[Music] good day everyone welcome to our first lesson today we will be discussing about composition stoichiometry let's start with the definition of the term stoichiometry so stoichiometry was coined from the greek word stoichion meaning elements and metron which means to measure and then they put the two terms together and then they came up with stoichiometry as a brief definition stoichiometry defines the mass to mole relationships in a chemical reaction leading to a quantitative analysis of any chemical reaction in other words stoichiometry deals with the conversion from one quantity to the other in a chemical reaction here are the list of the related concepts that we will be discussing in our study of stoichiometry atomic mass mole avogadro's number chemical formula percent compositions and chemical equations so this topics should just be a review for you so let's refresh your memory let's first start with atomic mass so atomic masses represent the average of the atomic masses of all known naturally occurring isotopes of an element so basically the atomic mass that you are reading from a periodic table that is the average atomic mass of all of the known naturally occurring isotopes of the elements so for example in hydrogen there are three known isotopes so they are the same element but they are just varying in their number of neutrons so we have the proton that's h1 we have duterium h2 and we have tritium h3 the listed atomic mass of hydrogen in the periodic table is the average of the atomic masses of all of these three naturally occurring isotopes for atomic mass we are using the unified atomic mass units signified by the letter u or daltons but you are more familiar with the units of grams per mole in the atomic mass so we are going to use grams per mole as well technically atomic mass is defined as 1 12 the mass of a single atom of carbon 12. i have previously mentioned that the atomic mass listed in the periodic table of elements is the average of the atomic masses of the naturally occurring isotopes of that element that's what they call relative abundance the abundance of the naturally occurring isotopes of a given element are the basis for the average atomic mass and it's very easy to compute it's as simple as taking the weighted average of the atomic masses of the individual isotopes so let's try it with this example chlorine has two naturally occurring isotopes that's chlorine 35 with atomic mass of 34.9689 eu and chlorine 37 with the atomic mass of 36.9659 u with relative abundance of 75.76 percent and 24.25 respectively what is the average atomic mass of chlorine let's try to solve this example so i have listed here our given so again for chlorine we have the two naturally occurring isotopes chlorine 35 and chlorine 37 with their atomic masses listed there as well as their percent abundance so how do we compute the average atomic mass of chlorine that is simply done by multiplying the percent abundance to the corresponding atomic mass of the isotope so we write it as 75.76 so we are simply converting it to a decimal notation multiplied by 34.9689 atomic mass units this this is for chlorine 35. plus 0.2424 or 24.24 percent multiplied by the atomic mass of chlorine 37 36.9659 atomic mass units and you simply enter that in your calculator then we will have the average atomic mass of chlorine so that is point seven five seven six times thirty four point nine eight nine six nine plus 0.2424 times 36.9659 that is 35.45 atomic mass units that is the average molecular weight of chlorine so if you take a look at a periodic table this is the atomic mass for chlorine that you will see because that's already the average atomic mass for all of the naturally occurring isotopes of urine okay you can do this with all of the other elements let's proceed we have here another problem silicon has three naturally occurring isotopes with atomic masses and relative abundance listed on the table below what is the average atomic mass of silicon now that you know that you will just be getting the weighted average of the atomic masses we can even do it without writing the formula you simply multiply the percent abundance to the atomic mass and then you simply add all of those products you can try that with your calculator next now let's define the mole because the concept of the mole along with its conversion with mass is very important in chemistry a unit of measurement defined as an amount of substance with the same number of particles as there are in 0.012 kilograms of carbon 12. that is a very technical definition of the mole and we will circle back to the concept of the mole later let's first define avogadro's number so avogadro's number is a physical constant with a value of 6.022 times 10 raised to 23. now what is the significance of this avogadro's number this is tied to the concept of the mole because one mole of something means that you have avogadro's number amounts of particles so let's say for example if you have one mole of eggs that means that you have six point zero two two times ten raised to twenty three pieces of eggs and that's a very very big number that is six zero two two followed by twenty zeros okay we use the concept of avogadro's number in conjunction with the definition of the mole to define how many atoms of a substance are there in a given mole of the substance so another example here if you have one mole of sodium that means that you have 6.022 times 10 raised to 23 atoms of sodium so do not forget about this conversion the conversion