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How to get atoms of something having % of mass?

  • Thread starter HCverma
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  • #1
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Homework Statement


Two oxides of a metal contain 27.6% and 30% of oxygen respectively. If the formula of the first oxide is MO, find that of the second?
Now my problem is how to find the atoms of O from the masses of 27.6% and 30.0 %?

Homework Equations




The Attempt at a Solution


I can see that in MO, there is one atom of O in the first oxide, So 27.6 = 1 oxygen atom?
 

Answers and Replies

  • #2
.Scott
Homework Helper
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This is just deaing with ratios. You don;t even need to compute the AMU's.
Set up your equations. Om and Mm are your AMUs.

So you have:
Om = 0.276 (Om+Mm)
and you are looking for an integer solution for On and Mn to:
OnOm = 0.3 (OnOm+MnMm)

Take it from there.
 
  • #3
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This is just deaing with ratios. You don;t even need to compute the AMU's.
Set up your equations. Om and Mm are your AMUs.
I don't understand the step above. Could you get it a little bit easier, please?
 
  • #4
.Scott
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Let's see if I can be more helpful without spoiling the answer...

First, let me change my nomenclature to avoid confusion.
I will represent the AMUs as ##Mass_O## and ##Mass_M##

So you have: ##Mass_O = 0.276 (Mass_O+Mass_M)## thus ##0.724 Mass_O = 0.276Mass_M)##

Start by forgetting about the integer issue.
So we're looking for ##OM_X## where X is the number of metal atoms in the molecule - and we are going to let that be a non-integer for the time being.
So we have: ##Mass_O = 0.3(Mass_O+XMass_M)## thus ##0.7Mass_O = 0.3XMass_M##

Now for you:
* compute the ratio ##Mass_M/Mass_O## from the 1st equation above;
* use that ratio to compute X in the second equation;
* you should recognize X as being pretty close to a simple integer ratio;
* apply the numerator and denominator of that simple ratio to ##O_dM_n##

One more hint:
The numerator and denominator are each larger than 3.

Also:
If you need more assistance, show me how far you got in attempting to follow those four steps I just provided.
 
  • #5
82
2
Let's see if I can be more helpful without spoiling the answer...

First, let me change my nomenclature to avoid confusion.
I will represent the AMUs as ##Mass_O## and ##Mass_M##

So you have: ##Mass_O = 0.276 (Mass_O+Mass_M)## thus ##0.724 Mass_O = 0.276Mass_M)##

.
How do you get the above equeation? As I know Mass(O) = 0.276 and Mass(M) = 100 - 0.276 = 0.724.
Why have you written '##Mass_O = 0.276 (Mass_O+Mass_M)## thus ##0.724 Mass_O = 0.276Mass_M)##'? it should be only ##Mass_O = 0.276 and ##Mass_M = 0.724 Could you simplfy youe equation please?
 
  • #6
.Scott
Homework Helper
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We know that the mass of the Oxygen is 27.6% of the total mass. Thus:
##Mass_O = 0.276(Mass_O+Mass_M)##

Expanding that out:
##Mass_O = 0.276Mass_O+0.276Mass_M##

Subtracting ##0.276Mass_O## from each side of the equation:
##Mass_O - 0.276Mass_O = 0.276Mass_M##

Thus:
##0.724Mass_O = 0.276Mass_M##
 
  • #7
82
2
We know that the mass of the Oxygen is 27.6% of the total mass. Thus:
##Mass_O = 0.276(Mass_O+Mass_M)##

Expanding that out:
##Mass_O = 0.276Mass_O+0.276Mass_M##

Subtracting ##0.276Mass_O## from each side of the equation:
##Mass_O - 0.276Mass_O = 0.276Mass_M##

Thus:
##0.724Mass_O = 0.276Mass_M##
Could you clarify me here How you get '##0.724Mass_O = 0.276Mass_M##' from '##Mass_O - 0.276Mass_O = 0.276Mass_M##'?
 
  • #8
.Scott
Homework Helper
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Could you clarify me here How you get '##0.724Mass_O = 0.276Mass_M##' from '##Mass_O - 0.276Mass_O = 0.276Mass_M##'?
It's the Distributive Property of Multiplication.

