Question

A .05438 g sample of a liquid consisting of only C, H, and O was burned in pure oxygen, and 1.039g of carbon dioxide and .6369g of water were obtained. What is the empirical formula of the compound?

Elementary analysis showed that an organic compound contained C,H,N and O as its elementary constituents. A 1.279g sample was burned completely as a result of which 1.60g of carbon dioxide and .77g of water were obtained. A separately weighed 1.625g sample contained .216g nitrogen. What is the empirical formula of the compound?

When solid ammonium dichromate is heated, it decomposes to Cr2O3(s) and N2 (g) and H2O (g). In a given experiment, the following data were obtained.
Crucible + ammonium dichromate. 33.622g
crucible + Cr2O3(s). 29.608g
What is the weight of the crucible?

Answer 1

Wait sorry i didnt finish typing yet haha

Answer 2

A .05438 g sample of a liquid consisting of only C, H, and O was burned in pure oxygen, and 1.039g of carbon dioxide and .6369g of water were obtained. What is the empirical formula of the compound?

Convert the values to moles:
\(CO_2\) will give you moles of carbon
\(H_2O\) will give you moles of hydrogen

you can figure out the moles of oxygen from what is left

Answer 3

post only question per thread, otherwise it's going to get confusing as heck

Answer 4

Water is added to 4.267 g of UF6. The only products are 3.730 g of a solid containing only uranium, oxygen and fluorine and .970 g of a gas. The gas is 95% fluorine, and remainder is hydrogen.

A) from this data, determine empirical formula of the gas.
B) what fraction of fluorine of the original comp is in solid and what fraction in the gas after the reaction?
C) what is the formula of the solid product

Answer 5

Ooops

D) write a balanced equation for the reaction between UF6 and H2O. Assume that the empirical formula of the gas is the true formula.

Answer 6

i can't help you if you post more than 1 question per thread, it's site policy and everyone has to adhere to it.

Answer 7

Oops sorry didnt know that hhaa

Answer 8

So what do i do now haha concerning the questions

Answer 9

@toxicsugar22 you have 2 other very smart people helping you, if you post comments about this again on someone elses thread i will be forced to report you.

Answer 10

A .05438 g sample of a liquid consisting of only C, H, and O was burned in pure oxygen, and 1.039g of carbon dioxide and .6369g of water were obtained. What is the empirical formula of the compound?


Convert the values to moles:
CO2 will give you moles of carbon
H2O will give you moles of hydrogen

you can figure out the moles of oxygen from what is left

Answer 11

So i need to convert from co2 grams to moles then to c grams and moles then o grams to moles?

Answer 12

The same concept applies to h2o too?

Answer 13

so CO2 grams to moles of CO2

then moles of C to grams of C

subtract from total mass of sample

do the same with H2O

you'll have mass of O

Answer 14

Yea i just realized that right now

Answer 15

then once you get all the moles of each element, divide all by the one with the least amount (this is to normalize)

then multiply them all by a factor (the same factor) to get them to whole numbers

Answer 16

. Wait for h2 that is. 1.0079 x2 bc its diatomic right

Answer 17

hm theres no H2, it's H2O

Answer 18

What i was doing was isolating the h 2 and c from both h2o and co2 and then i would just subtract that value from the .5438g sample

Answer 19

Would that work?

Answer 20

you should use H not H2, you'll end up having to do other stuff to compensate when you try to get the formula

Answer 21

it would, but you'll end up having to multiply by 2 somewhere after

Answer 22

Ok so i got .2246...g of O. Now should i use those values and convert them to mol and then divide those values by the smallest total and then get the coefficients for the formula

Answer 23

Did that make sense?

Answer 24

yes thats it!

Answer 25

Ok so i just divided the values by the smallest value which was o and got weird numbers so i have to multiply until its whole numbers right

Answer 26

yessss

Answer 27

i'm gonna work it out too see if we get the same thing

Answer 28

Is it c5h5o or did i need to... I dont know

Answer 29

C 0.28336 g, 0.024 mol

H 0.07076 g, 0.071 mol

O 0.18966 g, 0.012 mol


O \(\dfrac{0.012}{0.012}=1\)

H \(\dfrac{0.071}{0.012}=5.8972\approx 6\)

C \(\dfrac{0.024 }{0.012}=2\)

\(C_2H_6O\) ethanol

Answer 30

Wait why do u have. .18.. And .70... For o and h? I got different values

Answer 31

So i did have to multiply h by two afterwards

Answer 32

because you would've found moles of H2 not of H, so the formula would've been funky.

Answer 33

well it depends on what you did, not if you took H2 into account. But use H not H2, you're making things more confusing for yourself in the future

Answer 34

why are you helping him but know me i am so tired i have been on the computer for so long trying to get help and no one is helping me.you ave stated to help me but then left me midway to finish it on my own. So now, please HELP ME arronq. Report me if you like. You are being very unfair, you said you will help me if you get the time, and now you have time and are helping someone else.

The two that are helping me don't know how to find and help me with the question.

You even blocked me so I cant message you and you are forcing me to write on other peoples question

Answer 35

O i got the answer for that thanks. I will wait for the next one

Answer 36

open a new thread pl0x

Answer 37

Yea.

Answer 38

i opened a new thread guys. PLEASE HELP MEEEEEEEEEEEEEE

Answer 39

Hey how do u open a new thread haha on the ipad?

Answer 40

@toxicsugar22 i was replying to the poster of this question. as i have asked you before, do not spam other people's questions. I have reported you because you have 2 other very capable people looking at your question yet you're here asking for help.

Answer 41

i imagine you'd close this one, then click on ask question or something lol

Answer 42

NO ONE IS HELPING ME ARRONQ

Answer 43

please help me

Answer 44

i kept telling you that for a long time that no one is helping me. I want to be helped by you