"Atoms, Molecules, & Ions" | Chemistry with Educator.com

"Atoms, Molecules, & Ions" | Chemistry with Educator.com

hi welcome back to educator calm and today’s lesson in general chemistry is on atoms molecules and ions we’re going to go ahead and look at the lesson overview right now we’re going to get a brief introduction into atomic structure which is more of a historical background following this will get into what we mean by isotopes we’ll get introduced to the periodic table in this lesson followed by naming compounds basically ionic compounds those that are made from metals and nonmetals polyatomic ions hydrates and also we mean by molecular compounds we’re then going to get associated with the concept of the mole and how that can be used to determine what is called a percentage composition which is going to ultimately lead to the finale of this lesson which is on determining what is known as the empirical and molecular formulas alright by the beginning of the 20th century physicists had already known the following number one that opposite charges attracted and that like charges repel and number two a subatomic particle existed that was already that was negatively charged which had been called the electron commonly abbreviated e- and one of the first accepted models of the atom was called the plum pudding model and what the plum pudding model was the following it was essentially a sphere of positive charged and embedded in the sphere of positive charge were the subatomic particles the electrons and these were completely stationary if you will but there was no accounting for any other type of subatomic particles so once again this is called the plum pudding model so a sphere a positive charge that contain electrons scattered throughout we now get into one of the earliest dearies that was accepted to explain for atomic structure this was called John Dalton’s atomic theory John’s Dalton atomic theory contain the following premises number one was that matter was composed of individual units called atoms and number two that these atoms were the smallest unit possible and that they were indivisible that is we could not break them down any further okay so atoms were indivisible okay number three the third premise of John’s Dalton’s atomic theory was that atoms of one element we’re all identical to each other number four atoms of one element are going to be different than atoms of another element so that’s what we reference our chemical identity right okay so atoms of one element different then atoms of another element and finally the fifth and final premise was that when elements combined they can combine in small whole number ratios to form compounds so atoms combine in small whole number ratios to form compounds and this fifth premise here is going to be looked at again and that’s what we call the law of multiple proportions okay I’m going to go ahead and put an asterisk next to each of the premises which we actually find not to be so true today atoms were indivisible well that’s not true because we can divide an atom in a process called nuclear fission and

so premise 2 is not completely accurate and the next one is atoms of what element are all identical to each other we’re going to talk about this is that not all atoms of the same element are identical because atoms can exist in different forms which we call isotopes so that’s going to be coming on later on all right following John Dalton’s atomic dairy was one of the first major experiments on atomic structure and this is what we call ernest rutherford’s gold foil experiment what Ernest Rutherford third was the following he shot a beam of high-energy particles that were positively charged and he’s positive charged particles were called a where what we call alpha rays and this was shot at a very thin piece of gold foil now what was the expected result the expected result was the following that all of the alpha rays should have gone straight through why because the positives fair was mostly empty space and that the positive ray should have been attracted to the electrons that are embedded in a sphere so this is expected result due to plum pudding however the actual result was the following from the gold foil yes the majority of the Rays did indeed go straight through however what was determined was that some Ray’s actually got deflected at various angles and you would actually have a couple raised get deflected completely backward now this was highly unusual because this was not predicted by the plum pudding model instead what came about was the following implication and the implication was the following that matter and Adams contained a dense center of positive charge because it’s that it’s positive charge that was making the alpha rays deflect and this positive charge is what we today refer to as the nucleus and the nucleus is at the center of all atoms and is positively charged so too re summarize the implications of the gold foil experiment was that all atoms have a dense center that is positively charged which is called a nucleus inside the nucleus two subatomic particles exists the proton is positively charged at the neutron has zero charge number three nucleus is responsible for the mass of the atom that’s what gives a and Adam wait and finally electrons which were already known to have existed exist outside the nucleus and have negligible weight and so let’s go ahead and summarize these subatomic particles ok Neutron ok proton and electron now please make a note that these relative masses and the relative charges ok are indeed in relation to each other and so a neutron is going to be a sign a mass of one and a charge of zero a proton is assigned a mass of one and a charge of positive one and an electron has a negligible mass a mass of zero and a charge of negative one and of course we symbolize the neutron lowercase n proton p and electron is going to be e all right now another fallacy one fallacy in John Dalton’s atomic theory was isotopes was that excuse me atoms of the same element are all identical to each other but we’re going to learn that atoms of different of the same element can exist in different forms and these are called isotopes then isotopes are the same atoms on the same element and they differ only in their number of neutrons and so number of protons is going to be equal to the number of electrons and this is because of charge neutrality but the only difference is the number of neutrons so number of n differs and so now we’re going to enter introduce somebody call isotope format okay so for example carbon has two isotopes carbon-12 and carbon-14 now anytime you see the number immediately

