summarize this transcript? these are the answers to the ap chemistry packet that covers topics 1.1 to 1.3 so here you're looking at unit 1 at a glance again unit 1 is entitled atomic structure and properties and in this particular video we'll take a look at topic 1.1 moles and molar mass topic 1.2 mass spectroscopy also known as spectrometry of elements and topic 1.3 elemental composition of pure substances in the video description area there should be a link to the ap chemistry course and exam description and there also should be a link to this particular packet that goes along with the video all right topic 1.1 moles and molar mass so here are some essential knowledge statements that come from the ap chemistry course and exam description one cannot count particles directly while performing laboratory work thus there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes avogadro's number 6.02 times 10 to the 23rd particles per mole provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles or formula units of that substance expressing the mass of an individual atom or molecule in atomic mass units amu is useful because the average mass in amu of one particle atom or molecule or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams thus there is a quantitative connection between the mass of a substance and the number of particles that the substance contains so here are three substances magnesium mg carbon dioxide co2 and sodium chloride nacl the particles of a substance can be described as atoms molecules or formula units as shown in the following examples the molar mass of a substance can be determined or calculated from the atomic mass values on the periodic table so if you look on your periodic table and you find the elemental symbol for magnesium mg you could say that a single atom of magnesium would weigh 24.30 amu or atomic mass units but once i change that to mole a mole of magnesium weighs 24.30 grams and it contains avogadro's number of atoms a mole of carbon dioxide if you add up the atomic mass of one carbon and two oxygens the molar mass of co2 is 44.01 grams and that would contain avogadro's number of molecules again referring to co2 as molecules that is a covalent substance whereas since sodium chloride is not covalent but rather ionic its fundamental building blocks or particles is not considered a molecule but rather a formula unit so one mole of sodium chloride would have a molar mass from the periodic table when you add up the atomic masses of sodium and chlorine 58.44 grams per mole and again would contain avogadro's number of formula units okay let's take a look at some practice problems so here we have two problems one of them says to calculate the mass in grams of 0.08 moles of something called barium hydroxide ba parentheses oh2 and then the other problem says to calculate the number of moles of c4h10 that are present in 2.00 grams of c4h10 so we are going to need to use the periodic table to do these calculations so starting with number one i have to go find barium oxygen and hydrogen on the periodic table so i can calculate the molar mass so there are those three particular elements and again let's go ahead and do that math to calculate the molar mass so for barium hydroxide ba parentheses oh2 the total of the barium and the two oxygens and the two hydrogens added together is 171.346 grams per mole so we'll start with the number of moles and then we'll set up a conversion factor that will put mole on the bottom since that's the unit we're trying to cancel out and grams on the top so that would be like this 171.346 grams of barium hydroxide for every one mole if we do that math on our calculator we get 14 point again you can see several significant digits there five six four four one so how do we round off our final answer in this case we go with the measurement that we were given so 0.0850 has three significant figures so we're going to round off our final answer to three significant figures and that would look like 14.6 grams in our next problem we're essentially doing the reverse of what we just did in number one so instead of going from moles to grams we're going to go from grams to moles so we start off with calculating the molar mass of c4h10 again this comes from the periodic table so carbon has an atomic mass of 12.01 and there are four carbons in the formula hydrogen is 1.008 and there are 10 of those so 58.12 grams per mole that's my molar mass of c4h10 i write down the number i was given so 2.00 grams of c4h10 and now i want to put grams on the bottom and moles on top so i use my molar mass as a conversion factor to go from grams to moles i do that math on my calculator again i get a long number and i'm not sure let's say how to round it off until i look at the measurement that i was given so 2.00 grams has three significant figures so we're going to round off our final answer to three significant figures which would be written as 0.0344 moles of c4h10 okay let's take a look at our next two problems number three calculate the number of atoms of silicon si that are present in 35.0 moles of silicon and then number four calculate the number of moles of o3 which is ozone that are present in 4.3 times 10 to the 24 molecules of ozone so number three we are going from atoms to moles and in number four we're going from moles to molecules so because we're talking about a conversion factor that goes between number of moles and number of particles we need avogadro's number but we in fact do not need to use the periodic table because we're not converting from grams to moles like we were in the previous or moles of grams okay so in this particular set of problems we're just going to use avogadro's number so number three we write down 35.0 moles of silicon put moles on the bottom and in this case atoms are the particles so atoms on top so there's avogadro's number the number of atoms or particles in one mole we do that math and again rounding this particular number off to the number of sig figs that we need in this case 35.0 has three significant figures so that number will be expressed with three significant figures as 2.11 times 10 to the 25th power and then number four we start with our number of particles in this case molecules of ozone o3 we want to put particles on the bottom and moles on top so avogadro's number of molecules of ozone on the bottom and then one mole of ozone on the top we do this math by again doing division like in the last problem we were multiplying the avogadro's number now we're dividing by avogadro's number and again on our calculator we get this long number let's round it off based on the number we're given in this case 4.3 has only two significant figures so our final answer is written as 7.1 moles of ozone o3 all right let's take a look at the next two questions number five calculate the mass in grams of 8.2 times 10 to the 22nd molecules of chcl3 and the number six calculate the number of formula units of na2so4 present in 0.248 grams of that substance so number five we know that when we use a conversion involving avogadro's number that's going to be between moles and particles or vice versa so we're definitely going to need avogadro's number but since the final