A compound containing only carbon and hydrogen has a carbontohydrogen mass ratio of 11.89. Which carbontohydrogen mass ratio is possible for another compound composed only of carbon and hydrogen?
a) 2.50
b) 3.97
c) 4.66
d) 7.89
I'm asking for a friend, so a thorough explanation would be extremely helpful!
a) 2.50
b) 3.97
c) 4.66
d) 7.89
I'm asking for a friend, so a thorough explanation would be extremely helpful!
2 Answers
Trying to present a possible Answer
We know that the molar mass of carbon and hydrogen are
When Carbon Hydrogen mass ratio is 11.89
The ratio of no. Of atoms of C and H is given by
So empirical formula
Now from the list
a)
When Carbon Hydrogen mass ratio is 2.5
The ratio of no. Of atoms of C and H is given by
So empirical formula
Beceause the molecular formula of alkane with 5 Catoms in the molecule is
b)
When Carbon Hydrogen mass ratio is 3.97
The ratio of no. Of atoms of C and H is given by
So empirical formula
c)
When Carbon Hydrogen mass ratio is 4.66
The ratio of no. Of atoms of C and H is given by
So empirical formula
d)
When Carbon Hydrogen mass ratio is 7.89
The ratio of no. Of atoms of C and H is given by
So empirical formula
Since the molecular formula of butyne or butadiene
This is an example of the Law of Multiple Proportions.
If a carbontohydrogen mass ratio is
(Note that it does not mean it is the only mass ratio that can exist. Decimal multiples of the mass ratio are allowed, but do not guarantee a real compound.)
You are given the molar mass ratio
#M_X/M_Y = ("g X/mol")/("g Y/mol") = "g X"/"g Y" = m_X/m_Y#
Therefore, for some
#11.89/2.50 = 4.8# , not a whole number.
#11.89/3.97 = 2.99 ~~ 3# , so this is possible.
#11.89/4.66 = 2.55# , not a whole number.
#11.89/7.89 = 1.50# , not a whole number.
Having this ratio, we could identify the empirical formula:
#11.89 = (x cdot "12.011 g C/mol")/(y cdot "1.0079 g H/mol")#
If
If the ratio is

Choice
#(i)# overcomplicates things because it adds a step of multiplying everything by#3# to not end up with a fractional atom. 
Choice
#(ii)# gives#color(blue)("CH"_3)# as the empirical (and molecular) formula right away:
#m_C/m_H = (1 cdot "12.011 g/mol")/(3 cdot "1.0079 g/mol") = 3.97# and this correlates with the molecular formula
#"C"_2"H"_6# (ethane).
Note that in general, the mass ratio is not equal to the mol ratio.