MYTUTOR SUBJECT ANSWERS

443 views

1.5 g of hydrocarbon undergoes complete combustion to give 4.4 g of CO2 and 2.7 g of H2O. Given this data, what is the empirical formula of this hydrocarbon?

The first step here is to determine the mass of C in CO2 and the mass of H in H2O. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 4.40 g = 1.1999 g

For H:

(1.0079 x 2 / 18.0148 g) x 2.70 g = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: 1.1999 g / 12.011 g mol-1 = 0.0999 mol

H: 0.3021 g / 1.0079 g mol-1 = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case this is the number of moles of carbon:

C: 0.0999 mol / 0.0999 mol = 1

H: 0.2997 mol / 0.0999 mol = 3

We therefore have a ratio of 1 carbon atom to 3 hydrogen atoms, thus the empirical formula for this hydrocarbon is CH3.

Joshua H. GCSE Chemistry tutor, A Level Chemistry tutor, IB Chemistry...

10 months ago

Answered by Joshua, an A Level Chemistry tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

72 SUBJECT SPECIALISTS

£22 /hr

Daniel W.

Degree: Chemistry and Maths (Bachelors) - Leeds University

Subjects offered: Chemistry, Maths

Chemistry
Maths

“Me, Myself & I I am studying Chemistry and Maths at the University of Leeds and I am about to go into the second year of my degree. From quite an early age I realised Maths and Science are where my interests lie.  I volunteered as a ...”

£20 /hr

Christie M.

Degree: Medicine and surgery (Bachelors) - Newcastle University

Subjects offered: Chemistry, Maths+ 4 more

Chemistry
Maths
English Literature
Biology
.UKCAT.
-Personal Statements-

“Hi, I'm Christie, currently studying 'Medicine and Surgery' at Newcastle University! I aim to share my recent experience of gaining a place at medical school plus to assist individuals who find biology or chemistry a challenge.”

£22 /hr

Hannah W.

Degree: Biology (Bachelors) - Bristol University

Subjects offered: Chemistry, Music+ 5 more

Chemistry
Music
Maths
History
English Literature
English Language
Biology

“Hi I'm Hannah and I am a Biology student at Bristol University. I have always loved the sciences, especially Biology and Chemistry but I am also passionate aboutcreative subjects such as Music and English.  I completely understand how...”

About the author

Joshua H.

Currently unavailable: for regular students

Degree: Chemistry (Masters) - Sheffield University

Subjects offered: Chemistry, -Personal Statements-

Chemistry
-Personal Statements-

“Hello everyone!I am currently in my fourth and final year of a masters degree in chemistry at the University of Sheffield.I am here to help you with any problems you may have in chemistry, at any levels. University friends often co...”

MyTutor guarantee

You may also like...

Posts by Joshua

0.250 g of a hydrocarbon known to contain carbon, hydrogen and oxygen was subject to complete combustion and produced 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this hydrocarbon?

1.5 g of hydrocarbon undergoes complete combustion to give 4.4 g of CO2 and 2.7 g of H2O. Given this data, what is the empirical formula of this hydrocarbon?

24 g of Magnesium reacts with 16 g of Oxygen to produce 40 g of magnesium oxide. What mass of magnesium would you need to produce 10 g of magnesium oxide?

The relative formula mass of CaO is 56 and the relative formula mass of CO2 is 44. What is the mass of CaO that can be obtained from 200 g of CaCO3. CaO3 -> CaO + CO2

Other A Level Chemistry questions

Explain the trend in Ionisation energy when moving across a period and down a group

What are isotopes and how do they differ from each other?

Proton NMR Made Easier

How do buffers work?

View A Level Chemistry tutors

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok