n@ttyb
Member
i cannot get my head around it.
Has anyone got notes or help on this
Has anyone got notes or help on this
-
Looking for HSC notes and resources? Check out our Notes & Resources page
moles (1 Viewer)
- Thread starter n@ttyb
- Start date
priesty
formerly wm_abusef
Moles are pretty much the scale that is used to measure the amount of substance you have.
Moceular Mass is the number that you see on the periodic table under the names of each element for example Hydrogen is 1.008. This means that if you have 1.008grams of Hydrogen you have 1 mole of it.
Number of Moles = Mass/Molecular Mass That is pretty much the basic equation you would have covered in Prelim.
When working with Moles in Chemistry and trying to compare the amount of Moles of various substances etc consider it like Central Station. Yes, might sound wierd but say for example they give you a reaction and they give you one of the products in grams and they want you to find the number of grams of product produced. You would convert that original mass of the reactant to moles and using the molar ratio that is given by a balanced equation then calculate how many moles of product has been produced. Then rearrage that equation from above to find out the Mass (in grams) of product.
So to go anywhere you need to go via Moles.
This though is the basis of Moles and Molar Calculations, there is plenty more you will need to learn. Make sure you understand the basics and that way it'll be easy to learn the harder concepts. If you have anymore questions just shout!
Moceular Mass is the number that you see on the periodic table under the names of each element for example Hydrogen is 1.008. This means that if you have 1.008grams of Hydrogen you have 1 mole of it.
Number of Moles = Mass/Molecular Mass That is pretty much the basic equation you would have covered in Prelim.
When working with Moles in Chemistry and trying to compare the amount of Moles of various substances etc consider it like Central Station. Yes, might sound wierd but say for example they give you a reaction and they give you one of the products in grams and they want you to find the number of grams of product produced. You would convert that original mass of the reactant to moles and using the molar ratio that is given by a balanced equation then calculate how many moles of product has been produced. Then rearrage that equation from above to find out the Mass (in grams) of product.
So to go anywhere you need to go via Moles.
This though is the basis of Moles and Molar Calculations, there is plenty more you will need to learn. Make sure you understand the basics and that way it'll be easy to learn the harder concepts. If you have anymore questions just shout!
Forbidden.
Banned
A mole
e.g
1 mol oxygen =16g
you dont know ?????11!?
e.g
1 mol oxygen =16g
you dont know ?????11!?
n@ttyb
Member
yeh i understand avagadros number, but when i look at a question i just got wtf
P
pLuvia
Guest
Give an example of a question, then we can guide you through it
Jono_2007
Member
- Joined
- Jan 29, 2005
- Messages
- 139
- Gender
- Male
- HSC
- 2007
The mole method is something like this:
1. -- write balence chemical equation
2. -- given the information in the question, find the moles of the substance which you know something about, (usually you'll be given the mass of the substance.)
3. -- determine the mole ratio of the mol of the unknown substance:mol of known substance.
4. find the mol of unknown. by multiplying the mol ratio with the mol of known.
5. convert to appropriate units. (if the Q asks for the mass you convert to mass.)
n=m\M and n=N\NA and for gases n=V/MV and for solutions c=n/v
_JONO_
1. -- write balence chemical equation
2. -- given the information in the question, find the moles of the substance which you know something about, (usually you'll be given the mass of the substance.)
3. -- determine the mole ratio of the mol of the unknown substance:mol of known substance.
4. find the mol of unknown. by multiplying the mol ratio with the mol of known.
5. convert to appropriate units. (if the Q asks for the mass you convert to mass.)
n=m\M and n=N\NA and for gases n=V/MV and for solutions c=n/v
_JONO_
n@ttyb
Member
ok i never fully got the basics of moles and questions interelated with moles,so heres some questions to help me get the hang of it:
1)10ml of a 0.05 mol L solution of sulfuric acid was diluted by making 1000ml with distilled water. What was the pH of the resulting solution
( do i times 0.05 by 1000 and then work out the pH from there ?)
