Summer 2026 Quiz 2 Practice

0:02 Okay, so we are back uh for the solutions, my solutions to uh the pract practice exam for midterm 2, which is going to cover chapters 4 through six. So let's dive right in. All right. Determine uh for the isotope of calcium with an atomic mass of 44. Uh determine the number of protons, neutrons, and electrons. So if this particular isotope has an atomic mass of calcium, I'm going to look on my periodic table, which is behind me. Are you with me? Calcium is in group two. It's an alkaline earth metal. It's element 20. So calcium

0:50 is element 20 which means here it's atomic number also means uh the number of protons. So the number of protons P+ is going to be 20. And if it's a neutral species, it's going to have protons and electrons. So 20 and 20. If it's an isotope with an atomic mass of 44, that means it's P+ plus its neutrons equals 44. And if we know that the protons are 20

1:38 plus n = 44, we could subtract 20 from both sides to get the number of neutrons is going to be 24. So that 20 and 24 is 44. We can do the same thing for chlorine with an atomic mass of 35. So again, going to my periodic table. This time chlorine's on the other side of the periodic table. It's in group 17 or 7A. It's a H hallogen. Uh it's element 17. So chlorine is 17. And again this atomic number is the number of protons equals the atomic number.

2:25 And therefore our P+ is told to us at 17. And if it's a neutral species, we have the same number of electrons. And so then again, our total uh mass number is going to be the protons plus the neutrons. And so 35 equals the protons plus the neutrons. And since our protons are 17, then we can subtract 17 from both sides and we get the number of neutrons, oops, is um 15 and 17.

3:14 18. That's right. Oh, wait. That's not right. Yeah, that's right. Morning math. I got it. All right. Give the group number the common name and predict the ion formed by florine and potassium. So, a little bit of knowledge of our periodic table, right? So, florine is in group 17 or 7A. Those are the hallogens and those next to the noble gases they always form a -1 ion. So the group number for florine is

4:03 group 17 or 7A and we'll talk about the A group numbering later in the class. Uh these are called the H hallogens. and they always have a -1 ion. So for potassium is on the other side of the periodic table. That's element K. It's in group one or 1 A or 1 A. And we call these the alkali metals.

4:52 And not to be confused with group two next to it are the alkaline earth metals. So these are the alkali metals in group one and they always have a + one ion. These are always plus two ions in group two. All right. Copper has two isotopes. copper 63 and

5:40 copper 65 determine uh calculate the atomic mass. And so what we're going to do for both of these percentages is to get them into their decimal form. We're going to divide that 100 out that we multiplied by to get them in the percentage form and then multiply each of those percentages in their decimal form by each of the isotopic masses that comes from um the mass spectrometer. And from each of those, that's the contribution to the total atomic mass from each of those isotopes. The sum of those is the atomic mass. Let me show you how to do it. So, first we're going to get these back to their decimal form. So 69.15% / 100 = 0.6915

6:32 and 30.8 8 uh 5% divided by 100 equals 03085. And so now we're going to multiply let's see the copper 63 which is 62.929 amus times 0 69 6915 and we'll just set up the copper 65 which is going to be equal to 64.92 27 amus times 03085

7:24 62.929 43 and we'll have four significant figures 52 amus 164.927 * 30.85. Then that's 20.03.

8:16 And so then to find the atomic mass, we're just going to add those up plus 43 and we get 63.545 or 63.55. We look at the periodic table and copper is 63 546 or 55 amu. So it makes sense what we see on the periodic table as well. How many carbon atoms are in the following? So this is going to require a little bit of nomenclature. So calcium

9:02 is uh always calcium 2+ and carbonate is on our list of polyatomic ions. CO3 2 minus. So as a result, we're going to need just one of each of these so that the charges sum to zero. So this is going to be calcium CO3. Carbon dioxide is uh let's see. Dioxide is showing us there's no metals here. So, we're just looking at the prefixes. The lack of a monocarbon means that mono's been dropped. So, there's just one carbon and die is the prefix for two. There's two oxides oxygen. So,