following the hyphen this is what we call the mass number of the isotope and the mass number of the isotope is simply equal to the number of neutrons plus the number of protons and so when we write an isotope in a specific format it’s always going to have the following a design okay element symbol X a on top and Z on the bottom as a subscript okay so X is simply your element symbol K a is your mass number and Z is simply or atomic number the atomic number is strictly equal to the number of protons of the element okay so now a mathematical relation follows that if you take the difference of a minus C you’re going to get the following the number of neutrons plus the number of protons and then minus the number of protons and as you can see protons councils and so when you take the difference of a minus Z you simply get the number of neutrons everyone so for example if you take 12 6 and carbon ok we have 12 is equal to the number of neutrons plus the number of protons ok six is on the bottom and that’s the number of protons which is also equal to the number of electrons there 4a minus Z is equal to six and that’s equal to the number of neutrons and so carbon-12 as six protons six neutrons and six electrons however if you do the same analysis for carbon 13 we’re going to get six protons seven neutrons and six electrons so as you can see isotope strictly differ by the number of neutrons present inside the nucleus and now we get introduced to the periodic table well the periodic table is essentially organized into different rows and columns okay so each column is also referred to as a group okay each row is referred to as a period and typically there’s always a staircase that is embedded in all of our periodic tables anything to the left side of the staircase is what we call a metal and editing to the right side of the staircase is what we call a nonmetal now when we look at the periodic table you see always the following okay you see a number on top the number on top is what we call the element number and the element number is actually the atomic number number of protons okay the number on the bottom is the atomic mass of the element and finally right in the middle is your element symbol we’re going to talk an L more about different areas of the periodic table now we talk about metals and metals have the following characteristics they appear lustrous which is shiny okay they are good conductors of heat and of electricity they are malleable which means we can form thin sheets out of them and they are ductile which means we can pull them into fine wires nonmetals are going to be the opposite they’re going to be electrically and thermally insulated which is what we mean by poor conductivity in addition they are brittle which means that if we take a hammer and hit them they are going to crumble ends that are not too malleable okay so will crumble okay right in the middle of the border between metals and nonmetals that is what we find

semimetals okay semimetals are also known as metalloids and these are lie they border that staircase that we drew previously okay so border the staircase and because they are called semimetals and/or metalloids they are going to have properties of both they tend to be good conductors of electricity but court conductors of heat and they tend to be malleable and the typical examples are of course a silicon and germanium and we see these two being used in a lot of electrical devices and we call semi conductors all right now what we’re going to analyze necks are some of the major columns of the periodic table a group one is what we call the alkali metals such as lithium and sodium and potassium etc and group one metals tend to have the following characteristics that is their moisture sensitive which means they have a tendency to react with water and they can be explosive when they react of water group 2 or what we call the alkali earth metals okay such as calcium and magnesium and barium and these are harder than group 1 and less moisture sensitive okay group six are the halogens which are fluorine chlorine bromine and iodine and all of these occur naturally in nature diatomic Lee which means they occur in pairs so when someone says fluorine it’s really f2 and that’s and cl 2 kb r 2 and I 2 and the halogens are incredibly reactive finally a group number 8 D rightmost column of the periodic table or what we call the noble gases such as helium and argon and neon and unlike the group seven these are going to be monatomic that is they occur just as individual units and they are called noble gases because they are relatively inert and have a harder have a smaller tendency to react okay now that we’ve been introduced to the periodic tables and the elements that comprise let’s now talk about different types of compounds the first type of compound we’re going to study are called ionic compounds ionic compounds are formed between a metal and a nonmetal or they contain what are known as polyatomic ions and because of the strong electrostatic attractions that exists between opposite charges ionic compounds tend to have very high melting points and below are a couple of these polyatomic ions so Polly Atomics could be the following something like sealed 32 minus which is called carbonate ok or so4 2 minus which is called sulfate kno3 1 minus which is called night ray so you see why these are called polyatomic ions because it’s several items coming together to form what entity ok so those are what we call polyatomic now that is different than monatomic ions of course which are just formed from one unit something like CL minus 0 to minus al three plus etc ok some polyatomic vs monatomic and it turns out that we can actually use the periodic table to protect the ionic charge for main group elements and main group elements are going to be group 1 and group 2 and then group 6 group 7 and group 8 excuse me group 3 to 6 and then seven and eight and it turns out that all of the group ones ok tend to form a one plus charge all of the group 2 s10 the form of two plus charge group 310 the form of three plus charge we’re going to skip four and five because they tend not to form ionic compounds to such a large degree and so column 6 we’re going to get a 2 minus charge column 7 is going to be a 1 minus charge and group 8 because we say no choice because group a tardar noble gases and they tend not to react with other species now you noticed also very importantly

