Here we're going to discuss chapter 6 of the book introductory chemistry. This chapter is named the chemical composition. By the end of this chapter you will be able to convert between moles and the number of atoms.
Also to convert between grams and moles. conversion between grams and the number of atoms or molecules. Also, you'll be able to convert between moles of a compound and mole of a constituent element. In the same way, we're going to be able to convert between grams of a compound and grams of a constituent element.
You will be able to use the mass percent composition as a conversion factor. determine the mass percent composition from a chemical formula, also you will be able to determine the an empirical formula from experimental data, and finally you would be able to calculate a molecular formula from an empirical formula and the molar mass. So sodium is an important dietary mineral that we eat in our food. primarily as sodium chloride also known as a table salt sodium is involved in the regulation of body fluids and eating too much of it can lead to high blood pressure so that's why it's really important to control the amount of how much sodium we eat because it is necessary to to our work to to our body to work properly but also could be a little bit detrimental we ate too much so the FDA has some recommendations about how much sodium you can eat and you basically can eat 2.4 grams or 2400 milligrams of sodium per day the mass of sodium that we eat is not the same as the mass of sodium chloride that we eat how many grams of sodium chloride we can consume and still stay below the FDA recommendation for sodium.
The chemical composition of sodium chloride is given by this formula NaCl. Na for sodium and Cl for chloride. So as we can see there there is one sodium ion to every chloride ion. So it's a one to one that's why you have here you don't put it but you assume that there is a one and also here is a one. Since the masses of sodium and chlorine are different the relationship between the mass of sodium and the mass of sodium chloride is Not clear from the chemical formula alone That's why we need to calculate the amount of a constituent element in a given amount of a compound Information in a chemical formula along with an atomic and formula masses can be used to calculate the amount of a constituent element in a compound.
So for example, how much iron is in a given amount of iron ore? Or how much chlorine is in a given amount of a chlorofluorocarbon? So these questions could be answered by having all this information, the chemical formula, the atomic and formula masses.
With those information, we can calculate and we can look for the answer for these questions. So, some hardware stores sell nails by pound, which is easier than selling them by nail. So when you go to any hardware store, you look for 3 pounds of nails. You don't ask for 50 nails. This problem is similar to asking how many atoms are in a given mass of an element.
So it's more easy to ask for 8.25 grams for a specific mass instead of asking for thousands of atoms or of specific element or a thousand of molecules for a compound so that's why we use grams instead of specific amount of items a customer buys 2.6 pounds of medium-sized nails and a dozen of these nails weight 0.150 pounds how many nails did the customer buy So, to try to answer this kind of questions, we need to create what we call a solution map. This is going to give us like a big picture of how we can go from one point to another, from A to B. Okay, for example here, this is the map. We start with pounds of nails because that is the information that we have. He wants to buy 2.6 pounds of this, of nails, so we start with that.
Then we need to get to the number of nails. But in the middle, we have... we need to look for something that help us to change to transform the units of pounds to the number of nails and this is a definition that we have also here in the question but it says that one dozen of these nails weigh 0.150 pounds so this is a relationship that we can create one dozen over 0.150 pounds or we can also rearrange this equation put the denominator in the denominator and the denominator in the denominator to use it also remember that for each definition we always produce two conversion factors and we need to use the one that cancel the unit that we need to cancel for example this one would need to cancel the the pounds of nails so we're gonna use this conversion factor instead of the one that had the dozen in the denominator and the pounds in the denominator okay so that's why we are using this one here so by using this we're going to have we're going to change pounds to dozens to number of nails okay and then the dozen in one dozen we have 12 nails so now we can determine the number of nails so these are the two conversion factor that we need to use so this is point a this is point b and this is the intermediate point so we need to transformation between this two and then another factor to change from dozens to number of nails from here to here. So these are the two conversion factor. So we now convert the pounds to the number of nails by following this.
We start with 2.6 pounds of nails times one dozen of nails is equal to 0.15 pounds. So we cancel the pounds with pounds and we have here the units of dozen. Then we cancel this unit by using the second coefficient factor that we need to use To goes from dozens to number of nails and this will be 12 nails in one dozen We cancel this Dozen and dozen units and we have here the nails unit the ones that we need how many nails they do the customer buy So we have now the nails and we then multiply two point sixteen sixty times one times twelve divided by point one fifty pounds and this will be equal to 208 nails.
So the conversion factor for the first part is the weight per dozen nails. Okay so this is the definition that was in the problem and as a definition remember that we can create two conversion factors. We can put this in the denominator and this one in the denominator or this one in the denominator and this one in the denominator creating two conversion factor And remember that we need to use the one that the unit that we need to cancel is in the denominator. That's the conversion factor that we need to use. The second conversion factor for the second part was the number of nails in 1,000.
So 1,000 of nails is equal to 12 nails. That's another definition. And also from here, we can obtain two conversion factors, but we're just going to use the one that helped us to cancel.
the unit that we are interested to cancel and produce the other unit eventually that will be at the end to help us to find our unit for answer the question. Now with atoms we must use their mass as a way to count them. Atoms are too small and too numerous to count them individually. Even if you could see atoms and count counted them 24 hours a day, as long as you live, you would barely begin to count the number of atoms in something as small as a grain of sand.
