A compound with a molecular weight of 110.1 g

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Discussion Overview

The discussion revolves around determining the molecular formula of a compound with a molecular weight of 110.1 g, composed of carbon, hydrogen, and oxygen, based on combustion analysis results. The analysis includes calculations related to the moles of carbon dioxide and water produced from a known mass of the compound.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the moles of carbon dioxide produced from the combustion and expresses confusion about calculating moles of water.
  • Another participant confirms the calculation of moles of water and suggests deriving molar ratios for the reaction to find the molecular formula.
  • A participant clarifies that moles of water do not directly equate to moles of hydrogen, indicating the need for conversion to atomic hydrogen.
  • Further clarification is provided regarding the relationship between moles of water and moles of hydrogen in the reactant, leading to a calculation of hydrogen content in the hydrocarbon.
  • Participants discuss the need to perform similar calculations for carbon dioxide to ascertain the carbon content in the compound.

Areas of Agreement / Disagreement

Participants generally agree on the approach to calculating moles and deriving the molecular formula, though there is some confusion regarding the conversion between moles of water and moles of hydrogen. The discussion reflects a collaborative effort to clarify these calculations without reaching a definitive conclusion on the molecular formula.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly in the conversion of moles of water to moles of hydrogen and the overall stoichiometry of the combustion reaction.

Who May Find This Useful

Students and individuals interested in chemistry, particularly those studying combustion analysis and molecular formula determination.

priscilla89
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Homework Statement



A compound with a molecular weight of 110.1 g that contained only C, H, and O was analyzed by combustion. the combustion of a 5.19 g sample of the compound resulted in the formation of 12.4 g of carbon dioxide and 2.55 g of water. What is the molecular formulas of this compound?

2. The attempt at a solution

CO2

12.44
------ = .2818
44

But I'm confused on how to do this for H20
Is it this below?

2.55
-----
18

= .1416
 
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Thats correct for the moles of water, and for the hydrocarbon, 5.19g/110.1g/m = .0471 moles.
So now you can get the molar ratios for the reaction . And from there the formula for the hydrocarbon.
 
Last edited:


Moles of water is not moles of hydrogen. They are related, but not identical.
 


Were not looking for moles of hydrogen, we want to know how many hydrogens on the
reactant. So .1416 moles water/.0471 moles hydrocarbon = 3/1
So with three waters in the product we have 6 hydrogens on the hydrocarbon.
 


Sorry, my lack of clarity. I was not referring to the moles of molecular hydrogen, just the fact moles of water have to be converted to moles of atomic hydrogen. That means moles of water times two, that's what you did. We were talking the same all the time, obviously neither of us was entirely clear.
 


morrobay said:
Were not looking for moles of hydrogen, we want to know how many hydrogens on the
reactant. So .1416 moles water/.0471 moles hydrocarbon = 3/1
So with three waters in the product we have 6 hydrogens on the hydrocarbon.

Right, now I understand it. You would've to do the same for CO2::

CO2

.2818 / .0471 = 6

C6 H6 O2
 


Thanks for the help. I really do appreciate this

- Happy Holidays Everybody
 

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