How to get moles from MO having % of mass?

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  • #1
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Homework Statement


This is a question from a chemistry book
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?

3. The Attempt at a Solution

In the first MO, if the total amount is 100g, then O is 27.6g similarly,
in the second MO, if the total amount is 100g, then O is 30g respectively
1.725 mole of O is 27.6g.
in the MO, 72.4g of M left.
But my question is how to calculate the mole of M? because I don't know the molar mass of M
 

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  • #2
Ygggdrasil
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You have sufficient information to calculate the molecular weight of M. For 100g of MO, you calculated that you have 1.725 mol of oxygen. Given the chemical formula, how many moles of M do you have?
 
  • #3
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You have sufficient information to calculate the molecular weight of M. For 100g of MO, you calculated that you have 1.725 mol of oxygen. Given the chemical formula, how many moles of M do you have?
My question is how to calculate the mole of M? because I don't know the molecular mass of M
 
  • #4
Ygggdrasil
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What does the chemical formula MO tell you about the ratio of moles of oxygen to moles of M?
 
  • #5
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What does the chemical formula MO tell you about the ratio of moles of oxygen to moles of M?
If we break MO, we get 1 mole of O and 1 mole of O So O: M = 1:1
 
  • #6
Ygggdrasil
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Correct Can you use that information along with what you have previously calculated to determine the molar mass of M?
 
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  • #7
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Correct Can you use that information along with what you have previously calculated to determine the molar mass of M?
Yes molar mass of M = (100 -27.6) = 72.4g
So, 1 mol of M weighs 72.4g
 
  • #8
By that reasoning, molar mass of O is 27.6. Try again.

Here's a hint. With MO, you have 1 mol of M for EACH MOL of O. For your initial 100g of MO, how many mols of O do you have, how many mols of M do you have, what is the mass of that number of mols, and, therefore, what is the mass of a single mol of M?
 
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  • #9
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By that reasoning, molar mass of O is 27.6. Try again.

Here's a hint. With MO, you have 1 mol of M for EACH MOL of O. For your initial 100g of MO, how many mols of O do you have, how many mols of M do you have, what is the mass of that number of mols, and, therefore, what is the mass of a single mol of M?
When O is 27.6%, in 100g MO, M = 1 mole = 72.4g and O = 1.725 mole = 27.6g
when O is 30%, in 100g MO, M = 1 mole = 70g and O = 1.875 mole = 30g
but how to calculate 'what is the mass of a single mol of M?'
 
  • #10
symbolipoint
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Homework Statement


This is a question from a chemistry book
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?

3. The Attempt at a Solution

In the first MO, if the total amount is 100g, then O is 27.6g similarly,
in the second MO, if the total amount is 100g, then O is 30g respectively
1.725 mole of O is 27.6g.
in the MO, 72.4g of M left.
But my question is how to calculate the mole of M? because I don't know the molar mass of M
I have not yet analyzed the arithmetic carefully.

If you have TWO DIFFERENT metal oxides of the particular metal M, corresponding to two different percentages of oxygen element in the "molecule", then you should expect that one of your metal oxides may correspond to MO but the other will NOT be a one-to-one mole relationship between the Oxygen and the metal M.

You are merely looking for a ratio for the second compound which does not use very small whole numbers, but slightly or more larger whole numbers.
 
  • #11
symbolipoint
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Note, there are two compounds with M and O described. One is MO and the other is of unknown atom ratio of M and O. You found the correct moles oxygen of 100 g sample, for MO compound.

0.276*100/15.9994=1.725 moles of Oxygen.

This means, 1.725 moles M for 100 g of MO.
How many grams of M?
From 72.4 grams of the 100 g being M,
72.4/1.725=41.971 grams per mole, M.

You already know that 15.9994 grams per mole, O.

You next want to find the mole ratio for the unknown M and O compound which is 30 % Oxygen; and is 70 % M. Consider a sample of 100 grams.
Of this sample, 30 grams is Oxygen and 70 grams is M.
How many moles of each?
Can you then try to find a good whole number ratio for each element to match?

My values found of each M and O:
--
70/41.971=1.667818 moles M
--
30/15.9994=1.87507 moles O
--
These raw values not yet as whole values for the compound, M1.667818O1.87507
and you need to try to find a good equivalent of that which uses whole numbers.
 
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  • #12
symbolipoint
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Review post #11.
I tried this as a way to find best whole number values:

  1. Divide both values by 1.667818
  2. Multiply both values by 8

Finding something close to M8O9 .
 
  • #13
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Note, there are two compounds with M and O described. One is MO and the other is of unknown atom ratio of M and O. You found the correct moles oxygen of 100 g sample, for MO compound.

0.276*100/15.9994=1.725 moles of Oxygen.