answer is in grams we also need the periodic table and the calculated value for the molar mass to go between moles and grams and so i'm going to write the calculation for the molar mass of chcl so the atomic mass of carbon hydrogen and chlorine are what you see here so 12.01 1.008 and 35.45 now there are three chlorines in the formula that's why that's times three so 119.368 grams per mole that's the molar mass of chcl3 we're also going to need to use avogadro's number so here we have the given information 8.2 times 10 to the 22nd molecules of this substance chcl3 let's put molecules on the bottom and moles on top again using avogadro's number so avogadro's number on the bottom number of molecules in one mole and then let's use the molar mass that's going to be grams on top and then one mole on the bottom so here we pick up our calculator and we divide by avogadro's number we multiply by 119.368 and then we get this number on our calculator which if we round this off to two significant figures based on the 8.2 having two significant figures we're going to write this as not 16.2 or 16.3 but rather just 16 again two significant figures to match the value that we're given so i'm not going to be limited by avogadro's number or the molar mass from the periodic table those are not going to limit my significant figures in my answer but the number i was given in the problem that number has a certain number of significant figures that limits my answer all right on to number six molar mass for na2so4 again based on the atomic masses of sodium sulfur and oxygen so if i do that math 22.99 that's the atomic mass of sodium 32.06 the atomic mass of sulfur and then 16 the atomic mass of oxygen times 4. so we do all of that and we get 142.04 grams per mole that's the molar mass of na2so4 we'll start with the number we're given so 0.248 we're going to go from grams to moles so grams on the bottom one mole on top so there's the molar mass using that as a conversion factor and then now it's time to incorporate avogadro's number in our calculation so that would be one mole on the bottom and avogadro's number of particles in this case formula units on the top so if we divide by the molar mass multiplied by avogadro's number we get a number that needs to be rounded in this case 0.248 was the number we're given that's the measurement that has three significant figures so 0.248 three significant figures let's round off our final answer to 1.05 times 10 to the 21st formula units of na2so4 all right now we are going to move on to topic 1.2 mass spectroscopy of elements again also known as mass spectrometry and here are two essential knowledge statements from the ap chemistry course and exam description the mass spectrum of a sample containing a single element can be used to determine the identity of the isotopes of that element and the relative abundance of each isotope in nature the average atomic mass of an element can be estimated from the weighted average of the isotopic masses using the mass of each isotope and its relative abundance so in this first example for number seven from your packet we're looking at chlorine and on the left we see that there are two isotopes of chlorine so chlorine with a mass number of 35 and chlorine with a mass number of 37. the abundance that's like the percentage of that particular element as it would appear in a sample naturally 75.8 percent chlorine 35 24.2 chlorine 37. so on the right we're seeing a mass spectrum and again this particular diagram shows two bars so one on the x-axis at atomic mass or mass number of 35 for the chlorine-35 isotope and then a smaller bar on the x-axis you can see that's 37 so again mass number 37 on the y-axis we see percent abundance so number seven part a calculate the average atomic mass of cl so based on this information that would be to calculate the weighted average so the weighted average means we're going to multiply each mass number by its percent abundance that is written as a decimal so 35 times 0.758 plus 37 times 0.242 so that's the weighted average that we get from this data and our calculator would give us an answer of approximately 35.5 in part b of this question we're given a table with chlorine 35 and chlorine 37 and we're asked to fill in the number of protons and neutrons so if we take a look at the periodic table we can see that chlorine does in fact have an atomic mass that is approximately 35.5 so that answer to part a makes sense and we can see atomic number 17 that refers to the number of protons in the nucleus of chlorine so 17 protons for each isotope protons plus neutrons will give us the mass number so 17 plus 18 is 35 or you could do 35 minus 17 equals 18 and then 17 plus 20 equals 37. all right let's take a look at number eight so we're looking at a different element that says not chlorine you can see along the x-axis we have a mass number of either 121 or 123 and along the y-axis for the percent abundance you can see written above each bar 57.4 percent for the mass number 121 and 42.6 percent for the abundance of mass number 123. so to calculate the average atomic mass of this particular element we're going to do just like we did before take the mass number times the percent abundance that is written as a decimal so 121 times 0.574 plus 123 times 0.426 so we do that math and we get approximately 121.9 now to look on the periodic table to figure out what is the most likely identity of this element it's okay if the atomic mass of the element is not exactly 121.9 but is very close to that value let's take a look so again tin sn is only 118.71 so that's too small tellurium te is 127.60 that's too big so the only element that this could represent is atomic number 51 antimony sb with an atomic mass of 121.76 again that's the closest to the value that we calculated in part a so again to part b what is the most likely identity of this element it is going to be antimony element symbol sb okay let's take a look at number nine so we have a mass spectrum with mass numbers of 79 and 81 and we have the percent abundances that are listed so now we're not going to actually need to calculate the average atomic mass in this particular problem they come right out with what is the most likely identity of this element so if we look at the periodic table we're looking for an element that has an atomic mass that falls somewhere between 79 and 81. again these are almost exactly a 50 50 mixture of these two isotopes so should be pretty close to 80 for the atomic mass which would be the average of 79 and 81. let's take a look in the periodic table so selenium se has an average atomic mass of 78.97 so that's too small krypton kr is going to be about 83.80 that's too large so the only element that has an atomic mass that falls between 79 and 81 is bromine br and its atomic mass is 79.90 so bromine has 35 protons in its nucleus and has an average mass atomic mass of 79.90 so bromine is the answer to part a and then for part b again bromine has 35 protons in its nucleus so mass number minus atomic number would give us the number of neutrons so 79 minus 35 is 44 and 81 minus 35 is 46. so again 35 protons and 44 or 46 neutrons for the two isotopes of bromine okay moving on to number 10 this time we don't actually have a mass spectrum to look at but we are told that a certain element has two naturally occurring isotopes with mass numbers of 63 and 65 what is the most likely identity of this element let's take a look at the periodic table and see if we can find an element whose atomic mass falls between 63 and 65. nickel is around 58.69 zinc is around 65.38 so copper element cu the only one that would fall in that range between 63 and 65. so 63.55 is the average atomic mass of copper cu now to fill in the table with protons and neutrons again on the periodic table we know that copper has 29 protons in the nucleus again that would be the atomic number of copper 29 and then since the protons plus the neutrons equals the mass number 63 minus 29 will give us the number of neutrons and again that's either 34 or 36 for the number of neutrons again these are the number of neutrons so 29 plus 34 equals 63 29 plus 36 equals 65. now again we're not looking at a mass spectrum so we don't know the percent abundance of these two isotopes of copper so part c of this question which isotope of this element mass number 63 or mass number 65 is more abundant in nature justify your answer so on the periodic table the average atomic mass of copper is 63.55 so the isotope with mass number 63 should be more abundant in nature because the average the weighted average 63.55 is closer in value to 63 than it is to 65. so for example if it had been exactly a 50 50 mixture of these two isotopes then the average of 63 and 65 just a straight up average would be 64. since 63.55 is closer to the number 63 that means that the isotope with mass number 63 is more abundant okay let's take a look at number 11. so so far in the examples that i'm showing you for a mass spectrum it's been really simple in that there's only been two naturally occurring isotopes of the elements in the examples you've seen so far so what i wanted to do here with this example is to indicate that the process the calculation of the average atomic mass it can be a bit more complicated it's still the same principle but you just have more numbers to enter into your calculator so here number 11 if an element has several naturally occurring isotopes then here's how you would do the calculation part a based on the information above estimate the average atomic mass of the element to the nearest whole number and then use a calculator to determine the average atomic mass so if i had to estimate this it looks like it's going to be anywhere between 154 and 160 according to the data that we have for six different isotopes of this particular element so where would the average fall where would the weighted average fall looks like 154 is a very small amount it's about 2 percent and then all of the other isotopes are somewhere between 15 and about 22 25 so i'm going to estimate again just a rough estimate that the average atomic mass is probably somewhere around the middle like maybe around 157 or so so you might estimate that the average atomic mass of this particular element is approximately 157. let's go ahead and do the math so to do the math we're going to take the mass number times the percent abundance written as a decimal so it looks like this 154 times 0.0218 plus 155 times 0.1480 and then you just keep going 156 times its percent abundance as a decimal 157 158 160. so all of that into my calculator gives me an answer of four 157.0241 so again approximately 157 point zero two what is the most likely identity of this element again let's look on the periodic table and see can we find an element that has an average atomic mass that is approximately 157 so down there in the lanthanide series of the periodic table the elements between atomic number 58 cerium and atomic number 71 lutesium we see that an element with a symbol gd which is gadolinium has an atomic mass of 157.25 so that's going to be the most likely identity of this particular element so gadolinium gd with atomic mass 157.25 okay well moving on to our next topic the last topic in this particular video topic 1.3 elemental composition of pure substances so here are some essential knowledge statements from the ap chemistry course and exam description some pure substances are composed of individual molecules while others consist of atoms or ions held together in fixed proportions as described by a formula unit according to the law of definite proportions the ratio of the masses of the constituent elements in any pure sample of that compound is always the same the chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula so let's take a look at our first question here number 12. calculate the percent composition by mass of each element in glucose c6h12o6 so what you're doing here is you're using the skill that you already have acquired from topic 1.1 about calculating the molar mass so the molar mass of glucose c6h12o6 would be the atomic mass of carbon 12.01 times 6 the atomic mass of hydrogen 1.008 times and the atomic mass of oxygen 16.00 times 6. so grand total molar mass or formula mass is 180.156 in the case of the molar mass that would be grams per mole if this were a single molecule of glucose you could say amu or atomic mass units but the percent composition by mass what that means is that we have to take the part divided by the whole for each element in this particular compound so the percentage of carbon by mass the percentage of hydrogen by mass the percentage of oxygen by mass so for the percent carbon by mass that would be 12.01 times six that's the total mass of carbon and then divided by the formula mass or the molar mass 180.156 times 100 to convert that to a percent and we do that math and we get 40.00 percent carbon by mass now we look at hydrogen and then oxygen so for hydrogen it would be 1.008 times 12 that's the total mass of hydrogen in one unit of glucose again divided by the total mass times 100 6.71 hydrogen by mass and then for oxygen 16.00 times 6 that's the total mass of oxygen within the formula glucose divided by the total mass times 100 and we get 53.29 percent oxygen by mass so no matter where i get glucose a pure sample of glucose from any source should have these same percentages of carbon hydrogen and oxygen by mass so the law of definite proportion or constant composition says that glucose will always have these particular percentages of carbon hydrogen and oxygen any pure sample of glucose has that so moving on to number 13 in the packet now we have a different compound which is called erythrose and its chemical formula is c4h8o4 let's go ahead and do the same process that we did in number 12 for glucose so the total mass of erythros from the periodic table so four carbons eight hydrogens and four oxygens is 120.104 grams per mole or you could say amu's if you were talking about a single molecule of erythrose let's break it down for carbon hydrogen and oxygen just like we did in number 12. mass of carbon so 12.01 times 4 divided by the total mass 120 120.104 times 100. interestingly enough we get the exact same percentage of carbon that we did in number 12. and so you're going to see a little deja vu now we do the percentage of hydrogen by mass in this compound so 1.008 