also gas produced i dont get ( and i have my trials tomorrow :S)
2)0.66 g NO(g) was mixed with 0.72 g O3 (g)
what is the maximum volume of NO2(g) produced at 0 degreesC and 100kPa
1)10ml of a 0.05 mol L solution of sulfuric acid was diluted by making 1000ml with distilled water. What was the pH of the resulting solution
( do i times 0.05 by 1000 and then work out the pH from there ?)
also gas produced i dont get ( and i have my trials tomorrow :S)
2)0.66 g NO(g) was mixed with 0.72 g O3 (g)
what is the maximum volume of NO2(g) produced at 0 degreesC and 100kPa
tristambrown
Member
- Joined
- May 1, 2006
- Messages
- 85
- Gender
- Male
- HSC
- 2007
A mole is just a decent sized constant desriptor of a SET amount of atoms in any given substance so it can be compared against another.
For example we can compare one apple to one grape because we know what ONE of something we can see is (note even though they are one of something the apple is much heavier than the grape ... this can be likened to carbon and hydrogen ... when considering one mole of each we can see from the periodic table that the carbon is much heavier than the hydrogen).
Atoms are too bloody small to talk about one at a time so we compare a massive number of them (6.022*10^23 to be exact) at once but refer to this standard number of atoms as ONE mole....
this is how we get molar ratios - eg one (mole) heap of oxygen atoms compared to one (mole) heap of carbon atoms ...
and from that how we get comparison of fractions of that standard measurement the "mole"
eg: Half of one heap (0.5 moles) of oxygen atoms compared to one (0.5 moles) heap of carbon atoms (this example has a ratio of 1:1 in an amount of O.5 moles)
If the ratio called for by a chemical reaction involding oxygen and carbon was 1:2 O:C and we had .5 moles of oxygen atoms we would then need to have one whole (1) mole of carbon atoms to satisfy the reaction
when you are looking at things like 0.05 moles it really means 5/100 parts of a mole (that nice large standard scientists settled to make things easier)
Without the word mole and relatively small fractions we would know 0.05mol of any substance as 301100000000000000000000 atoms of something (now you see why the word mole is so handy)
(calc for that is 0.05*[6.022*10^23]
Looking at the q's posted:
"1)10ml of a 0.05 mol L solution of sulfuric acid was diluted by making 1000ml with distilled water. What was the pH of the resulting solution "
As it stands 10ml really is 10ml/10ml * 0.05 Mol/L = 0.05 Mol/l
Once you dilute 10 ml with 1000 ml of water you are essentially dividing 10ml worth of moles into 1000 .. so your ratio is now 10/1000 * 0.05 Mol/L =0.0005 moles/L
PH is calculated simply by entering -LOG(H+ Moles/L)
(LOG to base 10 that is - ie the log button not ln)
We can safely assume with Strong acids like HCL and H2SO4 that the acid Ionises completley and as such the molar H+ concentration is equivelent to the acid's concentration. THIS IS NOT TRUE FOR WEAK ACIDS LIKE CITRIC, ETHANOIC etc (see the dotpoint about comparing stong and weak acids).
so, assuming complete ionisation, the ph of the above solution = -LOG(0.0005) = 3.301029996 (~ Ph 3.3)
"also gas produced i dont get ( and i have my trials tomorrow :S)"
Gas produced is calculated by multiplying AMOUNT of moles by the VOLUME (IN LITRES) Specified by temperature (the volumes are on the data sheet which accompanies the HSC exam for both 0C and 25c at specified pressure).
at 25c ALL Gasses will occupy 24.79L of space per mole
data sheet
http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry_data_sheet.pdf
This is just simply extending the idea that we have 5/100 of one mole (which according to the data sheet occupies a set amount of space in litres)
So for the above question you have 10ml (which is not a litre so be careful) of 0.05Mol/L concentration solution
(note n moles/L really means n moles/1000ml)
So the amount of moles you have in that 10ml is
10ml * 0.05mol/1000ml = 0.0005 Moles
at 25c 1 mole of gas does actually occupy 24.79L
the volume 0.0005 moles would occupy is calculated by (number of moles) * (Volume one mole occupies at specific temp & pressure)
0.0005 * 24.79 = 0.12395L or 124.0mL
"2)0.66 g NO(g) was mixed with 0.72 g O3 (g)
what is the maximum volume of NO2(g) produced at 0 degreesC and 100kPa"
The reaction Eqn should be (it is also reversible but we wont take that into consideration is HSC Chemistry for a q like this)
NO(G) + O3(g) ---> NO2(g) + O2(G)
The ratio for this is all 1:1 as only one of everything is needed to balance the products on the right hand side.
