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An Unknown Compound

by haengbon
Tags: atomic, chemistry, element, weight
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haengbon
#1
Oct29-09, 11:14 AM
P: 38
1. The problem statement, all variables and given/known data

A compound containing element A and oxygen has a mole ratio of A:O = 2:3. If 8.0 grams of the oxide contains 2.4 grams of oxygen...

a) what is the atomic weight of A?
b) what is the weight of one mole of the oxide?
c) what is the theoretical weight of the oxide formed when 28 g of A is heated in excess oxygen?
d) what is the % yield if 38 grams was produced from 28 grams of A?

2. Relevant equations

....

3. The attempt at a solution

I didn't have any because I barely understood what to do :( can someone please guide me on how to solve this? It'll be of great help ^^
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symbolipoint
#2
Oct29-09, 11:22 AM
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P: 2,796
Start with the oxygen. How many moles of oxygen is equivalent to that 2.4 grams of oxygen from the compound?
haengbon
#3
Oct30-09, 10:09 PM
P: 38
so I'll convert the oxygen grams to moles right? :)

Borek
#4
Oct31-09, 04:29 AM
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P: 23,386
An Unknown Compound

Yes. Then try to find out how many moles of the oxide you have (look at the formula).
haengbon
#5
Mar13-10, 04:31 AM
P: 38
8.0 - 5.6 = 2.4 gO

2.4 gO x 1 mole O / 16 gO = 0.15 moles O

A:O = 2:3
A:0.15 = 2:3


[tex]\frac{A}{0.15}[/tex] x [tex]\frac{2}{3}[/tex]


3A = (0.15)(2)
3A = 0.30
A = 0.1 moles Oxygen

a)
AW = g/mole
AW = 5.6 gO / 0.1 moles O
AW = 56 g/mole

b)
2:3
A2O3
= (56)(2) + (16)(3)
= 160 g

c)
A + O2 [tex]\rightarrow[/tex] A2O3
4A + 3O2 [tex]\rightarrow[/tex] 2A2O3

1 mole A x 2moles A2O3 / 4moles A = 0.5

.5 moles A x 2 moles A2O3 / 4moles A = 0.25

LR: 0.25
= 0.25 x 160
= 40 g

d)
% = [tex]\frac{actual}{theoretical}[/tex] (100)

% = [tex]\frac{38}{40}[/tex] (100)

% = 95
Is this correct? I'm sure with my a & b but with c and d, I don't really get how that happened :( I asked for help in solving this, and this was the solution he gave.
Borek
#6
Mar13-10, 05:54 AM
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Quote Quote by haengbon View Post
3A = (0.15)(2)
3A = 0.30
A = 0.1 moles Oxygen
No - this is not "moles of Oxygen". This is number of moles of A in the sample.

[quote]AW = g/mole
AW = 5.6 gO / 0.1 moles O
AW = 56 g/mole[/quot]e

Here it becomes obvious - 56 g/mol is not molar of oxygen. It is molar mass of A.

160 g
Miraculously you get the correct answer, by ignoring incorrect information

Answers given for c & d are correct, but this is just a simple stoichiometry. You have a balanced reaction equation

4A + 3O2 -> 2 A2O3

that gives molar ratio, you know molar mass of A... I guess it can be symbolic A instead of some well defined element that throws you off. Would it be easier if I will tell you to solve c & d assuming A is the iron?

--
methods
haengbon
#7
Mar13-10, 09:46 PM
P: 38
Quote Quote by Borek View Post
No - this is not "moles of Oxygen". This is number of moles of A in the sample.

AW = g/mole
AW = 5.6 gO / 0.1 moles O
AW = 56 g/mole

Here it becomes obvious - 56 g/mol is not molar of oxygen. It is molar mass of A.
oops! I'm very sorry for that :)) the one in my paper was 0.1 moles of A :D

Miraculously you get the correct answer, by ignoring incorrect information
haha xD


Answers given for c & d are correct, but this is just a simple stoichiometry. You have a balanced reaction equation

4A + 3O2 -> 2 A2O3

that gives molar ratio, you know molar mass of A... I guess it can be symbolic A instead of some well defined element that throws you off. Would it be easier if I will tell you to solve c & d assuming A is the iron?
thank you! um, but the part that I'm really confused about is this one ...

4A + 3O2 -> 2 A2O3

if we take the balancing off..
A + O2 -> A2O3

Where did we get A + O2 ? Is this common knowledge? because all I understood was A2O3 since we had a mole ratio of 2:3 right? :D
Borek
#8
Mar14-10, 05:02 AM
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Not sure what your problem is. There are many ways of producing oxides, but direct reaction between the element and oxygen is the simplest one (even if technically not always possible). Even if it is not possible, you were told in the question that 38 grams of A2O3 were produced from 28 grams of A - no matter what the real reaction was, molar ratio of the A and A2O3 will be always the same.

Then A2O3 was given, O[ub]2[/sub] is a common knowledge, A is just the simplest way of approaching the problem.


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