times 8 total mass of hydrogen divided by total mass of erythros again that number should be familiar six point seven one percent hydrogen by mass and then finally with oxygen total mass of oxygen sixteen point zero zero times four divided by the total mass of erythros times a hundred fifty three point two nine percent oxygen by mass so if you're wondering why i chose the compounds in number 12 and number 13 there was a particular point that i wanted to make with these two compounds glucose and erythros and that is summarized by the work you're doing in number 14. so the empirical formula is going to be the simplest whole number ratio so when we take glucose c6h and we divide each of those subscripts by the greatest common factor in this case 6 we get c1h2o1 now even though erythrose is a very different chemical with different properties different formula when we take c4h8o4 and we divide each of those subscripts by the greatest common factor in this case four we get the exact same empirical formula so the point that i wanted to make is that we have two different compounds but because they have the same empirical formula then they are going to lead to a calculation that will give you the same percent composition by mass all right let's move on to number 15 in the packet now in this particular question we're actually going to go in the opposite direction so instead of being given a chemical formula and then calculating the percent composition by mass in number 15 we are told what the percent composition by mass is and we have to figure out the chemical formula specifically the empirical formula the one that has the lowest whole number ratio of elements so this is phosphorus and oxygen 43.64 phosphorus by mass 56.36 oxygen by mass so a good first step in this particular problem is to assume that the mass of this sample this compound that contains phosphorus and oxygen is 100 grams that way since a percentage is based on 100 we can just simply write down the number of grams of each element so we have 43.64 grams of the element phosphorus and we have 56.36 grams of the element oxygen well as you learned in topic 1.1 you can use the periodic table to go from grams to moles and so we have grams right now let's convert grams into moles because that way the mole ratio will give us the information that we need about the chemical formula so let's take the periodic table which has for phosphorous an atomic mass of 30.97 let's use that number to go from grams to moles now at this point i'm going to write that number down on my calculator i don't want to round it off to just 1 or 1.4 or even 1.41 it's actually better in calculations like this where you're trying to figure out the empirical formula it's better if you have extra digits on your calculator you want to really resist the temptation to start rounding these numbers off or not writing down the full number so i chose four significant figures in this problem the reason why you don't want to round it off too much is because you're looking for a ratio of moles of one element to the other and the more you round off your numbers the less accurate your values will be when you're trying to find that mole ratio so it's better to have extra precision in your answers than to just cut off the number at a couple of digits all right oxygen 56.36 grams of oxygen you might be wondering if you have good chemistry knowledge about oxygen being a diatomic element right we normally write oxygen as o2 you might be wondering do we have to use o2 in our calculations and the answer is no we're looking for individual ratios of elements o2 is a molecule it's not an individual atom so therefore it would be a single o so on the periodic table i'm going to write 16.00 i'm not going to write 32 because we're not talking about molecules of o2 we're talking about individual atoms of phosphorus and oxygen in this particular unknown compound so again be careful when you're going from grams to moles of individual elements that you ignore any elements that might be diatomic because in this particular case it's individual atoms not diatomic molecules all right i do this math 56.36 divided by 16 and i get again look at me not rounding off this answer too much i'm not going to call it 4 or 3.5 so i went ahead and chose 4 significant figures okay what do we do next to get the empirical formula of this compound we're looking for a simple whole number ratio between the moles of phosphorus and the moles of oxygen so how do we do that well because it's based on a simple whole number ratio of moles we're going to go ahead and take these two values of moles and we're going to divide them by the smaller of the two values so 1.409 divided by itself will give us 1 and 3.523 divided by 1.409 if you do that math on your calculator you get 2.50 so again now do we have simple whole numbers no we're not going to take 2.5 and just round it up to 3. what we have to do is take this empirical formula which again this is not correct and we're going to try to get simple whole numbers from the numbers that we have right now so instead of saying 1 and 2.5 let's just go ahead and double these values that way it keeps the ratio the same but we turn a 1 into a 2 and a 2.5 into a 5. so by doubling the ratio of 1 and 2.5 we get 2 and 5. and then that's it that's our empirical formula p2o5 is the empirical formula of this particular compound all right so to go ahead and summarize the procedure that we used in number 15 of this packet if you are given percent composition by mass or mass data for a certain compound so grams of each element the following procedure will help you to determine the empirical formula of the compound convert the grams or the mass of each element into moles again that would involve the periodic table then now that you have moles of each element divide each value of moles by the lowest value the lowest number now at this point you may already have whole numbers based on that step in the procedure but if not if you have numbers that end in 0.5 like we just saw in the last example then you could multiply by 2. or if for example your number of moles they end with .33 or .66 or 0.67 you may have to multiply by 3 to turn those values into whole numbers so again be careful with your data don't round off your numbers too early in the calculation as i said before but then whatever whole numbers you get those whole number values of moles then they will be used in the subscripts to write the empirical formula of that compound okay here's our next problem number 16. we are given the percentage by mass for the elements carbon hydrogen and oxygen let's convert each of these numbers from grams into moles so we'll assume that we have a hundred grams of this particular compound and that way the percentages can just be written as grams so we have grams of carbon let's divide that number by the molar mass of carbon which would be 12.01 we're converting from grams to moles again i'm not going to round that number off to four or 4.3 so i wrote that number with four significant figures a little extra precision is a good thing in the early stage of the calculation now we have the same calculation for hydrogen and for oxygen so 13.13 again