gen chem quest that are similar to the DAT

[reliefe123]

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1)what mass of carbon is present in .500mol of sucrose C12H22O6 (mm=342)
the answer is 72g??i keep getting .84 when i try and get the percent by mass of carbon!

2)a sample of a compound of xenon and flouirne contains molecules of a single type; XeFn, where n is a whole number. If 9.03x10^20 of these XeFn moleules have a mass of .311g , what is the value of n??
can someone explain it like im in the second grade please

3)what mass of carbon is present in 1.4x10^20 molecules of sucrose C12H22O6 (mm=342) the answer is 3.3 x10^-2...again i get close but no cigar...

4)a 2 g sample of an unknown metal,M, was completely burned in excess O2 to yield .02224 mol of the metal oxide, M2O3. what is the metal...answer is Sc

if you can please help me and explain like im 10...thanks

 

[UCLAzy]

for 1, pretty easy. for 1 mol, there would be 12*12.01 g of C, so for .5 mol there would be (12*12.01)/2 = 72g

 

[karrotcake]

#1 I did like this:

How much sucrose you currently have = 0.5mol(342g/1mol) = 171g C12H22O6

How much C is in one sucrose? 12mol C (12g/1mol) = 144g C per sucrose. So 144g C / 342g sucrose = 0.42 = sucrose made of 42% C

How much C (mass of C) is in the amount of sucrose you currently have? 42% of 171g = 71.82g ~ 72g

 

[karrotcake]

actually UCLazy's way is way easier lol.

 

[UCLAzy]

3 is the same concept as 1, except you have to use avagadro's number to find the number of moles. so there are (1.4x10^20/6.022x10^23) = 2.324x10^-4 mols

then using the same method as 1, we have (12x12.01)(2.324x10^-4) = .033

i left units out but you can hopefully figure it out

 

[UCLAzy]

for 2, you approach same way as 3, except go a little further

so you use avagadro's number again to go from molecules to mols

(9.03x10^20)/(6.022x10^23) ~ .0015mol of the unknown compound

so you know the atomic mass of Xe = 131.29 and you know the molar mass of F is 18.998 so you set up the equation

[(131.29g/mol)+(18.998x)]*[.0015] = .311

solve for x and you get 4

 

[UCLAzy]

4 is alot simpler than it looks too.

the key is to note that the product is M2O3, which means that 2 will be the reactant M's coefficient in the balanced rxn

from there, you say that the mols of M are 2x the mols of M2O3, or .04448mol

since you have the grams of M and the mols of M, you can find the molar mass of M by

(2g)/(.0448mol) = 44.96g/ mol, which corresponds to Sc on the periodic table

 

[UCLAzy]

again..i dont like to usually put units because i know what they are already, but if you get confused, just put the units in on any of these problems and you'll see how they cancel to make sense

when i see a problem i dont know how to do, the first thing i do is see what units are given and what units are needed. then you just need to find an equation to cancel the right units and you can plug in the numbers at the end

 

[reliefe123]

again..i dont like to usually put units because i know what they are already, but if you get confused, just put the units in on any of these problems and you'll see how they cancel to make sense

when i see a problem i dont know how to do, the first thing i do is see what units are given and what units are needed. then you just need to find an equation to cancel the right units and you can plug in the numbers at the end

thanks bro...i get it now...1 and 3 where super easy i was trying to find the percent by mass by taking the total C and dividing by the MM of the compound...but i just needed the mass of C and did not realize that....as for 2 and 4...i get the gist of it...but i dont think i would be able to answer those on test day...i hate stoichiometery