This means, 1.725 moles M for 100 g of MO.
I don't understand the point above 'This means, 1.725 moles M for 100 g of MO.' Could you simplify it, please? As I know 1.725 mole of O weighs.276g, so, how did you get 1.725 moles of M? it should be 1 mole of M and 1 mole of M weighs 72.4g. Please explain it. I am confused.
 
  • #14
symbolipoint
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Note, there are two compounds with M and O described. One is MO and the other is of unknown atom ratio of M and O. You found the correct moles oxygen of 100 g sample, for MO compound.
I don't understand the point above 'This means, 1.725 moles M for 100 g of MO.' Could you simplify it, please? As I know 1.725 mole of O weighs.276g, so, how did you get 1.725 moles of M? it should be 1 mole of M and 1 mole of M weighs 72.4g. Please explain it. I am confused.
I reemphasised that the compound MO, in regard to a.m.u.'s, has 1 a.m.u. of M and 1 a.m.u. of O.

To say this a little differently, 1 mole of MO has 1 mole of M and 1 mole of O.
 
  • #15
symbolipoint
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This is the key question to handle:
You have sufficient information to calculate the molecular weight of M. For 100g of MO, you calculated that you have 1.725 mol of oxygen. Given the chemical formula, how many moles of M do you have?
You would arrange the rest after finding the formula weight for the element M.
 
  • #16
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I reemphasised that the compound MO, in regard to a.m.u.'s, has 1 a.m.u. of M and 1 a.m.u. of O.

To say this a little differently, 1 mole of MO has 1 mole of M and 1 mole of O.
Ya, I understand it that 1 mole of MO has 1 mole of M and 1 mole of O. But still, I don't understand what you mean to say by ' 'This means, 1.725 moles M for 100 g of MO.' Could you simplify your point, please?
 
  • #17
Ygggdrasil
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Perhaps we can try to rephrase the question.

A 100 g sample contains an unknonwn number of moles of MO. The sample is 27.6% oxygen. How many moles of oxygen are present in the sample?

Once you figure that out you then should be able to answer how many moles of MO are present in the sample and how many moles of M are present in the sample.
 
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  • #18
You found that 100 grams of MO had 1.725 mol O. But by molecular formula, for each mol of O, you have one mol of M. Right? Formula of MO means one mol of each per molecule. Consider if you had started with 200 grams of MO instead of 100, and and found that you had 3.45 mol O instead of 1.725 mol. Would you expect that it must also have exactly 1 mol M? Molecular weight of M is fixed; as the mass of M changes, the number of mols must also change, and you must determine that number of mols based on the molecular weight.

If you have 1.725 mol O, you must have, identically, 1.725 mol M. (Also, identically, 1.725 mol of MO).

You also found that your 100 g MO contained 72.4 g M. Therefore, you have 1.725 mol M with a mass of 72.4 g; the gram-molecular weight is (72.4g/1.725 mol), or 41.971 g/g-mol.
 
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  • #19
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Perhaps we can try to rephrase the question.

A 100 g sample contains an unknonwn number of moles of MO. The sample is 27.6% oxygen. How many moles of oxygen are present in the sample?

Once you figure that out you then should be able to answer how many moles of MO are present in the sample and how many moles of M are present in the sample.
1.725 moles of O present in the sample. 1.725 moles of MO present in the solution. So 1.725 moles of M present in the sample
1.275 M + 1.375 O2 = 1.275 MO
 
  • #20
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You found that 100 grams of MO had 1.725 mol O. But by molecular formula, for each mol of O, you have one mol of M. Right? The formula of MO means one mol of each per molecule. Consider if you had started with 200 grams of MO instead of 100, and found that you had 3.45 mol O instead of 1.725 mol. Would you expect that it must also have exactly 1 mol M? The molecular weight of M is fixed; as the mass of M changes, the number of mols must also change, and you must determine that the number of mols based on the molecular weight.

If you have 1.725 mol O, you must have, identically, 1.725 mol M. (Also, identically, 1.725 mol of MO).

You also found that your 100 g MO contained 72.4 g M. Therefore, you have 1.725 mol M with a mass of 72.4 g; the gram-molecular weight is (72.4g/1.725 mol), or 41.971 g/g-mol.
I got it now.
Now my new questions are
1. 27.6g of O = how many atoms of O?
2. 72.4g M = how many atoms of M?
 
  • #21
I got it now.
Now my new questions are
1. 27.6g of O = how many atoms of O?
2. 72.4g M = how many atoms of M?
You know how many g-mols of each (1.725). And you (should) know Avagadro's number, which is the number of atoms per g-mol. With that information, it is left as an exercise for the reader....

But I don't see that this gets you closer to the answer for your original question, which can be restated "what is the molecular formula for an oxide of M and O that contains 30% O by weight."

That is, find "a" and "b" such that MaOb contains 30 wt% (0.3 mass fraction) O.