that the positive charges would are which are what are called cations okay these are going to be formed by metals and that the negative charges or what are called anions and these are all non metals okay oh I’m sorry I’m leaving out group five everybody group five is of course this is going to be a three- charge okay so you have your cations and then you’re an eye on so once again cations are going to be formed by metals and anions are going to be formed by nonmetals now we have to learn how to name ionic compounds such as nacl such as ba f2 and so we need to now go over the systematic procedure for doing so okay so we’re going to first always you’re going to name the cation first followed by the ni on and the anion will end in ide so something like nacl ok so the cation is simply going to be sodium and Cl ok is usually chlorine but now we’re going to become chloride and so the correct name of nacl is simply sodium chloride very simple okay you do not use prefixes mono die trying cetera to denote subscripts in a formula so something like for example baf too ok let’s go ahead and name it so that’s going to be a barium followed by floor ride ok and you do not say die floor I know no difluoride so for example again we ignore the prefixes for certain transition metals you’re going to use a Roman numeral in parentheses to indicate the metal charge so if we what iron two plus we’re going to note denote it as I and Roman numeral two and iron three-plus cation is going to be Fe roman numeral three and again you should definitely ask your instructor as to which transition metals he or she would like you to know for sure where you have to use a parentheses to indicate charge finally you can also have a hydrate and a hydrate is when you have water attached so for example cocl 2 dot 5 h2o so we’re going to use the prefixes followed by the word hydrate so five using the Greek prefix would be Penta and so we would just say that this is the pentahydrate cowbell of a cocl2 okay let’s go ahead and put everything together and do a couple of examples okay so al2o3 okay fecl3 see you I to dot 3 h 2 o ok so we’re going to name the cation first so al 2 o 3 that’s going to be aluminum you followed by oxygen becoming oxide ok fecl3 so Fe is one of those we need to know so i earn something in parentheses and then chlorine becomes chloride now we have to come up with the charge ok so each chlorine is a 1 minus charge and there’s three of them giving me a three- overall and so for iron iron is going to be 3 plus here so giving us iron roman road 3 chloride in the end so don’t forget for the ionic compounds the net charge should equal zero to maintain charge neutrality ok right here we’re going to have copper copper is one of those you need a Roman numeral for followed by hydrate final followed by I’ll died and then we have three waters which becomes try hydrate and here iodine is a 1 minus charge and there’s two of them giving negative 2 overall and so copper must be a positive 2 to balance it and so I’m going to indicate that in the ribbon in the parentheses as Roman numeral now we got to do it the other way so before I gave you the formula and you’re asked to provide the name now let’s go ahead and write the name given write the formula given the name so let’s go ahead and look at the following examples okay so

