This discussion is now closed.
Check out other Related discussions
- Can someone please provide me a solution to this
- Moles question
- Mole Calculations.
- Chemistry Question Help please
- Acids and bases help
- Chemistry question
- chemistry limiting reagents question aqa alevel
- Chem alevel help
- AQA Chemistry Chapter 2 Practice Questions HELP
- NSAA Chemistry help
- Is finding the moles enough to decide?
- I don't understand mark schemes
- Chemistry A Level Titrations Help
- Amount of substance q help
- Volume of gas equations
- AQA A Level Chemistry
- Chemistry alevel with dyscalculia
- Help chemistry
- chemistry help
- moles and volume
I would do it algebraically 
The other way to do it is to say 16.3% by mass is 16.3/18 and 83.7/460.5 (the mass compound without water)
gives ratios 0.9055 and 0.1817, division by the smallest => x = 5
it's very much like the question 'a hydrocarbon is 53.5% carbon, 9% hydrogen the remainer oxygen calculate the empirical formula' except you're dividing by molecules rather than atoms
edit: typo!
The other way to do it is to say 16.3% by mass is 16.3/18 and 83.7/460.5 (the mass compound without water)
gives ratios 0.9055 and 0.1817, division by the smallest => x = 5
it's very much like the question 'a hydrocarbon is 53.5% carbon, 9% hydrogen the remainer oxygen calculate the empirical formula' except you're dividing by molecules rather than atoms
edit: typo!
EierVonSatan
I would do it algebraically
The other way to do it is to say 16.3% by mass is 16.3/18 and 83.7/460.5 (the mass compound without water)
gives ratios 0.9055 and 0.1817, division by the smallest => x = 5
it's very much like the question 'a hydrocarbon is 53.5% carbon, 9% hydrogen the remainer oxygen calculate the empirical formula' except you're dividing by molecules rather than atoms
edit: typo!
The other way to do it is to say 16.3% by mass is 16.3/18 and 83.7/460.5 (the mass compound without water)
gives ratios 0.9055 and 0.1817, division by the smallest => x = 5
it's very much like the question 'a hydrocarbon is 53.5% carbon, 9% hydrogen the remainer oxygen calculate the empirical formula' except you're dividing by molecules rather than atoms
edit: typo!
Ah. The 'other way' method is assuming that you have 100g of the chemical right?
So then n =m/Mr can be applied, giving you the molar ratio.
sonic23
Ah. The 'other way' method is assuming that you have 100g of the chemical right?
So then n =m/Mr can be applied, giving you the molar ratio.
So then n =m/Mr can be applied, giving you the molar ratio.
No you don't need to assume a certain mass - you have the percentages.
Methane is 75% carbon by mass, 25% hydrogen
75/12 = 6.25 and 25/1 = 25
25/6.25 = 4
CH4
Related discussions
- Can someone please provide me a solution to this
- Moles question
- Mole Calculations.
- Chemistry Question Help please
- Acids and bases help
- Chemistry question
- chemistry limiting reagents question aqa alevel
- Chem alevel help
- AQA Chemistry Chapter 2 Practice Questions HELP
- NSAA Chemistry help
- Is finding the moles enough to decide?
- I don't understand mark schemes
- Chemistry A Level Titrations Help
- Amount of substance q help
- Volume of gas equations
- AQA A Level Chemistry
- Chemistry alevel with dyscalculia
- Help chemistry
- chemistry help
- moles and volume
Latest
Trending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these productsTrending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these products
