STOICHIOMETRY- whoever gets it plz help-urgent!!!!gag (1 Viewer)

[Templar]

Ionic compounds only have empirical formula since they are a continous lattice. I'm not sure what you meant by the other parts of the question.

 

[n a + a l i e]

i meant molecular compounds, do they display molecular formula?
n also would there b any rule 4 covalent compounds as well?

 

[xxxx8888]

i meant molecular compounds, do they display molecular formula?
n also would there b any rule 4 covalent compounds as well?

what covalent compounds.

 

[Dreamerish*~]

i meant molecular compounds, do they display molecular formula?
n also would there b any rule 4 covalent compounds as well?

I don't understand what you're trying to ask.

 

[n a + a l i e]

i mean as in molecular formula- like for instance the carbon compounds, they display molecular formula- im kind of finding this hard to explain but yeh, do u u kno what i mean by molecular formular, also what is it that u dont get about my question, is it the terms because im kind of finding it hard to simply myself. soz

 

[n a + a l i e]

n also, covalent compounds r referred to as molecular compounds if thts any help but im not sure y..if thts any help =)

 

[Dreamerish*~]

Lol, ok, what was the question again?

 

[n a + a l i e]

molecular compounds, do they display molecular formula?
n also would there b any rule 4 covalent compounds as well?

 

[Dreamerish*~]

molecular compounds, do they display molecular formula?
n also would there b any rule 4 covalent compounds as well?

Um, yes. Molecular formulae are displayed by ... how should i put this ... molecular compounds?

Rules for finding the empirical formula? For naming them? Or something else?

 

[n a + a l i e]

molecular formula is formula that displays the exact no of atoms for a particular compound,

 

[DraconisV]

1. What mass of silver carbonate precipitates when 25 mL of 0.2 M silver nitrate solution is added to 30 mL of 0.150 M sodium carbonate solution?

Ok, I have done this question. I now have came up with the answer of 1.38g of Silver Carbonate formed. Can someone please check this for me. and IF i'm wrong can they please shwo me how its done? but im preety confident.

DraconisV

 

[jennylim]

we have a song for empirical formula.

percentage to mass,
mass to mole,
divide by small,
multiply till whole.

it works. like just take the percentages and assume there are 85.15 g of carbon or whatever. then work out how many moles are there (85.15/12.01) for all of them, then work out the ratio by dividing all of them by the smallest number of moles there.

 

[Dreamerish*~]

Ok, I have done this question. I now have came up with the answer of 1.38g of Silver Carbonate formed. Can someone please check this for me. and IF i'm wrong can they please shwo me how its done? but im preety confident.

DraconisV

Sorry to say, that's incorrect. I triple checked to make sure I didn't make a mistake. I'm sorry for taking so long to notice the question. You have probably already figured it out but anyway, the equation is:

2AgNO_3(aq) + Na₂CO_3(aq) → Ag₂CO_3(s) + 2NaNO_3(aq)

We have 0.025 x 0.2 = 0.005 moles of AgNO₃.
We have 0.030 x 0.15 = 0.0045 moles of Na₂CO₃.

You'd probably look at it and think that AgNO₃ is in excess, but if you look at the balanced equation, the molar ratio of AgNO₃ to Na₂CO₃ is 2:1.

This means that every mole of Na₂CO₃ will react with 2 moles of AgNO₃.

That would mean we need 0.0045 x 2 = 0.009 moles of AgNO₃, but we only have 0.005 moles. Hence Na₂CO₃ is in excess.

The molar ratio of AgNO₃:Na₂CO₃:Ag₂CO₃ is 2:1:1, which is 0.005:0.0025:0.0025 as we have only 0.005 moles of AgNO₃.

Molar mass of Ag₂CO₃ is 107.9 x 2 + 12 + 16 x 3 = 275.8 g

The mass of the Ag₂CO₃ is then 275.8 x 0.0025 = 0.69 (2 d.p).

 

[DraconisV]

You'd probably look at it and think that AgNO₃ is in excess, but if you look at the balanced equation, the molar ratio of AgNO₃ to Na₂CO₃ is 2:1.

You are Correct their. It now makes sense, oh my god i cant believe i made such a stupid mistake ahh, ive been spending to much time with one of my chem buddies(lol, yearlies result:34%) ahhh

 

[Dreamerish*~]

You are Correct their. It now makes sense, oh my god i cant believe i made such a stupid mistake ahh, ive been spending to much time with one of my chem buddies(lol, yearlies result:34%) ahhh

You have to watch out for these things. People who write up chemistry exams can be very very sneaky.

Always balance your equations, and double check your calculations. Also, use some logic - say if you have 10 mL of AgNO₃ and 10 mL of NaCl reacting and your mass of the AgCl precipitate is 30000 g (ok, exaggeration. ), then obviously there's something wrong. Don't just scribble down numbers. Look at them and ask yourself if they look right.