##Mass_O - 0.276Mass_O = 0.276Mass_M##
##1.0Mass_O - 0.276Mass_O = 0.276Mass_M##
##(1.0 - 0.276)Mass_O = 0.276Mass_M##
##(0.724)Mass_O = 0.276Mass_M##

What is your grade level? You should have had this in High School Algebra class.
 
  • #9
82
2
It's the Distributive Property of Multiplication.

##Mass_O - 0.276Mass_O = 0.276Mass_M##
##1.0Mass_O - 0.276Mass_O = 0.276Mass_M##
##(1.0 - 0.276)Mass_O = 0.276Mass_M##
##(0.724)Mass_O = 0.276Mass_M##

What is your grade level? You should have had this in High School Algebra class.
I passed high school but I am sorry to say, I am weak at math.
 
  • #10
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I passed high school but I am sorry to say, I am weak at math.
Then this is something you should rectify, sooner rather than later. As you should are seeing, your weakness in elementary algebra is preventing you from being proficient in parts of chemistry, particularly in balancing chemical reaction equations.

You should think about bolstering your math skills. One site that might be helpful is khanacademy.org. They have lots of videos in algebra, chemistry, and many more subjects.
 
Last edited:
  • #11
82
2
Then this is something you should rectify, sooner rather than later. As you should are seeing, your weakness in elementary algebra is preventing you from being proficient in parts of chemistry, particularly in balancing chemical reaction equations.

You should think about bolstering your math skills. One site that might be helpful is khanacademy.org. They have lots of videos in algebra, chemistry, and many more subjects.
Thanks a lot.
 
  • #12
82
2
Let's see if I can be more helpful without spoiling the answer...

First, let me change my nomenclature to avoid confusion.
I will represent the AMUs as ##Mass_O## and ##Mass_M##

So you have: ##Mass_O = 0.276 (Mass_O+Mass_M)## thus ##0.724 Mass_O = 0.276Mass_M)##

Start by forgetting about the integer issue.
So we're looking for ##OM_X## where X is the number of metal atoms in the molecule - and we are going to let that be a non-integer for the time being.
So we have: ##Mass_O = 0.3(Mass_O+XMass_M)## thus ##0.7Mass_O = 0.3XMass_M##

Now for you:
* compute the ratio ##Mass_M/Mass_O## from the 1st equation above;
If you need more assistance, show me how far you got in attempting to follow those four steps I just provided.
the ratio is 181: 69
 
  • #13
.Scott
Homework Helper
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861
Yes.
That is the first ratio. That's step 1.
2.6232

Now you should be able to use ##Mass_M/Mass_O = 2.6232## with the equation ##0.7Mass_O=0.3XMass_M## to solve for ##X##.
 
Last edited:
  • #14
82
2
Yes.
That is the first ratio. That's step 1.
2.6232

Now you should be able to use ##Mass_M/Mass_O = 2.6232## with the equation ##0.7Mass_O=0.3XMass_M## to solve for ##X##.
Hw to solve X? X = 0.7 Masso / 0.3 Massm but here Massm / Masso = 2.6232. How to put the value in that equeation?
 
  • #15
.Scott
Homework Helper
2,464
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Hw to solve X? X = 0.7 Masso / 0.3 Massm but here Massm / Masso = 2.6232. How to put the value in that equation?
It will be a bit easier for us to work with ##Mass_O/Mass_M## than ##Mass_M/Mass_O##.
So let's start there: ##Mass_M/Mass_O = 2.6232##
Invert each side of that equation: ##Mass_O/Mass_M = 1/2.6232 = 0.38122##

Now back to this equation: ##0.7Mass_O = 0.3XMass_M##
What do we have to do to the left side of that equation to change it from ##0.7Mass_O## to ##Mass_O/Mass_M##?
Whatever you do to it, you have to do the same to the other side of that equation.
For example, if I added 99 to ##0.7Mass_O##, I would have to add 99 to ##0.3XMass_M## to keep both sides of the equation equal.
If I did that, I would have: ##0.7Mass_O + 99 = 0.3XMass_M + 99##. But that doesn't do us any good because we are looking to get ##Mass_O/Mass_M## on one side.

So what operation should we perform? Hint: It's not an addition or subtraction.

Also: I noticed that @Mark44 has made a suggestion in reply #10. Please take it to heart.
 

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