barium phosphate phosphide okay ammonium phosphate ok so now barium phosphide okay barium is BA ok phosphide is phosphorus which is P now what I like to do is I like to put the charges up in the top just for my own purposes so barium is two plus phosphorus is three minus and what I always tell my students to do is then like I use the crossing rule basically the if you cross the charges they become the subscripts for each of the other elements and so ba is going to get if subscript of 3 and phosphorus it’s going to get a subscript of two and you can convince yourself to make sure that the charges cancel three of the barium times to positive plus two of the phosphorus times 3 minus and yes they do council 20 ok next one is ammonium phosphate monium is nh4+ okay phosphate is p 0 for k ammonium is a 1 plus charge philosophy is a three- charge let’s go ahead and accounts and cross them again okay and when we cross them we’re going to get nh4 three and then p 0 for willful one and let’s go ahead and make sure that they all work out so three of the ammonium times a plus one charge plus one of the phosphates and a three- charge and yes it does council 20 every 10 so that’s some nice properties on ionic nice sample problems on naming ionic compounds the other type of Calvin we’re going to learn about now is called a molecular compound and molecular compounds are not a metal and nonmetal but instead they’re composed from two nonmetals something like co2 something like CEO and of course h2o itself now to name a molecular compound the names the rules are going to be very similar with a cup which is one main exception we’re going to name the element first element first by its entire name and then the second element has the IDE suffix and this time we do pay attention to the subscripts we’re going to use the Greek prefixes mono di tri tetra etc to determine what the subscript is however you will MIT model for the first nonmetal so something like co2 name the first element first which is carbon okay the second element is oxygen becoming oxide and there’s two of them which is how we come up with carbon dioxide okay something like n2o5 okay so there’s two nitrogens which is die and the first element gets the full name so dinitrogen and then five oxide or just pet oxide so actually naming a molecular compound is always easier than naming an ionic compound why because you don’t have to worry about charge now on to what we call the concept of the mole when you go to the supermarket and you purchase eggs they are always in either a half-dozen or he doesn’t use very hard and very seldom that you’re going to purchase just an individual egg it’s convenience its convenience for us to refer to eggs as a dozen and so for chemistry we have the equivalent of our does it and the mole is our chemist as dozen for a convenience when you go into lab you don’t measure just the weight of for example one atom you don’t deal with an individual atom it’s not practical so what the mold is all about is that it goes from the micro scale to a more macro to a macro scale and it’s more practical and basically our a dozen which is what we call the mole and the mole is abbreviated is simply mo l is equal to the following 6.022 times 10 to the 23rd units of that of whatever you are measuring so this is equivalent to saying one dozen is equal to 12 units of whatever you are measuring ok so again our dozen the chemistries does is what we call the mole and this mole is probably this numbers you have to memorize i’m pretty sure and six point 0 to 2 times 10 to the 23rd it’s what we call Avogadro’s

number ok I’ll go goggles number now the mole we’re going to find is essential unit and it can connect the following quantities the mo we can go and calculate individual units the mole we can also get what are the grams now when we go from mold two units we are going to multiply this way by Avogadro number 6 point 0 to 2 times 10 years we need their units for every one mole but when we go to the opposite direction we’re going to divide ok we’re going to divide by 6 point oh we’re going to divide by Avogadro number mol can also be used to go to mass and so when we go to mass we can actually multiply the mole by a rate conversion factor of gram for every one mole and when we go to the opposite direction we’re going to divide by 1 mole we’re going to divide by grams so we can see that mole is truly a central unit and the conversion factor of gram per mole that’s what we call the molar mess and this is equal to the grams that one mole of a substance ways now we’re going to come back to this quite heavily and we’re going to do a lot of us example problems utilizing the bowl now where do we get this molar mass from then well on the periodic table okay so on the periodic table atomic masses remember it’s the number below the element symbol are also called molar masses and so when you look at c-6 12.01 on the periodic table this bottom number here is the molar mass of carbon and so one mole of carbon weighs exactly 12.01 grams that’s how you write it out now that’s going to be a conversion factor so how many moles are in twenty point five point seven grams of sodium so you say 20 5.7 grams of sodium times something over something and that’s going to give us moles of sodium alright so remembering remembering our dimensional analysis now grams is going to go downstairs of sodium and moles of sodium is going to go upstairs so we just look up the value and that’s going to be 22.99 grams of sodium for every one mole of sodium giving you your answer in units of moles of sodium let’s now go ahead and utilize Avogadro’s number how many atoms are in one point two moles of carbon okay so one point two moles of carbon times something over something is going to give us our answer in units of atoms of carbon okay so to get cancelled moles of carbon goes downstairs and then atoms of carbon goes upstairs and so we know from our flow chart with the mole that this is involving Avogadro’s number so one will go downstairs and six point 0 to 2 times 10 to the 23rd atoms goes upstairs to give us our answer in units of atoms of carbon alright so whenever you see the word atoms or molecules that’s pretty much a good giveaway that you’re going to be using Avogadro’s number as your conversion factor so again always keep a mind into that especially if the word moles is mentioned to all right now that we’ve been introduced to what the mole can help us get we’re going to find something else that the mole can be used to calculate what’s called molar mass of not only an atom but also of a compound let me say that again the mold can be used to calculate the molar mass of a compound excuse me the molar mass of an element is really provided to us on the periodic table and basically the molar

