Solving Unknown Compound: A:O = 2:3, 8.0g O, 28g A

[haengbon]

Homework Statement

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?

Homework Equations

...

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 ^^

 

[symbolipoint]

Start with the oxygen. How many moles of oxygen is equivalent to that 2.4 grams of oxygen from the compound?

 

[haengbon]

so I'll convert the oxygen grams to moles right? :)

 

[Borek]

Yes. Then try to find out how many moles of the oxide you have (look at the formula).

 

[haengbon]

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


\frac{A}{0.15} x \frac{2}{3}


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 \rightarrow A₂O₃
4A + 3O₂ \rightarrow 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)
% = \frac{actual}{theoretical} (100)

% = \frac{38}{40} (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]

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.

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]

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]

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.