9:50 this is C O2 or C12. Let's see. This is already in its molecular formula. So we see how many sodiums, hydrogen's, carbons, and oxygen are here. And then magnesium cyanide. Is there a metal? Yes, it's magnesium, which is always 2 plus. And it's an excuse me, it's always 2 plus because it's an alkaline earth metal in group two or 2 a. Theide is a polyatomic ion. We look for our sheet. It's CN minus. And so as a result to have two plus balance out with negative charges, we're going to need two minus. So we're going to need two of these

10:36 and we get Mg CN2. So now looking at the actual question which is asking how many carbon atoms are in each in this species we have one calcium one carbon and three oxygens. So that three only applies to the oxygens. So we should have just one carbon. Carbon dioxide we have one carbon and two oxygen. So one carbon here as well. Here we have three oxygen, one carbon, one hydrogen and one sodium. So one carbon here as well. Here we have one magnesium and we have two carbons and

11:23 two nitrogens because there is CN minus CN minus. There's two carbons here. So we're going to do the same thing for iron 3 sulfite. Uh but now we're looking for oxygen atoms. But let's find the molecular formula first. So iron 3 sulfate means that there is an iron 3+ and sulfite is on our polyatomic list SO3 2 minus. And so to pair up iron 3+ and sulfite, we're going to need that 3 22 23 relationship. We're going to need two of the 3+ to get 6 plus total and three of

12:08 the 2 minus to get 6 minus total. And so that's going to be Fe2 and we'll need an open parentheses for the SO3 which will there'll be three of sodium hypocchlorite. Sodium's always one plus. So we have this sodium 1 plus and hypocchlorite is CL minus. So to have sodium hypocchlorite together it's just Na We don't need any parentheses for this.

12:56 Uh so this is manganese 3 carbonate. So we have two manganesees and three carbonates. Carbonates are always 2 minus. CO3 2 minus. That means this is 6 minus over here. which has to equal 6 plus over here. So each of these manganesees must be 3+. And then aluminum oxide. Aluminum is a metal that's always 3+ and oxide is always 2 minus. So we're going to need two of these and three of these. So aluminum oxide is Al23 so that the charges sum to zero. 6 + 6 minus these have to add to zero.

13:45 So now that we know the molecular formulas, we can look to try to see what are the um how many of each oxygen atoms are present. So here we have three sulfites. So that means that there's SO3 2 minus SO3 2 minus there's three of these and each one of them has three oxygens. So there is nine oxygens, three sulfers and two ions here and iron 3 sulfate. But here for sodium hypocchlorite there's one sodium, there's one chlorine and one oxygen. Here there's two manganesees and then there's three of these carbonates. Each

14:31 carbonate again has three oxygens. So there's three carbons and nine oxygens and then there's two aluminums and three oxygens here. So this question is definitely a nomenclature question as is this one. Circle the ionic compounds. ionic compounds are going to exist when there is a metal or there is a polyatomic ion. So the question is there a metal carbon sulfur not a metal but going down this is copper and oxygen. So copper is a transition metal Cu. And so this is an ionic compound copper 2 oxide. All right. So here we have a typo. Uh

15:17 this should be NH4 NOO3. Uh so if you marked this as not ionic because that's nonsense. That would make sense. But as NH4, this is supposed to be NH4 plus and NO3 minus. two polyatomic ions. This is an ionic species. Uh phosphorus triric chloride. So uh is there a metal? No, no metals here. But this is PT platinum. So this is platinum 4 oxide. And this is ionic. Here is NH3. I don't think this is a typo. Uh this is nitrogen and hydrogen. Ammonia. No metals. So this is not ionic. carbon and oxygen. So there is a

16:03 metal cobalt which is uppercase C lowerase O. That's not what this is. This is uppercase C uppercase O. So this is carbon and oxygen both non-metals. But K is potassium. It's an alkali metal in group one. So K with I this would be ionic. This is potassium iodide. So this is the way that we solve a a problem like this. This is nickel with four chlorines. Ni is nickel. It's a transition metal. So this would be ionic. Carbon, florine, and chlorine. These are all non-metals. And I don't see any polyatomic ions here even with typos. So

16:49 this is not ionic. This is molecular. But Fe is iron. So another transition metal right in the middle. This is iron and nitrate. A polyatomic ion. So this is a ionic species. Is there a metal? Carbon. Chlorine. No metals. Not ionic. Is there a metal? No. This is just helium. Not ionic. Nitrogen. Uh triiodide. Nitrogen and iodide. Remember N capitalized I is not a metal. and lowercase I is nickel. That'd be something different. This is not ionic. And argon is also helium and argon are noble gases. They don't form

17:35 any compounds uh including ionic compounds. All right. So we're just going to work through the nomenclature and the formula masses for this table. So L I F is there a metal? Yes. Lithium. Do we always know lithium's charge? Yeah, it's one plus. The all the alkali metals are always one plus in their charge. And so we don't have to write the charge in um parentheses. This is just lithium fluoride. And if we add up the lithium's 6.94 amus and fluorides 18.998, we get 25.93.

18:25 So it confirms bromine trilouide. So the presence of these prefixes tell us there's no metals and bromine and florine aren't metals. We just write the species there. So it's one bromine Br and F3Fs. We just drop the mono bromine. It's just bromine triloride. And so to get the formula mass, we get the formula or the mass for bromine which is 79.904 from the periodic table plus the uh fluorides which is 3 * 18.998.

19:12 So that's the florine, that's the bromine, and our answer is 136.898. All right. So, is there a metal? Yes. Magnesium. It's in group two. It's a alkaline earth metal. Do we always know its charge? Yeah, it's always 2 plus. So, we don't have to write it in parenthesis. We just write the name of the metal, magnesium. And then there's a polyatomic ion O

19:57 minus, which is hydroxide. We just put the name of hydroxide. We don't change anything to it. And what this two means here is that there's two hydroxides. So this species is really an Mg2+, an O minus, and another O minus. So to get the mass, we need one magnesium, two oxygens, and two hydrogens. So all of that together, magnesium is 24.305. 305 plus the oxygens which are 15.999

20:43 amus* 2 plus the hydrogen which is 1.08 O8 amus * 2 equal. So this is the hydrogen, the oxygen, nesium. I'll write it up here on the top. Not very much. Let's see. 24.305 305 58 319

21:36 Okay, nitrous acid. So, we have an acid here. That means that it's going to come with an aquous symbol after the formula. It's uh doesn't start with hydro, which means it's not a binary acid. It's an oxo acid. An oxo acid that ends in o. So o is the suffix that comes when our original oxoanion ended in it. So this is an oxoanion that um was made from something that originally was called nitrite. Nitrite acid and nitrite is NO2 minus.

22:23 And an acid is just something where H+ is the cation and we always have as many H+es as we need to balance out the charge. So in this case, nitrite is just one minus. So we just need one H+. So it's H N O2. And we can double check that 47.013 is two oxygen, one nitrogen, and one hydrogen. All right. Is there a metal? No, just carbon and florine. So we're going to do those prefixes. uh and the prefixes if the first one is mono monoarbon 4 which is tetra tetra florine and then the suffix eyid but we're going to drop the mono we're going to just drop it just carbon

23:13 and then tetra florine but then the suffix eyid lauride carbon tetra fluoride and let's see 1 * 12.011 amus and 4 * 18 998 amus is 88.003 O3 HF aquous. So it's aquous and it has H

