This was reposted because the original was posted on another student's thread.

Question

An oxide of carbon contains 27% carbon. What is its empirical formula?
I don’t understand this question can you please help

0 0
posted by Raida
today at 5:03pm
Answer ID 1814044

Answer 1

Well, oxide of carbon means it has the formula C_xO_y

assume 1000 grams
270 grams C, 730g O
moles C= 270/12=22.5
moles O= 730/16=46.6
divide both numbers of moles by the lowest, so the ratio is CO2

Answer 2

atomic mass of carbon = 12
atomic mass of oxygen = 16

12 * moles of carbon / (12 * moles of carbon + 16 * moles O) = 0.27
12 c/(12 c + 16 o) = 0.27
12 c = 3.24 c + 4.32 o
4.32 o = 8.76 c
o/c = 2.03
looks like about 2 O for each C
How about CO2 ?

Answer 3

It's a carbon oxide, so carbon and oxygen only

Assume 27% equals 27g of carbon and 100-27=73, is oxygen.

27g*(1 mole/12.01g)=2.24 moles

73g*(1 mole/16.00g)=4.56 moles

4.56/2.24= around 2

So, C==> 1 and O ==> 2

CO2

Answer 4

Three different ways to look at the problem but all three are correct...........

Answer 5

I just ejected a loud gas LAUGH OUT LOUD -- Bob did you hear it? If your office stinks it might be because of that...

Answer 6

Dude come on, don't impersonate Ms. Sue. You forgot the period after "Ms" so you are obviously a fake.