- Forums
###### chem a level help

Watch

2 months ago

1 mole of a hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192g of oxygen. Work out the formula of the hydrocarbon.

i cant find a recent thread of this question.

one answer says:

'CnH2n + X O2 ==> n CO2 + n H2O

balancing oxygen, we have X= 3n/2

192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4

and the hydrocarbon is C4H8'

why have they done 3n/2?

another answer says:

'192g of oxygen = 12 mole

bCnH2n + aO2 -> xCO2 + yH2O

For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.

:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?

thanyou

i cant find a recent thread of this question.

one answer says:

'CnH2n + X O2 ==> n CO2 + n H2O

balancing oxygen, we have X= 3n/2

192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4

and the hydrocarbon is C4H8'

why have they done 3n/2?

another answer says:

'192g of oxygen = 12 mole

bCnH2n + aO2 -> xCO2 + yH2O

For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.

:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?

thanyou

Reply 1

2 months ago

Original post by miac0328

1 mole of a hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192g of oxygen. Work out the formula of the hydrocarbon.

i cant find a recent thread of this question.

one answer says:

'CnH2n + X O2 ==> n CO2 + n H2O

balancing oxygen, we have X= 3n/2

192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4

and the hydrocarbon is C4H8'

why have they done 3n/2?

another answer says:

'192g of oxygen = 12 mole

bCnH2n + aO2 -> xCO2 + yH2O

For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.

:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?

thanyou

i cant find a recent thread of this question.

one answer says:

'CnH2n + X O2 ==> n CO2 + n H2O

balancing oxygen, we have X= 3n/2

192g of oxygen is 6 moles, so 3n/2 = 6 ==> n=4

and the hydrocarbon is C4H8'

why have they done 3n/2?

another answer says:

'192g of oxygen = 12 mole

bCnH2n + aO2 -> xCO2 + yH2O

For every 3 moles of oxygen there are 2 of hydrogen and one of carbon.

:. bCnH2n has 8 moles of hydrogen and 4 of carbon'

why have they said theres 12mol? isnt there 6moles?(192/32, to get 12 they wouldve done 192/16 but thats not right so then they wouldve got 6 mol but that gives c2h4 and thats not right?

thanyou

There must be the same number of oxygens on both sides of the equation:

On the LHS, only O2 has oxygen atoms in it and since the reaction coefficient (the big number in front of the formula) is X and there are two oxygen atoms in each O2 molecule, then there must be 2X oxygens on the LHS.

On the RHS, CO2 and H2O both have oxygens in them and using their reaction coefficients, there must be 2n + n = 3n oxygens.

So we have 2X = 3n ==> X = 3n/2

I agree, 192 g of O2 is 6 mol of oxygen (since O2 has a relative mass of 32 g/mol) and therefore the correct continuation would be to equate 6 with 3n/2 and solve for n, which you can use to deduce the formula of the hydrocarbon.

Original post by TypicalNerd

There must be the same number of oxygens on both sides of the equation:

On the LHS, only O2 has oxygen atoms in it and since the reaction coefficient (the big number in front of the formula) is X and there are two oxygen atoms in each O2 molecule, then there must be 2X oxygens on the LHS.

On the RHS, CO2 and H2O both have oxygens in them and using their reaction coefficients, there must be 2n + n = 3n oxygens.

So we have 2X = 3n ==> X = 3n/2

I agree, 192 g of O2 is 6 mol of oxygen (since O2 has a relative mass of 32 g/mol) and therefore the correct continuation would be to equate 6 with 3n/2 and solve for n, which you can use to deduce the formula of the hydrocarbon.

On the LHS, only O2 has oxygen atoms in it and since the reaction coefficient (the big number in front of the formula) is X and there are two oxygen atoms in each O2 molecule, then there must be 2X oxygens on the LHS.

On the RHS, CO2 and H2O both have oxygens in them and using their reaction coefficients, there must be 2n + n = 3n oxygens.

