Rod's Chemistry 3/4 Questions Thread

[Rod]

When 1.193g of a hydrocarbon compoound was completely burnt in pure oxygen to produce 4.039 g of c02 and 0.826 g of water. What is the empirical formula?

It's a multi choice and I have no idea.

Know I need to find the mole of the hydrocarbon. Could start of finding mole of c02 and water, but then what do I do?

thanks

[Rishi97]

When 1.193g of a hydrocarbon compoound was completely burnt in pure oxygen to produce 4.039 g of c02 and 0.826 g of water. What is the empirical formula?

It's a multi choice and I have no idea.

Know I need to find the mole of the hydrocarbon. Could start of finding mole of c02 and water, but then what do I do?

thanks

I got the answer CH. If that's right, let me know and I'll show you what I did. If it's wrong, I'll try again cause I was unsure of one step.

[Blondie21]

I got the answer CH. If that's right, let me know and I'll show you what I did. If it's wrong, I'll try again cause I was unsure of one step.

My answer was CH aswell.

m(CO2)=4.039g
n(CO2)=0.0918mol
n(C) = 0.0918mol

m(H2O)= 0.826g
n(H2O)=0.0458mol
n(H)=0.0918mol

Empirical Formula
                                   C                                   H
                         0.0918/0.0918       :         0.0918/0.0918
                                                    1:1

.:. CH is the formula

[Rod]

My answer was CH aswell.

m(CO2)=4.039g
n(CO2)=0.0918mol
n(C) = 0.0918mol

m(H2O)= 0.826g
n(H2O)=0.0458mol
n(H)=0.0918mol

Empirical Formula
                                   C                                   H
                         0.0918/0.0918       :         0.0918/0.0918
                                                    1:1

.:. CH is the formula

Ohh I get it, thanks. Ill check if it's right and let you guys know soon, but it looks as if it's the only way. We didn't use the mass of the hydrocarbon but ;\.

Thanks Rishi and Blondie

[Blondie21]

Ohh I get it, thanks. Ill check if it's right and let you guys know soon, but it looks as if it's the only way. We didn't use the mass of the hydrocarbon but ;\.

Thanks Rishi and Blondie

n(c)=0.0918 x12
m(c) = 1.10g

m(h)= 0.0918g

1.10 + 0.0918 = 1.19g

.. Which is the mass of the hydrocarbon!

[Rod]

n(c)=0.0918 x12
m(c) = 1.10g

m(h)= 0.0918g

1.10 + 0.0918 = 1.19g

.. Which is the mass of the hydrocarbon!

Yay! thanks heaps guys

[Rod]

Hey guys I've got a couple of more mole questions, sorry

Someone precipitated the nickel present as a compound Ni(C4H7N2O2)2.        A sample of ore was 4.248 g, it yielded 1.525g of precicpitate, what is the nickel content of this ore expressed as a percentage by mass?

What I'm thinking:
Mole of ore: 4.248/288.7 =0.0147
Mole of Ni: 0.0147

Precipitate = 1.525g, n(Ni) = 0.0147.. Not sure what to do next.



CaC03 ---------> CaO + CO2

a 1.839g sample of mixture of calcium carbonate and calcium oxide was heated and a mass loss of 0.572 g was recorded. What is the percentage by mass of calcium oxide in original sample?

What I'm thinking:
Not sure with this one ;\

And one more:
11g of a hydrocarbon was completely burnt in air and 18g of water was produced. What is the emperical formula of the hydrocarbon?

N(h20= 18/18 = 1
n(h) = 2x1 = 2 mol

m(h) = 2x1 = 2

so m(carbon of hydrocarbon) = 11-2 = 9

9/12 = 0.75 mol
so n(c) = 0.75
n (h) = 2

[Rishi97]

Hey guys I've got a couple of more mole questions, sorry

Someone precipitated the nickel present as a compound Ni(C4H7N2O2)2.        A sample of ore was 4.248 g, it yielded 1.525g of precicpitate, what is the nickel content of this ore expressed as a percentage by mass?

What I'm thinking:
Mole of ore: 4.248/288.7 =0.0147
Mole of Ni: 0.0147

Precipitate = 1.525g, n(Ni) = 0.0147.. Not sure what to do next.

Hey Rod
First of all, don't be sorry about asking questions. It's good that you are clarifying things you are unsure about
Here's what I would and if I'm wrong, sorry
n(Ore) =0.0147
n( Nickel) 0.0147 (Great start)
m( Nickel) = 0.0147 x 58.7
                = 0.86289 g
(% of nickel) = 0.86289 / 1.525
                   = 0.5658295082 x 100 =  56.58 %

[Rishi97]

CaC03 ---------> CaO + CO2

a 1.839g sample of mixture of calcium carbonate and calcium oxide was heated and a mass loss of 0.572 g was recorded. What is the percentage by mass of calcium oxide in original sample?[/b]

What I'm thinking:
Not sure with this one ;\

Having a complete guess with this one:
Since calcium carbonate + calcium oxide= 1.839
Heating it would get rid of the water including the oxygen. 1.839-0.572 = 1.267g
1.267/1.839 x 100 = 68.9 %
This is a complete guess as even I'm not sure why I did certain steps It would be a easter miracle if I got it right. lol

[Rishi97]

N(h20= 18/18 = 1
n(h) = 2x1 = 2 mol

m(h) = 2x1 = 2

so m(carbon of hydrocarbon) = 11-2 = 9

9/12 = 0.75 mol
so n(c) = 0.75
n (h) = 2

Good start. First of all, is the answer C₂₄H₄O ? The answer doesn't look right so let me know if it's not and I'll try again

[nhmn0301]

Hey Rod
First of all, don't be sorry about asking questions. It's good that you are clarifying things you are unsure about
Here's what I would and if I'm wrong, sorry
n(Ore) =0.0147
n( Nickel) 0.0147 (Great start)
m( Nickel) = 0.0147 x 58.7
                = 0.86289 g
(% of nickel) = 0.86289 / 1.525
                   = 0.5658295082 x 100 =  56.58 %

[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =7.299%.

[Rishi97]

I don't think you should divide the mass of nickel to the mass of precipitate, cause they are asking for percentage by mass in the ore sample. hence  mass of 0.86289 g of nickel in ore is roughly around 20.31%.

Oh really? I always thought percentage by mass was dividing by the total amount. Also, why would they give you that extra info if you don't need it?
Ah well, we'll just wait for Rod's answer

[Rod]

Hey guys, thanks for all your help.

The answer to the first question is 7.3%, second question is 29.3% and third C3H8

[Rod]

[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =0.07299%.

So is this what you did?

n(Ni(C4H7N2O2)2 ) = 0.0052823

n(Ni) = 0.005283

m(Ni) = 0.005283 x 58.7 = 0.31 g

Hence 0.31/4.248 x 100 = 7.3 % And that's correct. Thanks pal.

Now, how about the other two ?

[Rishi97]

[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =7.299%.

oh yeah...makes more sense now. Sorry for the wrong answer Rod and nhmn0301