Question

Heres the question

Answer 1

@Callisto @shivam_bhalla

Answer 2

M O2
Mass 2.0 (2.8-2.0=0.8)
No. of mole 0.05 0.8/(16x2) =0.025
Mole ratio 2 1
So, M : O2 = 2:1
2M + O2 -> 2MO
Right?

Answer 3

it is MO only

Answer 4

The oxide is MO, but since I wrote it in the equation form, I needed to balance it, so I wrote 2 before it, meaning that for 2 moles of M reacting with 1 mole O2, 2 moles of MO will be produced.

Answer 5

okay :)....can u do this method with the method (u told in the previous)

Answer 6

What do you mean?

Answer 7

xM+(y/2)O2---> MxOy

Answer 8

Sorry, I'm not the one who used that method :|

Answer 9

u know it anyway :)

Answer 10

Hmm... I don't know :|
xM + (y/2) O2 -> MxOy
no. of mole of O2 used = (2.8-2.0)/(16x2) = 0.025
So, y = 0.025 (2) = 0.05
no of mole of M used = 2.0/40 = 0.05
So, x = 0.05 and y = 0.05
So, x=y , x:y = 1:1
So, empirical formula for metal oxide is MO ._.
I prefer the method of comparing the mole ratio to this...

Answer 11

"no. of mole of O2 used = (2.8-2.0)/(16x2) = 0.025
So, y = 0.025 (2) = 0.05"

can u explain this part?

Answer 12

no. of mole = mass / molar mass => got it?

Answer 13

yeah i know thaht, the calculation part?

Answer 14

mass of O2 = 2.8-2.0
molar mass of O2 = 16.0x 2

Answer 15

just curious , can you tick all?

Answer 16

then the y part

Answer 17

That's the part I'm not quite sure ...
I thought no. of mole of O2 = y/2
But later (just now) I realised that it's just the mole ratio...

Answer 18

hm...this method is not applicable then?

Answer 19

I don't know. I rarely use this method in doing such question.
Usually, I use it when I do question related to combustion of hydrocarbons.
I'm sorry!

Answer 20

hm okay...anyway thanks alot :D

Answer 21

Let's fetch help from @Kayne :)
I hope she will notice this when she goes online

Answer 22

I get MO

Answer 23

Equation
\[M+O_2->M_xO_y\]
You know that all the mass of M reacted will be found in M_xO_y,
So you just substract the 2.8g with 2g from MO and get the grams of oxygen, which is 0.8g of oxygen,
Then, use empirical formula,

M O
mass(g) 2 0.8
moles(mol) 0.05 0.05
Ratio 1 1

MO

Answer 24

@Callisto ..he is using O---> not O2

Answer 25

Okay, I'll retake chemistry (if possible)

Answer 26

what do u mean?

Answer 27

I'll retake chemistry -> I'm really bad in chemistry that I think I need to study it again (but probably I end up doing the same question using the same method)

(if possible) -> Only if I need to retake my secondary 6 and do the exam again, or in university, I choose chemistry-related subjects. (That's not very true, since I'll ask my teacher about this as soon as possible)

Actually, I'm confused with his equation. Did he hide the '2'?
Sorry I really haven't learnt doing that question in that way...
*cooking again* brb

Answer 28

You have to use O only not O2 because you cant balance the equation

Answer 29

*I really dislike google :S*

the balanced equation should be
\[M + \frac{y}{2}O_2 -> M_xO_y\]
in that case, y/2 is like ''being hidden'' as 2 in O2 is cancelled when multiplying (y/2), which gives y O
So, in the calculation, 2 is not taken into account.
I don't know if this explanation is correct (and probably not correct), but this is the only way I can persuade myself so far.

Answer 30

Sorry for confusing you all with this method.. :S

xM+(y/2)O2--->MxOy

x mol of M give 1 mole of MxOy
(40x)g of M give (40x+16y) of MxOy
2g of M give [((40x+16y)/40x)x2.0]=2.8g of MxOy

40g of M =1 mole of M
2g of M=2/40 = 0.05

x= 0.05
Multiplying by 20,
x=1

replacing x=1 in the equation:
[((40x+16y)/40x)x2.0]=2.8g
y=1

Hence, MO

Answer 31

wow ur back..good that u came before my exam..:D

Answer 32

@zaphod if my method is confusing.. maybe you should use that of @callisto...
either way.. you'll get the answer..

Answer 33

http://www.twiddla.com/845236
can u come here{to the link}...can u explain that method clearly, coz i know the method is reliable..Please if u dont mind

Answer 34

actually, I'm kinda busy right now.. and I'll have to leave in a moment.. please use the two examples on which we've all worked on.. you'll get the pattern..

Answer 35

hm alright..thanks anyway:)

Answer 36

sorry:/.. good luck for your exams..;)

Answer 37

thanks:)