Question

a certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. at 120 deg Celsius & 750 mmHg, 1.00 L of the gaseous compound has a mass of 2.30g. what is the molecular formula of the compound? PLEASE HELP! THANKYOU!

Answer 1

i really can't understand this. please help me. :)

Answer 2

I use to be able to do these in my sleep give me a second I might still have notes on this

Answer 3

yay. thankyou so much!

Answer 4

Ok first thing you do is set each percentage to grams, and ignore the other stuff so,

C -> 64.9g -> 64.9g/12.01g/mol = 5.4mol -> 5.4mol/1.35mol = 4
H -> 13.5g -> 13.5g/1.00794g/mol = 13.39mol ->13.39mol/1.35mol = 10
O ->21.6g -> 21.6g/15.9994g/mol = 1.35mol ->1.35mol/1.35mol = 1

Oxygen Has the lowest amount of moles, thus we divide everything by it

Thus we have an empirical formula of

C4H10O

Answer 5

\[C_4H_{10}O\]

Answer 6

https://en.wikipedia.org/wiki/Methoxypropane

Answer 7

:)

Answer 8

I use the formula

\[Moles = \frac{Grams}{MolecularMass}\]

Answer 9

*used

Answer 10

other than that is just about memorizing the steps :)

Answer 11

If you have any questions feel free to ask

Answer 12

thankyou so much. i'll start to analyze it. :D

Answer 13

1. find the mass of C, H and O respectively:
m(C)= 2.3 x 0.649 m(H)=2.3 x 0.315 m(O)=2.3 x 0.216
= 1.4927g =0.3105g =0.4968g
n(C)=m/M n(H)=m/M n(O)=m/M
=1.4927/12.01 =0.3105/1.008 =0.4968/16.00
=0.1243 mol =0.308 mol =0.03105 mol

so ratio for C : H : O = 0.1243 : 0.308 : 0.03105
= 4 : 10 : 1

so the empirical formula is C4H10O

Answer 14

you dont really need to do that extra step Kystal but we are both correct in this case :)

Answer 15

You found the empirical formula, but the question asks for the molecular formula.

Answer 16

I don't have time to help you through that, but someone else here should be able to. Remember to always fulfill the question!

Answer 17

Hint: You'll need to use PV=nRT.

Answer 18

Haha. it's okay but thanks. :)

Answer 19

@Xishem what am i exactly looking for here? please help. :)

Answer 20

You'll need to find how many moles are in the sample given: 2.30g, 1.00L at 120C and 750mmHg using PV=nRT.

Then you'll need to find a molar mass, and choose the appropriate multiplier for the empirical formula so that the molar mass matches up.

The best I can do is help you in about 7 hours. If your question hasn't been answered by then, I'll be glad to.

Answer 21

@Australopithecus please help. thanks. :)

Answer 22

PV = nRT

P = 750mmHg = 0.987atm
n = moles
V = 1.00L
T = 120 + 273 = 393.2K
R = 0.082 Gas Constant

so

(0.987)(1.00) = n(0.082)(393.2)
n = 0.987/((0.082)(393.2))
n = 0.0306mol
now figure out the molecular mass of the compound

0.0306mol = 2.3g/x

x = 2.3g/0.0306mol
x = 75.2g/mol

With the empirical formula we have determine its molecular mass

Molecular mass of C4H10O
is 74.1216

so I would say it is the correct formula based on the fact I did a lot of rounding and such.

Answer 23

Also I kind of confirmed it is right with that wikipedia link :)

Answer 24

I had a feeling I had to do something like this but it has been a long time

Answer 25

thanks. one question: is molecular formula = molecular mass / formula mass???

Answer 26

I dont quite understand what you mean

Answer 27

Although, you can always use,

molecular mass of calculated molecular formula/molecular mass of empirical formula

when this ratio equals 1 you have the correct formula, if it equals 0.5 you need to double the atoms in the formula etc till you get to 1

Answer 28

well until the ratio gives you 1

Answer 29

ugh

Answer 30

oh. that some kind of answer my ques.

Answer 31

\[moles = \frac{grams}{molecularmass}\]

Answer 32

is what I used to solve for the molecular mass in the second part of the question