balance an equation for the reaction between acetic acid and sodium bicarbonate to get this relationship. Then
    1 g NaHCO3 mol NaHCO3 mole-to-mole ratio
    from balanced equation

    mol CO2 cc CO2
    ...so you've cut a tough problem into two easier ones: the g-to-mol conversion of NaHCO3, and the mol-to-cc conversion of CO2. The first part is easy; you can always convert grams of a substance to moles using the substance's molecular weight. You know a relationship between moles of gas and volume of gas, too: the ideal gas law. But since the problem doesn't say anything about temperature and pressure, you'll have assume specific values for these; P = 1 atm and T = 298 K are reasonable conditions. Using the ideal gas law, you can get a mole-to-volume conversion factor at this temperature and pressure.
  4. Do the math. Set up a series of conversion factors so that units cancel to ultimately give you cc CO2.
  5. Check the answer. You know that a mole of gas will occupy about 22.4 L at STP (and you're not far off from STP). Every mole of bicarbonate produces a mole of carbon dioxide. So multiplying moles of bicarbonate by 22.4 L should give you a rough estimate in liters.

You can follow the same general procedure to answer the second question. This time, no gases are involved.

A mixture of cyclopropane gas and oxygen is used in an anaesthetic. Cyclopropane contains 85.7% C, and 14.3% H by mass. At 50.0 C and 0.984 atm pressure 1.56g cyclopropane has a volume of 1 L. What is the empirical and molecular formula of the cyclopropane?

Heather

Heather,

First, use the element mass percents to obtain the empirical formula of the cyclopropane.

  1. Write the mass percent as a mass (mass percent is just mass per 100 g sample). So you have 85.7 g C and 14.3 g H in a hundred-gram sample of cyclopropane.
  2. Convert each of these masses to moles.
  3. Which of the elements is present in the smallest molar amount? Divide moles of C and moles of H by the smallest molar amount. You now have two mole ratios.
  4. Round the mole ratios to the nearest whole number (or simple fraction). These are the subscripts in the empirical formula. They should be whole numbers.

You can see an example of an empirical formula calculation in General Chemistry Online's section on compounds.

You're halfway done. To find the molecular formula, you need to find the molecular weight of the compound. The ideal gas law will be useful here.

  1. Calculate the number of moles of gas from the given pressure, volume, and temperature.
  2. The molecular weight of the gas is grams of gas (1.56 g) divided by moles of gas.

Finally, to find the molecular formula:

  1. Find the empirical formula weight by adding up the weights of the atoms in the empirical formula.
  2. Divide the molecular weight by the empirical formula weight. This number tells you how many times the empirical formula is repeated to make the molecular formula.

    For example, if your empirical formula was OH, and the molecular weight was 34, the empirical formula would be repeated 34/17 or 2 times to make the molecular formula. So the molecular formula would be H2O2.

I hope this helps.

What steps do I need to take to solve this problem?

The brominated alkane ethylene dibromide, or 1,2-dibromoethane is a common but controversial fumigant and is also an anti- knock addditive in gasolines. It is prepared by the action of bromine on ethene gas. How many liters of ethane at 35°C and 3.00 atmospheres are required to prepare 500.0 kg of ethylene dibromide, assuming the reaction goes to completion?

-Ryan Cramer

This is a reaction stoichiometry problem, since you're trying to relate an amount of one substance to an amount of another through a chemical reaction. Here is a general strategy for solving such problems:

  1. Pick out the target. You want liters of ethane.
  2. List the given information. You have 500.0 kg of ethylene dibromide. The ethane used to make it is at 35°C and 3.00 atm. Your strategy is to convert kg of ethylene dibromide into L of ethane gas:

    500.0 kg C2H4Br2 --> L C2H6(g)

  3. Search for relationships that connect the given information with the target. This is usually the most difficult step. It helps to work backwards from the target towards the given information. Also, whenever an amount of one substance is being used to obtain an amount of another, you need a mole-to-mole ratio that relates the two substances. Knowing that lets you break the problem into three easier pieces:

    500.0 kg C2H4Br2 --> mol C2H4Br2 --> mol C2H6 --> L C2H6(g)

    The kg C2H4Br2 to mol C2H4Br2 conversion will require the molar mass of C2H4Br2. The mol C2H4Br2 to mol C2H6 conversion requires a balanced chemical equation for the reaction (although, if you look at the formulas, you'll realize that every mole of C2H4Br2 formed requires one mole of C2H6). Finally, to compute L C2H6(g) from moles of C2H6, you'll need to think about how to compute the volume of a gas given the moles, pressure, and temperature of the gas...
  4. Check your answer. Before you pick up your calculator, decide roughly how many of liters of ethane you expect to get. Whenever you can, you should use a different method to check your calculation.

Could you, please, show and explain how to do the following problem? I kind of have an idea, but not sure. Please, help!!!!!

  1. How many molecules are present in 475 mL of CO2(g) at STP?
  2. What is the volume occupied by 3.50 x 1024 SO2 molecules at STP?
  3. Also, please, tell me the relationship between molecules of a compound or atom and the number of moles.

