Lab 10: Molar Volume of a Gas & the Percent KClO3 in a Mixture

Only help on the Problems in the lab manual are available at this time.
PROBLEMS (page 98)

1. 2.037 g of a mixture of NaClO3 and NaCl (inert) was heated until all of the NaClO3 has decomposed. The mass of the residue was 1.610 g. Calculate the % NaClO3 in the mixture.

__ NaClO3 (s) --> __NaCl (s) + __O2(g)

NaCl(s) --> NaCl(s)

This problem is very similar to the problems in Lab 4. That's where hydrated salts were heated and the water was driven off. It also had NaCl as an inert component of the mixture. The chemical equation is given so that you can balance it. That's needed to know the ratio of the decomposition products compared to the original NaClO3. In the first equation, we can see that Na and Cl are already balanced. However, there are 3 oxygen atoms on the left side and only 2 on the right side. So that will require us to double NaClO3 and triple O2. That will make the oxygen atoms balance.

2NaClO3 (s) --> 2NaCl (s) + 3O2 (g)

The approach to solving the problem is to realize that the grams lost by the mixture is from the loss of oxygen gas. So we had 2.037 g and ended with 1.610 g. That's a loss of 0.427 grams of oxygen gas. That can be changed to moles of oxygen gas. Then using the balanced equation, we can find the moles of NaClO3 that must have decomposed. The moles of NaClO3 is then converted to grams of NaClO3. That will be a fraction of the original mixture, which leads to our percent by weight. The question didn't specify % by weight, but we will assume that's what is wanted.

 Start Mass of mixture Starting Mixture decomposition products Final masses 2.037 g NaClO3 Heat --> O2 0.427 g of O2 gas escapes NaCl from NaClO3 1.610 g in test tube inert NaCl original NaCl

 A B C D E F G H I J K L M N 1 Mass of O2 escapes mass to moles mole ratio 2:3 from equation to get moles NaClO3 turn moles to grams NaClO3 using its molar mass divide by total grams of mixture to get fraction of NaClO3 x 100 to get % % NaClO3 2 0.427 g 1 mole O2 2 mole NaClO3 106.44 g 100 = ?? % 3 32 g 3 mole O2 1 mole 2.037 g 100

To find percent, they say to multiply by 100, but you are actually multiplying by 100/100. You just don't divide by 100 because you see "/100". You simply turn that "/100" into the % sign. "%" means "per 100", so that means there is a 100 still in the denominator.

2. In the above experiment (problem 1) a student collected 343 mL of gas by water displacement at 23 °C and 731 torr. Calculate the molar volume of O2 at STP from these results. Also calculate the percent error in the student's value.
First we assume that the 343 mL of gas is O2, coming from the decomposition of NaClO3. If this 343 mL were at 0°C and 760 torr, then at standard conditions, it would still be 343 mL. However, 23°C is warmer than standard conditions and 731 torr is less pressure than standard conditions. So you need to decide if the changing to STP conditions will cause the volume to go up or down. If temperature goes from 23°C to 0°C, then the gas molecules are moving slower and they bang against with less force causing them to shrink more (less volume). To find out by how much we have to turn Celsius temperatures into Kelvin to get them on the proper scale. 0°C is 273K and 23°C is 296K. Now we see their perspective sizes. To make the volume go down, it makes sense to use the fraction of 273K/296K.
Going from 731 torr to 760 torr means the gas will feel more pressure, causing the molecules to squeeze together more (less volume). The exact amount depends on a fraction based on the starting and final pressures. So a fraction of 731 torr/760 torr, will reduce the volume.
Molar volume is "Liters per mole". So we need to take the volume in liters and divide by moles. The volume of 343 mL adjusted to STP will find the volume (Liters). Problem 1 gives us the grams of O2. So we start with finding liters per gram, but then change the grams of O2 to moles. See rows 5 to 7.
 A B C D E F G H I J 1 Volume of gas Adjust for going to an increased pressure (Boyle's Law) Adjust for going to a lower temperature (Charles Law) Volume of O2 gas at standard conditions (760 torr, 273K) 2 343 mL 731 torr 273 Kelvin = ??? mL 3 760 torr 296 Kelvin 4 5 O2 STP volume from I2 Divide by mass of O2 from Problem 1 Turn grams O2 to moles O2 using molar mass of O2 cancel milli Molar Volume of O2 is Liters per mole as calculated from this experiment 6 =I2 mL 31.9989 grams 0.001 = ??? Liters 7 0.427 g 1 mole milli 1 mole

The final answer is in the ballpark of the expected 22.4 liters per mole that all ideal gases are suppose to have. To get percent error this value, subtract 22.4 from cell H6 and then divide by 22.4. Then multiply by 100
Experimental value - 22.4 (true value) x 100 = % error.
22.4

3. How much KClO3 must be decomposed to provide 400. mL of oxygen gas collected at STP?

__KClO3 (s) --> __KCl (s) + __O2 (g)

The balancing of this equation is just like the one in Problem 1. Instead of sodium chlorate (NaClO3, we have potassium chlorate (KClO3).