So we have 2X = 3n ==> X = 3n/2

I agree, 192 g of O2 is 6 mol of oxygen (since O2 has a relative mass of 32 g/mol) and therefore the correct continuation would be to equate 6 with 3n/2 and solve for n, which you can use to deduce the formula of the hydrocarbon.

i just dont really understand why o2 is 2x?

why cant we call it x?

Reply 3

2 months ago

Original post by miac0328

i just dont really understand why o2 is 2x?

why cant we call it x?

why cant we call it x?

We are looking at the number of oxygen atoms in XO2. If there are X molecules made up of two atoms, then there are X x 2 = 2X oxygen atoms

- TSR Study Together - STEM vs Humanities!
- Swapping a levels
- A level options??
- Grade 6 in maths to do bio and chem a level
- A level decisions
- A Levels
- Thinking about dropping A-level Physics (pre-med student)
- chemistry alevel mock
- A level choices
- finding AS Chemistry very hard to the point of my college stepping in
- Medicine
- Biology A-Level Question Resources ...
- A-level choices
- A Level Economics overview
- A level Chemistry
- What A levels should I choose for Dentistry?
- A level subject choices
- chem and bio a level advice
- A levels for Dentistry?
- A level chemistry

- a premed miniblog
- AQA GCSE Spanish Writing Higher Tier 4H (8698/WH) - 13th June 2023 [Exam Chat]
- Official: University of Birmingham A100 2024 Entry Applicants
- My head girl application
- Further maths in one year
- I’m overthinking about this guy.
- I'm a student at LIS, ask me anything!
- Architecture Applicants 2024
- Official The London Interdisciplinary School Applicant Thread for 2024
- Official: Hull York Medical School A100 2024 Entry Applicants
- Presentations and DSA
- How long for Tesco right to work check?
- CTAM: Count to a million (Part 79)
- Space EPQ Ideas
- Survey
- does anyone do this a level combo?
- KCL Dentistry 2024 Entry
- Mst in English (all streams)
- Isaac Senior Physics Challenge 2024
- Accenture Degree Apprenticeship 2024

- Abbreviations and acronyms for Personal statement
- is epq useful at all?
- IDEA random coding
- Family and mental health
- Official Cambridge Postgraduate Applicants 2024 Thread
- Are tattoos lower class?
- 2023/24 Ask a UAL Student
- Motivating myself - my GYG blog :)
- GCSE English Literature Study Group 2023-2024
- a level functions help pls !!
- Airbus degree apprenticeships 2024
- Has anyone studied in Turkey?
- teaching while in sixth form
- Official University College London Applicant Thread for 2024
- Bath Spa University - Primary PGCE 2024
- University of Chester First Deposit
- Improving Vet Student? | 3rd year GYG blog
- KPMG Audit Apprenticeship 2024
- Degree apprenticeships 2024
- Official Politics and/or International Relations Applicants Thread 2024

- chemistry limiting reagents question aqa alevel
- A level chemistry help please
- The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete combustion is
- Need Help on an Electrochem Q
- reactivity
- chemistry olympiad round 2
- Chemistry olympiad
- OCR AS chemistry 2023
- MCQ from OCR A-level chemistry
- Catalyst: iodide and peroxodisulphate ions

- chemistry limiting reagents question aqa alevel
- A level chemistry help please
- The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete combustion is
- Need Help on an Electrochem Q
- reactivity
- chemistry olympiad round 2
- Chemistry olympiad
- OCR AS chemistry 2023
- MCQ from OCR A-level chemistry
- Catalyst: iodide and peroxodisulphate ions

The Student Room and The Uni Guide are both part of The Student Room Group.

© Copyright The Student Room 2024 all rights reserved

The Student Room and The Uni Guide are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: Imperial House, 2nd Floor, 40-42 Queens Road, Brighton, East Sussex, BN1 3XB