isaac physics help

Complete combustion of compound N occurs in a stoichiometric ratio of 1:6 with oxygen gas. Complete combustion of 4.2g of compound N produces 13.2g of carbon dioxide and 5.4g of water. find the molecular formula of N.

please help me with this question as i cannot work it out! at first i thought it was CH4 but it was wrong so. if anyone can provide working too thta would be great!

Reply 1

Nitrotoluene

If the exercise also gives the final result, could you share it with us? Thanks in advance.
Here are some ideas I'd like to share with you: The Chemistry Forum encourages users to try questions independently before asking for help. Mistakes are acceptable as they are part of the learning process.
Sharing your attempt allows helpers to give targeted advice.
Failure to do so may delay receiving an answer, so it's important to try first.

Kind regards.

Reply 2

user3009

Original post by Nitrotoluene

that is all the question gives.

Reply 3

Alex_c10

Original post by user3009

Complete combustion of compound N occurs in a stoichiometric ratio of 1:6 with oxygen gas. Complete combustion of 4.2g of compound N produces 13.2g of carbon dioxide and 5.4g of water. find the molecular formula of N.
please help me with this question as i cannot work it out! at first i thought it was CH4 but it was wrong so. if anyone can provide working too thta would be great!

Hi, I know this is a week late but still hope this helps.
Since it is complete combustion that means it must have reacted with O2. To find how much oxygen you add 13.2+5.4 then take away 4.2, which I found out to be 14.4g. I then found the number of Moles in O2, CO2 and H20. Since the ratio of N to O is 1:6 you divide the number the moles of oxygen by 6.
My working gave me 0.075 mol of N, 0.45 mol of O2 and 0.3 mol of both CO2 and H2O. You then divide by 0.075 to give the ratio of 1:6:4:4.
I then used this to get the equation - N + 6O2 --> 4CO2 + 4H2O. I then worked out from this what the formula of N was.
Please correct me if I did anything wrong.