Chemistry How Accurate Should Experimental Data Be?

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The discussion revolves around the hydrogenation of aldehydes to form alcohols, specifically focusing on the calculation of mass percentages and molar masses. Participants analyze the addition of 4.45% mass of hydrogen to an aldehyde, leading to the formation of alcohol. Algebraic expressions are used to derive the molar mass of the aldehyde, with calculations indicating a theoretical error margin of only 2%, which is deemed acceptable in experimental data. The conversation emphasizes the importance of understanding the relationship between the molar masses of aldehydes and alcohols during this process. Overall, the discussion highlights the challenges and learning opportunities in experimental chemistry calculations.
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Homework Statement
The mass of an alcohol by the way of hydrogenation of a coresponding aldehyde is 4.45% higher than the mass of the aldehyde. Find the mass of the aldehyde
Relevant Equations
Aldehyde + H2 = alcohol
Ethanal + H2 = ethanol
Aldehyde + H2 = alcohol
Ethanal + H2 = ethanol
This is all that I have done so far
 
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4.45% of the mass of the aldehyde is the mass of what you are adding to the aldehyde.
All that's left is the Algebra.
 
.Scott said:
4.45% of the mass of the aldehyde is the mass of what you are adding to the aldehyde.
All that's left is the Algebra.
In hydrogenation, we add only hygrogen
So meaning we are adding 4.45% mass of the aldehyde worth of hydrogen. Let me try
 
.Scott said:
4.45% of the mass of the aldehyde is the mass of what you are adding to the aldehyde.
All that's left is the Algebra.
Let's say x=100
X+H2 = 100+4.45
100+4.45=104.45
This is what I got. I'm sort of lost here

I sort of get this question as have an x g of aldehyde
Add 4.45% of x g of hydrogen to it and we get 4.45% g higher alcohol.
But I'm stuck with the algebra
 
Let's say that X is the mass of the aldehyde.
X+mass(H2) = what function of mass(alcohol)?
 
.Scott said:
Let's say that X is the mass of the aldehyde.
X+mass(H2) = what function of mass(alcohol)?
X+2=(4.45/100) x (X)+X
Is this ok?
We could say x=100
100g + 4.45g= 104.45g
 
Last edited:
Assuming you used 1 mole of hydrogen (what was its mass?) what has to be the molar mass of the aldehyde?
 
The mass referred to here is molar mass. Going from alcohol to aldehyde represents a loss of two hydrogens. Let X be the molar mass of the aldehyde and X+2 be the molar mass of alcohol.

The ratio of (X+2)/X = what?

molar mass alcohol / molar mass aldehyde = what? in this example...
 
chemisttree said:
The mass referred to here is molar mass. Going from alcohol to aldehyde represents a loss of two hydrogens. Let X be the molar mass of the aldehyde and X+2 be the molar mass of alcohol.

The ratio of (X+2)/X = what?

molar mass alcohol / molar mass aldehyde = what? in this example...
0.0455
When that is multiplied with 100 we get 4.45%
 
  • #10
Almost. The value of that ratio is 1 + 0.0445 or 1.0445

Do you see that if the numerator is greater than the denominator the result must be > 1?

Solve for X
 
  • #11
chemisttree said:
Almost. The value of that ratio is 1 + 0.0445 or 1.0445

Solve for X
Yes because it's an increase right?

X+2/x = 1.0445
X+2=1.0445x
1.0445x-x=2
0.0445x=2
X=2/0.0445
X=44.94
 
  • #12
That molar mass corresponds to which aldehyde? Consider only C,H&O in the formula.
 
  • #13
chemisttree said:
That molar mass corresponds to which aldehyde? Consider only C,H&O in the formula.
The lowest formula of aldehyde is HCOH
Mass of HCOH= 30
44.94/30=1.498
 
  • #14
Nope. Keep trying. This formula wt is not 30... it is ~45.
 
Last edited:
  • #15
chemisttree said:
Nope. Keep trying. This formula wt is not 30... it is ~45.
CnH2n+1CHO
So CH3CHO mass = 44
44.94/44=1.02
 
  • #16
Close enough!
 
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  • #17
chemisttree said:
Close enough!
Close enough?
What was the error here?
 
  • #18
There was no error on your part. Sometimes the data we get from an experiment isn’t perfect, isn’t exact. The data from this example was only off from theoretical by 2%. That’s usually good enough. Close enough!
 
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  • #19
chemisttree said:
There was no error on your part. Sometimes the data we get from an experiment isn’t perfect, isn’t exact. The data from this example was only off from theoretical by 2%. That’s usually good enough. Close enough!
Thank you very much
You teaching technique is awesome. I really learned a lot.
PS. There is one more question that I have posted could you take a look at that too please

Here is the link
https://www.physicsforums.com/threa...f-a-hydrocarbon-using-combustion-data.990753/
 
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