Welcome to a mad scientist production of… Mole Conversions, in which we will take a look at how to convert from mass and particles to moles, using dimensional analysis! First let’s define the mole: it’s a NUMBER! A very large, incomprehensibly large number. But we use it because on the every day level we use incomprehensibly large amounts of particles, which in general includes atoms, molecules, ions, and formula units. So on the macro, or every day level, normal amounts of material are often close to molar amounts. For example, 8 oz of water is 7.6 × 10²⁴ H₂O molecules, a little more than 12 moles of water molecules. And a cup of sugar is a bit less than 1 mole of sucrose molecules. In the laboratory we use molar amounts all the time. For example, 75.9 grams of acetone, something you might use in an experiment, is 1.31 moles of acetone. Everyday, or macro, amounts of substances are composed of vast amounts of particles, and the mole allows us to express those very large amounts using convenient numbers; AND, more importantly, to express the amounts in grams so that it can be measured directly. Here is the essential nugget for understanding the mole’s existence. There are many applications in chemistry that you will encounter through the year that require knowing the amount of particles. But particles cannot be counted directly! We can however measure the mass of particles. And so we can connect mass and particles by using the mole. And this allows us to determine specific amounts of particles. The mole represents an amount of particles we would use on a human scale. So how can we connect these three measurements mathematically? This video will look at mathematically converting from mass to mole and mole to mass, and from mole to particle and particle to mole. Each one of these four conversions requires a single calculation. And so you might hear to them referred to as single step conversions. Let’s first look at the two conversions on the left, mole to mass and mass to mole conversions. To do that we need to first understand what molar mass is. In other words, how are the amounts of particles related to the mass of particles? And second, we’ll see how to set up the calculation to convert from one unit to another. We need the periodic table to answer that question. We define it as the mass of one mole of particles given by the periodic table masses. And we’ll keep in mind what a mole represents. We can see on this periodic table that the mass of each element is given at the bottom of each element’s box. That is very significant, because 1 mole of any element has the mass given by the periodic table in grams. Let’s give that statement meaning with some examples. So a mole of silicon atoms has a mass of 28.09 grams, a mole of phosphorus atoms has a mass of 30.97 grams, and a mole of sulfur atoms has a mass of 32.07 grams. What about molecules? For example one mole of NF₃, or nitrogen trifluoride has a mass of 71.0 grams, the mass of 1 N and 3 F added together. So this is how molar mass is defined. Let’s look at our examples. 1 mole of sulfur is 32.07 g, so the molar mass of sulfur is 32.07 g per mol. 1 mole of NF₃ is 71.0 g, so the molar mass of NF₃ is 71.0 g/mol. Now let’s take a look at how we can use molar mass to convert from mols to mass or mass to mols. For example: How many grams are in 6.58 mol silicon? We are given moles, and want to get to mass, so this is a mole to mass calculation. The connection of moles and mass will always be found on the periodic table. We find that silicon’s molar mass is 28.1 g per mole, rounded to the nearest tenth. In first determining this calculation you might intuitively just multiply the two together. This would in fact get you the right answer, But if we look at WHY that simple calculation gives the correct answer, then we will have the tools to solve more difficult problems. So let’s change the structure slightly to give us a better look at how the units are set up, which is the dimensional analysis approach. the molar mass is 28.1 g for every 1 mole of silicon. We can also arrange the first term as a fraction to more clearly see what will happen to the units. Comparing to the original calculation we see that with dimensional analysis we ensure the unit we start with is canceled, leaving us with the unit wanted. Mathematically we multiply what’s on top, divide by what’s on bottom. The answer tells us there are 185 g Si in 6.58 mol Si. Let’s use this set-up for the next problem, something slightly more complicated. How many mols are in 48.4 g of Copper I carbonate, Cu₂CO₃? We are given mass and want to get to mols, so this is a mass to mole conversion. We need the PT again as with any problem involving mass. Here we have two coppers, 1 carbon, and 3 oxygens, giving a molar mass of 187.2 g/mol, rounded to the nearest tenth. We’ll start with the amount given in the problem, multiply by a fraction with the given unit on bottom and