Question

An organic compound contained C 66.7%, H 11.1%, 0 22.2% by mass. Its relative formula mass was 72. Find:
a) the empirical formula of the compound
b) the molecular formula of the compound. (Ar : H = 1, C = 12, 0 = 16)

Answer 1

Answer:

Find the area of a rectangular field in hectare whose sides are:

70 m 9dm and 125m

Answer 2

CN2H4O

Explanation: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

Change the percentages to grams. If there were 100 grams

C= 20 grams H = 6.66 grams N = 47.33 grams O = 26 grams.

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

Find the simplest mole ratios.

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