from mole to the number of particles so when we say particles here we can be referring to atoms we can also be referring to molecules and we can also be referring to other types of particles okay next let's define molecular weight so molecular weight is a very important conversion factor in chemistry because molecular weight gives you the avenue in converting from mass to moles or vice versa the definition of the molecular weight is the amount of mass of a substance divided by its number of moles okay and this is where the unit grams per mole is derived based from the definition of molecular weight the mass translates to grams and then the mole translates to mole as the units okay the unit dalton is equivalent to the unit grams per mole so now that we have defined mole avogadro's number and molecular weights we can now define the relationships between mass mole and number of atoms we will now be drafting a chart that relates mass to mole to number of atoms so that you will not forget we have three quantities that is mass mole and the number of atoms if we want to convert from mass to mole or mole to mass we are going to use the molecular weight as our conversion factor the molecular weight can be obtained from the periodic table of the elements when we want to convert from moles to number of atoms or vice versa we are going to use avogadro's number 6.0 times 10 raised to 23. now there is no direct conversion from mass to number of atoms if you want to convert from mass to number of atoms what we need to do is to first convert mass to mole and then convert mole to number of atoms so it's a two-step process okay so do not forget this chart let's try some examples okay first one how much does 3.5 moles of aluminum weigh so let's try to solve that you are given 3.5 moles of aluminum and we are asked how much does it weigh so that implies that we want to convert the moles of aluminum to grams of aluminum and how do we do that if you go back to our chart earlier if you want to convert from moles to mass we must use the molecular weight of the substance and our substance in this case is aluminum so let's look for the molecular weight of aluminum i have here a periodic table of the elements and you see here aluminum has a molecular weight of 26.9 grams per mole so let's write that the molecular weight of aluminum is 26.982 grams per mole now this molecular weight is an important conversion factor for us to convert from moles to mass let's use the unit factor analysis method for our conversion we start with the given that's 3.50 moles of aluminum and since your unit here is moles that means in our next factor moles should be in the denominator and that makes grams our unit in the numerator so that means that we will multiply the molecular weight to the number of moles to obtain the mass we get 3.5 times 26.982 94.437 grams of aluminum now the given is three significant figures so we must only retain three significant figures in the answer that's 94.4 grams aluminum so that is how we use the molecular weight in converting from mass to mole okay next okay next problem how many moles are there in 200 grams of osmium so the given is 200 grams osmium we want to know the number of moles so let's let's try to solve this we are converting from grams to mole so we still need the molecular weight as our conversion factor so let's look for the molecular weight of osmium so from from our periodic table osmium is 190.23 grams per mole 190 points 23 grams per mole and then we use the same unit factor analysis method we start with the given 200 grams of osmium and this time grams is our unit so we need to have grams in the denominator of the factor for it to cancel so grams below moles at the top that means that we are dividing by the molecular weight of osmium 190.23 grams is to one mole of osmium so grams of osmium cancels we are left with the unit moles of osmium we can now compute for the number of moles of osmium that is 200 divided by 190.23 and assuming that the two zeros in 200 are significant we retain three significant figures that is 1.05 moles of osmium next problem how many atoms are there in one gram of carbon so based on this given we are now converting from mass to number of atoms if you consult our chart earlier there is no direct conversion from mass to number of atoms so we need to convert first from mass to mole and then from mole to number of atoms so let's write the given we have one gram of carbon and we want to know how many atoms are there in this sample okay so our plan is again from mass we first convert this to mole using the molecular weight because the molecular weight is the conversion factor from mass to mole and then from mole to the number of atoms we use avogadro's number as our conversion factor okay we start with a given one gram of carbon first unit refers to the molecular weights and if we go back to our periodic table the molecular weight of carbon is 12.011 grams per mole and since grams is our unit for the given that means that grams should be in the denominator and mole is in the numerator so we are dividing with the molecular weight 12.011 grams of carbon is to one mole gram of carbon cancels next factor we want to convert from mole to the number of atoms so we are using the avogadro's number as our conversion factor avogadro's number states that there are 6.022 times 10 raised to 23 atoms for every mole of the substance so we can use this relationship in our last factor so in our previous factor mole is in the numerator that