0:02 in this video we're continuing to learn
0:05 about quantitative chemical analysis
0:07 here we're going to describe the
0:09 fundamental aspects of gravimetric
0:10 analysis and we're going to perform
0:13 stoichiometric calculations using
0:16 typical gravimetric data so graphometric
0:18 analysis is the type of analysis where
0:21 we have a sample that is our analyte we
0:23 subject it so some sort of treatment
0:27 that causes a change or reaction and we
0:28 can then separate it from other
0:31 components of the sample if we make Mass
0:33 measurements of this
0:36 separated component then we can find out
0:38 information about the original sample
0:41 right how much of it we have or what its
0:44 concentration would be so as one example
0:47 we could precipitate something from a
0:49 solution get the mass of that
0:51 precipitate and then be able to use
0:53 stoichiometric calculations to figure
0:55 out how many moles of a given component
0:58 were precipitated
1:00 and then we can use that information to
1:02 figure out for example the concentration
1:04 of the original unknown so in this
1:07 example we have a solid mixture that
1:10 contains magnesium sulfate that solid
1:12 mixture is then dissolved in water and
1:15 treated with an excess of barium nitrate
1:18 and this causes a precipitation of
1:20 barium sulfate so we can see that our
1:23 original component that's of interest is
1:26 magnesium sulfate when it reacts with
1:29 the barium nitrate the barium will
1:32 precipitate out with the sulfate of our
1:34 original component
1:36 we have our balanced chemical equation
1:38 where we see the reaction of the
1:39 magnesium sulfate with the barium
1:42 nitrate to form a our precipitate which
1:45 we can see here is the solid and our
1:48 spectator ions magnesium and nitrate so
1:50 the question is what is the
1:53 concentration or mass percent of the
1:56 magnesium sulfate that is in the mixture
1:58 so in gravimetric analysis we're going
2:01 to use the mass of the product from our
2:04 reaction right this is the mass of the
2:06 barium sulfate precipitate that was
2:08 formed from the reaction so that's where
2:10 we're going to start from that we'll be
2:12 able to use the molar mass to calculate
2:15 the moles of barium sulfate
2:17 if we know the male the moles of barium
2:20 sulfate we can use the stoichiometric
2:21 factor from our balanced chemical
2:24 equation to calculate the moles of
2:27 magnesium sulfates that were reacted
2:29 from that we can use the molar mass to
2:31 calculate the mass of magnesium sulfate
2:34 that we started with and then finally
2:37 using the sample Mass right the samp the
2:39 mass of the solid mixture we can
2:41 calculate what percentage of it was
2:44 magnesium sulfate so our flow diagram
2:46 tells us to start with the mass of the
2:50 barium sulfate so this was
3:00 we have a massive barium sulfate but we
3:02 would want to know the number of moles
3:05 so that we can use our stoichiometric
3:07 Factor so we're going to use the molar
3:09 mass of barium sulfate to calculate
3:13 moles of barium sulfate and this is 233
3:15 0.43 grams
3:17 for every one mole
3:25 taking this product grams will cancel
3:28 and we'll have moles of barium sulfate
3:30 so now we can use our stoichiometric
3:32 factor from our chemical equation we
3:36 should see that for every one mole of
3:40 barium sulfate that is produced
3:53 all right so now we know how many moles
3:57 of magnesium sulfate we reacted we want
4:00 to find the mass percent of magnesium
4:02 sulfate so we'll want to know what Mass
4:04 this corresponds to
4:06 so we'll use the molar mass of magnesium
4:10 sulfate this is one
4:14 120.37 grams of magnesium sulfate for
4:23 doing this calculation we find that we have
4:30 0.3181 grams of magnesium sulfate that
4:32 were reacted
4:35 now our question is asking what is the
4:38 mass percent of magnesium sulfate in our
4:40 original mixture so there could have
4:43 been other ionic compounds present in
4:45 the solid mixture we're not exactly sure
4:47 what the other components were but we
4:49 know that only a portion of it is
4:51 magnesium sulfate so to find our mass
4:53 percent we're going to want to take the
5:02 and compare this to the sample it's
5:08 the mass of the mixture
5:10 and to calculate a mass percent you'll
5:12 want to multiply by a hundred
5:14 so here let's plug in the numbers that
5:17 we have the mass of magnesium sulfate
5:19 that we found is
5:26 and our denominator the mass of the
5:29 entire mixture is
5:32 0.4550 grams
5:35 multiplying this by a hundred percent to
5:38 get our percentage we find that the mass
5:41 percent of magnesium sulfate is
5:45 69.91 another type of gravimetric
5:48 analysis is combustion analysis so in
5:51 this case the elemental composition
5:53 right how much of each element you have
5:56 for hydrocarbons and other types of
5:59 compounds may be determined using
6:02 combustion analysis in this case a wage