this is individual elements individual atoms of hydrogen so it's not h2 it's not the diatomic molecule that we think about in terms of its substance you might have hydrogen gas for example it's h2 this is individual atoms of hydrogen so divided by 1.008 that came from the periodic table 13.03 i'm using the atomic masses for each element as the molar mass of each element so 34.73 grams of oxygen again not divided by 32 but 16. so individual molar masses of these elements with no subscripts so now i'm staring at three numbers which are 4.31 sorry 4.341 moles of carbon 13.03 moles of hydrogen and 2.171 moles of oxygen so the next step in the procedure is to find the lowest value in this case 2.171 and we're going to go ahead and divide each of these values of moles by the smallest value so divided by 2.171 we get 2 divided by 2.171 we get 6 and obviously divided by itself we get 1. so those values 2 6 and 1 are already whole numbers there's no extra step of multiplying by 2 or 3. so what i have right now is an empirical formula c2 h6 and then o1 or just o the one is understood as a subscript okay so our empirical formula was c2h6o let's move on to number 17 in the packet a pure sample of tin so elemental symbol sn with a mass of 6.18 grams is burned in air until the tin is completely converted into a tin oxide so a tin oxide compound would contain both tin and oxygen the mass of the tin oxide is equal to 7.85 grams determine the empirical formula of this tin oxide compound so we start with 6.18 grams of tin it is burned in the air which means it's going to react with oxygen in the air and it produces 7.85 grams of tin oxide now i'm not writing a balanced chemical equation for this process i'm writing the elements in words because we don't know the exact formula of the tin oxide so therefore even though i know that tin is sn i know that oxygen gas is written as o2 i don't know what the chemical formula of this particular tin oxide compound is so the difference in mass between 7.85 grams and 6.18 grams that difference is 1.67 grams that's how much oxygen reacted with the tin in this experiment so while this looks like a different problem in the previous examples we had the percentage of each element by mass so percent composition now that we know from processing this information that we have grams of tin sn and then grams of oxygen in this case i'm going to write this as individual atoms of o we can go ahead and convert grams to moles like we have been doing in the other problems number 15 and 16. so grams of tin 6.18 grams divided by the atomic mass of tin on the periodic table 118.71 that gives me the number of moles of tin again i'm going to try to keep a little extra precision so i'm not going to write 0.05 or 0.052 i went ahead and i chose three significant figures here but if you had four or five significant figures that would be fine too again a little extra precision is fine in this stage of the calculation but don't round off your numbers because you might throw away important precision that will help you get to those simple whole number ratios of moles again like i said in the previous examples we have individual oxygen so yes there is oxygen in the air which is o2 but i want to know the ratio of atoms of tin to atoms of oxygen in this particular compound so that's why i'm not dividing by 32 i'm not talking about o2 i'm talking about individual atoms of o within this tin oxide compound so 1.67 divided by the atomic mass of o 16.00 again i'm not going to call that 0.1 i'm going to call that .104 so at least three significant figures so what's the next step in the process we have moles of 10 and moles of oxygen or o divide by the lower of the two values in this case we're going to divide by .0521 so now we have a 1 for 10 and we have a 2 for oxygen so those simple whole number values now become the subscripts so our chemical formula of this tin oxide compound from our data and our calculations is sno2 all right let's move on to number 18. number 18 starts off looking very similar to the other types of problems where we're trying to figure out the empirical formula of an unknown compound but then there's a little bit of a twist in parts b and c of this question where we actually make the connection between the empirical formula and the molecular formula of a compound okay so part a determine the empirical formula of compound x so we know that compound x consists of the elements carbon hydrogen and nitrogen we have a 15.00 gram sample of this compound and if you break it down by grams of each element we have 9.81 grams of carbon 1.37 grams of hydrogen and 3.82 grams of nitrogen so procedure is going to look like the previous problems where we have grams of each element let's go from grams to moles so divide by the molar mass of carbon which would be 12.01 and i'm going to write this number down as moles of carbon again i'm not rounding too much so i have three significant figures 0.817 moles of carbon moving on for hydrogen 1.37 divide by the atomic mass or the molar mass of h so 1.008 from the periodic table and i get 1.36 moles of hydrogen again not h2 just h and then for nitrogen again nitrogen is a diatomic element but in this case we're talking about individual atoms of c h and n so i'm not going to divide by 28.02 but rather 14.01 the atomic mass or the molar mass of individual n so ch and n this again that number gives me if i do that math 0.273 so i kept some precision at least three significant figures in my number of moles which of these three values is the smallest so 0.273 so i'm going to divide each of these value of moles by the smallest value so divide .817 by 0.273 and i get 3. and then for hydrogen i divide by 0.273 and i get 5 and obviously divide it by itself is 1. so 3 5 and 1 those are my subscripts in the empirical formula c3 h5n now the empirical formula is not necessarily the same as the molecular formula so in part c of this question we're going to have to get the molecular formula for compound x and we don't know what that is yet but we do know that the empirical formula which is the lowest simple whole number ratio of moles is c3h5n let's move on to part b of this question in part b it is determined that a 25.0 gram sample of compound x contains 9.11 times 10 to the 22nd molecules calculate the molar mass of compound x in units of grams per mole now you might be thinking molar mass don't we already know the chemical formula c3h5n well again that's the empirical formula which is not necessarily the same as the molecular formula so at this point we're not using our answer to part a to do this calculation we're given grams and molecules so i'm going to go ahead and write down the numbers i'm given which is 25 grams 25.0 grams and 9.11 times 10 to the 22nd molecules now what i did in starting this particular problem or setting up the calculation is i put grams on the top and i put molecules on the bottom if i'm paying close attention to what they're looking for molar mass is not grams per molecule but rather grams per mole so what i'm thinking is that i would like there to be a conversion factor that changes grams per molecule into grams per mole i don't want the grams to go away in fact i want