With nails, we use a dozen as a convenient number in our conversation. A dozen is too small to use with atoms. We need a larger number because atoms are so small.
We can't use the... you know the word one dozen of atoms but honestly that's too too small as compared with it with the males so in chemists our dozen is called the mole the definition of mole is 1 mol is equal to 6.022 times 10 power of the 23rd particles. So 1 mol is equal to this amount of particles. And those particles could be atoms, could be compounds.
1 mol of nails will be 6.022 times 10 to the 23rd power nails. 1 mol of cows is 6.022 times 10 power of the 23rd cows. So this is the definition as the same way as one dozen is equal to 12 particles. So one mole is equal to 6.022 times 10 to the power of 23rd. This number is known as the Avogadro's number.
One mole of anything is 6.022 times 10 to the power of 23rd units of that thing. This number is called Avogadro's number named due to Amadeo's Avogadro. that was the ones that define it and establish this number so one mole of marbles correspond to 6.022 times 10 to the power of 23rd marbles the same way one mole of sand grains corresponds to 6.022 times 10 to the power of 23rd sand grains so one mole of atoms ions or molecules generally makes up objects of reasonable sizes for example 22 real copper pennies contains about one mole of copper atoms so one mole of copper atoms can be represented by 22 pennies also two large helium balloons contain approximately one mole of helium atoms so that's the amount of helium that you need to have one mole of helium atoms The size of the mole is a measured quantity.
The numerical value of the mole is defined as being equal to the number of atoms in exactly 12 grams of pure carbon-12. This definition of the mole establishes a relationship between the mass, the grams of carbons, and the number of atoms, the Avogadro's number. So that's where comes all the definition of the Avogadro's number. This relationship allows us to count... atoms by waning them So let's convert moles to numbers of atoms convert 3.5 moles of helium to the number of helium atoms So here we have in this problem the data that the first information is that we have 3.45 moles of helium Okay, and we need to find the atoms of helium So we need to look for something that help us to go from moles to atoms.
So we create solution map we need to we need a coefficient factor to goes from the molts of helium to atoms and we have one definition that is the Avogadro's number that goes from moles to number of particles remember that Avogadro's number is a relationship between one mole and number of particles in this case those particles are the atoms okay so it's not like it's one mole and then six point zero twenty two times ten power of thirty third atoms all the time no one mole is equal to that number of particles and in this case the atoms are those particles so we have one mole of helium is equal to 6.022 times 10 power of 33rd atoms of helium so this is the conversion factor that we need and because we start with moles of helium we put the one mole in the denominator so we can then cancel that and have the atoms of helium and the denominator. So here we have the solution 3.5 moles that is in the problem. The definition that we, I mean the conversion factor that we can obtain from the definition of mole, we cancel the mole of helium and then we have the atoms of helium.
So 3.5 times the Avogadro's number divided by 1 is equal to 2.1 times 10 to the power of 24th number of atoms of helium. the same thing we can convert from atoms to moles this is basically the reverse way to use the Avogadro's number so what we have first is the atoms of silver so we need to go from atoms of silver to moles of silver and basically the same we need to use a solution map but now it goes from atoms from particles to moles and as we can see here the conversion factor that we need is the Avogadro's number, the definition of the Avogadro's number here. So one mole basically of silver is equal to 6.022 times 10 power to the 23rd atoms of silver.
So in this way, by using this, we can go from the atoms to the mole. So 1.1 times 10 to the power of 22 atoms of silver divided by the Avogadro's number that would give us the number of moles of silver. That is 1.8 times 10 to the power of minus 2, the moles of silver.
Now, these pictures have the same number of nails. Both of them we have one dozen of large nails, we have one dozen of small nails. The weight of one dozen nails change for different nails because that size of each nail.
This picture have the same number of atoms. The weight of one mole of atoms change for different elements. So we have here the same number of atoms as here but here the mass of this is 32.07 gram while the mass of this is 12.01 gram but the number of atoms is the same in both.
So the atomic mass unit is defined as 1 12th of the mass of a carbon 12 atom. The molar mass, as of any element, is the mass of one molar atom of the element is equal to the atomic mass of that element expressed in atomic mass unit. One copper atom has an atomic mass of 63.55 atomic mass unit.
one mole of copper atoms has the mass of 63.55 grams of copper the molar mass of copper is 63.55 grams of by mole okay so this basically is the molar mass of copper and this one is the atomic mass so when we use the atomic mass we're going to use that units of atomic mass unit and when we're looking for the molar mass We're going to use the 63.55 grams per mole, those units, because this one is going to give us the relationship between grams or mass and moles. So, the mass of one mole of atom of an element is a smaller mass. The mass of one mole of atoms change for different elements. For example, 32.07 grams of sulfur is equal to one mole of sulfur.
and the number of our gathered atoms of sulfur. 12.01 grams of carbon is equal to one mole of carbon and also the same name the same number of atoms and 6.94 grams of lithium is equal to one mole of lithium and also the same number of atoms. This is equal to if we can think about a dozen. A dozen of cows, a dozen of mice, and a dozen of whales. You have a dozen of each one, but each group is going to have a different weight because of the weight of the individual animal.
