Answer to Question #168832 in General Chemistry for Michael Parker

Question

A compound contains 66.7% carbon, 11.1% hydrogen and 22.2% oxygen by mass. Its relative formula mass is 72. What is the empirical formula and the molecular formula of the compound?

Answer 1

1. Find the percentage of Oxygen by adding all of the given percentages then subtracting from 100

20.0+6.66+47.33=73.999=74 100−74=26

1. Change the percentages to grams. If there were 100 grams
2. C= 20 grams H = 6.66 grams N = 47.33 grams O = 26 grams.
3. Change the grams to moles by dividing by the molecular mass of the elements

Moles Carbon = 20/12=1.66moles C

Moles Hydrogen = 6.66/1=6.66moles H

Moles Nitrogen = 47.33/14=3.38molesN

Moles Oxygen = 26/16=1.625moles O

1. Find the simplest mole ratios.
2. Since Oxygen is the smallest divide all the other nu=mber of moles by the moles of Oxygen

Carbon ratio = C/O=1.66/1.625=1.02or1:1

Nitrogen ratio = N/O=3.38/1.625=2.04or 2:1

Hydrogen ratio = H/O=6.66/1.625=4.09or 4:1

So the compound has a ratio of 1 C: 2 N : 4 H : 1O for the empirical formula

The mass of one empirical formula is 62 grams per mole. This is slightly higher than the experimental molecular mass of 60 grams but is within experimental error.

So the compound most likely has a formula of C

N2H4O