the grams to stay exactly where they are but i would like to get rid of the molecules i would like to get rid of the molecules and replace that with per mole so i want molecules to go on top and i want moles to go on the bottom now is the time for avogadro's number as you've already learned there are avagadra's numbers 6.02 times 10 to the 23rd particles in this case molecules of a substance in one mole of that substance so if i do this calculation 25 divided by 9.11 times 10 to the 22nd times 6.02 times 10 to the 23rd i get on my calculator approximately 165 grams per mole again going with three significant figures based on the 25.0 and all the other numbers 165 grams per mole if you watching this video are thinking well is there another way to solve this problem besides the way i set it up absolutely so what you could do is you could analyze the information in part b and you could say well i don't know how to convert 25.0 grams into moles because i'm not really sure what the chemical formula again not the empirical formula but the molecular formula of compound x is so let's say for a moment we don't know what to do with 25.0 grams have we learned in chemistry how to convert molecules to moles yes absolutely so you could have written down the number of molecules and then again use avogadro's number to go from molecules to moles if i just did that i'd be dividing by avogadro's number again trying to get rid of molecules and go to moles and if i did that on my calculator 9.11 times 10 to the 22nd divided by avogadro's number i would get a value of approximately 0.151 moles but then i could go back to the 25.0 grams which i did not use yet in this calculation and i could say all right grams per mole which means grams divided by moles so 25.0 grams divided by the number of moles that i just calculated using avogadro's number if i did that math i would definitely still get the same answer that i got earlier which is 165 grams per mole so we're going to move on now to part c of this question which involves taking our answers to part a and part b and kind of putting them together in a certain way so in part a we had an empirical formula in part b we have a molar mass so in part c based on your answers to parts a and b determine the molecular formula of compound x so in part a we had an empirical formula of c3h5n again that's the simplest whole number ratio for those elements for the number of moles in part b we had approximately 165 grams per mole so how can we combine these two quantities or this information to get the molecular formula well if you take a moment and look at c3h5n that's three carbons five hydrogens and one nitrogen that number the formula mass of the empirical formula is not 165 it's about 55 55.08 on my calculator so how are those two numerical quantities related to each other well the ratio of the actual molecular formula which is the number of atoms of each element in that particular compound the ratio of the molecular formula to the empirical formula is going to be a whole number so for example c3h5n that's the lowest whole number ratio if i multiply it by 2 it would be c6h10 n2 but i could multiply that empirical formula by three or by four by five whatever and i would turn the empirical formula into the molecular formula so 165 is the molar mass 55 or 55.08 that is the empirical formula mass let's take the larger number divided by the smaller number and see what happens if we do that math 165 divided by 55.08 we get 3. so what does this tell us that the empirical formula should be multiplied by 3 and so 3 times 3 is 9 5 times 3 is 15 1 times 3 is 3 so the actual molecular formula is not c3h5n but rather times 3 c9 h15n3 and that would have a molar mass of 165 grams per mole so the answer to part c is c9 h15 n3 okay let's take a look at our next question so another way to determine the empirical formula of a compound instead of just being given the percent composition or the number of grams of each element in that compound is to use data from a combustion experiment now we saw this briefly with the tin converting that into the tin oxide earlier but if i take a compound that contains carbon and hydrogen and i burn it all the way to completion in the presence of excess oxygen gas then all of the carbon will be converted into carbon dioxide co2 and all of the hydrogen will combine with oxygen and convert it into h2o water so if the compound contains more than just carbon and hydrogen let's say it contains maybe nitrogen or sulfur then other gases may be formed oxides of nitrogen or oxides of sulfur so in number 19 we are given a data table and it says that there's a sample of a compound that contains carbon hydrogen and oxygen but we don't know the chemical formula of that compound all we know is it contains c h and o and it's burned completely in oxygen and o2 and it produced the sample produced 10.99 grams of carbon dioxide and it also produced 6.00 grams of water when it was burned completely so part a determine the mass of carbon present in 5.00 grams of the compound now we're not taking 5.00 grams and doing any math with that and the reason why we're not doing any math with that is because we don't know what the chemical formula is so we can't figure out how much carbon is present in 5.00 grams on its own what we are doing is looking at the co2 so to remind you on the periodic table the molar mass of co2 as calculated from the atomic masses of carbon and oxygen is 44.00 grams per mole so how much carbon is present in co2 that's what's going on in the calculation for part a so 10.99 grams of co2 was taken directly from the table and then now using the calculations from the periodic table i'm going to convert from grams of co2 into grams of carbon so how do i do that well at least i want to put grams of co2 on the bottom and i want to put grams of c grams of carbon on the top so this is how i did it the 44.01 that's the total mass the molar mass of co2 how much carbon is in that particular quantity is 12.01 as you can see from the calculations in the upper left corner of the slide so now i'm going to do the math of converting grams of co2 into grams of carbon so i do that math and i get approximately 3 grams i get 2.99908 i'll round that off to four significant figures at least for right now so 2.999 again i'm trying to be as precise as possible and not just cut off a number to a single significant digit so four sig figs in my mass of co2 four sig figs in my mass of carbon is there another way that i could have done this calculation that involved one mole as in converting from grams to moles absolutely we're still going to get the same answer but watch how you might have done it this way you could have taken the grams of co2 and first converted that into moles moles of co2 well the problem with just doing that much is that you have to get to grams of carbon so what might the next step be in converting moles of co2 into let's say grams of carbon as you might have done it like this there's one mole of carbon for every one mole of co2 because there are no subscripts with the carbon in this chemical formula for carbon dioxide so one mole of carbon is present in one mole of co2 now you have to go back to