So this is the same thing. We have the same number of atoms for each of these three different elements. And that amount of atoms is going to give a different weight for each of the elements.
The lighter the atom, the less mass in one mole of that atom. Okay, so if you have a lighter atom that's going to have the less mass because it's the lightest. So that's why you can see that here.
Okay, the sulfur is a bigger one so it's going to have a bigger mass for the same number of atoms. So let's convert between grams and moles. And grams and moles...
conversion factor or a definition that we talked before that is the molar mass okay so that's the conversion factor basic that we need to use when we wants to go from mass to moles or grams okay remember that units of mass the most common is grams so calculate the number of moles of carbon in point 58 grams of diamond so what we have is that first information point 58 grams of diamond We need to look for the moles of carbon, so we can create this solution map that goes from grams of carbon to moles of carbon. And here we can see that our conversion factor now. is the molar mass of carbon the ones that can relate mass and mole because we know that one mole of carbon is equal to 12.01 grams of carbon this comes from the periodic table okay remember that the atomic mass can be transformed to the molar mass by using the Avogadro's number so that's the relationship that we're going to use in this case and that will be our conversion factor so We start with 0.58 grams of carbon, and we need to put the grams in the denominator as we can see here. So there's going to be basically these two factors that we're going to use, 0.58 of both data, divided by 12.01 grams of carbon, and we have the moles of carbon in the denominator.
That's why there's the unit for our result. And here we have how many moles we have in 0.58 grams of carbon. That is 4.8 times 10 to the power of minus 10. to moles of carbon.
So what about if we want to know the number of atoms that we have in.58 so we have this and this one we have a mole so what we need to use to change from moles to number of atoms of carbon to number of particles. We need to use then the Avogadro's number. Perfect.
So and now we can create another map here okay solution map we need to start from the mass of carbon goes through the moles of carbon and then finally to the number of carbon because we don't we don't have right now a conversion factor that can goes directly from mass of carbon to number of atoms we need to go first through the moles of carbon so to do this from map from the mass of carbon to the moles we use the molar mass And to transform from the moles to the number of atoms or number of particles, we need to use the Avogadro's number. So now we add this part to the one solution before that we have in the previous slide. OK, so now we have the Avogadro's number here.
We now can cancel the moles of carbon. And now we have the atoms of carbon here. So 0.58 grams of diamond is equal to 2.9 times. 10 to the 22nd atoms of carbons.
So converting now between grams and the number of atoms. How many aluminum atoms are in an aluminum can with a mass of 16.2 grams? So we have mass here, and these two go to atoms.
But we don't have a specific... conversion factor that goes from mass to atoms but we can go from mass to moles and then from moles to atoms okay that way we can create our solution map here that goes from the mass of aluminum to the moles of aluminum and from moles of aluminum to the number of particles or number of atoms in this case of aluminum so to go from mass to moles we can use the molar mass to go from moles to number of atoms we need to use our definition of the Avogadro's number. So we need to look for the molar mass of aluminum because we are looking for the atoms of aluminum and the initial data is 16.2 grams of aluminum.
So we use our molar mass of aluminum and we can then solve a problem by using 16.2 grams of aluminum divided by the molar mass because remember that we need to put the mass in the denominator so that way we can cancel the mass of aluminum of the sample okay and now we have here moles of aluminum so we need to cancel this so that's why we need to use in this way the Avogadro's number we need to put the moles in the bottom because we can then cancel this to this if we use the other way we put the moles on the top and then the the atoms of aluminum our result will be mole squared divided by atoms So that's not what we're looking for. So that's why we need to invert and have it like this. So that way we can cancel the unit of moles and we will have here the unit of aluminum atoms. So that's why 16.2 grams of aluminum at 3.62 times 10 to the power of 23 atoms of aluminum. counting now the molecules by the gram.
For elements, the molar mass is the mass of one mole of atoms of that element. For compounds, the molar mass is the mass of one mole of molecules formula units if we are talking about ionic compounds of that compound ionic compounds do not contain individual molecules we convert between the mass of a compound and moles of the compound and then we calculate the number of molecules or formula units for from the moles so the molar mass of a compound in grams per mole is numerically equal to the formula mass of the compound in atomic mass units or ionic compounds. The formula mass for a compound is the sum of the atomic masses of all of the atoms in a chemical formula. So let's now convert between grams and moles of a compound. So let's calculate the mass in grams of 1.75 moles of water.
So we need to go from moles to mass. The only conversion factor that we have is the molar mass. So we need to look for the molar mass of water.
Because we need to go from moles to grams. And the only conversion factor that has both units is the molar mass. So we have 1.75 moles of water who wants to determine the mass of water. So we need to goes from mole to mass and what the conversion factor is going to help us to determine this one will be the molar mass Okay, that is 18.02 grams of water in one mole of water.