means in the next factor it should be in the denominator moles carbon and our numerator is atoms of carbon and using the value of avogadro's number we write 6.022 times 10 raised to 23 atoms of carbon per 1 mole of carbon the number of moles of carbon cancels and we can now compute for the number of atoms we compute as 1 divided by 12.011 times 6.022 times 10 raised to 23. answer in three significant figures 5.01 times 10 raised to 22 atoms of carbon okay so this is how we convert from mass to number of atoms if you're given number of atoms and you want to convert it to mass you simply follow the same path but in reverse okay last problem how much does a single oxygen atom weigh so i will be leaving this problem to you as an exercise so that you can practice in this one we are converting from atoms so you have one atom you want to convert it to mass okay so it's just the reverse of our previous example let's proceed next concept that we will be discussing is chemical formula chemical formula are very important in the study of chemistry because these are used to represent elements and compounds in an equation and chemistry is all about chemical reactions so you need to learn about chemical formulas chemical formula are governed by the law of definite proportions just to refresh you know the law of definite proportion states that for every compound there exists a fixed ratio between its constituent atoms and if you were to change that ratio you will end up with a different compound okay so that's the law of definite proportions the molecular weight of a compound is simply the sum of the atomic masses of each atom in a compound so you will just be adding atomic masses for example determine the molecular weight of water water has a chemical formula of h2o that means that it has two atoms of hydrogen and one atom of oxygen so if you wanted to so if we want to determine the molecular weight of water we need to know the atomic masses of hydrogen and oxygen so let's try to solve that you're given is h2o again you have two atoms of hydrogen and one atom of oxygen let's look at the periodic table for the molecular weights for oxygen the molecular weight is 15.999 grams per mole and for hydrogen the molecular weight is here 1.008 grams per mole okay so there's our given now we need to multiply that molecular weight with the number of atoms present in the compound so for oxygen there's only one so we need to multiply 15.999 by one and for hydrogen there's two so we need to multiply 1.008 by 2. so we have 15.999 for oxygen and for hydrogen we have 2.016 next it's just a matter of adding these two to get the molecular weight of water the molecular weight of water is simply 15.999 plus 2.016 that is 18.015 grams per mole so we have determined the molecular weight of water okay so that is how you compute for the molecular weight of a compound you must know its chemical formula because without the chemical formula we cannot know the number of atoms in the compound okay let's try the next one determine the molecular weight of sucrose so sucrose is table sugar this is the sugar that you add to your coffee and to your other drinks its chemical formula is c12h22o11 let's determine its molecular formula we will just be repeating our process from earlier so we list the atomic masses of the individual elements we multiply them by the number of atoms in the compound and then add so let's start with oxygen that is 15.999 multiplied by eleven hydrogen that is 1.008 multiplied by 22 and for carbon that is 12.011 multiplied by 12 and we simply multiply first for oxygen 15.999 times 11 175.989 next for hydrogen that is 22 22.176 and lastly for carbon 144.132 and then get the sum we get the molecular weight of sucrose that's 144.132 plus 22.176 plus 175.989 that's 342.297 grams per mole so that is how we compute for molecular weights of compounds okay so if you are if you are performing stoichiometry using compounds this is how you determine its molecular weight let's proceed next is percent composition this tells us the elemental composition of a compound in terms of percentage by weight in order for you to determine the percent composition you must first determine its molecular weight because the molecular weight factors in the calculation of the percent composition let's try this example determine the percent composition of the elements in aluminum sulfates okay the first question is what is the chemical formula of aluminum sulfate you notice that aluminum sulfate there was given as a proper name and it was not given as a formula so it's up to us to determine the molecular formula of aluminum sulfate based from your previous learnings aluminum as a cat ion has a charge of positive three and sulfate is an an ion with a formula of so4 with a charge of two minus so if we're going to apply the crisscross method we drop the signs and then the three goes for the sulfate and then the two goes to the aluminum our formula becomes al2 and then so4 3 this is the chemical formula of aluminum sulfate now we can determine the percent composition of the elements we have three elements here aluminum sulfur and oxygen the process of obtaining the percent composition starts with determining the molecular weight of the compound so let's first determine its molecular weights for oxygen that's 15.999 multiplied by 12 so this is 4 times 3 okay remember that the subscript 3 applies to the entire sulfate and ion you will need to multiply 3 to both the sulfur and the oxygen next is sulfur let's look for