2KClO3 (s) --> 2KCl (s) + 3O2 (g)

Practice Problem for #3. These calculations are useful in survival types of scenarios. For example, let's say you supervise a mining operation and want the workers to carry around an oxygen source of KClO3 that could provide 10 minutes of oxygen. A person needs about 140 milliliters of oxygen per minute, so that is 14000 mL of oxygen for the 10 minutes. How many grams of KClO3 must be decomposed to provide 14000 mL of oxygen gas at STP? Also, what would be mL volume of the KClO3 powder?
This problem is easier because the conditions are at STP. We know that 1 mole of a gas at STP is 22.4 liters. So we can use that to convert the 14,000 mL into moles of O2 (columns D & E below). Then using the balanced equation, we change that to moles of KClO3 (columns F & G). Then it's easy to go to grams of KClO3. Then using looking up the density of KClO3, we can get its volume.

 A B C D E F G H I J K L M N O P Q 1 mL of O2 cancel milli Liters>moles O2 Use balanced equation moles KClO3> g grams KClO3 density Volume 2 14000 mL 0.001 1 mole O2 2 moles KClO3 122.55 grams = 51.1 grams 1 mL = 21.8 mL 3 milli 22.4 liters 3 moles O2 1 mole 2.34 g
The first answer is 51.1 grams of KClO3 needed to provide 10 minutes of oxygen for breathing. The 51.1 grams of KClO3 only takes up 21.8 mL of volume. That's only 1.5 tablespoons. The KClO3 could be heated (decomposed) with two AA batteries powering a heating filament. So the whole unit would be about the size of a ballpoint pen.
4. The volume of a sample of gas that weighs 0.324 g is 295 mL at 26°C and 732 torr. Calculate the density of the gas at STP.
Density of gases is normally written in grams per liter (grams divided by liters. You are given the grams and the millilters, so the current density is 0.324 g/295mL = 1.098 g/L. However, it wants the density at STP. So there is an adjustment of the volume when going from 26°C to 0°C and 732 torr to 760 torr. Like in Problem 2, you decide if these changes raise or reduce the volume (mass doesn't change). That guides you on how to set up the fraction. Also, like before, temperature's always needs to be converted to the Kelvin scale first. So 0°C is 273K and 26°C is 299K.
 A B C D E F G H I 1 Volume of unknown gas Adjust for going to a higher pressure (gas contracts) Boyle's Law Adjust for going to a lower temperature (gas contracts) (Charles Law) Volume of gas at STP (0°C and 760 torr) 2 295 mL 732 torr 273 K (50°C) = ??? mL 3 760 torr 299 K (0°C) 4 5 Mass of unknown gas Divide by volume from H2 cancel milli Density in grams per liter at STP 6 0.324 g milli = ??? grams 7 ??? mL 0.001 1 Liter
Knowing the density of a pure gas at STP is the first step in knowing its molar mass, which can help identify the gas. By multiplying your answer times 22.4 liters per mole, the liters cancel and you get grams per mole, which is the molar mass. In this problem, molar mass turns out to be 27.975, so the gas appears to be nitrogen (N2).
5. What is the molar mass of a gas if 0.528 g of it occupies 457 mL when collected at STP.
This problem is easier than Problem 4. There are no adjustments due to temperature or pressure. So it's just rows 5, 6 and 7 in the above spreadsheet plus the multiplication of 22.4 liters per mole as mentioned above. Hint: The molar mass is close to the gas, acetylene, which is used in acetylene torches for welding.
 A B C D E F G H I 1 Mass of unknown gas Divide by volume cancel milli Convert (per liter) to (per mole) using 22.4 L = 1 mole molar mass of unknown gas 2 0.528 g milli 22.4 L = ??? grams 3 457 mL 0.001 1 mole mole
These gas problems are a little different due to the fact that we are normally changing grams to moles, but in this case we are changing the volume to moles using the fact that 22.4 liters equals one mole for gases.