How can we found this 18.02? grams of water Well, we need to look for each of the atomic mass of the atoms that compose the compound. And for example, the molar mass for hydrogen is 1. We have in the molecular formula 2 hydrogen, so we multiply 2 by the atomic mass of hydrogen. And for oxygen, we just have 1 oxygen here, so that's why we multiply 1 times the atomic mass of oxygen. This is 1 times 16. So, this is going to be equal to 18.02.
grams per mole. This is the molar mass okay for water. Now we can use this to convert from 1.75 moles of water to the mass of water.
We have the conversion factor here. We need to use this one instead of the one that has the mole in the denominator and the mass in the denominator. So here we can cancel the mole of water and we have in the denominator the units of mass, the unit of gram of water, that is what we're looking for. So when we multiply this, we're going to have a total of 31.5 grams of water. So converting between number of molecules and mass of a compound.
Remember, now because we're talking about a compound, we're talking about molecules. Okay, so in the other example, we're talking about elements so we can calculate number of atoms. Eventually then we can calculate also the number of atoms if we want to determine from a molecule.
But for now we want to calculate the number of molecules when we have a mass or the data of the mass of a compound. So here is what is the mass of 4.72 times 10 to the power of 24 molecules of nitrogen oxide. So the data that we have is for the the number of molecules is this one for NO2 we want to find the mass of NO2 that represent this amount of molecules so we need to go from molecules to moles remember that we don't have a specific conversion factor that goes from molecules to mass the only thing that we know that has something related with mass is the molar mass that is mole and mass so molecules is particles In this example, the molecules is the number of particles.
So when we talk about particles, that has to make a click in our mind. It's the number of particles, Avogadro's number. So we need to go from number of molecules to the mole by using the Avogadro's number. And then from moles to mass using the second coefficient factor that is the molar mass. Okay, so we want to go from particles to moles.
Avogadro's number from mole to mass is going to be the molar mass. The same way is the opposite. If we want to go from mass to moles, we need to use the same. kind of conversion factor that we can obtain from the molar mass and from moles to number of particles we need to use a conversion factor that can come from the avogadro's number so here we need to use determine the molar mass of no2 and by doing that we need to calculate it in this case is 46.01 grams of nitrogen oxide divided by one mole of no2 So how can we do that?
Here we have our first relationship, okay, the ones that we know before from Avogadro's number. And this one is to obtain the molar mass of NO2. We have one nitrogen and two oxygen from our chemical formula here, NO2. So we need to look for the atomic mass of nitrogen and multiply by one the atomic mass of oxygen and multiply by two.
So the atomic mass of nitrogen is 40.1 times one is this one. while the atomic mass of oxygen is basically 16.0 times 2. So that means that our molar mass of NO2 is 46.01 gram times mole. So now we need to go, going back to the problem, we need to go from molecules to mass. So first we change our molecules to moles by using the Avogadro's number, from which a factor, we can cancel here the... molecules and then this moles can be cancelled by the units of moles that we can find in our conversion factor from the molar mass so we can cancel here the moles of NO2 and we have the mass of NO2 that is 365 grams of NO2 in 4.78 times 10 to the power 24 NO2 molecules So chemical formulas as a conversion factor, we can also use those chemical formulas to go from number of molecules to number of atoms, as I mentioned before.
So here, for example, we have a three leaf clover analogy. How many leaves of unforting clovers? So if each clover have three leaves, that means that we need to go from number of clovers. We're going to use the same. solution map, number of clovers to number of leaves.
We have the conversion factor that we can find by just looking, ok, the clover. So we then can use 14 clovers divided by 1 clover multiplied by 3 leaves. This is a conversion factor from here.
And now we can have 42 leaves, ok, we can have 42 leaves from 14 clovers. So the same way we can use our chemical formulas. The formula for carbon dioxide means that there are two oxygens of atoms per one CO2 molecule. So we have, in other words, two atoms of oxygen per each molecule of CO2. So this is a conversion factor.
We can use this as the denominator and this as the denominator or vice versa. It all depends on what is the question and what we're looking for. The same way, 12 dozens of oxygen atoms are in one dozen of molecules of CO2, because it's basically twice the number of CO2 of oxygen.
So the same thing happens if we are talking about moles. Two moles of oxygen are in one mole of CO2. So the conversion factor comes directly from chemical formulas.
eight legs in one spider okay for each spider we're gonna find always eight legs in a chair we have four legs and in one molecule of hydrogen we have two atoms of hydrogen so all of these are conversion factor using basically these types of example so we can understand a little bit more that conversion factor from the molecules so converting between moles of a compound and moles of a constituent element. Find the number of moles of oxygen in 1.7 moles of calcium carbonate. So we start with this value. This is what we have in the problem.
We have 1.7 moles of calcium carbonate, and we need to determine the number of moles of oxygen. So we need to look for a conversion factor that can relate. the number of moles of oxygen and the number of moles of carbon calcium carbonate. And that will be from our basically chemical formula. In our chemical formula we can see here that we have three moles of oxygen per each mole of calcium carbonate.
So the solution map will be moles of calcium carbonate to moles of oxygen. We have three moles of oxygen by each mole of calcium carbonate. This is our conversion factor. Okay, so we have 1.7 mole of calcium carbonate.
times this will give us the moles of oxygen and here we have 5.1 moles of oxygen so in 1.7 moles of calcium carbonate we have 5.1 moles of oxygen so here we are using the chemical formula the information from the chemical formula to create a conversion factor if we want to transform from moles of the compound to moles of a constituent element or vice versa if we want to start with moles of a constituent element to find how many moles of a compound can be produced it's the same thing we need to use a conversion factor that we can create or obtain from the chemical formula. So let's do another example here we need to convert from grams to a compound to a grams of a constituent element. So let's find the mass of sodium in 50 grams of sodium chloride. So we start with the mass of sodium chloride here.
Okay, I want to find out the mass of sodium itself. So that means that maybe we need to go from the mass of sodium chloride to the number of moles of sodium chloride. then from that point we can determine the number of moles of sodium and finally we can get to the mass of sodium. So here we have basically our solution map.
We start with the mass of the compound because that's what we have in the problem, 50 grams, the mass of sodium chloride. So from the mass we can go to the moles using the molar mass of the compound sodium chloride that is 58.44. grams over one mole and we use this rearrange of that molar mass so we can cancel the units of mass of sodium chloride.
Now we have the moles of sodium chloride so we can go from moles to sodium chloride to moles of sodium by using the conversion factor that we can obtain from the chemical formula because we can see here that in one mole of sodium chloride How many moles of sodium we have in the chemical formula? Just one. So that's another conversion factor that we can use to go from the moles of sodium chloride to the moles of sodium.
And finally, we can go from the moles of sodium to the mass of sodium using the molar mass for sodium. Okay, that is 22.9 grams of sodium over one mole. We put here the moles of the denominator so we can cancel the moles of sodium. So in other words... The solution will be 50 grams, okay, of sodium chloride, and we can substitute all this value here in this equation, multiply 15 by 22.9, and then divide it by 58.44, and that will be 5.9 grams of sodium, okay?
So because 1 times 1 times 21, this will be 22.9, and then the denominator of 58.4 times 1 times 1 will be 58.4, so we can multiply 15 times 22.9, divide it. by 58.44 and that will be the mass of sodium that we can obtain from 15 grams of sodium chloride so for each 15 grams that we ate eat every day of sodium chloride from those 50 15 grams 5.9 grams are from sodium so as we saw before there is a mole relationship from the chemical formula here we have one mole of CCL4 tetrachloride of carbon tetrachloride carbon okay and here we have four moles okay that we can obtain from the one mole of CCL4 so the relationship inherent in a chemical formula allows us to convert between moles of the compound and moles of a constituent element and vice versa so that's how we can use the chemical formula as a conversion factor As we can see here, one mole of CCL4 can produce four moles of CCL, of Cl chloride. And what about if we... We need to look for the relationship using the CCl4 of moles of carbon. So if we need to determine the moles of carbon, the conversion factor from here would be 1 mole of CCl4.
We'll produce 1 mole of carbon because here in the chemical formula, we see that for each CCl4, we have just 1 carbon. And for each CCl4, we have 4 moles of carbon. chlorine now talking about compounds that have chlorine synthetic compounds known as chlorofluorocarbons CFCs CFCs are destroying a vital compound called ozone in Earth upper atmosphere CFCs are chemically inert molecules used primarily as refrigerants and industrial solvents in the upper atmosphere Sunlight break bonds within CFCs resulting in the release of chlorine atoms.
And there's the situation. Chlorine atoms react with ozone and destroyed it by converting from ozone to O2. Ozone is O3 and O2 is oxygen. The finding of ozone over populated areas is dangerous because ultraviolet light can harm living things and induce skin cancer in humans. because the ozone protects us against the ultraviolet light from the sun.
Most developed nations banned the production of CFCs on January 1st in 1996. The CFCs still lurk in older refrigerators and air conditioning units and can leak into the atmosphere and destroy the ozone. Now, upper atmospheric ozone is important. as I mentioned before because it acts as a shield to protect life and on earth from harmful ultraviolet light so Ozone is basically around all around the earth Okay, those molecules of all three and the UV is absorbed by those molecules and avoid To impact directly to the earth and also affect us Now the Antarctic ozone hole area from 1980 to 2012 is right here in this graph. This is basically the years and this is the size of the hole of the hole in the ozone close to the arc the darkest blue colors indicate the lowest ozone level so that means that this area is less protected than the rest of the world now in in the previous slide we mentioned that in 1996 a lot of nation banned the production of cfcs and we can see here that in from 1995 on there is like like a flat up there because um there is no production of CFCs. So that's why it's important to understand and learn about the different compounds with chlorine and the impact in our lives.
Let's talk now about the mass percent composition of compounds. This, also known as the mass percent of an element, is the element's percentage of the total mass of the compound. So for example, we have the mass percent of element X. is the mass of x in a sample of the compound divided by the mass of the sample of the compound times 100. So basically the mass percent of the element x is the mass of x divided by the total mass of the compound. So 0.358 grams sample of chromium reacts with oxygen to form 0.523 grams of metal oxide.