mass is simply the net sum of all mo of all molar masses of each element and so the molar mass of carbon dioxide is simply equal to the molar mass of carbon plus the molar mass of oxygen plus the molar mass of oxygen because we have two oxygens so that’s going to be equal to 12.01 grams plus 16.00 grams plus 16.00 grams giving us a grand total of forty 4.01 grams and so one mole of co2 weighs exactly forty 4.01 grams having said that we can also use molar mass as a conversion factor for a compound so how many grams are in one point two moles of carbon dioxide one point two moles of co2 times something over something it’s going to give us our answer in units of grams of co2 ok so the mole is going to go downstairs and the grams is going to go upstairs and right there in the previous problem we calculated this conversion factor our molar mass of cl2 anytime you want to relate mass and molds it’s always molar mass to be a conversion factor and so it’s going to be one on the bottom is going to be 44.0 one on top giving us our answer in units of grams of cl2 now the next application of the moles what we call percentage composition and so percentage composition is given to us to be the following okay so percent composition of an element tells us basically the parts of the element divided by total parts x 100 it basically tells us your relative amount that a specific element makes up of the compound so this equation that we actually use is going to be then be the total mass of the element divided by the molar mass of the compound all of that times 100 so let’s go ahead and answer this question how many grams of carbon dioxide are contained in 60 5.1 grams of sealed to so this problem is strictly a wanting to know how much carbon is going to be contained in so much cl2 so let’s go ahead and first calculate the percentage composition of carbon in CL 2 so we’re going to say 12.01 grams because there’s only one carbon / the entire molar mass of co2 which is 40 4.01 grams times 100 and that’s going to give us 20 7.3 percent so carbon dioxide is only 20 7.3 percent carbon the majority is carp is oxygen now that tells me that any sample I hold of co2 in my hand 27.3 percent of that is carbon so we’re going to use percentage as our conversion factor and so we’re going to say point two seven three times the 60 5.1 grams of co2 is going to give me grams of carbon and we’re going to get 17.8 grams as our answer so once again anytime you see a problem that relates the amount of an element contained in a certain amount of compound you want to use percentage composition as your tool finally we’re now going to see how we can determine what’s known as an empirical and molecular formula an empirical formula is the simplest whole number ratio of a formula so for example co2 this is an exec one to two ratio of carbon to oxygen and that’s my smallest ratio I cannot go smaller than it however let’s go ahead and look at one that’s maybe c2h6 in this example I have a ratio of two to six of carbon to hydrogen but you notice that i can divide i can divide by a lowest common

denominator is common multiple which is two and i’m going to get 1 to 3 ratio instead therefore 123 is actually my smallest ratio and so that’s going to be ch3 which is my empirical formula so what we’re going to do is we’re going to determine how to calculate the empirical formula and the empirical formula is determined from elemental analysis basically you have an unknown sample and you put it through you perform what’s called elemental analysis tests on it and after you perform this analysis you’re going to get a printout of the percentage that a certain element make something your unknown so you’re going to get X percent of for example carbon you’re going to get y % for example oxygen you’re going to get z % for example hydrogen so these percentages again are only to be used for empirical formulas they may or may not necessarily give you the actual formula only relative amounts are reported now if you want to get the actual formula which is known as the molecular formula you can determine it but only if the molar mass is already provided for you let’s go ahead and look at a typical example okay so compound used as an additive for gasoline to help prevent engine knock has the following percent composition by mass 71.6 5% chlorine 24.2 seven percent carbon and 4.0 seven percent hydrogen the molar mass is known to be 98.9 six grams per mole determine both the empirical and molecular formulas all right now the very first step is to get your percentages into grams okay so get percentage to grams and after we get everything intagram then we can go to our central unit from the chapter which is moles again whenever you’re in doubt get the moles now the nice thing about percentages is that percentages always mean out of 100 so if we assume 100 gram of compound the percentages automatically become grams and so we have 71.6 five grams of chlorine 24.2 seven grams of carbon and 4.0 seven grams of hydrogen again assuming 100 gram of compound okay so now we’re going to go ahead and get the moles and this is something we know how to do by now so when we go ahead and get two moles we’re going to get two point zero two moles of chlorine we’re going to get two point zero two moles of carbon and we’re going to get 4.0 for moles of hydrogen okay step one is done okay now step two step 2 we’re going to take our molds i’m going to the divine we’re going to divide by the smallest most present okay so / smallest moles present everything and we’re going to get a nice whole number that is going to become our subscripts are named part in our empirical formula okay so the result is the empirical formula subscript so when we take chlorine and / the smallest number which is 2 point 0 2 we’re going to get one carbon / 2 point 0 to get one and four hydrogen for point 04 / 2.02 we’re going to get to giving us our empirical formula of ch2 co k now what happens if you don’t get a perfect whole number what if you get one point five comma 1 point 5 comma 2 well you know dalton’s law of multiple proportion tells us that we cannot combine elements in other than a small whole number ratio so basically any time you have a decimal number well we’re just going to then multiply everything by the same factor

