5

Researchers useda combustion method to analyze a compoundusedas anantiknockadditivein gasolie A 9.394-mg sample of the compoundyielded 31154 mgof carbon dioxide and...

Question

Researchers useda combustion method to analyze a compoundusedas anantiknockadditivein gasolie A 9.394-mg sample of the compoundyielded 31154 mgof carbon dioxide and7.977 mg ofwater inthe combustion (mass of €= 12.01,H=1.008,0-16 mg/mmole) (a) Calculate the percentage composition of the compound. (6) Determine its empirical formula.OUQJi Gxljbl a CHsb.C_HscCsH4

Researchers useda combustion method to analyze a compoundusedas anantiknockadditivein gasolie A 9.394-mg sample of the compoundyielded 31154 mgof carbon dioxide and7.977 mg ofwater inthe combustion (mass of €= 12.01,H=1.008,0-16 mg/mmole) (a) Calculate the percentage composition of the compound. (6) Determine its empirical formula. OUQJi Gxljbl a CHs b.C_Hs cCsH4



Answers

Combustion analysis of a 13.42-g sample of equilin (which contains only carbon, hydrogen, and oxygen) produces 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g>mol. Find its molecular formula.

Because the problem gives us the exact masses of each element in the compound. We don't have to determine an arbitrary mass or percentage. Things is an easier place to start, so we can go right ahead and determine how many moles of each element of president a compound. So starting with carbon, we know that we have, ah, 0.23 to 2 grams and we can go ahead and determine how maney moules there are by dividing Been the molar mass. Ah, we cross out our grams here leaving as with the mole unit that we desire. And that leaves us with 0.193 five moles of carbon moving on to oxygen. We know that we have 0.3091 grams that we're gonna again to fighting by the molar mass getting the mole. You know that we want we end up with 0.19 31 moules of oxygen moving on the hydrogen. We know we have 0.5848 grams again dividing by the molar ass. We have 0.5 eat moles of hydrogen so we can go in here we know we're gonna write in the values that we have. They were gonna divide by the smallest value we have, which is that 0.1 nine three value. Which leaves us with I see one h three one for a final empirical formula of C H 30

Okay, So for this question, they want us to find the empirical formula of the unknown compounds. So whenever they ask you to find the Imperio formula with the molecular formula, all you have to do is think of a like as this. So out of 100 grams of the unknown compound eggs, carbon makes up 92.26 grams of it and hydrogen makes up 7.74 grams of it, Right? So now we can, since we know the groundwork and convert these into most by so 7.74 grams divided by one grand. Because that is the multi when hydrogen gives you 7.74 mose and we do so many 2.26 of 12 grounds and we get 7.69 mold. So now what we do is we divide by the lowest number, which is 7.697 point 69 7.74 divided by 7.69 is about one and some points. Excited about it by 7.69 is also one. So we know that the, uh, Melco formula is C h. And this is also D, uh, empirical formula, because you can't simplify this down any further

We'll have to find the molecular formula of equally in. We have 13.42 grams of Equalling and equaling is composed of carbon, hydrogen and oxygen were told upon combustion. We have 39.61 grams of CO two that is formed and 9.1 grams of H 20 that its forms all of the carbon in equal lean is present in the carbon dioxide so we can find out the mouths of carbon Here, Uh, that is present. So let's go, um, 44 grams of co two in one more and then in one more, Uh oh, to there's one more of carbon and then one more carbon 12 10 grams. Working the suit, we'll find that we have 10.8 zero grams of carbon present. And let's do a similar conversion for the H 20 18.0 grams Permal in one mol of H. 20 They're two moles of hydrogen and one more of hydrogen Muller. Massive one gram and working this out, we'll get one point 00 grams of hydrogen. So to find the mass of oxygen present in Equalling will be the 13.42 grams minus the 10.80 grams of carbon minus the 1.0 grams of hydrogen, and we'll find that we have 1.62 grams of oxygen present. So now let's calculate the empirical formula, starting from 10.80 grams of carbon, one gram of hydrogen and 1.62 grams of oxygen. Convert to moles get 0.9 and 1.62 divided by 16. I would give us 0.10125 by 0.101 This would give me nine, and this would give me 10. So my empirical formula here would be C nine h 10 old. And let's find the empirical mass. What should be equal to 134 grams per mole and R N would be molar mass over empirical mass. 268.34 grams per mole is the molar mass given in the question, divided by 134.0 grounds for moves. The empirical mass we just calculated, and we get a whole number ratio to their former molecular formula will be two times C nine h 10 0 to give me a molecular formula that's equal to C 18 h 2002