Bear in mind that At 0c and 100kpa (according to data sheet) 1 mole of gas will occupy 22.71 L
First find How many moles of everything we have
According to the periodic table:
Oxygen weighs 16g
Nitrogen weighs 14g
we have .72 of O3
1 mole of O3 weighs 3* the weight of oxygen, so 3 * 16g = 48 grams
to find how many moles we have simply divide the weight we have by the weight ONE mole should weigh - ie .72/48 = 0.015 moles
Find NO's molar ratio
NO weighs weight of N + weight of O = 16g + 14G = 30 g
Moles NO = .66g/30g = 0.022 moles
Now this is a trick q because NOT all can react as we do not have an equal molar ratio to make all the NO react
We have only .015 moles of O3 which will react to donate 0.015 moles of O to the NO to make NO2 leaving 0.015 moles O2
(NOTE that even though O3 splits into O2 and O there is still 0.015 of ALL the Atoms there .... ie there is 0.015 *O + 0.015 * O + 0.015 * O .. which in this equation works out to be 0.015 * O2 and 0.015 * the O in the NO)
From the original 0.022 moles of NO there will be (0.022 - 0.015=0.007) 0.007 moles of UNREACTED NO left
So the MAXIMUM amount of NO2 that can be procduced (in the 1:1 ratio that the reaction describes) is 0.15 (the molar quantity of the limiting reactant, the O3) will be 0.015 Moles of NO2
At Oc and 100kpa we know from the standard data sheet that 1 mole of gas will take up 22.71 L .
Again we dont have a mole we only have 0.015 moles so we will multiply our part mole by the volume of a whole mole to find the volume our part mole occupies
ie 0.015*22.71L = 0.34065L = 340.65mL
(*proof that this works is that part of a whole * one whole = original part of a whole
if we have 1/4 an apple * 1 apple we get 1/4 of that apple
here we have 0.015 of a mole * 1 mole's volume = 0.015 mole's volume)
For example we can compare one apple to one grape because we know what ONE of something we can see is (note even though they are one of something the apple is much heavier than the grape ... this can be likened to carbon and hydrogen ... when considering one mole of each we can see from the periodic table that the carbon is much heavier than the hydrogen).
Atoms are too bloody small to talk about one at a time so we compare a massive number of them (6.022*10^23 to be exact) at once but refer to this standard number of atoms as ONE mole....
this is how we get molar ratios - eg one (mole) heap of oxygen atoms compared to one (mole) heap of carbon atoms ...
and from that how we get comparison of fractions of that standard measurement the "mole"
eg: Half of one heap (0.5 moles) of oxygen atoms compared to one (0.5 moles) heap of carbon atoms (this example has a ratio of 1:1 in an amount of O.5 moles)
If the ratio called for by a chemical reaction involding oxygen and carbon was 1:2 O:C and we had .5 moles of oxygen atoms we would then need to have one whole (1) mole of carbon atoms to satisfy the reaction
when you are looking at things like 0.05 moles it really means 5/100 parts of a mole (that nice large standard scientists settled to make things easier)
Without the word mole and relatively small fractions we would know 0.05mol of any substance as 301100000000000000000000 atoms of something (now you see why the word mole is so handy)
(calc for that is 0.05*[6.022*10^23]
Looking at the q's posted:
"1)10ml of a 0.05 mol L solution of sulfuric acid was diluted by making 1000ml with distilled water. What was the pH of the resulting solution "
As it stands 10ml really is 10ml/10ml * 0.05 Mol/L = 0.05 Mol/l
Once you dilute 10 ml with 1000 ml of water you are essentially dividing 10ml worth of moles into 1000 .. so your ratio is now 10/1000 * 0.05 Mol/L =0.0005 moles/L
PH is calculated simply by entering -LOG(H+ Moles/L)
(LOG to base 10 that is - ie the log button not ln)
We can safely assume with Strong acids like HCL and H2SO4 that the acid Ionises completley and as such the molar H+ concentration is equivelent to the acid's concentration. THIS IS NOT TRUE FOR WEAK ACIDS LIKE CITRIC, ETHANOIC etc (see the dotpoint about comparing stong and weak acids).
