Empirical Formulas

Hey Guys

I am a bit stuck with my Chemistry H/W, it’s not particularly hard but I need to know if I am going the right way, so please help me out with these two questions. (btw, I am doing As Chemistry with OCR). The questions are on empirical Formula masses.

For the first one, could u check if I am going the right way:

Q.
3.53g of Fe reacts with Cl to form 10.24 g of Iron Chloride. Find the empirical formula mass of Iron Chloride.

A.

Fe + Cl --> Iron Chloride
3.53g + X --> 10.24

So x much be 6.71g
Now I divided by the Relative Atomic Masses, which gave the following moles:

0.06 moles + 0.19 moles --> Iron Chloride

Which means the Empirical Formula must be FeCl3
Is that correct?
Out of interest, I thought Cl was diatomic so would it not react as Cl2

I am lost with this one:

Q2.

An organic compound X, which contains Carbon, Hydrogen, and Oxygen only has the Mr of 86. When 0.43g of X is burned in excess oxygen, 1.10g of Carbon dioxide, and 0.45g of water are formed.
Find the empirical and molecular formulae of compound X

A. :eek:

I’d find it most helpful if someone can explain how to do it, rather than list an answer so that I’ll be able to do questions like this in the future.

Thanks in advance :biggrin:

Reply 1

EierVonSatan

21

The first question looks good :yy:

For the second question you can calculate how much carbon and hydrogen there was from CO2 and H2O...then work out the rest :smile:

Reply 2

cpchem

14

The first question is all fine. Chlorine is diatomic, so you'd have 3/2 Cl₂ per iron.

With the second one, this is something called combustion analysis. The mass of carbon dioxide allows you to work out how much carbon is present (what proportion of the mass of carbon dioxide is carbon?), similarly the mass of water lets you work out how much hydrogen is present. The missing mass is clearly oxygen (there's no other way to work that out, because you're burning the sample in oxygen). From the masses, you can calculate the moles, and the empirical formula is the simplest proportion of C to H to O. You might then need to multiply that by some factor to achieve the mass you're after, and the right molecular formula.

Hope that helps!

Reply 3

tazarooni89

20

You're correct with the first question. Chlorine gas is diatomic, but that doesn't mean you can't get FeCl3.

"FeCl3" doesn't mean one atom of iron and three atoms of chlorine, it means three atoms of chlorine per atom of iron. I hope that makes some sense :s-smilie:

Q2. First work out the mass of O2 that's reacting

X + O2 ==> H2O + CO2
0.43g 1.12g 0.45g 1.10g

Divide each reactant/product's mass by its Mr

0.005 0.035 0.025 0.025

Divide everything by the smallest value (in this case, 0.005)

1 7 5 5

So your equation is X + 7 O2 ==> 5 H2O + 5 CO2

Now if you just count up the atoms on both sides of the equations, you'll see that X contains 10 Hydrogen Atoms, 5 Carbon Atoms, and 1 Oxygen Atom. (which is correct because it has a molecular mass of 86)

So you can say that the molecular formula is C5 H10 O

Since 5, 10 and 1 have no common factors, you can't cancel down the formula, so the empirical formula is also C5 H10 O

Reply 4

New Username

OP

10

tazarooni89

You're correct with the first question. Chlorine gas is diatomic, but that doesn't mean you can't get FeCl3.

"FeCl3" doesn't mean one atom of iron and three atoms of chlorine, it means three atoms of chlorine per atom of iron. I hope that makes some sense :s-smilie:

Q2. First work out the mass of O2 that's reacting

X + O2 ==> H2O + CO2
0.43g 1.12g 0.45g 1.10g

Divide each reactant/product's mass by its Mr

0.005 0.035 0.025 0.025

Divide everything by the smallest value (in this case, 0.005)

1 7 5 5

So your equation is X + 7 O2 ==> 5 H2O + 5 CO2

Now if you just count up the atoms on both sides of the equations, you'll see that X contains 10 Hydrogen Atoms, 5 Carbon Atoms, and 1 Oxygen Atom. (which is correct because it has a molecular mass of 86)

So you can say that the molecular formula is C5 H10 O

Since 5, 10 and 1 have no common factors, you can't cancel down the formula, so the empirical formula is also C5 H10 O

ohh...that is quiet genius way of working it out...I'd never have thought of that.

I just got one doubt on the diatomic bit, in Q2 we assume that it is gona be 02 (the reactant) as it is diatomic,(i can see sense in that as 02 is present in the atmosphere as diatomic gas), so does that mean I should also consider the Cl in the first equation to be diatomic? :rolleyes:

edit: so in the first question, should the mole of Chlorine be 0.095?

Reply 5

tazarooni89

20

I don't think it matters whether you consider it as diatomic or not. You'll either end up with 3Cl or 3/2 Cl2 which are both the same thing.

Same for the second question, you'll either end up with 14O or 7O2. I only assumed it was diatomic because...well...it is :P

Reply 6

New Username

OP

10

tazarooni89

I don't think it matters whether you consider it as diatomic or not. You'll either end up with 3Cl or 3/2 Cl2 which are both the same thing.

Same for the second question, you'll either end up with 14O or 7O2. I only assumed it was diatomic because...well...it is :P

haha...I prefer to take them as diatomic most of the time :smile:

Anyways...thanks a lot for your help. :biggrin:

Also..thanks everyone, for your help :smile:

Reply 7

mathieua0001

4

11 years have passed and schools are still setting the same questions; i struggled like you did but this thread really helped ,so thanks.

Reply 8

username5303988

13

Another year passed, and another student, late at night pondering on a question that these people pondered over 13 years ago. Yet in another 13 years more students will ponder on the same questions, and will look at us, and think the same...