And this problem, we're going to find the molecular formula for the compound that we have analyze using combustion analysis. So in order to do this, we're going to determine the percent composition of each element in this compound. So we have obtained two products, we have obtained carbon dioxide and water and from these products we can obtain the mass of the carbon and the hydrogen in the sample originally. So we're going to take the mass of the current dioxide converted two g. In order to do that, we know there are 1000 mg in one g. So divide the number of milligrams by 1000. Move the decimal place three places to the left and we get 30.561 g of carbon dioxide. Then we are going to compare two moles of carbon dioxide using the molar mass of carbon dioxide, we have the same units on the top and bottom. They will cancel out and then we will get moles of carbon dioxide, then cover two moles of carbon. Knowing that in one mole of carbon dioxide we have one mole of just carbon Because the subscript on the carbon is one. And then we need to convert from moles of carbon to the massive carbon using the atomic mass of carbon. Then we're able to get the mass carbon obtains so that's the master current in the sample and then from the water we obtained, we do the same exact thing except now we're just going to use the molar mass of the water And then in one formula unit of a show, we have two moles of hydrogen. So we're going to use that mole ratio instead. And then we get the mass of the hydrogen. So now we can find the percent composition by mass of each element compared to the original sample of the compound. So we take the mass of the carbon divided by the mass of the sample. Making sure units are both in grams, Multiplied by 100 and we get 60% carbon. Then we do the same for the hydrogen and we get 13.3, hydrogen. Then after that, since the only other element in this compound is oxygen, we can find the percent composition by mass of oxygen in the compound. So to do that we do 100 minus 2% composition of carbon plus percent composition of hydrogen. We end up with 26.68% of oxygen. Then from this we can find the malls of each element in the compound. So we start with .6266g of the sample and multiplied by 60g of the element or the Massive element over a 100g of the samples for carbon would be 60g carbon Over 100g of the sample and then grams of sample cancels out. And then we would use the Atomic mass of carbon to convert from massive carbon, two moles of carbon and then we end up with molds of the carbon. So we do the same exact thing with the hydrogen except now we're going to use the percentage of the hydrogen instead percent of the carbon and also the atomic mass of the hydrogen to get moles of the hydrogen. And then we're just going to do the same thing with the oxygen as well. So we end up with multiple oxygen. So now that we have the moles of each element in the compound, we can now find the smallest number hole, the smallest whole number ratios between these amounts. So what we're going to do is we're going to divide by the smallest number of moles. So we're taking the moles of carbon, divide by the moles of oxygen, which is the smallest number out of them all. And then we get up three and then we're going to divide bowls of action by itself to get one, we get moles of hydrogen divided by most of oxygen to get eight. And this is also an approximation. So just round off your numbers. Then This gives us the empirical formula. So these are going to just be the subscript. So we get C301 H8. Then we need to find the empirical formula math. So we're just going to Calculate the molar mass of the empirical formula. We got 60 g per mole. Then we're going to divide the molar mass were given by the empirical mass and we end up eating three. So we have to multiply all the sub scripts in the empirical formula by three. And once we do that we obtained the molecular for.


