How to get atoms of something having % of mass?

• HCverma
In summary, Homework Equations: - The first oxide of a metal contains 27.6% and 30% of oxygen respectively. - To solve for On and Mn, set up equations and forget about the integer issue. - X in the second equation is close to a simple integer ratio. - Apply the numerator and denominator of that simple ratio to ##O_dM_n##.

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 %?

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?

This is just deaing with ratios. You don;t even need to compute the AMU's.

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.

.Scott said:
This is just deaing with ratios. You don;t even need to compute the AMU's.
I don't understand the step above. Could you get it a little bit easier, please?

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.

.Scott said:
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?

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##

.Scott said:
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##'?

HCverma said:
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##

.Scott said:
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##

I passed high school but I am sorry to say, I am weak at math.

HCverma said:
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:
.Scott
Mark44 said:
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.

.Scott said:
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

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:
.Scott said:
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?

HCverma said:
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.

1. How can I determine the number of atoms in a substance based on its mass percentage?

To determine the number of atoms in a substance based on its mass percentage, you will need to use Avogadro's constant (6.022 x 10^23) and the molar mass of the substance. The equation for calculating the number of atoms is: Number of atoms = (mass percentage / 100) x (mass of substance in grams) / (molar mass of substance in grams)

2. Can the mass percentage of a substance change over time?

Yes, the mass percentage of a substance can change over time due to factors such as evaporation, chemical reactions, and decay. It is important to take into account any changes in mass percentage when conducting experiments or analyzing data.

3. How do I convert between mass percentage and other units of concentration?

Mass percentage is typically expressed as a percentage by mass (g/g) or sometimes as parts per million (ppm). To convert between these units and other units of concentration, such as molarity (mol/L) or molality (mol/kg), you will need to use the molar mass of the substance. The equations for converting between these units can be found in most chemistry textbooks.

4. Can the mass percentage of a substance be greater than 100%?

No, the mass percentage of a substance cannot be greater than 100%. This percentage represents the proportion of the substance's mass in relation to the total mass of the mixture or solution. Therefore, it cannot exceed 100%.

5. How does the mass percentage of a substance affect its properties?

The mass percentage of a substance can greatly affect its properties, as it determines the amount of that substance present in a mixture or solution. For example, a higher mass percentage of a solute in a solution will result in a more concentrated solution, leading to differences in properties such as boiling point and density. Additionally, different mass percentages of a substance can also impact its reactivity and stability.