The Mole? Homework help!

[iNerd]

Hey all; we are being taught the mole concept at Year 10 and the book has zero information on it and the teacher fails. If someone could answer the following question with explanations it would be greatly appreciated! I know the basics like n = m/M and molar mass but yeah...confused!

Q: What is the mass of one mol of i) ethane, ii) hexene

A: Ethane: 30grams per mole. Hexene: 84grams per mole.

Q: Determine the amount (no. of mol) in 10 grams of hexene.

A: n = m/M... n = 10/84...n = 5/42 mol...n = 0.119mol....Therefore 0.12mol in 10grams of hexene.

Q: How many mole of carbon atoms in 10 grams of hexene

A: ?

Q: How many hydrogen atoms in 10 grams of hexene

A: ?

Q: How many mole of carbon dioxide is produced if 10 grams of ethane is burnt given the equation 2C₂H₆ + 7O₂ >>> 4CO₂ + 6H₂O

A: ?

[98.40_for_sure]

I wouldn't recommend just blindly applying n=m/M, understanding is better
Molar mass is just the "mass of one mole" of something

Q: What is the mass of one mol of i) ethane, ii) hexene
A: Ethane: 30grams per mole. Hexene: 84grams per mole.

So this is just the basic definition of Molar mass, which you grab from your periodic table. Ethane = CH3CH3 = C2H6 = 24+6 = 30

Q: Determine the amount (no. of mol) in 10 grams of hexene.
A: ?

Ok so now you want mol. You know that mass is g, molar mass is g/mol. Therefore mass divided by molar mass will get you the unit mol. 10g divided by 84g/mol = 0.12mol

Q: How many mole of carbon atoms in 10 grams of hexene
A: ?

In the previous question, we worked out that n(hexene or C6H12)=0.12
So now you're asked to find n(C), since there is 6 C's in hexene, n(C)=n(C6H12)x6 = 0.12x6 = 0.72mol
no. of carbon atoms = 0.72 x 6.02x10^23

Q: How many hydrogen atoms in 10 grams of hexene
A: ?

This is the same as last question, except this time it's 0.12x12, since there is 12 H's

Q: How many mole of carbon dioxide is produced if 10 grams of ethane is burnt given the equation 2C2H6 + 7O2 >>> 4CO2 + 6H2O
A: ?

This question requires mole ratios. The question is asking about CO2 in relation to C2H6(ethane).
From the already balanced equation, the ratio is 4:2
So work out n(ethane)=10/30=0.33mol
So now using mole ratio, you know that n(CO2)=0.33 divided by 2, times by 4 = 0.66mol

Hope that helps a bit

[iNerd]

Ehh yeah...um...for the hydrogen ATOMS question...dusn't that mean I should multiply it by Avagrado's number? The rest makes sense; +1 karma

[98.40_for_sure]

Ehh yeah...um...for the hydrogen ATOMS question...dusn't that mean I should multiply it by Avagrado's number? The rest makes sense; +1 karma

Oh yeah shit sorry, i have small eyes, can't read properly
If it says atoms/particles, then mol x avocado
The formula is no. of particles = n x na i think, where na is avocado number

[iNerd]

Ehh yeah...um...for the hydrogen ATOMS question...dusn't that mean I should multiply it by Avagrado's number? The rest makes sense; +1 karma

Oh yeah shit sorry, i have small eyes, can't read properly
If it says atoms/particles, then mol x avocado
The formula is no. of particles = n x na i think, where na is avocado number

ah of course! N=nNa; yeah got it thanks

EDIT: looked over the ratio Q and I don't get how you got the right answer lol. The mole of ethane is 10/60 isnt it as there is a 2 infront of the ethane thus making the molar mass 60grams? Apparently the method is that 10/60 = 1/6. Then you do 1/6 times 4 = 4/6 = 2/3 = 0.66mol. Two different methods; im confused

[iNerd]

*BUMP*

Q: How many mole of carbon dioxide is produced if 10 grams of ethane is burnt given the equation 2C2H6 + 7O2 >>> 4CO2 + 6H2O
A: ?

