# Converting Between Grams and Moles

in this video we're going to learn how to solve problems like these where we have to convert back and forth between

### grams

and## moles

what is the mass in### grams

of 4.30## moles

of aluminum people from some other countries might pronounce this aluminium so in this problem we're talking about### grams

and## moles

so the first thing we want to ask is what is the molar mass of aluminium the molar mass is going to tell us how much one mole of aluminium weighs in### grams

so to figure this out we look upaluminium on the periodic table and we check out this number here okay this number tells us that the molar mass of aluminium is twenty six point nine eight

### grams

per mole which means that one mole of aluminium weighs 26 point nine eight### grams

now that we have this information I first want to show you how we can solve this problem but just kind of thinking through it okay we're looking for the mass in### grams

of four point three## moles

of aluminium basically what the four point three## moles

ofaluminium weigh this is basically a multiplication problem let me show you what I mean okay look at this information here okay we can say that if we had one mole of aluminum it would weigh twenty six point nine eight

### grams

okay how much would two## moles

of aluminium weigh what we take twenty six point nine eight which is how much one mole weighs and we'd multiply it by two okay and we get this answer how much would three## moles

way we take twenty six point nine eight which is how much one moleweighs and we'd multiply it by three and get this answer okay you get the idea right now in this problem we're talking about how much 4.30

## moles

way once again we know that this is how much one mole weighs we don't have just one mole we have 4.30## moles

so let's take this number and multiply it by four three zero that is going to give us 116 rounded to three significant figures I haven't worried about significant figures for the numbers up here okay so basically all I'mdoing here is just thinking through the problem keeping in mind that the molar mass tells me how much one mole weighs and all I have to do is just multiply how much one mole weighs by however many

## moles

I have to kind of sum up what I'm doing here I could write the simple rule that says that when we're going from## moles

to### grams

## moles

to### grams

what we want to do is multiply## moles

by the molar mass okay here's how we think through it now I want to show you how we can solve the sameproblem with conversion factors because a lot of times teachers and textbooks use conversion factors instead of this kind of like thinking through in method okay so here we're starting with 4.30

## moles

of aluminum we get that from the problem here now let's write some conversion factors I want to take this and I want to multiply it by a conversion factor that's going to get rid of## moles

and is going to convert## moles

into### grams

okay to write conversion factors we often want to look atsome sort of an equation or equivalence or relationship okay so I can take this information about the molar mass and I can write it sort of as an equation here as a relationship right which is basically saying that if we have one mole of aluminum well that means we have twenty six point nine eight

### grams

of aluminum check it out we have equal sign and something on either side this is like a perfect example of the sort of relationship that we can write a conversion factor with okay there are twoconversion factors we can write with this relationship okay the first is going to have this on top and this on the bottom okay one mole of aluminum over twenty six point nine eight

### grams

of aluminum okay or we can flip it we can do this on the top and this on the bottom okay so on the top twenty six point nine eight### grams

of aluminum and then on the bottom is going to be one mole of aluminum okay both of these conversion factors are totally cool they're totally fine but we only want to useone of them and the one that we're going to want to use is the one that gets rid of

## moles

of aluminum from up here it's on the top here so in order to cancel out## moles

of aluminum is going to be on the bottom of the conversion factor that we want to use okay it's on the top here but it's on the bottom here so this is the conversion factor we want to use okay so get rid of this one for now bring this over and now we have## moles

of aluminum on the top cancels out## moles

of aluminumon the bottom and we're left with

### grams

of aluminum which is exactly what we want okay so now the math that we do is this times this divided by one which is going to give us 116### grams

of aluminum rounded to three significant figures because there are three significant figures in here and we don't worry about the number of significant figures in one here and check this out the math that we're doing 4.30 times twenty six point nine eight is exactly the same math that we did whenwe're just thinking through it right because dividing something by one mole dividing something by one doesn't change its value so whether you do the thinking through it method or you do the conversion factor method you're doing the exact same math right you're just taking the

## moles

and you're multiplying them by the molar mass and this tells us how much 4.30## moles

of aluminum would weigh how many## moles

in 127 point five### grams

of sodium chloride NaCl as before we'retalking about

## moles

and### grams

so the first thing I want to ask is how much does one mole of sodium chloride weigh in### grams

this is just another way of saying what is the molar mass of sodium chloride to figure this out I look at the formula and I look at the elements that make it up okay I've got na and I've got CL here so I look them up on the periodic table sodium and chlorine and these numbers here are the molar masses of the two different elements now I want to add the up okay I lookat the chemical formula here and this tells me that I have one sodium atom so I'm going to do one times twenty two point nine nine which is the molar mass of sodium and then I have one CL so I'm going to do one times thirty five point four five which is the molar mass of chlorine add these together and I get fifty eight point four four which tells me that the molar mass of NaCl sodium chloride is fifty eight point four four

### grams

per mole or it means that one mole of sodium chlorideweighs fifty eight point four four

### grams

that is how much one mole of sodium chloride weighs okay so we have a certain amount of### grams

of sodium chloride and we want to know how many## moles

is that here's how we think through okay based on this information here about the molar mass we know that if we had 58 point four four### grams

that would be one mole okay but we don't have fifty eight point four four### grams

we have a hundred twenty seven point five### grams

okay this is more than just onemole the question that we want to ask is how many fifty eight point four fours are there in 127 point five in other words how many times does 58.4 four go into 120 seven point five this is a division problem okay we're going to do a hundred and twenty seven point five divided by the molar mass we're going to find out how many of these are in this and that's going to tell us how many

## moles

we have when we do the math rounding to four significant figures we get two point one eight two## moles

we can sum up what I did here by saying that when we're## converting

from### grams

to## moles

we take### grams

and we divide by the molar mass so that's what I'm doing here taking the### grams

and dividing by the molar mass to see how many molar masses fit in the number of### grams

that we have now for the conversion factor approach I'm starting here with one hundred twenty seven point five### grams

of sodium chloride and I want to multiply this by conversion factor that will get rid of### grams

sodium chloride and will take me to

## moles

of sodium chloride so to do that I'll take this information about the molar mass and I'll rewrite it as an equivalence or a relationship like this okay I got one thing on either side of the equal sign which makes it really easy to turn this into a conversion factor here are the two conversion factors that I can write with it okay one has## moles

on the top and this on the bottom the other has### grams

on the top and## moles

on the bottom the one that Iwant to use here is a conversion factor that is going to cancel out

### grams

sodium chloride which is on the top here so I want it to be on the bottom over here I'm going to use this one so now I've got gram sodium chloride on the top gram sodium chloride on the bottom they cancel out leaves me with## moles

of sodium chloride and the math that I'll do is this times this divided by this and get to point one eight two## moles

sodium chloride and check out that the final answer is rounded tofour significant figures because there are four here and there are four here and I don't worry about the one here because it's part of the definition now I just want to point out the math I'm doing here is exactly the same as the math in the think it through method I take one twenty seven point five and multiplying up by one doesn't change at all so I'm still just doing one twenty seven point five divided by fifty eight point four four which is exactly the math that I did up

here which just shows when we're going from