grams because they want the mass of carbon in grams so we could use the periodic table and write 12.01 grams of carbon per mole now it's the exact same calculation but it's in three steps so if you would prefer to do a problem a calculation like what i just showed you on the bottom part of that slide if that makes more sense to you go for it so the top version is just a ratio of grams of carbon to grams of co2 and then the bottom three-step sequence converts grams to moles of a compound converts moles of a compound to moles of an individual element within that compound and then back to grams using the periodic table okay with answered part a let's take a look at part b of this question we talked about carbon part b is let's talk about the mass of hydrogen present so again we're not using 5.00 grams in our calculation we're going to go ahead and figure out how much water was produced so 6.00 grams of water is produced and again the molar mass of water as you i'm sure already know would be just two hydrogens and one oxygen so 18.016 that is the molar mass of water so how much hydrogen is present in water is going to be part of our calculation if i write down 6.00 grams of water which i got from the table that i was given and i want to convert grams of h2o into grams of h so watch how i do that again be careful that you don't miss an important step so how much hydrogen is present in water looks like this so in the previous example it was 12.01 for carbon and 44.01 for co2 but within the water molecule within the water formula there's two h's in that formula so that's why this ratio is not 1.00 it's the total mass of hydrogen in the formula h2o if i do that math i get again i'm not going to round this off too much i'll say at least three sig figs so 0.671 grams of hydrogen if you would rather first convert grams to moles you could do that again grams to moles is a good first step but then we have to remember that there are two moles of hydrogen atoms for every one mole of water molecules and then if i go to grams of hydrogen that would be 1.008 grams of h per mole i'm still gonna get the same answer i'm gonna get 0.671 but if you prefer that bottom three-step sequence if that makes more conceptual sense to you to go grams to moles and then remember that there are two h's in every h2o then you're gonna get the exact same answer okay so let's move on to part c we've got the grams of carbon the grams of hydrogen let's see what it says in part c determine the mass of oxygen present in 5.00 grams of the compound so at the top of the slide you can see that the answer to part a was 2.999 grams of carbon and the answer to part b is 0.671 grams of hydrogen well there's a 5 gram sample 5.00 gram sample of a compound that contains three different elements c h and o we already know the mass of c and h so all we really have to do is subtract the answers from part a and b from the total mass of the compound the 5.00 and if we do that math so 5.00 minus 2.999 minus 0.671 when you add and subtract measurements we go with the least number of decimal places so i'm going to go ahead and with two decimal places i'm going to call that 1.33 grams of oxygen and again that's coming from just subtracting the answer to part a and b from the total mass of the sample that was burned all right almost done part d will be the traditional empirical formula calculation that you've seen before we're going to convert the grams of each element into moles so based on my answers to part a b and c i have 2.999 grams of carbon 0.671 grams of hydrogen and 1.33 grams of oxygen let's convert each of these elements from grams to moles as you've seen before so divide by either 12.01 or 1.008 or 16.00 so this is my calculation of moles of carbon so that's 0.2497 moles of carbon and then let's move on to the hydrogen let's convert from grams to moles so 0.671 grams of hydrogen divide by the molar mass of h 0.666 moles of hydrogen of h and then finally with the oxygen 1.33 grams of o divide by the molar mass of o 16.00 do that math and we get 0.0831 moles of o so now we have moles of c moles of h and moles of o to get the empirical formula we're going to divide each of these mole values by the smallest value and let's see what happens so divide by point zero eight three one for carbon we get three divide by point .0831 for hydrogen we get eight and obviously dividing by itself we get one so three eight and one those are the subscripts in the empirical formula so the empirical formula of this compound is c3 h8 o okay the last three problems in this packet have to do with ratios of moles but not exactly from element to element but involving water so it says another type of situation that involves mass and mole ratios involves a substance that is known as a hydrate a hydrate is a substance in which water molecules are included in the chemical formula these substances are often ionic compounds in which water molecules are bonded to the ions in the crystal structure so a hydrate salt can be heated to remove the water through evaporation and then you would form what is known as an anhydrous salt where the word anhydrous just means without water so here you can see in the table on the left two examples of anhydrous salts so we have copper roman numeral two sulfate cuso4 and then calcium chloride cacl2 on the right we see examples of hydrate salts of these compounds so copper roman numeral 2 sulfate pentahydrate so there are 5 moles of water for every 1 mole of cuso4 calcium chloride dihydrate there are two moles of water for every one mole of cacl2 so that dot just indicates that the water is coordinated and included in the chemical formula but that dot does not mean to multiply it just means that there are five or in this case two five additional moles of water are included in the chemical formula so let's see how we can take a look at some example problems that involve hydrates number 20 a sample of this copper roman numeral two sulfate pentahydrate cus04.5 h2o has a mass of 25.00 grams part a calculate the mass of cuso4 present in this 25.00 gram sample so how much cuso4 is present is going to be a number that is less than 25 grams because it's going to include both the cuso4 and the water so how much cuso4 is present well let me go ahead and start writing down the molar mass calculations so on the periodic table the atomic mass of copper 63.55 atomic mass of sulfur 32.06 atomic mass of oxygen 16.00 and there are four of them so 159.61 that's the molar mass of the anhydrous form of this compound without the water if i take 159.61 and then i add 5 moles of water to that calculation so 18.016 that's the molar mass of water h2o times 5 because there are 5 moles of water in the formula for the hydrate i get a number which is 249.69 that's the molar mass of the copper sulfate pentahydrate compound so 25.00 grams of this hydrated compound if i want to convert that to grams of just cuso4 i can use the numbers that are right here on the slide so 159.61 grams of just cus 04 for every 249.69 grams of the entire hydrate compound so that ratio will be set up as a conversion factor so i can get grams of cuso4 and i do that math and i get rounding off to four significant figures here 