so, assuming complete ionisation, the ph of the above solution = -LOG(0.0005) = 3.301029996 (~ Ph 3.3)
"also gas produced i dont get ( and i have my trials tomorrow :S)"
Gas produced is calculated by multiplying AMOUNT of moles by the VOLUME (IN LITRES) Specified by temperature (the volumes are on the data sheet which accompanies the HSC exam for both 0C and 25c at specified pressure).
at 25c ALL Gasses will occupy 24.79L of space per mole
data sheet
http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry_data_sheet.pdf
This is just simply extending the idea that we have 5/100 of one mole (which according to the data sheet occupies a set amount of space in litres)
So for the above question you have 10ml (which is not a litre so be careful) of 0.05Mol/L concentration solution
(note n moles/L really means n moles/1000ml)
So the amount of moles you have in that 10ml is
10ml * 0.05mol/1000ml = 0.0005 Moles
at 25c 1 mole of gas does actually occupy 24.79L
the volume 0.0005 moles would occupy is calculated by (number of moles) * (Volume one mole occupies at specific temp & pressure)
0.0005 * 24.79 = 0.12395L or 124.0mL
"2)0.66 g NO(g) was mixed with 0.72 g O3 (g)
what is the maximum volume of NO2(g) produced at 0 degreesC and 100kPa"
The reaction Eqn should be (it is also reversible but we wont take that into consideration is HSC Chemistry for a q like this)
NO(G) + O3(g) ---> NO2(g) + O2(G)
The ratio for this is all 1:1 as only one of everything is needed to balance the products on the right hand side.
Bear in mind that At 0c and 100kpa (according to data sheet) 1 mole of gas will occupy 22.71 L
First find How many moles of everything we have
According to the periodic table:
Oxygen weighs 16g
Nitrogen weighs 14g
we have .72 of O3
1 mole of O3 weighs 3* the weight of oxygen, so 3 * 16g = 48 grams
to find how many moles we have simply divide the weight we have by the weight ONE mole should weigh - ie .72/48 = 0.015 moles
Find NO's molar ratio
NO weighs weight of N + weight of O = 16g + 14G = 30 g
Moles NO = .66g/30g = 0.022 moles
Now this is a trick q because NOT all can react as we do not have an equal molar ratio to make all the NO react
We have only .015 moles of O3 which will react to donate 0.015 moles of O to the NO to make NO2 leaving 0.015 moles O2
(NOTE that even though O3 splits into O2 and O there is still 0.015 of ALL the Atoms there .... ie there is 0.015 *O + 0.015 * O + 0.015 * O .. which in this equation works out to be 0.015 * O2 and 0.015 * the O in the NO)
From the original 0.022 moles of NO there will be (0.022 - 0.015=0.007) 0.007 moles of UNREACTED NO left
So the MAXIMUM amount of NO2 that can be procduced (in the 1:1 ratio that the reaction describes) is 0.15 (the molar quantity of the limiting reactant, the O3) will be 0.015 Moles of NO2
At Oc and 100kpa we know from the standard data sheet that 1 mole of gas will take up 22.71 L .
Again we dont have a mole we only have 0.015 moles so we will multiply our part mole by the volume of a whole mole to find the volume our part mole occupies
ie 0.015*22.71L = 0.34065L = 340.65mL
(*proof that this works is that part of a whole * one whole = original part of a whole
if we have 1/4 an apple * 1 apple we get 1/4 of that apple
here we have 0.015 of a mole * 1 mole's volume = 0.015 mole's volume)
Last edited:
n@ttyb
Member
hey thanks !