the wanted unit on top. The molar mass gives the numerical relationship between mols and mass: There is one mole for every 187.2 grams. Since grams cancel and we are left with mols, which is the wanted unit, we know it is set up correctly. The fraction used to convert from one unit to another is called a conversion factor. The math is top divided by bottom, and in correct sig figs, there is 0.265 mol Cu₂CO₃ in 48.4 g of Cu₂CO₃. Looking at the two problems side by side, we can see that the conversion factors, in this case the molar masses, can be used in either configuration—mols on top or bottom, grams top or bottom, to suit the needs of the problem. Mols over grams coverts grams to moles; grams over mols converts mols to grams. The units can be reciprocals because their relationship is derived from a mathematical equality. This is dimensional analysis. If you need a more detailed review of dimensional analysis, click here. Let’s look at mole – particle conversions. Here we have mole to particle and particle to mole conversions. Instead of the mass – mole relationship given by molar mass, we will use the mole – particle relationship given by Avogadro’s number, N sub A. So one mol of sulfur is Avogadro’s number of sulfur atoms. And one mole of NF₃ is 6.02 x 10²³ molecules of NF₃. Let’s keep Avogadro’s number in a safe place for reference. Our next example is how many atoms are in 7.3 mols of Nickel? We begin with the amount given in the problem, multiply by a fraction, the conversion factor, with unit given on bottom, unit wanted on top. The numerical relationship between atoms and mols is given by Avogadro’s number. The mols cancel, we are left with the desired unit, atoms. Note that particles, such as atoms or molecules, can be represented with just the symbol rather than writing out the terms atoms or molecules. In this last of the four single step conversion problems, we are given particles, in this case molecules of carbon dioxide, and asked to find moles. As before we begin with the amount given in the problem, multiply by a conversion factor with unit given on bottom, unit wanted on top. Fill in the numerical relationship, 1 mol of CO₂ molecules for every 6.02 x 10²³ CO₂ molecules, molecules cancel and we are left with moles. Again, if we compare the two problems side by side, we see that moles or particles can be on top or bottom, depending on the needs of the calculation. This converts molecules to moles, this gets us moles to atoms. They can be reciprocals because they derive from mathematical equalities. So knowing the previous four conversions gives us access to converting to and from mass and particles. The connection between the two as we have seen is through moles. A mass to particle conversion is actually a mass to mole to particle conversion. And a particle to mass conversion is actually a particle to mole to mass conversion. Let’s look at two examples: How many molecules are in 25.3 grams of acetone? You may want to turn off the video here and work out the problem given what you learned in the first part of the video… So we are given mass and want to find an amount of particles. So this is a mass – particle conversion, which as we know requires a mass—mole—particle conversion. We start with what is given, multiply by a conversion factor with unit given, grams, on bottom and moles on top. We look at the periodic table to give us the mole—mass relationship. 3 carbons 6 H and one O gives a molar mass of 58.0 g per mole, rounded to the nearest tenth, which we use to fill in the conversion factor. The next step is converting to molecules. The first calculation gives us moles. So now moles go on bottom, and finally molecules on top. We know the relationship between moles and particles, and that goes in the equation. Moles cancel and we are left with molecules. Multiply everything on top, divide by everything on bottom. There are 3.46 x 10²³ molecules of acetone in 25.3 g of acetone. Last problem, what is the mass of 2.91 x 10²¹ atoms of zinc? Again, stop the video to see if you can work it out yourself… Here we are starting with particles and want to end up with mass. The conversions are from particle to mole to mass. Let’s turn that around to make it easier to follow. Write down what’s given, multiply by the mole particle relationship to get moles, multiply by the molar mass to get mass of zinc. We use zinc’s periodic table mass in the conversion factor. and the calculation gives 0.316 g Zn has 2.91 x 10²¹ atoms of zinc. These mole conversion problems follow a specific pattern that we can summarize on a map. Converting mass to moles we use the conversion factor of 1 mol over the molar mass. Particles to moles uses 1 mol over Avogadro’s number. Moles to particles uses Avogadro’s number divided by 1 mole. And mole to mass we uses molar mass per mole. If you’d like to apply mole conversions to stoichiometry, click here. And that is it for mole conversions! Seeya!