imagine trying to count every grain of sand on a beach or every star in the sky just as vast and incomprehensible is the world of chemistry where we think about tracking and measuring individual particles like atoms molecules and formula units this is where the concept of the mole comes in originating from the mind of Amadeo avagadro the mole provides us a way to bridge the gap between the microscopic world of particles and the macroscopic world of measurable quantities like a baker who counts their pastries in sets of dozens a chemist will count particles in sets of moles where one mole represents 6.02 * 10 23rd particles in this video we'll explore some of the many applications of the mole in chemistry from understanding empirical formulas to determining concentration and using mole ratios of gases recall that a molecular formula shows us the number of each type of atom present in a particular compound for example one glucose molecule contains six atoms of carbon 12 of hydrogen and six of oxygen prevalently bonded together as we noted earlier in this video it's virtually impossible to work with track or even measure individual atoms like carbon hydrogen and oxygen or even one molecule like glucose however ever we can work with and measure one mole of glucose which weighs a more practical 18018 G one mole of glucose molecules would contain 6 moles of carbon atoms 12 moles of hydrogen atoms and 6 moles of oxygen atoms although the molecular formula gives us a complete accounting of the numbers of each type of atom present in the compound it could be useful to understand the concept of empirical formulas the empirical formula represents the simplest ratio of atoms of each element present in the compound in the case of glucose the 6 to 12 to 6 ratio simplifies to 1 to 2: 1 empirical formulas provide fundamental information about the composition of chemical compounds and play a key role in various aspects of chemistry from stochiometry and reaction analysis to substance identification and qualitative analysis before diving into empirical formula calculation let's quickly take a look at the concept of percent composition using glucose our table outlines how we can calculate percent composition for a compound we already know we take the moles of atoms of each element times their mass and divide that by the total mass of the compound to find the mass percent composition of that element in the compound we can total up our values to check our work the total masses of the atoms of each element add to the mass of glucose and the % composition of each element adds to 100% in the lab we can use the mole concept along with percent composition to determine an empirical formula of an unknown compound here's a sample problem a compound contains 43.7% of phosphorus and 56.3% of oxygen by mass what is the empirical formula of this compound remember the goal is to find the lowest whole number mole ratio of oxygen and phosphorus in the sample to do this we'll follow three steps first we'll assume that we have a 100 G sample of the compound since more ratios within a compound are independent of how much of the compound we're actually using we could start with any mass of sample starting with 100 G is mostly out of convenience since our percentages would directly translate into Mass we would have 43.7 G of phosphorus and 56.3 G of oxygen in our 100 G sample from here here we'll determine the number of moles of each element present we'll take the mass of each element in the sample and divide it by the respective M Mass which we find using the periodic table finally we'll take these M values and find the lowest whole number ratio this last step can feel the most complicated a simple trick is to divide each value by the lowest moles you have available in this case that would be the 1.41 moles of phosphorus doing this gives us a 1 to 2.5 Mo ratio between phosphorus and oxygen we're not quite done yet since this is still not a whole number ratio mathematically the closest whole number to 2.5 is 5 which we get by multiplying the value by two this means we'll multiply both mole values by two giving a 2:5 mole ratio between phosphorus and oxygen and our empirical formula of p25 we can go one step further from the empirical formula if the M mass of a compound is known for example a compound containing carbon and hydrogen is determined to have the empirical formula ch2 if the mass of the compound is known to be 98.21 G per mole what is the molecular formula of the compound to relate an empirical formula back to a molecular formula we'll first need to determine the mass of the empirical formula We'll add together the M masses of each atom of each element based on the values found on the periodic table then we'll compare the empirical Mass to the molecular mass if we divide the molecular mass by the empirical Mass we see that it is 7times the size this means the mole ratio and therefore each subscript in the molecular formula must be seven times that of the empirical formula this gives a molecular formula of C7 h14 as you've seen now and will continue to see throughout your study of chemistry mole calculations are important in fact one can say that the mole is the currency of chemistry when we think about the forms in which substances are found in a normal chemical reaction the states typically include pure solids pure liquids aquous Solutions and gases you'll need to know how to derive the number of moles of a substance in any of these states this mole conversion diagram addresses the most basic mole calculations which are used when a substance like an element or compound is present in the solid state let's address how to calculate moles if a substance is found as a pure liquid or as an aquous solution liquids are typically measured by volume therefore the mass of a pure liquid can be determined from its density once the mass of a liquid is known it's easy to determine the moles present which we can see by adding this to our diagram density will act as our bridge between volume and mass of a pure substance to find mass we'll take the density time volume and to find volume we'll divide a substance's mass by its density for example if a 5.50 CM Cub sample of methanol is used in an experiment and the density of methanol is found to be 07866 G per cm cubed at 25° how many moles of methanol were used since we're converting from the volume of a pure liquid into moles we first convert from volume into Mass using density then into moles using the substance's mass with the density equation we can plug in our known density and volume to find the mass of methanol used in our sample thear mass of methanol is 32.4 G per mole meaning that 1 mole of methanol weighs 32.0 G we'll divide our substance's mass by its mass to find