to get everything in whole numbers and so when we do that we can multiply 1.5 x 2 and everything by 2 giving us nothing about whole numbers three 3 & 4 so if that happened then those numbers would become the empirical we get c3h for co 3 all right now to get the molecular formula you simply do the following you take the molar mass of the molecular formula which is given to you and you divided by the molar mass of your empirical formula and that is always going to give you a nice whole number and when we do this we’re going to get 98.9 6 divided by our molar mass of our empirical formula which is ch 2 CL and we actually get 48.4 68 and when we do this we get approximately 2 you take that whole number and you multiply the subscripts of the empirical formula by this and so I get our molecular formula of c2h4 CL 2 which is the formula of the actual compound you can always double check your work you can add up the molar mass of c2h4 cl2 and we’re going to get very close to ninety eight point nine six alright let’s go ahead and summarize okay atoms are composed of a central nucleus that is positively charged protons and neutrons reside within the nucleus and electrons are outside we saw that isotopes of the same element differ only by their number of neutrons and we’ve went through specific rules for naming ionic and molecular compounds we also saw that the mole is a central unit that allows for conversion between number of atoms and molecules and from s and finally the mole is a central unit that is required for several types of different problems including percentage composition and the empirical and molecular formula problems alright let’s go ahead and look at some sample problems you have one milligram of the theam metal reacting with molecular flowing gas and the resulting fluoride salt as a mass of three point seven three milligrams determine the empirical formula of lithium fluoride well you have 1.00 milligrams of lithium and after it reacts with the molecular flowing grass gas which is f2 you get three point seven three milligrams of your lithium fluoride well if i start with only one milligram of lithium and i wind up with 3.73 milligrams of compound why did my mass get heavier the mascot heavier because it reacted with fluorine fluorine came on board and so we can actually determine the mass of fluorine and it’s just a difference 3.73 mg minus 1 point 0 0 mg and that’s going to give us two point seven three mg which is the mass of fluorine reacted okay so we got all of our masses now now let’s get it raining into molds and so two point seven three milligrams of fluorine is going to become one point four four times ten to the negative 4 moles of fluorine 1.00 milligrams of lithium is going to become one point four four times ten to the negative 4 moles of lithium now that we have all of our moles we can divide by the smallest number present and for this problem they are identical and so when we divide everything we’re going to get Li 1 and F 1 for our empirical formula or just LIF and this edge sir actually makes sense because we know that lifts theam is a one plus charge and that flooring is a 1 minus charge and so indeed the charges due balance each other out and now moving on to simple problem too how many atoms of carbon are present in 2.67 kilograms of C 686 so in this case we are asked about an element within a certain compound well that sounds a lot like percentage composition so let’s go ahead and calculate the percentage composition of carbon in c6h6 ok so the percentage composition is equal to the total mass of carbon there’s six of them by the molar mass chill point 0 1 divided by the total mass with C 686 which is 6 by 12 point 0 1 plus the six

hydrogen’s by 1.00 eight and all of that times 100 so when we do this we get 92.3 percent so c6h6 is mostly carbon ninety-two percent carbon ok so remember what we did last time we’re going to take our percentage composition point 9 23 and we’re going to multiply it by the total mess two point six seven kilograms and that’s going to give us the 2.4 six kilograms of carbon in this specific sample now the question is asking for atoms of carbon dough so somehow we have to go from kilograms to atoms well remember anytime you see the word atoms or molecules it’s going to be via andava gajeel number so our first step is to go from kg just a regular g and then from g the central to the central unit which is moles and then of course moles on two atoms and we can go ahead and do this so we’re going to say 2.46 kilograms of carbon times something over something so g goes on top kg on the bottom that’s going to be 10 to the third g over 1kg and now on to moles so x something over something geez goes downstairs to get council moles goes upstairs to get carried through to the final answer and that’s molar mass and so you look up the molar mass of carbon which is 12.1 grams for every one mole and then finally times something over something giving us our answer in atoms of carbon so that’s Avogadro’s number so mole on the bottom and avocados on top which is Adams here that’s one mole on the bottom and 6.022 times 10 to the 23 atoms on top and you should get an answer of 1.2 three times ten to the twenty six atoms of carbon okay so I want to thank you for your attention and this concludes our lecture on atoms molecules and ions and thank you for using educator com

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