The mass percent of chromium will be when all react together and produce this will be point will be basically the mass of chromium divided by the mass of the metal oxide so it will be 0.358 grams divided by 0.523 grams this is the mass of the chromite as we can see here chromium the mass of the chromium and this is the total mass of the metal oxide times 100 this will be equal to 68.5 percent of chromium in that oxide. So that's how you can determine the mass percent of an element. So using mass percent composition as a conversion factor, we can use mass percent composition as a conversion factor between grams of an element and the grams of the compound.
So if we have for example, a sample of sodium chloride, 39% of sodium That means that we have 39 grams of sodium in 100 grams of sodium chloride. Because the composition of sodium in that sample of sodium chloride is 39% so that means that 39 of the 100 part are from sodium while the 61% is from chloride so By using the sodium Relation we can find out a conversion factor Okay, about 39 grams of sodium for every for each 100 gram of sodium chloride and that will be the conversion factor obtained from the mass percent composition so here we have the mass percent composition of chloride 39 so we can have two conversion factor obtained from that percent because we can say that we have 39 grams of sodium in 100 grams of southern chloride or 100 grams of following chloride over 39 grams of sodium so these fractions are conversion factor factors between the grams of sodium and the grams of sodium chloride in a sample of 39% sodium in sodium chloride so as we mentioned at the beginning of this video the FDA recommends to adults to consumes 2.4 grams of sodium per day so how many grams of sodium chloride can you consume and still be within the FDA guidelines So sodium chloride is 39% of sodium by mass, as we mentioned before. So if we have, for example, if we need, for example, 2.4 grams of sodium, okay, and we have this conversion factor that says that for each 100 grams of sodium chloride, we have 39 grams of sodium because the sample is 39% sodium in sodium chloride, and we want to determine the grams or mass of sodium chloride. We can use this conversion factor, okay? And we know that we need to go from the mass of sodium to the mass of sodium chloride by using the conversion factor of the mass percent composition.
So we can multiply 2.4 times 100 and divide it by 39. And this will give us the mass of sodium chloride that we need to eat every day. okay to obtain the amount recommended of Sodium per day by FDA So if we wants to eat 2.4 grams of sodium we need to really eat 6.2 grams of sodium chloride Okay in our diet in a way that we can consume the levels that FDA recommended for adults So based on the chemical formula the mass percent of element chlorine in compound CCl2F2 is as follow the mass percent of Cl will be 2 times the molar mass of chlorine because we have 2 moles of chlorine there divided by the molar mass of the compound CCl2F2 times 100 okay so because we have 2 moles of that element then we need to multiply twice the molar mass of that element And that will give us the mass of that element in this molecule, in this compound. And we then divide it by the molar mass of the whole compound times 100. And that will give us the mass percent for chlorine.
The same thing will be with fluorine because we have two. If we need to calculate the mass percent for fluorine, it will be two times the molar mass of fluorine divided by the molar mass of CCl2. And if this will be...
the mass percent of carbon will be just one time the molar mass of carbon divided by the molar mass of the whole um compound so let's do an example calculate the mass percent of chlorine in c in c2cl4 f2 that is free on 114. so we have the compound we need to calculate the percent mass of chlorine so we have how many chlorines per molecule of C2Cl4F2 we have 4 so that means that we need to multiply the molar mass of chlorines times 4 and then divided by the molar mass of C2Cl4F2 so as we have here the percent of chlorine will be 4 times the molar mass of chlorine divided by the molar mass of the compound times 100 that will be the equation that we need to solve. So in other words, basically we have the mass percent of an element x as we mentioned before It will be the mass of element X in one mole, divided by one mole of the compound. So here we have in the situation of 3114, the molar mass of chlorine is 35.45 times 4. It will be 141.8.
This is the molar mass for chlorine and the amount of mass for chlorine in the compound. And the molar mass of the compound will be 2 times the atomic molar mass of carbon, 4 times the molar mass of chlorine, and 2 times the molar mass of fluorine. So this is the amount of mass due to fluorine, this one due to chlorine, and this is due to carbon. That all of them will give us the molar mass of the compound.
That is 203.8 grams. Okay, so this is the molar mass for the compound, and this is the mass, basically, that we have from chlorine in this compound. So this will be 141.8, that is this, divided by 203, that is the molar mass for the compound, times 100. So we have a total mass percent for chlorine of 69.58%.
so that's the percent of fluorine I mean of chlorine and this compound if we would like to determine the percent of fluorine it will be the molar mass of fluorine times two and then that move that product you put that product here divided by this amount because it's the same compound so you need to use the same molar mass of the compound times 100 and that will give you the molar mass for I mean the mass percent of the floor so now that's we're talking about fluoride fluoride fluoride strengthens to an animal which prevents tooth decay too much fluoride can cause teeth to become brown and spot a condition known as dental fluorosis extremely high levels can lead to skeletal ross's the scientific consensus is that like many minerals, fluoride shows some health benefits at certain levels, about 1 to 4 milligrams a day for adults, but can have detrimental effects at higher levels. So basically this is what is recommended to have every single day. Adults who drink between 1 and 2 liters of water per day would receive the beneficial amounts of fluoride from the water.