24:00 in this leadoff position. So this is an acid. And then the question is does it have oxygen in its um annion? Nope. Just florine. So it's a binary acid. Binary acids start with the prefix hydro. Then the root annion which is florine with the suffix iick. So this is hydric and then the word acid. Hydrofluoric acid. And then it's just the 1 times the 1.008 08 for the hydrogen and the florine 18.998

24:45 amus 20.06. All right, moving on. So, more nomenclature. Just want to keep practicing, practicing, practicing this. So, is there a metal? Yes. We always know it's charged, which is why there's no Roman numerals in the parentheses. Calcium is always 2 plus. Sulfate is a polyatomic ion. Sulfide is S2 minus, but that's not what we're

25:32 talking about here. We're talking about sulfate, which is SO4 2 minus. And so we just need one and one of the calcium and the sulfate because we want our charges to sum to zero. So two plus and 2 minus. So it's just calcium S4. And the formula mass is Uh let's see one calcium times and calcium is 40.078.

26:21 And then sulfur we just have 1 * 32.06 amus. and oxygen * 4 which is 15.999 amus 136.134 All right. Chromium. Is it a metal? Yes. Do we always know it's charged? No. That's why it's written right here. This

27:08 is chromium 2+. This two shows the charge on the one single metal ion. And chlorite is part of that chlorate per chlorate. Chlorite hypocchlorite oxo annion series. and chloride is CL2 1 minus. So for this 1 minus um annion to b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b out with the chromium 2+, we're going to need two of these. So the formula is going to be chromium and then open parentheses the chlorite Cl2 minus. We can omit the charge. It's

27:55 um assumed, but we'll need the two here. So, for those of us just keeping track, that's going to be one chromium 2 plus, Cl2 minus, and Cl2 minus. All of these things are grouped together. So, when we're looking for the mass, we're going to have a chromium. There's one of those. We're going to have chlorines. There's two of them. And oxygens. There's four of them. The chromium on the periodic table element 24 is 51.996.

28:42 The chlorine is 35.45 45 and the oxygen is 15.999 amus. So 1 chromium 51.996 plus 35.45 * 2 plus to that 15.999 * 4. So I get 186.892 892. Is there a metal? Yes. Do we always know its charge? No, we don't. So, we need to put the charge of

29:28 titanium next to it. Titanium is element 22. It's a transition metal. It's not an alkaline metal, which is always + one. An alkaline earth metal, which is always plus two. It's not silver, which is always + one, or cadmium. and zinc, which is always plus two, and it's not aluminum, which is always plus three. Those metals, we know titanium is not one of them. And so, as a result, we're going to put it in in the um parenthesis. So, we'll put titanium parenthesis. And this titanium is with two oxygens, which is always 2 minus. The noble gases are always zero. The H hallogens in group 17 are always

30:13 negative -1 when paired up with a metal. The group 16 oxygen and sulfur are negative -2. And when paired up with a metal, nitrogen nitride is always -3. So these are -2. So as a result here we have a -4 charge which has to be balanced out by one titanium metal. So that titanium metal must be + 4 plus4 and plus minus. So four in Roman numerals is IV titanium 4 oxide. All right. Dulfur dioxide. So is there a metal? No. And we know

31:00 that because of those prefixes. So d is the prefix for two. So, this is just a sulfur and two of them and an oxygen and two of them. Dulfur dioxide works for me. For the masses, we've got sulfur, which we're going to need two of them, which is 32.06 amus each. And oxygen, we only need two of them, which is 15.999 amus each. Oh man, and I forgot the titanium. It's right in front of me. Well, while I'm there, titanium is and we'll need one of those, which is 47

31:45 867 amus. And oxygen* 4, which is 15.999 amus. So now in my calculator, I'm going to add up one titanium and four oxygens here. And then I'm going to add up separately two sulfers plus two oxygen here. So let's pluck those numbers in. Sorry for this mistake.

