[prncess23]

We are doing a general unknown. According to one of the tests it states that if the unknown solution dissolves/soluble in HCl, it means its an amine. When I told my teacher this and why I don’t have a NO2 mass in my combustion analysis sheet, she said that all you need is on the sheet she gave me.

Given
Unknown compound: 1.075g
CO2: 2.625g
H2O: 1.075g
Molecular weight: 72.104g

I did this…but the empirical formula doesn’t look right.

1) determine the mass of carbon, hydrogen, and oxygen in 1.575 g of compound.
carbon: 2.625 g x (12.01 / 44.01) = 0.7163 g
hydrogen: 1.075 g x (2.016 / 18.02) = 0.1203 g
oxygen: 1.075 – 0.7163 – 0.1203 = 0.2384 g

2) convert to moles.
carbon: 0.7163 / 12.01 = 0.0596 mol
hydrogen: 0.1203 / 1.008 = 0.1193 mol
oxygen: 0.2384 / 16 = 0.0015 mol

3) divide each molar amount by the lowest value to obtain small integers
carbon: 0.0596 / 0.0015 = 40
hydrogen: 0.1193 / 0.0015 = 80
oxygen: 0.0015 / 0.0015 = 1

The empirical formula of the substance is C₄₀H₈₀O.

[UG]

Are you absolutely 100% certain that is it an amine?

[UG]

And where did the number 1.575g pop out from? 

I did this…but the empirical formula doesn’t look right.

1) determine the mass of carbon, hydrogen, and oxygen in 1.575 g of compound.
carbon: 2.625 g x (12.01 / 44.01) = 0.7163 g
hydrogen: 1.075 g x (2.016 / 18.02) = 0.1203 g
oxygen: 1.075 – 0.7163 – 0.1203 = 0.2384 g

[prncess23]

the 1.575g came from the information sheet our teacher gave me which is specific for each person's unknown.

according to the HCl test...it is supposed to me an anime. But...wheni  asked my teacher...it seemed like i didn it wrong and none of the unknowns contain an amine...they just contain C, O and H's.

[UG]

I thought the unknow compound was 1.075g.  Which one is it? 

Given
Unknown compound: 1.075g
CO2: 2.625g
H2O: 1.075g
Molecular weight: 72.104g

[prncess23]

yes u are correct. the unknown is 1.075g. The 1.575 g was a typo.
Thank you!

[UG]

You made a mistake in your calculation:

oxygen: 0.2384 / 16 = 0.0015 mol

The answer should actually be 0.0149 ~ 0.015 moles. Now you should be able to work out the emperical formula.