Okay, so one of the principal sources of fluoride is the water fluoride is often added to water as sodium fluoride now what is the mass percent composition of fluorine in sodium fluoride how many grams of sodium fluoride should be added to 1500 liters of water to fluoridate it at levels of 1 milligram of fluoride per liter okay so this question must be answered we get a work with them by using all this convection factor we can do that by using conversion factors so let's talk now about the empirical formulas from mass percent composition here we have chemical formula okay and here we have the mass percent composition and empirical formula give us give us only the smallest whole number ratio of each type of atom in a compound Not the specific number of each type of atom in a molecule So this give it just a ratio of a compound. Okay by by looking them at the element level The molecular formula is always a whole number multiple of the empirical formula for example The molecular formula for hydrogen peroxide is h2o2 and it's empirical formula is H. Oh, okay. So here we have the molecular formula and this is the empirical formula from this molecular formula and sometimes we have a situation where the molecular formula is the same as the empirical formula as for example in water H2O. H2O is the molecular formula but also is the empirical formula.
So the molecular formula is equal to the empirical formula times n where n is 1, 2, 3, etc. and is equal to 2 for hydrogen peroxide because here we see that hydrogen peroxide is twice okay empirical formula of HO so that's why the N is gonna be equal to 2 for hydrogen peroxide now we can calculate empirical formula from experimental data we're going to use the example of water to give us to tell you how how can you do that with the compose a sample of water in the lab and find that it produced three grams of hydrogen and 24 grams of oxygen. By using this data how can we determine the empirical formula? Okay so we created the composition of water here and we observed that we have 30 I mean 3 grams of hydrogen and 24 grams of oxygen.
So how many moles of each element are formed during the decomposition of water? The first step to determine The empirical formula, if we have as initial data a mass, we need to transform that mass, okay, that data with unit of mass to moles, okay? So if we have 3 grams of hydrogen, how can we change from grams to moles?
Using what? We can use the molar mass. The units of molar mass, remember that is grams per mole. So we can go from the... 3 grams of hydrogen and divide it here by the molar mass of hydrogen.
1.01 grams of hydrogen is equal to 1 mole. And by this, we have the moles of hydrogen. The same thing we can do with oxygen.
We can multiply the mass of oxygen times 1 mole and then divide it by 16.0 grams of oxygen. This is the molar mass of oxygen. This is the molar mass of hydrogen.
And it's flipped because we need to that have those units of gram in the denominator so that we can cancel the units of gram here from hydrogen and also from oxygen so here we have that we can produce three moles of hydrogen okay and we can produce 1.5 moles of oxygens so that means that in this sample the ratio basically goes from three moles of hydrogen for every 1.5 moles of oxygen so we can create and we can write But it's known as a pseudo formula for water. It will be H3O1.5. Okay. But as we have been said before, we need to have basically whole numbers subscript.
Okay. We can't have decimals. So, but if we can look here, 1.5, we multiply by 2. Okay. But basically, sorry, we need to divide this by the smallest subscript.
Okay. So, we divide this by the smallest subscript. And we have 1.5 divided by 1.5, 3 divided by 1.5.
And from here, from this division, we have the subscript for each element. And we can find out that the molecular formula is H2O. So our empirical formula for water, which in this case also happens to be the molecular formula, is H2O.
So that's the way that how we can determine or we can calculate the empirical formula by using data. of the mass of each of the elements that create that compound. We have before, as we mentioned here, the decomposition of water producing 3 grams of hydrogen and 24 grams of oxygen.
So by using this, we can calculate, we can find out the empirical formula. Once again, we have the mass of the element. We divide that mass by the molar mass of each element, and that will give us the moles.
So we can write a pseudo-formula, okay, as we wrote here, H3O1.5, but we need to have whole numbers. So to do that, we divide this 2 by the smallest one. In this case, it's 1.5. So 1.5 divided by 1.5 is 1. 3 divided by 1.5 is 2. So H2O is basically, in our case, the empirical formula as well as the molecular formula. So, these are some rules or some steps to determine the empirical formula from experimental data.
First, write down as given the masses of each element present in a sample of the compound. If you are given a mass percent composition, assume a 100 gram sample and calculate the masses of each element from the given percentage. Then convert each of the masses, this is step one, to moles.
as I mentioned before, by using the appropriate molar mass for each element as a conversion factor. So that way, when you use the molar mass, and multiply by the mass you will have the moles. Then you can write a pseudo formula for the compound using the moles of each element as a subscript.
And then you need to divide all the subscript in the formula by the smallest subscript. If the subscripts are not whole numbers multiply all by the subscript by a small whole number. As I'm going to show in the next slide table. to arrive at a whole number subscript.