the molecular weight of sulfur so sulfur is 32.06 multiply that with three and then lastly we have aluminum aluminum is 26.982 times 2 and then we multiply so that is 15.999 times 12 that's 191.988 next first of all for 32.06 times three 96.18 next for aluminum 26.982 multiplied by two that's 53.964 and then we add to determine the molecular weight of aluminum sulfate so that's 53.964 plus 96.18 plus 191.988 so the molecular weight is 342.132 grams per mole now we will be solving for the percent composition for the percent composition we are just going to divide the molecular weight per element to the molecular weight of the entire compound so let's say for example for oxygen we just need to compute the percent composition as 191.988 this is the total atomic mass of oxygen in aluminum sulfate divided by the total molecular weight of aluminum sulfate 342.132 and that is times 100 to make it percent so our answer is 56.12 percent okay we do the same for sulfur so the total atomic mass of sulfur is 96.18 you divide that with the molecular weight 342.132 times 100 to make it percent that's 28.11 and then lastly for aluminum the total molecular rate is 53.96 divide that 53.964 divide that with 342.132 times 100 we get 15.77 percent and if you did everything correctly if you total the percent composition of the individual elements you should arrive with 100 percent to know that you are correct so let's check we have 56.12 plus 28.11 plus 15.77 the total is 100 percent we know that we performed the correct computations now if we obtain 99.99 as the total or 100.01 that's okay because that might be due to rounding off error okay so that is how we compute for the percent composition of the compound we have here a do it yourself you simply follow the same solution but this time you are to determine the percent composition of the elements in calcium phosphate you also have three elements calcium phosphorus and oxygen you follow our example earlier and you can find the percent composition of calcium phosphate okay do this as your practice next next we have empirical and molecular formula so these are two related topics these are two related topics that are connected with the percent composition since the elements in a compound exist at a fixed proportion which is dictated by the law of definite proportions you can use percentage composition data to determine the empirical formula in our previous slide you are given the chemical formula and you are to solve for the percent composition this time around you will be given the percent composition and we will manipulate those numbers for us to determine the chemical formula and chemical formulas come in two types the empirical and the molecular so what's the difference the empirical formula is the simplest positive integer ratio of atoms present in a compound and it's not the same as a molecular formula so the empirical formula is a simpler version of the molecular formula the molecular formula lists the complete atomic composition of a compound for you to be able to differentiate between empirical and molecular formula here are some examples and by the way please do take note that empirical and molecular formulas can sometimes be the same not all of the time okay first example you can have an empirical formula of no2 but also have a molecular formula of n2o4 if you notice we simply multiplied no2 by 2 so that became n2 and then o became 4. another example you can have an empirical formula of co2 and a molecular formula of c2o4 again we multiplied this by two and then last example you can have an empirical formula of h2o and also a molecular formula of h2o so this is a case where in the empirical and molecular formula are the same let's try to solve some problems regarding empirical and molecular formula a compound contains 21.6 percent sodium 33.3 percent chlorine and 45.1 percent oxygen determine the empirical formula of the compound so notice that you're given here is the percentage composition which we have determined earlier but this time we don't know the chemical formula of the compound we need to determine the empirical formula let's solve for that our given is we have 21.6 percent sodium 33.3 percent chlorine and 45.1 percent oxygen okay now for this types of problems remember that this composition is in terms of mass or this is percent by mass for our first step in the solution we are going to assume that we have 100 grams of the compound so you might be asking why did i choose to assume that i have 100 grams of the compound this is for simplicity reason if you multiply 100 by 21.6 the answer is 21.6 grams so it's not complicated to determine the mass of the elements if you assume that you have 100 grams of the compound so in theory you can assume any type of number in your initial assumptions let's say for example you want to assume that there's 42.67 grams of your compound that's okay but you would be adding another layer of calculations because you need to multiply 42.67 with 21.6 percent in easy terms by assuming 100 grams of the compound that means that we have 21.6 grams of sodium we have 33.3 grams of chlorine and we also have 45.1 grams of oxygen okay that's the beauty of assuming 100 grams now we have to convert this to moles we have three atoms so we need three molecular weights for sodium for chlorine and for oxygen let's convert first for sodium let's take a look at the molecular rate of sodium that is 22.99 so we divide by the molecular weight to convert it to mole the molecular weight of chlorine