32:48 Is there a metal for this next one? Yes, silver. Do we always know silver's charged? Yeah, it's a transition metal, but we always know silver is always going to be one plus. So, um, no parentheses here. This is just silver. Just write the name. And then this is a polyatomic ion. After the metal, we have a non-metal, more than one non-metal. So that's polyatomic. And it has to be an ion. Uh this is CN minus. Has to be an ion because metals only form ionic compounds. And so uh when we see more than one non-metal, we should look at the polyatomic ion sheet. And CN minus

33:33 is cyanide. And so we're going to go one silver AG which is uh 1078682 one carbon which is 12.011 011 amus and one nitrogen which is 14007 amuse

34:30 133.88. All right. And then sodium sulfide. Is there a metal? Yes. Na. Uh, we always know the charge. It's one plus. And then sulfide. Not sulfate. Not uh sulfite. Sulfide as I mentioned up here and I crossed off is when sulfur loses like oxygen next to the H hallogens. Oxygen sulfur and group 16 always lose two electrons to make 2 minus and then would become sulfide. And so for there to be a 2 minus charge here it means that we're going to need to lose two electrons.

35:17 So we're going to need two sodiums with the one sulfide. So sodium sulfide is Na2 Sulfide and sodiums. So Na* 2 is 22.9 89 and sulfur * 1 is 32.06 amus. So grand total of 78.038.

36:05 All right. Hope for all of our sakes that those match up. Let's continue on. Just chugging right along here. the following questions. We're considering CH4, C2H6, and C2H4. How many moles of CH4 are in uh C4? Uh 45.55 g of CH4 or methane. So, we're going to need the molar mass of CH4, which is one carbon and four hydrogens. So, 12.011 loss 4 * 1 16.043

36:55 g per mole. Um, and so we're going to start with what we know, which is 45.55 g of CH4. We want grams to cancel and we want moles to come out. And according to our periodic table and Avagadro's number for methane there is 16 043 g for every one mole. Grams cancel with grams and we divide the numbers that are on the bottom. So 45.55

37:42 / with four significant figures which come from the measurement we get 2.839 moles of CH4. How many grams are in molecules? Well, we can't go straight from molecules to grams, but what we can do is go from molecules to moles. Chemists always work in moles and from moles to grams. Now, from molecules to moles is always the same thing. Avagadro's number that there is 6.022* 10 23rd things in a mole. uh in this case molecules

38:31 in one mole and to go from grams to moles we need the molar mass which for C2H6 is two carbons 12.011 * 2 plus 6 hydrogens 1.008 * 6 so 30.07 07 g per mole for C2H6 30.07 g per mole. So once we know the molar mass of C2H6 and use Avagadro's number, we can find out how many grams of C2H6 are in 1.00 * 10 25th molecules. Let's start with what we know, which is 1.00 00 * 10 25th

39:18 molecules of C2H6. We want molecules to go away and we want moles to come out. Chemists always work in moles. And according to Avagadro's number, there is 6.022 * 10 23rd molecules in every one mole of everything including C2H6. But we want grams of C2H6. So for every 1 mole of C2H6, let's be specific. There is we measured 30 point or calculated 07 g for every one mole. So now molecules cancel with molecules and moles cancel with moles and grams comes out

40:03 g of C2H6. And so we take 1 * 10 the 25th and divide it by Avagadro's number and times it by the molar mass. So let's give it a whirl. 1* Watch your parentheses when adding these things or dividing these things up. Unless the calculator makes a mistake, not you. But if the number is really really big, it should be about 16 moles. 16.6 moles. anything that's crazy different than that, check the parentheses in your calculator. Um, that's often the problem. So, I get uh with three sigfigs 499

40:51 g. All right. How many grams of C2H4 is in molecules? Uh so this is very similar except we have C2H4 now. So we're going to do the same thing. We're going to go from molecules to moles using the same Avagadro's number. And now from moles to grams, but this time we have C2H4. So that's 12.0 011 * 2 + 1.08

41:37 * 4. So it's only 28.054 when we're talking about C2H4. So let's see how it makes a difference. 1.0* O * 10 the 25th molecules of C2H4. We want molecules to go away. One mole of anything in this case C2H4 is Avagadro's number 6.022 * 10 23rd. And now we have our different molar mass which says that for one mole

42:22 of C2H4 is 28.054 gram. So divide by Avagadro's number multiply by the molar mass. And so we get 460 with three sigfigs. 466 g of C2H4.

43:08 And how many molecules are in C2H4 uh of C2H4 are in 25.55 gram. So now we're going from grams to molecules. We just have to go in the exact opposite way through moles. Chemists always work in moles. So, we're going to use the molar mass, which is C2H4 is still 28.054 g per mole. And for moles to molecules, we'll use Avagadro's number. So, start with what we know, which is how many grams of C2H4, which is 25.55 g. And we want to convert to moles. So

43:53 we'll put grams on the bottom, moles up top. And there is 28.054 g in every one mole. And then we want to go from moles to Avagadro's number. So we have about a mole of C2H4. So we should have about Avagadro's number of molecules. But let's see. So for every one mole of anything including C2H4, there is 6.0 022 * 10 23rd molecules. So 25.55 / the molar mass which is 28.054 and then times Avagadro's number being

44:39 careful of my parentheses in my calculator. Don't trip me up. I should have 5 484* 10 the 23rd molecules. Cool. One more part to this question which is part E. What's the mass percent of hydrogen in the following compounds? CH4, C2H6, and C2H4. So, mass percent of hydrogen is going to be the mass of hydrogen over the mass

45:27 of everything or ball then times 100. So, easy enough. We just write the mass of the hydrogen which is for this one four hydrogens times 1.08 g per mole over the same thing which is 4 * 1.08 g per mole plus that carbon which is 1 * 12.011 g per N times 100

46:22 25.1 just 25.13%. Can do the same thing for C2H6. So now we have six hydrogens. So 6 * 1.08 g per mole. So 6 hydrogens over those same same six hydrogens. 6 * 1.08 g per mole plus the two carbons. 2 * 12.011 g per mole* 100. So 6 * 1.08 08 plus.

47:18 So I get 20.11%. And then finally C2H4. So four hydrogens over everything. So that's 4 * 1.08 g per mole over everything which is 4 * 1.08 g per mole plus the carbons plus 2 * 12.011 n* 100. So I get

48:16 14.37%. So the hydrogen by mass is decreasing as we go. Alrighty keep going. All right. Number 10. Determine the empirical formulas. So, in this these questions, we're always going to be looking to go from the mass to the moles, then compare the moles, divide by the smallest number, and then scale up if needed. So, in this species, we've got 10.260 g of carbon and 2.296 g of hydrogen. So, let's start with what we know.

49:01 10.260 260 g of carbon and 2.296 g of hydrogen. So, we're going to convert to moles. So, there is 12.011 g of carbon in every one mole of carbon. There is 1.008 O8 oops 8 g of hydrogen in every one mole of hydrogen. So we have 0.854

49:48 moles. Let's see 42. Oh man, there's five two. Oh, bless me. 85422 moles of carbon and 2.296 / 1 2 314 And so we're going to put our carbon and then 0.8 5422

50:35 and then our hydrogen in a preliminary formula 2.314. We'll divide by the smallest number of moles which is the carbon 0.85422. This will become 1 0.85 8 5422. This will by definition become a number more than one because every atom has to be more than one and so I get uh 2.7. So C H

52:17 Oh, I think that there's a mistake here. I see this. If I check my math again, 2.296 divided by 1.08. I must have typed it in wrong. 2 277. So, sorry about that. Going to make it a little bit different here. This number is correct. But then I should get 2

53:04 667 something like that. And so then with this we're going to multiply these aren't whole numbers. So we'll multiply this by three and we get C3 H8. Sorry about that. A little bit of math error on my part. Calculator error. Let's see. A molecule is decomposed in the lab and found to have this much carbon and hydrogen. Let's find the empirical formula. Do it right this time, Michael. All right. 0.41 41 59 g of carbon and 0.13