So, for example, there's a fraction of subscript that you're going to have is 0.1, you multiply by 10, this is 0.2, you multiply by 5, 0.25, multiply by 4, 0.33, multiply by 3. So, if your subscript is any of this, this is the factor that you need to multiply to have eventually a whole number. Follow an example. A 3.24 gram sample of titanium reacts with oxygen to form 5.40 grams of the metal oxide. What is the empirical formula of the metal oxide?
So we need to know the empirical formula for the metal oxide that is produced by titanium with oxygen. So what we have initial is the mass of titanium, the sample of titanium, and the mass of... the metal oxide so we need to find out the empirical formula you cannot convert mass of metal oxide into moles because you don't know the chemical formula for this metal oxide yet so and that's what you need eventually that's what you need to to find out okay so what we can do first of all you're given the mass of the initial titanium and the mass of the oxide But we need to determine the mass of oxygen. How can you think that we can determine the mass of oxygen if this is what we have? So remember the law of conservation of mass?
That can be destroyed and created. So we have here that the total mass of the reaction of titanium with oxygen is 5.40. So if we subtract 3.24 from 5.40... 5.40...
0.40 what we're going to have is the mass of oxygen because this metal oxide is the combination of the mass of oxygen plus the titanium so if we subtract the mass of the titanium we're going to have the mass of the oxygen so the difference of this will be the mass of oxygen that combined with the titanium will produce the metal oxide so to find the mass of oxygen we need to subtract the mass of titanium from the mass of the metal oxide that will be 5.40 minus 3.24 and this is the mass of oxygen. So now that we have the mass of oxygen what we need to do eventually we need to convert that mass in moles and then divide it by the smallest number okay once we created the pseudo empirical formula okay we divide it by the smallest and then we can have the empirical formula. So now we can calculate molecular formulas for compounds from empirical formula and for molar masses.
So once we have the empirical formula, we can determine the molecular formula for that compound. But we need to have the molar mass to do that. So the molecular formula is always a whole number multiplied.
of the empirical formula we need to find the end in the expression okay remember that the molecular formula is equal to the empirical formulas time n so we need this value of n we can find n in the expression of molar mass equal to the empirical formula molar mass times n and will be equal to the molar mass divided by the empirical formula molar mass Okay, so if we have the empirical formula, okay, we can find out the molar mass of that empirical formula, and somehow in the problem, they must provide information about the molar mass. So we can divide the molar mass of the compound divided by the molar mass of the empirical formula, and that will give us the n. So we can multiply to the empirical formula and eventually have there the molecular formula for that compound. calculating the molecular formula for compounds in this we're going to do the example of fructose So find out the molecular formula for fructose with the empirical formula CH2O.
And the molar mass is 180.2 grams per mole. So the molecular formula is a whole number multiple of the CH2O. This is the empirical formula.
We need to calculate the n value. So we can multiply that n value for each of the subscript of this empirical formula to create the... molecular formula for fructose so for fructose the empirical formula mass will be 1 times 12.01 plus 2 times 1.01 plus 16 because we have one carbon two hydrogen and one oxygen as we can see here see one carbon two hydrogen one oxygen so we need to multiply one by the molar mass of carbon Two, molar mass of hydrogen.
One, molar mass of oxygen. So that's what we have here. One, molar mass of carbon.
Two, molar mass of hydrogen. One, the molar mass of oxygen. And this will be equal to 30.03.
This is the molar mass for the empirical formula. And they gave us the molar mass for the compound. So now we can divide the molar mass for fructose. Divided by the molar mass for the empirical formula and this will be equals to 6 so the n is equal to 6 So that means that we need to multiply each of the subscript of the empirical formula for fructose times 6 So the critical is CH2O will be 6 times 1 C 6 C times 2 H 12 6 times 1 O 6 this now is the molecular formula of our So from the empirical formula, we can calculate or evaluate or find out the molecular formula of a compound. So calculating molecular formulas for a compound, use the molar mass, which is given, and the empirical formula molar mass, which you can calculate by the empirical formula.
To determine the n, that is the integer by which you must multiply the empirical formula to get the molecular formula. Then multiply the subscript in the empirical formula by n to arrive to the molecular formula. So this is how you can find out the molecular formula by using the empirical formula. So let's review our chapter 6. The mole constant of the mole is a specific number. 6.02 times 10 to the power of 23rd particles, okay?
That allows us to easily count atoms or molecules by weighing them. One mole of any element has a mass equivalent to its atomic mass in grams. One mole of any compound has a mass equivalent to its formula mass in grams.
And the mass of one mole of an element or compound is its molar mass. About the chemical formulas and chemical composition, the chemical formulas indicate the relative number of each kind of element in a compound. These numbers are based on atoms or moles.
By using the molar masses, the information in the chemical formula can be used to determine the relative masses of each kind of element in a compound. And the total mass of a sample of a compound. can be related to the masses of the constituent elements contained in the compound. And finally the empirical and molecular formulas from lab data. We can refer to the relative masses of each kind of element within a compound to determine the empirical formula of the compound.
If the chemist also knows the molar mass of the compound, he or she can also determine its molecular formula. And this will be all with Chapter 6, Chemical Composition from the Introductory Chemistry by Dr. Trow.