I don't understand  :buck2:

[Russ]

Convert 10g to mol and then use a stoichiometric ratio to go from how much ethane you have, to how much carbon dioxide you will produce.

So n = m/M to work out how many mole of ethane there is.
Then work out a stoich ratio: what you WANT to know / what you ALREADY know

Multiply.

[iNerd]

Convert 10g to mol and then use a stoichiometric ratio to go from how much ethane you have, to how much carbon dioxide you will produce.

So n = m/M to work out how many mole of ethane there is.
Then work out a stoich ratio: what you WANT to know / what you ALREADY know

Multiply.

yeah so can you do that step by step please because wouldn't the molar mass be 60grams as the number 2 is infront of the equation so 10/60 = 1/6mol. This is contrary to what 99.95_for_sure said so yeah i'm lost

[|ll|lll|]

*BUMP*

Q: How many mole of carbon dioxide is produced if 10 grams of ethane is burnt given the equation 2C2H6 + 7O2 >>> 4CO2 + 6H2O
A: ?

I don't understand  :buck2:

Lemme try
n(C2H6) = 10/(12x2 + 6x1) = 1/3
n(CO2) = 1/3 x 2 = 2/3

Answer: 2/3 moles
Mass of CO2 produced is actually 2/3 x (12 + 16x2) = 29.33g

[iNerd]

*BUMP*

Q: How many mole of carbon dioxide is produced if 10 grams of ethane is burnt given the equation 2C2H6 + 7O2 >>> 4CO2 + 6H2O
A: ?

I don't understand  :buck2:

Lemme try
n(C2H6) = 10/(12x2 + 6x1) = 1/3
n(CO2) = 1/3 x 2 = 2/3

Answer: 2/3 moles
Mass of CO2 produced is actually 2/3 x (12 + 16x2) = 29.33g

This is what I dont get...so the ethane has a molar mass of 30grams not 60? the 2 in front of it doesn't matter? And where did you get 1/3 times 2 from? There is a 4 infront of the carbon dioxide? kill me

[Russ]

yeah so can you do that step by step please because wouldn't the molar mass be 60grams as the number 2 is infront of the equation so 10/60 = 1/6mol. This is contrary to what 99.95_for_sure said so yeah i'm lost

Ah okay. No, you ignore the 2 out the front of the equation, that's only for balancing. The molar mass is JUST the molecule, you don't multiply it by anything after calculating it.

What cherylim23 posted is right.


edit: the "2" is from the stoich ratio

ratio = what we want / what we know
what we want = carbon dioxide = 4
what we know = ethane = 2

ratio = 4 / 2 = 2

[iNerd]

yeah so can you do that step by step please because wouldn't the molar mass be 60grams as the number 2 is infront of the equation so 10/60 = 1/6mol. This is contrary to what 99.95_for_sure said so yeah i'm lost

Ah okay. No, you ignore the 2 out the front of the equation, that's only for balancing. The molar mass is JUST the molecule, you don't multiply it by anything after calculating it.

What cherylim23 posted is right.


edit: the "2" is from the stoich ratio

ratio = what we want / what we know
what we want = carbon dioxide = 4
what we know = ethane = 2

ratio = 4 / 2 = 2

At last...the big number doesn't matter...stupid balancing...thanks man

[|ll|lll|]

This is what I dont get...so the ethane has a molar mass of 30grams not 60? the 2 in front of it doesn't matter? And where did you get 1/3 times 2 from? There is a 4 infront of the carbon dioxide? kill me

It's not a molar mass! It's a RELATIVE molar mass, which is in comparison to 1/12 of a Carbon atom.
It's like comparing and ratios.

It's like saying, if 2 units of paperclips weigh 50gram, and there are 4 times as many paperclips as there are for pencils, how heavy would the pencils weigh
So you have to compare the individual masses of paperclips and individual masses of pencils (that's the relative molar mass) and then the total weight in all (which is 50g for paperclips). That's the whole compound in comparison.

Get what i mean?

If you're interested, mol = conc x vol as well.

Oops, editted. Thanks jason

[jasoN-]

^ n=Cv