53:55 96 g of hydrogen. Convert to moles. 12.011 g in every one mole carbon. There is 1.008 g for every one mole of hydrogen. 0.0346 0346 moles of carbon and.13 96 0.138

54:42 5 moles of hydrogen. We'll put them in their preliminary formulas. C 0.0 0 3 4 6 H 0.1385 and we'll divide by the smallest number which is carbon and we get CH4 and there's no need to multiply methane The last two have the added step of the molecular formula. So not just the

55:28 empirical formula, the lowest common ratio, but actually what's there. So to do this, we're going to find the empirical formula just the same. And then we're going to um to scale that uh up to the molecular formula using the molecular mass. Let me show you how. So in this case, we have percentages of carbon, hydrogen, and oxygen. But that's easy. What we're going to do is we're just going to assume for this case that there is a 100 g sample. And as a result, if there's a 100 g sample, then 65.45% carbon would be 65.45 g of carbon. It would also then be 5.492

56:15 g of hydrogen and 29.06 g of oxygen. And all of those, just like the percentages add up to 100%, those grams would add up to 100 g. So, by choosing 100 g in a percentage problem like this, it makes it much easier for me. And so, I'm going to convert to moles. There's 12.011 gram in every one mole. of carbon. There is 1.008 g. In every one mole of hydrogen, there is 15.999 g in every one mole of oxygen. So, this is all about ratios, which is

57:01 why I get to choose that starting mass as long as the percentages, the ratios are the same, and they are. So let's see in this case I would have 65.45 / 12.011 5.45 moles of carbon five.45 Four, five moles of hydrogen 1

57:47 8 2 moles of oxygen. We can put our preliminary formula C 5.45H 5.45 45 1.82 and divide by the smallest number and we get C3 H3 O. So almost there. This is our empirical formula and we need to see how much to scale that to get to our molecular formula. So we need our empirical formula mass. So three carbons, three

58:33 hydrogens and one oxygen which is 55.056 g per mole. And so we compare that to our molecular uh formula um formula mass which is 112.128 g per mole over 55.056 g per mole supposed to be a five. I don't know what happened there. And so when we divide those, we see that there's a factor of two, which means that this entire empirical formula needs

59:20 to be scaled by two. And so we end up getting C6 H6 O2 as our molecular formula. All right, let's try one more. Carbon, hydrogen, oxygen by percentages. Let's find the molecular formula. So, we are going to have a 100 g sample. We're just going to choose that. So, that would make us have 39.342 g of carbon, 8.25 g of hydrogen, 52.44 g of oxygen.

1:00:07 There's 12.011 011 gram of carb. For every one mole of carbon, there is 1.008 g. For every one mole of hydrogen, there is 15.999 g. For every one mole of oxygen 39 342 Right.

1:00:55 Uh significant figures aren't as important in a question like this because our end game here is rounding. But we should keep the significant figures so that we're able to see what the correct answer is when it comes up. So now we'll put our prelim 3.275 H185 3.275 and we'll divide by 3.275 275.

1:01:43 So it's C H 2.5. Even in an empirical formula, we can't have fractions of an atom. So we're going to multiply these by two. That's supposed to be a two, which gets us to our empirical formula of C2 H5O2. And so to get to the molecular mass, we need to know what the empirical formula mass is. So 12 two carbons plus five hydrogens plus two oxygens

1:02:30 is 61.06 g per mole. So again the ratio between the molecular 122.120 mass g per mole and the empirical formula mass which is 61.06 g per mole again in this case is 2. And so we are going to multiply our empirical formula by two to get C4 H104 as our molecular formula. Okay. Well, I hope that helps you uh in your studying for the um quiz. Uh best of luck

1:03:15 with uh all of your preparation.