the moles of methanol in the sample this works because methanol is a pure liquid which allows us to use density to relate volume and mass for an aquous solution we'll need to take a different approach recall the basics of solutions which are composed of a solute dissolved in a solvent which is generally water thus the term aquous in an aquous solution of say sodium chloride we have a substance that is composed mostly of water the solvent with sodium and chloride ions dispersed uniformly throughout as the because an aqu solution of sodium chloride is mostly water the determination of the number of moles of sodium chloride present requires a method different from what we previously used we must incorporate the concept of solution concentration into our calculation solution concentration or marity is defined as the moles of solute per volume of solution mathematically we represent this as a lowercase Cal n / V that is moles ided by volume of solution in decimet cubed marity will therefore have the units of moles per decim cubed from this equation we can see that in order to determine the number of moles present in an aquous solution we must rearrange the equation to make moles equal to marity time volume We'll add this to our mole diagram as well marity access the bridge between the volume of solution and moles to convert from solution volume into moles we would multiply volume by concentration in the other direction converting from moles into solution volume we would divide moles by concentration remember using a diagram like this can help you build mental maps that guide you through unit conversions if you haven't done so already I would highly suggest adding something like this to your notes in an easily accessible area let's take a look at a problem that involves an aquous solution in which it's necessary to determine the number of moles and explain why we can't use density in that calculation a 350 CM Cub sample of glucose solution has a concentration of 0.0750 moles per decim cubed what is the mass of glucose in this sample since we're converting from volume into Mass we'll need to convert from volume of solution into moles first using marity and then from moles into Mass using molar mass this can sometimes be a place of high confusion many students at this point may try and solve this problem using density so why can't we use density here if we calculate mass using density we need to identify what we're actually taking the mass of say we know the density of the glucose solution and after multiplying it by the solution's volume we find that it has a mass of 3535 G that seems like a lot and it is because that's the mass of the entire solution meaning it's the mass of mostly water finding this would be like placing the solution on a scale after adjusting for the weight of the beaker this is perfectly fine if we want to know the mass of a pure liquid but if we're trying to find the mass of just the glucose dissolved in the solution we'll have to reference the concentration of the glucose in the solution remember that we use moles as a way to count molecules and bulk having concentration measured as marity gives us a ratio of CED molecules to volume of solution if we plug in our numbers making sure to convert volume into decimet cubed we can use this ratio to find the total moles of glucose in the solution sample we've used concentration to separate information about the glucose from the solution we'll then compare the moles of glucose to its mol mass that is the mass of one mole of glucose and determine its total mass in solution our answer here is much smaller than before which should make sense Solutions are most L water so the amount dissolved will usually be much smaller than the total mass the final of the most common States in which matter will appear in a chemical equation is the gas State because particles of a gas are widespread separated from one another and constantly moving the pressure temperature and volume of a sample of gas all contribute to the determination of the number of moles present in the sample this concept is covered in Greater detail with the idog gas law and idog gas law calculations but if temperature and pressure are held constant throughout a chemical reaction the volume in which the gases are contained will be directly proportional to the number of moles of gas present this is stated by avagadro's law which states that equal volumes of gases under identical conditions will have the same number of particles this was discussed in a previous video along with the definition of the mole in line with this observation we can use the mo relationship found in the balanced chemical equation to predict the moles or even volume of a gas produced or consumed within a reaction we can do this so long as we know at least one of the gas's molar or volume measurements beforehand and that the temperature and pressure remain constant let's quickly take a look at this using the reaction equation below for the ha process if 5.5 moles of hydrogen are reacted with excess nitrogen how many moles of ammonia NH3 will be produced if the temperature and pressure are held constant mol amounts remain proportional as temperature and pressure do not change meaning we can use our 3:2 mole ratio between hydrogen and ammonia as our conversion factor this ratio corresponds with how these compounds will be consumed and produced and tells us that 3.7 mol of ammonia will be made when 5.5 mol of hydrogen are allowed to react in excess nitrogen similarly we can do this with volumes of gases if 10 decim cubed of nitrogen reacts with hydrogen what volume of hydrogen gas is needed if the temperature and pressure remain constant instead of using moles we can represent the reaction ratios within the equation using volume this allows us to solve the problem just as before where we can use our reaction ratio between nitrogen and hydrogen to find the volume of gas needed we see that for every 1 decim cubed of nitrogen that's reacted we need 3 decim cubed of hydrogen so 10 DM Cub of nitrogen gas will require 30 DM cubed of hydrogen gas in summary the mole is used as a way to count large quantities of particles and as such there are many practical applications for the mole in chemistry for example we can break down and identify molecular and empirical formulas using mole ratios within compounds we could also find the mass concentration and volume of a pure substance and of a solution using either density or marity and and we can even use avagadro's law to find quantities and volumes of gases using ratios within chemical equations each of these represents a practical lab skill for IB chemistry students making it essential to not only understand the concept of the mole but also its many applications