Homework questions thread

[soccerboi]

Hi everyone, feel free to post any holiday hw/hw questions that u hav on this thread and we can all help each other out!

Here are some problems that i need help with:

1. A compound of carbon and hydrogen is found by analysis to contain 92.26% of carbon. This substance is a gas at 200 degrees celcius, 101 300 Pa pressure, and under these conditions has a density of 2.012 g dm^-3. calculate the molecular formula of the compound. Ans is C6H6

2. When 1g of a compound, known to contain only carbon, hydrogen, and oxygen, is burned in air, 1.910g of carbon dioxide and 1.173g of water are formed. The vapour of this compound is 1.64 times as dense as nitogen under the same conditions of temperature and pressure. what is the molecular formula of the compound? Ans is C2H6O

3. 2g phosphorous is burned in air and forms 4.582g of an oxide. This quantity of the oxide is found to react with exactly 1.744g of water to give a compound of hydrogen , phosphorous, and oxygen. Determine the empirical formula of the oxide and that of the other compound. Ans is P2O5;H3PO4

[Hamdog17]

Ok so
Per 100g of substance 92.26g is carbon
Therefore n(C)=92.26/12=7.688 mol
n(H)=(100-92.26)/1=7.74 mol
:. n(H):n(C). 7.688:7.74.   roughly 1:1
PV=nRT.     n=m/M
:.  PV=(mRT)/M
Transpose for M    :. M=(mRT)/PV.   However density=d=m/V
M=(dRT)/P
M=(2.012*8.31*(200+273))/(101.3)=78.069 g/mol
Therefore 78.069/(12+1)=6.   Therefore 6 units of CH
Therefore molecular formula is C6H6

[soccerboi]

Thank you so much

[soccerboi]

Could someone tell me the formula for iron(II) ammonium sulfate hexahydrate.

[dc302]

Is this what you're looking for? http://www.chemblink.com/products/7783-85-9.htm

[soccerboi]

yes, thanks.

[samad]

q. 2: find the empircal formula as per usual. you get C₂H₆O.
Now, let density of the compound be and density of nitrogen be .


since ,

now,

deriving from the general gas equation:


cancel pressures and temperatures on either side since they are equal in each case



hence

knowing this, we can cancel out the n's and v's from expression (1)

hence

the molar mass of the compound is 1.64*M(N2)
=1.64*28
=46

compare to the molar mass of the emperical formula (46 as well)

hence molecular formula is one unit of C2H6O. i.e the same

[Hamdog17]

So for q 3,
m(P)=2g. n(P)=m/M=2/31=0.06452mol
n(H2O)=1.744/18=0.09689mol
m(Oxygen in oxide)=4.582-2=2.582g
n(O)=2.582/16=0.16138mol
:. n(P):n(O).  0.06452:0.16138.  1:2.5
:. 2:5. 
:. The empirical formula of the compound is P2O5

n(P2O5)=0.5*n(P)=0.5*0.06452=0.03226mol
:. n(P2O5):n(H2O). 0.03226:0.09689.  1:3.
Now form a chemical equation (we know the substance formed involves H,O and P.):
P2O5(s) + 3H2O(l) ---> 2H3PO4 (aq)
Hence the second substance is H3PO4.

[soccerboi]

Thank you everyone,i really appreciate how u guys are willing to spend ur own time to help me out

[soccerboi]

In regards to Q3, where did the 0.5 come from? (n(P2O5)=0.5*n(P)=0.5*0.06452=0.03226mol)

[pi]

In regards to Q3, where did the 0.5 come from? (n(P2O5)=0.5*n(P)=0.5*0.06452=0.03226mol)

P2O5

Hence, the no. of moles of P2O5 must be half that of P

[soccerboi]

Q4. 20.2g of a crushed ore, which contains iron(III) oxide, is reacted with excess carbon monoxide. 11.2g of iron results from 100% reduction of the metallic oxide. Calculate the percentage of iron oxide in the original ore sample. Ans is 79.3%

And also, a quick question: Is HCl the formula for hydrochloric acid or hydrogen chloride or is it for both?
Thanks

[Aurelian]

Q4. 20.2g of a crushed ore, which contains iron(II) oxide, is reacted with excess carbon monoxide. 11.2g of iron results from 100% reduction of the metallic oxide. Calculate the percentage of iron oxide in the original ore sample. Ans is 79.3%

And also, a quick question: Is HCl the formula for hydrochloric acid or hydrogen chloride or is it for both?
Thanks

HCl (aq) is hydochloric acid, hydrogen chloride is generally HCl (g); the state is the defining feature here.

As for your other question;

Q4. 20.2g of a crushed ore, which contains iron(II) oxide, is reacted with excess carbon monoxide. 11.2g of iron results from 100% reduction of the metallic oxide. Calculate the percentage of iron oxide in the original ore sample. Ans is 79.3%

REACTION: Fe2O3 (s) + 3CO (g) ---> 2Fe (s) + 3CO2 (g)

CO is in excess, therefore Fe2O3 is the limiting reagent.

n(Fe)produced = m(Fe)/M(Fe) = 11.2g/55.9gmol-1 = 0.200mol
n(Fe):n(Fe2O3) = 2:1
Therefore, n(Fe2O3) = 1/2 x n(Fe) = 0.100mol
m(Fe2O3) = M(Fe2O3) x n(Fe2O3) = 159.8gmol-1 x 0.100mol = 15.98g = 16.0g

m(crushed ore) = 20.2g
% iron (II) oxide = 16.0/20.2 x 100 = 79.3% to 2 sig figs

EDIT: Just noticed that I did iron (III) oxide... yet you get the same answer... Lol. Weird. Anyway method is the same. Just use the below equation instead;

FeO (s) + CO (g) --> Fe (s) + CO2 (g)

EDIT 2: Just did it with the above equation and you get like 71.2%, so I'm thinking this question actually means iron (III) oxide anyway

[soccerboi]

Thanks Aurelian, and yeh i just checked the question, its actually iron(III) oxide, I mistyped lol...

[Bismuth]

My school loves being tricky with HCl on SACS/exams. For example, a question on a SAC would be a substitution reaction between ethane and chlorine gas. You'd state the conditions that are needed for the type of reaction to occur, write a structural equation for the reaction and everything would be straight forward even for the not so diligent of students.

Now, they throw in the question that hopefully distinguishes the masses. "Provide a systematic name for each of the products". A lot of students who see the H-Cl that they wrote above quickly scribble down "hydrochloric acid". However, the reaction takes place in a gaseous state and the products that are produced are also gaseous, so the correct answer is "hydrogen chloride gas". \

The same also applies for all hydrogen halides (such as HF), so be weary.