Similar Solved Questions

5 answers
Inifnecdin Tcll le <ortictej Idt cdis nc Meo EEtcuj (5 VjLaLfp Catnt eannlC 9anecrt Lltoncs 6Kahcettin 6aAT? "ralibujr oltta > Wite @an?ZimnaHarudMRound Your Intnuct ) ! JeenJpio-asSaC 270antzs#? cljcasJre hralintacto -Tmeinan 39 F4S> Kalnd { vJlue t0 2 AnlcerFrobuubieTconcn Zdenadnc Ont49mpieMAMIiTc? IRoltra ae c cecituihiet|bricen %6674 38 Eoums ' Rourd { Vi a6hmaininepaocubataSeen
Inifnecdin Tcll le <ortictej Idt cdis nc Meo EEtcuj (5 VjLaLfp Catnt eannlC 9 anecrt Lltoncs 6 Kahcettin 6a AT? "ralibujr oltta > Wite @an? Zimna Harud MRound Your Intnuct ) ! JeenJ pio-as SaC 270 antzs#? cljcasJre hralintacto - Tmeinan 39 F4S> Kalnd { vJlue t0 2 Anlcer Frobuubie Tconc...
5 answers
13. (5) A mixture of acetone (A) and methanol (M) was prepared at 57.20C and at. PA 786 torr: A) Sketch a graph of PA Vsapa indicating Raoult's Law behavior: You should also mark the location of PA on your Add a dashed line indicating the behavior in the real acetone-methanol solution the A-M interactions are relatively weak compared to A-A interactions: assUMIDG
13. (5) A mixture of acetone (A) and methanol (M) was prepared at 57.20C and at. PA 786 torr: A) Sketch a graph of PA Vsapa indicating Raoult's Law behavior: You should also mark the location of PA on your Add a dashed line indicating the behavior in the real acetone-methanol solution the A-M i...
5 answers
8-2Momentum of a System of Particles; Conservation of Linear Momentum 10 A 90.0 kg fullback attempts to score a touchdown by diving over the goal line. When he is at the goal line and moving horizontally at 6.00 m/s, he is met in midair by an opposing 110 kg (240 Ib) linebacker; who is initially moving at 4.00 m/s in the opposite di- rection. The two meet in head-on collision at the goal line. Does the fullback cross the goal line?
8-2 Momentum of a System of Particles; Conservation of Linear Momentum 10 A 90.0 kg fullback attempts to score a touchdown by diving over the goal line. When he is at the goal line and moving horizontally at 6.00 m/s, he is met in midair by an opposing 110 kg (240 Ib) linebacker; who is initially mo...
5 answers
In the manufacture of a product, fixed cost per week is RM4OOO. Fixed costs are costs such as rent and insurance, that remain constant at all levels of production during given time period. If the marginal-cost function is: dc 0.000001(0.002q2 dq 254)+0.2 Where C is the tolal cost (in RM) of producing q units of products per week Find the total cost function of producing q units of products per week 14 marks]
In the manufacture of a product, fixed cost per week is RM4OOO. Fixed costs are costs such as rent and insurance, that remain constant at all levels of production during given time period. If the marginal-cost function is: dc 0.000001(0.002q2 dq 254)+0.2 Where C is the tolal cost (in RM) of producin...
5 answers
Match each labelwith the appropriate target and answer the two questions below; Activation enerbyActivated complexReactants! -100 } ~200 250 -ProductsBond breakingBond formingProgress of reaction
Match each labelwith the appropriate target and answer the two questions below; Activation enerby Activated complex Reactants ! -100 } ~200 250 - Products Bond breaking Bond forming Progress of reaction...
4 answers
IonIGiven the following data in the form (z, f(z)): (1.10,1), (1.20,3) , (1.28,0), (1.30,5). Using Simpson'$ rule to approximate f f(r) dr, we getereded cut ofSelect one: f3 f(z) dx 0.6 S f(z) dr 0.16 6: f(z) dr 0.16| IE Nr) dz ? 0.6 Vh f(z) dr 0.4 JH fkz) dr 012ag questicn
Ion IGiven the following data in the form (z, f(z)): (1.10,1), (1.20,3) , (1.28,0), (1.30,5). Using Simpson'$ rule to approximate f f(r) dr, we get ered ed cut of Select one: f3 f(z) dx 0.6 S f(z) dr 0.16 6: f(z) dr 0.16| IE Nr) dz ? 0.6 Vh f(z) dr 0.4 JH fkz) dr 012 ag questicn...
5 answers
1 7 5 99$ 1 5 8 3 ] 1 1 L [ 1L 1 1
1 7 5 99$ 1 5 8 3 ] 1 1 L [ 1 L 1 1...
5 answers
30_ How much would you need to invest now at 3.&%/a compounded semi-annually to provide S1500 every 6 months for the next 5 years?
30_ How much would you need to invest now at 3.&%/a compounded semi-annually to provide S1500 every 6 months for the next 5 years?...
1 answers
For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then describe and sketch the graphs using polar graph paper. $$r=\frac{2}{2-3 \sin \theta}$$
For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then describe and sketch the graphs using polar graph paper. $$r=\frac{2}{2-3 \sin \theta}$$...
5 answers
%6 28T8 M 81 % 6 & 45 + m % 4 19 & (K: V " 3) Lct IP 6 {9 5I 4 % & 3 2 5 & 3 3L 61 3 28 % 2 2 site Fill tlle' inL L [YHItatixu [P Cun-iuct It TulluwM cudigDES ricriuitinl ["TWtAtULet M AA3B 32BC JADF {IUI (utS3n4e LL Ie At CAlz codeehe Menge Mul apply che initinl ptItutation the ILEYage M
%6 28 T8 M 81 % 6 & 45 + m % 4 19 & (K: V " 3) Lct IP 6 {9 5I 4 % & 3 2 5 & 3 3L 61 3 28 % 2 2 site Fill tlle' inL L [YHItatixu [P Cun-iuct It TulluwM cudig DES ricr iuitinl ["TWtAtU Let M AA3B 32BC JADF {IUI (utS3n4e LL Ie At CAlz codeehe Menge Mul apply che initinl ...
5 answers
Exercise 3.3.21. Let A and B be sets. Suppose that A # B. Suppose that E is a set such that Ax E = Bx E. Prove that E = 0_
Exercise 3.3.21. Let A and B be sets. Suppose that A # B. Suppose that E is a set such that Ax E = Bx E. Prove that E = 0_...
1 answers
Express the equation in exponential form. $$ \ln (x+1)=2 \quad \text { (b) } \ln (x-1)=4 $$
Express the equation in exponential form. $$ \ln (x+1)=2 \quad \text { (b) } \ln (x-1)=4 $$...
5 answers
Question 54 of 72The coenzyme(s) used in fatty acid synthesis is (are)NADH and NADPHNADPHFADHz and NADHNADHFADHz:4oounEAnnpnvach polisyMacBool
Question 54 of 72 The coenzyme(s) used in fatty acid synthesis is (are) NADH and NADPH NADPH FADHz and NADH NADH FADHz: 4ooun EAnn pnvach polisy MacBool...
5 answers
H: Ifa triangle three vertices are A (1.1) B(S 5) and C(4.6) Determine the following a AB AC BC b. Prove triangle AKC is 4 right triangle [2C] Caleulate Ihe arca of the triangle. [2K |
H: Ifa triangle three vertices are A (1.1) B(S 5) and C(4.6) Determine the following a AB AC BC b. Prove triangle AKC is 4 right triangle [2C] Caleulate Ihe arca of the triangle. [2K |...

-- 0.022795--