is 35.45 grams is to 1 mol and the molecular weight of oxygen is 15.999 grams per mole let's compute for these numbers for sodium 21.6 divided by 22.99 you can use four decimal places so this is 0.9395 for chlorine that's 33.3 divided by 35.45 we get 0.9394 and for oxygen 45.1 divided by 15.999 we get 2.8189 so these are all in moles moles sodium moles chlorine and moles oxygen now what's the next step we now have the number of moles of each of your elements we need to divide them by the lowest value that we obtained out of the three moles that we obtained the lowest is 0.9394 you divide everything by 0.9394 now we can get the number of atoms in the compound so for the first one 0.9395 over 0.9394 is very very close to 1. so we will label that as 1. for chlorine it's also 1. now for oxygen that's 0.2189 divided by point nine three nine four and we get a value of three so for this types of uh computation it's okay to round it up to the nearest whole number okay so this numbers that we have obtained this are already the counts of the atoms in the compound so that means that the empirical formula of the compound is now n a 1 c l one o three this is our empirical formula okay the determination of empirical formula for any compound follows the same procedure as long as you are given the percentage composition so again from the percentage composition you assume that you have 100 grams of the compound so that you can get the masses you convert the masses to moles and then you divide all of the moles with the lowest value that you obtain and that gives you the count of the atoms for the empirical formula let's try the next example vitamin c or ascorbic acid was found to contain 40.91 carbon 4.59 hydrogen and 54.50 oxygen determine its empirical formula so we are going to follow the same procedure okay the given 40.91 percent oxygen 4.59 percent hydrogen and 54.5 percent oxygen okay so next we assume that you have 100 grams of the compound and doing so we obtain 40.91 grams of carbon 4.59 grams of hydrogen and 54.5 grams 54.5 grams of oxygen now we can convert this to moles so the molecular weight of carbon is 12.011 for hydrogen that is 1.008 and for oxygen that is 15.999 let's compute 40.91 divided by 12.011 we get 3.406 moles of carbon next for hydrogen 4.59 divided by 1.008 we get 4.553 moles of hydrogen and lastly for oxygen 54.5 divided by 15.999 we get 3.406 moles of oxygen okay after we have obtained the number of moles of the elements we divide them by the lowest value that we obtained and that is 3.406 okay so for carbon that is one for oxygen that is also one for hydrogen that's four point five five three divided by three point four zero six that's one point three three seven in this case 1.337 cannot be rounded down to 1 and it cannot be rounded up to 2. what we're going to do is we need to multiply all of those obtained numbers with an integer such that the result would also be an integer or a whole number okay what can you multiply to 1.337 to make it a whole number or to make it very very close to a whole number let's try two so if you multiply this by 2 you get 2.67 so that's not a whole number so let's go back if we multiply this by 3 we get 4.01 that is close enough to a whole number which is 4. now we know that we need to multiply three you multiply three to everything so that is one times three and one point three three seven times three and one times three so our results would be this is three this is four and this is three thus our empirical formula for ascorbic acid is now c 3 h 4 0 3 ok so let's proceed with the next item from other experiments it was shown that the molecular weight of ascorbic acid is 176.14 grams per mole determine its molecular formula so this is a continuation of the previous item now that we have the empirical formula of of ascorbic acid the question is how do we determine its molecular formula i have listed the empirical formula of ascorbic acid from our previous calculations and it was stated in the problem that the molecular weight of ascorbic acid is 176.14 grams per mole okay for the determination of the molecular formula we need to know two values first is the molecular weight of the compound and second is the empirical weight of the compound the empirical weight is simply the molecular weight of the empirical formula of the compound so the empirical formula c3h4o3 we determine the empirical weight as oxygen 15.999 times 3 plus hydrogen 1.008 times 4 plus carbon 12.011 times 3. and the empirical formula is 88.062 grams per mole now we are going to divide the molecular weight with the empirical weight in order for us to obtain an integer so that is 176.14 this is the molecular weight divide that by the empirical weight 88.062 what do we have 176.14 divided by the empirical weight our integer is 2. what do we do with this integer you multiply this integer with the empirical formula to get the molecular formula so therefore the molecular formula of ascorbic acid is equal to two times the empirical formula of c3h4o3 we get a molecular formula of c6h8o6 okay so that is how we determine the molecular formula of a compound that's the end of this lesson for our next lesson we will be dealing about reaction stoichiometry or performing stoichiometric calculations for chemical reactions and that will be for next week so if you have any questions with this lecture you can always message me in course messages in blackboard okay that's it for this lesson i hope you've learned something and as always keep safe [Music] you