Question #4cd61

2 Answers
Oct 5, 2016

Answer:

#C_2H_5#

Explanation:

The equation for the oxidation or burning of a hydrocarbon is

#C_nH_n + O_2 = CO_2 + H_2O#

To find the number of moles of Carbon in the original molecule divide the grams of Carbon Dioxide by the molar mass of Carbon Dioxide.

# 1 xx C # = # 1 x 12# = 12
+#2 xx O# = # 2 x 16#= + 32
= 44 grams/ mole

16.4 / 44 = .373 moles of Carbon atoms used.

To find the number of moles of Hydrogen in the original molecule divide the grams of water by the molar mass of Water then multiple the answer by two because there are two atoms of Hydrogen in one molecule of water.

# 2 xx H # = 2 xx 1# = 2 # 1 xx O # = 1 xx 16# = +16
= 18 grams /mole

8.37/ 18.0 = .465 moles of water x 2 = .930 moles of Hydrogen.

Now find the ratio of Hydrogen to Carbon by dividing the number of moles.

.930/ .373 = 2.49 H/C round this off to 2.50 Hydrogen to Carbon.

Since it is not logical to have half an atom multiple the ratio by 2

# 2.5/1 xx 2/2 = 5/2 # There are five Hydrogen to 2 Carbons This is the empirical formula. ( It may not be the molecular formula)

Oct 5, 2016

Let the molecular formula of HC be #C_xH_y#

And the balanced equation of the combustion reaction of the HC in oxygen is

#C_xH_y+(x+y/4)O_2(g)->xCO_2(g)+y/2H_2O(l)#

So by this equation the stochiometric ratio of #CO_2# and #H_2O# produced on combustion is
#=(xmol)/(y/2 mol)=(44xg)/(18/2yg)=(44x)/(9y)#
#"Where " 44g/"mol"" is the molar mass of "CO_2 and 18g/"mol" " is the molar mass of "H_2O#

But by the given data this ratio is

#(16.4g)/(8.37g)=164/837#

So equating these two we get

#(44x)/(9y)=164/837#

#=>x/y=9/44xx164/837=0.4=2/5#

Since the ratio of number of atoms C and H in the HC molecule is #2:5# then we can easily say the Empirical formula of HC is #color(red)(C_2H_5)#