15.98 grams of just cos 04 without the water all right part b of this question calculate the mass of water present in this 25.00 gram sample well again it should be a number that's less than 25 grams so based on the same calculation i did for part a there's 159.61 grams of cuso4 for every 249.69 grams of the total hydrate compound what about the water well the water would be 18.016 times five that's how much water is present in the entire compound so 18.016 times 5 would give me the grams of water and then the 249.69 is the total molar mass of the hydrated compound so that math as a conversion factor would give me grams of water in my answer and i get approximately 9.02 or 9.019 so you might be thinking well did i really have to do that kind of calculation in part b if i already knew in part a that there was 15.98 grams of copper sulfate cuso4 couldn't i just subtract that from 25 and the answer is yes so another way perhaps a much easier way to calculate the answer to part b is just to subtract your answer to part a from the total mass of the sample so 25 minus 15.98 gives me the mass of water and 9.02 grams okay let's take a look at number 21. we're still talking about a hydrate compound but this time it's not the copper sulfate compound it's a calcium chloride compound this is calcium chloride dihydrate calculate the percentage of water by mass in this count calcium chloride dihydrate cacl2.2h2o well let's go to the periodic table calcium has an atomic mass of 40.08 chlorine has an atomic mass of 35.45 and there are two of them so just that math would give me 110.98 it's the molar mass of the anhydrous calcium chloride water has a molar mass of 18.016 and there are two of them in the chemical formula so grand total 110.98 plus the two moles of water is 147.012 grams per mole so the percentage of water by mass would be the mass of water divided by the total mass of the hydrate compound so we have 18.016 times 2 for the total mass of water within this compound and 147.012 for the total mass of the entire hydrate compound so that ratio times 100 would give us the percentage of water by mass if you do this math you get 24.51 percent water by mass all right on to number 22. we are given a data table it says in a certain experiment a sample of a hydrate of magnesium sulfate the formula for magnesium sulfate is mgso4 and then since we don't know exactly what the formula of the hydrate compound is we have dot n h2o where n is going to be a whole number it's going to be one two three four five etc that compound the hydrate is heated in order to remove all of the water from the sample so the empty container is 25.356 grams then they put the hydrate salt in that container and it weighs 28.418 grams before it's heated and then they heated it to evaporate all of the water from the hydrate sample and they heated it three times and the mass of the container and the sample after each heating is recorded in the table eventually once all of the water has been removed then we have mgso4 the anhydrous salt so let's take a look at the different parts of this question part a explain how the data indicates that all of the water has been removed from the hydrate salt in this experiment so notice that the mass is going down after each heating well take a look closely at the data after the second and the third heating and what you'll notice is that they are essentially the same number they're only different by .001 grams so at that point we can assume that all of the water has been removed if you take a look at the third heating which is again for the precision of the scale is essentially the same number it's only different by .001 grams because the mass does not change significantly between the second heating and the third heating as in like those values are essentially the same now we can assume that the mass is constant the mass is not changing so therefore all of the water must have been removed in this experiment all right let's take a look at part b calculate the mass of the hydrate salt that was used in this experiment so that would be the difference between the first two values in the table 28.418 is the mass of the container and the hydrate salt before it was heated 25.356 is the mass of the empty container so that difference in grams 3.062 that's the mass of the hydrate salt all right let's take a look at part c calculate the mass of water that was removed from the hydrate sample in this experiment so the mass of the container and the hydrate salt before heating was 28.418 grams and at the end of the experiment after all of the water was removed the mass of the container in the sample was 26.852 grams so that difference in mass represents how much water was lost from evaporation and that would be 1.566 grams of water all right let's take a look at part d determine the value of n in the formula mgso4 dot nh2o so n represents the ratio of moles of water to moles of mgso4 that means we're going to need to take our answers that we have so far in our data table and turn grams into moles for the chemical substances involved here so how many moles of water and how many moles of mgso4 that's what we're looking at so the mgso4 the anhydrous salt would be the difference between our answers to b and c 1.496 grams or you could get it directly from the table and you could say well at the end of the experiment the mass was 26.852 [Music] just subtract the empty container and that difference 25.356 subtracted from 26.852 you would get the same answer so now we know how many grams of mgso4 are present at the end of the experiment the anhydrous mgso4 let's turn that number into moles so the value of n represents a mole ratio of moles of water divided by moles of this anhydrous salt so using the periodic table 1.496 grams of mgso4 let's turn that into moles so divide by the molar mass 120.36 and we get notice how i'm not going to round this off too much so 0.01 0.012 it's better to have some extra precision in your calculations at this point so i'm not gonna be tempted to call that point zero one definitely not point zero one two i'm gonna keep going so there's my number of moles and then for the water 1.566 grams of water divided by its molar mass again i'm not gonna round that off to point zero nine see i might think the ratio is nine to one if i said .01 and .09 but that's incorrect so better to have some extra precision in your values especially when you're in the middle of a calculation so how am i going to get my answer for the value of n it's a mole ratio so we're going to take moles of water so 0.08692 divided by moles of the anhydrous salt .01243 let's see what we get on the calculator and we get 7. so the answer to this particular question the value of n is 7 and if you wanted to write the formula of the hydrate mgs04.c h2o this would be called magnesium sulfate heptahydrate if you were curious it's also found in epsom salts which are kind of an over the counter remedy for like soaking sore muscles things like that okay so seven was the answer to part d and i believe that this particular part of this question represents the last question in the packet so again i hope that the answers to this particular packet were helpful thanks for watching