Conversions.

Conversions:

In any given gas law problem you will need to solve, one of the necessary steps will most likely be converting between units. For example, in the sample problem given in Charles’s Law, the final answer for V2 was 4.92 L, but what if instead the question asked you for the volume of the gas to be expressed in mL?

Problem: A combustion reaction produces 6.48 kg of carbon dioxide gas. Convert this amount into moles of the gas, and then into milliliters.

Okay, so the amount of gas is given to you in kilograms and the first part of the question is asking you to convert that into moles. To go from mass to moles, you use a conversion factor. Because you are trying to get to moles, you know that the conversion should have 1 mol in the numerator and the mass in grams should go in the denominator so that the units will cancel out.

I mol/grams of gas

But first you need to change the amount of gas in kg to grams so that the conversion can be used. To do this, you will need to know the relationship between kilograms and grams.

I kilogram = 1000 grams Knowing this, you can put it into a conversion factor to convert kg to g. Because you are already in kg and trying to get into g, kg will have to go on the bottom (so that it cancels):

1000 g/1 kg

Now use it to change 6.48 kg of carbon dioxide gas into grams by multiplying it…

6.48 kg CO2 * (1000 g/1 kg) = 6480 g

The amount in grams is 6480! Now you can change it into moles.

For this, you’ll need another conversion. This conversion has to relate grams of the gas to 1 mole.

The first step is to figure out the atomic mass of your gas. This can be done by looking at the periodic table. The name carbon dioxide tells you all you need to know. CO2 gas is made from one atom of carbon and two atoms of oxygen so to get the atomic mass, simply add the masses of the atoms together.

Atomic mass C = 12.01 g Atomic mass O = 16.00 g * 2 = 32 g

12.01g + 32 g = 44.01 g CO2

Now you can write the conversion for grams to moles! It should have 1 mole on the top, atomic mass on the bottom.

1 mol/44.01 g CO2

Simply multiply the mass in grams from the reaction by this to get the amount in moles…

6480 g CO2 * (1 mol/44.01 g CO2) = 147.2 mol CO2

The first part of the question is complete!

Now to change it from moles to mL… All you have to do is convert it from moles to liters and then to mL. You can do this by multiplying the value in moles by the conversion: 22.4 L/1 mol. (One mole of any ideal gas at STP will always occupy 22.4 liters).

147.2 mol CO2 * (22.4 L/1 mol) = 3297.28 L CO2

Finally change it from liters to milliliters. The relationship between liters and milliliters is that 1000 mL is equal to 1 L. This can be expressed as 1000 mL/1 L. Just multiply your answer in liters by this to get it in mL (the denominator will cancel out with the liters)…

3297.28 L CO2 * (1000 mL/1 L) = 3297280 mL

All parts of the problem have been solved! :]]

To go from an amount in mL to kg, the steps are essentially the same. Just remember that the units you want always go on the top because the denominator always cancels.

Example:

3297280 mL CO2 * (1 L/1000 mL) = 3297.28 L CO2

3297.28 L CO2 * (1 mol/22.4 L) = 147.2 mol CO2

147.2 mol CO2 * (44.01 g CO2/1 mol) = 6480 g CO2

6480 g CO2 * (1 kg/1000 g) = 6.48 kg CO2

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