6:04 sample of the compound is heated to a
6:06 high temperature under stream of oxygen
6:09 right we know that oxygen is required in
6:11 order to combust or burn a sample so the
6:14 combustion results in the formation of
6:17 products we know for hydrocarbons which
6:20 contain carbon and hydrogen our products
6:22 are always going to be carbon dioxide
6:25 and water as shown in the chemical
6:27 equation on the slide in the diagram
6:30 that's shown here right our combustion
6:33 takes place in the furnace but then the
6:35 products of our reaction are swept
6:37 through into separate pre-weighed
6:39 collection devices so the first device
6:43 that you see here is a water absorber
6:45 meaning that it's going to absorb the
6:47 water that is the product of our
6:49 combustion reaction
6:52 other substances would continue to flow
6:54 through to the next pre-wave chamber
6:56 where we have a carbon dioxide absorber
6:59 and so any carbon dioxide that is
7:00 produced in our combustion reaction
7:03 would be absorbed now from these
7:05 pre-weighed collection devices we're
7:07 able to determine how much carbon
7:11 dioxide and how much water is produced
7:13 by our combustion reaction and with this
7:16 information we can calculate how many
7:19 moles of hydrogen for example are in the
7:21 water that's collected or how many moles
7:24 of carbon are in the carbon dioxide that
7:26 has been collected and once we know how
7:29 many moles of carbon or hydrogen we have
7:32 we can then use that information to
7:34 determine the empirical formula of your
7:37 unknown compound so let's look at an
7:40 example of a combustion analysis in this
7:43 question polyethylene a hydrocarbon
7:45 polymer is used to produce food storage
7:48 bags and many other flexible plastic
7:53 items a combustion analysis of a mass of
7:55 this sample yields
7:59 .00394 grams of carbon dioxide and
8:03 .00161 grams of water so using this
8:05 information we want to determine the
8:08 empirical formula of polyethylene
8:10 now we have a flow diagram to show us
8:13 what our approach should be we can see
8:14 that we're going to start with the mass
8:17 of carbon dioxide and the mass of water
8:19 that result from our combustion analysis
8:22 we're going to then use the molar masses
8:25 of these two products to determine the
8:27 moles of each of them so we'll determine
8:29 the moles of carbon dioxide and the
8:31 moles of water that have been produced
8:34 using the molecular formula of carbon
8:36 dioxide we can then calculate how many
8:39 moles of carbon have been collected in
8:41 the combustion analysis and how many
8:43 moles of hydrogen have also been
8:46 collected once we have moles of these we
8:48 can then use the molar ratio to
8:51 determine the empirical formula we're
8:52 first going to figure out how many moles
8:54 of carbon we have collected in our
8:58 combustion analysis they collected
9:05 according to our flow diagram we're
9:07 going to use the molar mass of carbon
9:09 dioxide to calculate how many moles of
9:11 CO2 that is
9:14 the molar mass of carbon dioxide is
9:22 grams of CO2 are going to cancel so
9:25 we'll have moles of CO2 and our flow
9:26 diagram says we're going to use the
9:29 stoichiometric factor in our molecular
9:31 formula for CO2 to convert this to moles
9:34 of CO2 now I can see from the formula
9:38 for every one mole of CO2 I only have
9:45 doing this calculation I calculate that
9:46 I have
9:50 8.95 times 10 to the minus 5th moles of carbon
9:52 carbon
9:54 now let's do the same thing for hydrogen
9:57 we can figure out the moles of hydrogen
10:00 by using the mass of water that was
10:01 collected in the combustion analysis
10:03 this is
10:11 next I'll use the molar mass of water to
10:14 convert this to moles of water the molar
10:17 mass of water is 18.02
10:25 grams of water are going to cancel and
10:26 my next step is to use the molecular
10:28 formula of water to write a
10:31 stoichiometric factor relating moles of
10:34 H2O to moles of hydrogen now based upon
10:37 the formula H2O I can see that I have
10:40 two moles of hydrogen for every one mole
10:42 of water doing this calculation allows
10:45 me to find that I have
10:49 1.79 times 10 to the minus fourth moles
10:52 of hydrogen collected from my combustion analysis
10:54 analysis
10:56 now the last last step to find my
10:58 empirical formula is to take the ratio
11:00 of the number of moles we learned
11:02 previously that when we do this we want
11:04 to divide by the smallest number of
11:06 moles which we have found for moles of
11:08 carbon so I'm going to set my ratio as
11:12 1.79 times 10 to the minus fourth moles
11:14 of hydrogen
11:17 divided by the smaller number which is
11:21 8.95 times 10 to the minus fifth moles
11:23 of carbon
11:25 doing this calculation I find that the
11:28 ratio is actually 2. so this means that
11:31 I have two moles of hydrogen for every
11:34 one mole of carbon in my empirical
