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A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

Answer 1

Naomi Aronson

Analysis of combustion!

#157mg * (10^-3g)/(mg) * (CO_2)/(44.01g) approx 3.57*10^-3mol#

#42.8mg * (10^-3g)/(mg) * (H_2O)/(18.02g)* (2H)/(H_2O) approx 4.75*10^-3mol#

The largest molar mass divided by the smallest molar mass yields...

#C_1.33H#

Inconvenient... multiply by three, causing the subscripts to be integers:

#C_4H_3#

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Answer 2

Caleb Adams

The empirical formula of the compound is CH2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.