Can you find an algebraic formulation of the molecular weight of the oxide that allows you to calculate "a" and "b"? Or can you set up a table varying "a" and "b" with the resulting weight fraction of O? If you set it up in Excel, you should include the case where a=1 and b=1 with the result of 0.276 mass fraction O.
 
  • #22
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You know how many g-mols of each (1.725). And you (should) know Avagadro's number, which is the number of atoms per g-mol. With that information, it is left as an exercise for the reader....

But I don't see that this gets you closer to the answer for your original question, which can be restated "what is the molecular formula for an oxide of M and O that contains 30% O by weight."

That is, find "a" and "b" such that MaOb contains 30 wt% (0.3 mass fraction) O.

Can you find an algebraic formulation of the molecular weight of the oxide that allows you to calculate "a" and "b"? Or can you set up a table varying "a" and "b" with the resulting weight fraction of O? If you set it up in Excel, you should include the case where a=1 and b=1 with the result of 0.276 mass fraction O.
I got a solution to the problem above. Please let me know whether the solution is correct or wrong.
In the first oxide, oxygen = 27.6,
Metal = (100 - 27.6) = 72.4 parts of mass.
As the formula of the oxide is MO, it means 72.4 parts y mass of metal = 3 atoms of metal ( I don't know how to get 3 atoms from 72.4 parts)
in the second oxide, oxygen - 30.0 parts by mass. and metal = (100 - 30) = 70 parts by mass. But 72.4 parts by mass of metal = 3 atoms of metal
So, 70 parts by mass of metal = (3/72.4) X 70 atoms of metal = 2.90 atoms of metal
Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen (Again I don't know how to get 4 atoms from that 27.6 parts)
So, 30 parts by mass of oxygen = (4/27.6) X 30 atoms of oxygen = 4.35 atoms of oxygen
Hence, the ratio of M : O in the second oxide = 2.90 : 4.35 = 1 : 1.5 = 2 : 3
Therefore, the formula of metal oxide is M2O3
The problems are mentioned below from the solution above
1. As the formula of the oxide is MO, it means 72.4 parts y mass of metal = 3 atoms of metal ( I don't know how to get 3 atoms from 72.4 parts)
2. Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen (Again I don't know how to get 4 atoms from that 27.6 parts)
Here I don't know the above problems. Please help me solve these problems.
 
  • #23
symbolipoint
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Did my posts #11, 12, 14 help you at all? In case i made mistakes, can anyone point to what my mistakes were?
 
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  • #24
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Did my posts #11, 12, 14 help you at all? In case i made mistakes, can anyone point to what my mistakes were?
Yes. your posts helped me a lot to learn the concept. If you check my solution above and help me solve this problem in another way provided above, and help me understand the below concept from the solution above, It will be very helpful for me.
The problems are mentioned below from the solution above
1. As the formula of the oxide is MO, it means 72.4 parts y mass of metal = 3 atoms of metal ( I don't know how to get 3 atoms from 72.4 parts)
2. Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen (Again I don't know how to get 4 atoms from that 27.6 parts)
Here I don't know the above problems. Please help me solve these problems.
 
  • #25
Did my posts #11, 12, 14 help you at all? In case i made mistakes, can anyone point to what my mistakes were?
I thought your posts were perfectly clear, but did not resolve the OPs confusion between 1 amu of M, and 1 amu FOR EACH amu of O (or MO, for that matter).

1. As the formula of the oxide is MO, it means 72.4 parts y mass of metal = 3 atoms of metal ( I don't know how to get 3 atoms from 72.4 parts)
2. Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen (Again I don't know how to get 4 atoms from that 27.6 parts)
Here I don't know the above problems. Please help me solve these problems.
I do not know how to get "3 atoms" or "4 atoms," either. They certainly aren't contained in the 1:1 ratio between M and O in the original oxide.

What you are looking for is an oxide of MO with a formula MaOb, such that O comprises 30 mass percent of the compound.

Post 11 shows how to get the relative (non-integer) values of a and b, starting with an arbitrary mass of 100g. Post 12 pretty much gives the answer you are looking for....

Alternatively, what you can do is is calculate the MW of an arbitrary oxide, MaOb. For example, one mol MO is comprised of one mol M and one mol O. Thus, the molecular weight of MO (the original oxide) is 15.9994 + 41.971 = 57.970. The weight percent O contained in the oxide is the mass of oxygen contained in one mol divided by the mass of one mol of the oxide. Mass of 1 mol O is 15.9994. Mass of 1 mol MO is 57.970. Mass fraction O in MO is 15.9994/57.970 = 0.276. Repeat for values of a and b to find the combination that yields a mass fraction closest to 0.30.

In the case of M2O3, the mass fraction of O is (3 * 15.9994) / [(2 * 41.971) + (3 * 15.9994)] = 47.9982 / 131.940 = 0.3638, which is not the answer you are looking for.
 
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