-Julia

Allow me to introduce you to the mole. Click on him to learn the answer to your last question first. Then return here.


To find the number of molecules in a gas sample, given the pressure, volume, and temperature of the sample:
  1. Figure out how many moles of CO2 gas you have. Hint: assume the CO2 is an ideal gas; you know the temperature and pressure (it's at STP) and you know the volume, so you should be able to compute the number of moles using the ideal gas equation of state.
  2. Use the mole-to-molecule conversion you learned above to compute the number of molecules.

To find the volume of a gas sample, given the number of molecules, the pressure, and the temperature:

  1. Convert molecules of gas to moles of gas, using the mole-to-molecule conversion.
  2. Assuming the gas sample is ideal, you can easily compute the volume of the gas using the ideal gas equation of state because you know the temperature, pressure, and moles of gas.

For more about Avogadro's law, look here.

How do I find the densities of oxygen , methane and carbon dioxide gases at s.t.p.? -Najam Mirza

Density is mass over volume. To solve the problem, you have to find the mass of a sample of unit volume. Here is a general strategy for getting a quick and dirty estimate of gas densities:

  1. Find the molar volume. At STP, one mole of an ideal gas will occupy 22.414 L. At other temperatures and pressures, the molar volume of an ideal gas V/n = RT/P. If you can't assume the gas is ideal, you'll need to either look up the molar volume in a handbook or use a real-gas equation of state to get it.
  2. Find the molar mass. You know the mass of one mole of each gas from its molecular weight.
  3. Density is mass over volume, so dividing the molar mass by the molar volume gives you the density. Gas densities are somewhere around 1 g/L at room temperature and pressure- about 1/1000th the density of liquids.

    Explain why air of low humidity has a higher density than air of high humidity at the same temperature.

    -Jamal

    There is a lot of empty space between gas molecules. The molecules are very far apart, relative to their sizes, and they neither attract nor repel each other much. So distances between molecules aren't affected by the type of molecules in the sample.

    That implies that the number of molecules in a given container won't change, when you change only the type of gas filling the container. (This is one way to state Avogadro's Law).

    Here is the same very small container, filled with dry air on the left, and wet air on the right. The drawings aren't to scale; the distance between molecules is much, much larger under normal conditions. Notice that in both pictures, there are 10 molecules in the container.

    The molecular weights of O2 (in red), N2 (in blue) and H2O are about 32, 28, and 18 atomic mass units, respectively. Substituting a water molecule for either an oxygen or a nitrogen molecule, then, will decrease the total mass of air in the container. Since the volume of the container is the same in both cases, the density of water-containing air must be less than the density of dry air.

    Temperature and pressure affect the distance between molecules, so in making this comparison, we've assumed that both the dry and wet air are at identical temperatures and pressures.

    When you open a bottle of soda does the air go in or out of the bottle and what is the volume of the air released or allowed in?
    mackenzie altland

    Stretching a balloon over the top of the bottle and popping off the cap under the balloon will reveal that gas leaves the bottle; the balloon most definitely expands. So gas will rush out of the bottle, but this gas is not air; it's mostly carbon dioxide saturated with water vapor, but other gases are present as well.

    Computing the final volume of the gas after the top is removed requires additional information. The gas is released in two stages:

    1. The gas originally trapped in the head space has higher than atmospheric pressure and will expand very quickly out of the bottle.
    2. Gases dissolved in the soda will be released as well, but more slowly. Notice that these gases stay in the soda until the bottle is uncapped. You don't see bubbles until the cap is off. This is because the solubility of gas in the soda is proportional to the pressure of the gas in the head space (an assertion sometimes known as "Henry's Law"). The soda bottler adds high pressure CO2 to the head space to ensure that soda will dissolve more CO2. If the pressure in the head space drops when the cap is popped off, the solubility of the gas in the bottle also drops. The soda is now supersaturated with carbon dioxide, and the excess gas is released slowly over the next hour in streams of bubbles.

    Let's just focus on the gas that is released quickly when the bottle is uncapped. To estimate the final volume of the gas, we need to have the following pieces of information:

    • The pressure drop. The gases in the head space have a pressure of approximately 2 atm, and after uncapping, the pressure will fall to about 1 atm.
    • The temperature drop. When the gas expands very quickly, it has to expend its own energy to push back the atmosphere. Additional energy will be required to overcome intermolecular attractions. Most real gas molecules weakly attract each other, and when the gas expands rapidly, additional energy is required to pull the molecules away from each other. The internal energy of the gas plummets, and so does its temperature. If the soda comes out of the fridge at 5°C, the temperature drop for the expansion of the headspace gases is about 40°C. He assumes that the expansion occurs so quickly that no heat flows between the gas and the surroundings. (Such expansions are referred to as adiabatic expansions).

    One may then use the combined gas law for a quick and rather dirty estimate of the final volume:

    V2/V1 = (P1/P2) * (T2/T1) = (2 atm/1 atm) * (248 K/278 K)

    so just after the cap is popped the gases in the head space will leave the neck of the bottle, expanding to about 1.8 times their original volume. The volume of the original head space is required if you want to know the volume of gas that actually exits the bottle.