If #"68.9 mg"# of safrole, #("C"_x"H"_y"O"_z)_n# combusts, and #"187 mg"# of #"CO"_2# and #"38.3 mg"# of #"H"_2"O"# form, what is the empirical formula of the hydrocarbon? If its molar mass is #"162 g/mol"#, what is the molecular formula?

Well, there are two ways to do it. I got a molecular formula of #"C"_10"H"_10"O"_2# for safrole. What is the empirical formula of safrole?

• You can solve for the mols of #"C"# and of #"H"# in the hydrocarbon, and then use conservation of mass to find the mass of #"O"# (and consequently the mols of #"O"#) in the hydrocarbon.
• Or, you can solve for the mass of #"C"# and #"H"#, and find their respective mass percents in the hydrocarbon. The remainder is the #"O"# mass percent, and the known molecular mass then gives the molecular formula first.

METHOD 1

#0.187 cancel("g CO"_2) xx cancel("1 mol CO"_2)/(44.01 cancel("g CO"_2)) xx cancel("1 mol C")/cancel("1 mol CO"_2) xx ("12.011 g C")/cancel("1 mol C")#

#=# #"0.0510 g C"#, and along the way, we got #ul"0.00425 mols C"#.

#"0.0383 g H"_2"O" xx cancel("1 mol H"_2"O")/(18.015 cancel("g H"_2"O")) xx cancel("2 mol H")/cancel("1 mol H"_2"O") xx ("1.008 g H")/cancel("1 mol H")#

#=# #"0.00429 g H"#, and along the way, we got #ul"0.00425 mols H"#.

By conservation of mass, the mass of #"O"# in the hydrocarbon is:

#m_(zO) = m_(C_xH_yO_z) - m_(xC) - m_(yH)#

#= "0.0689 g" - "0.0510 g C" - "0.00429 g H"#

#=# #"0.0136 g O"#

And its mols would then be:

#0.0136 cancel"g O" xx "1 mol O"/(15.999 cancel"g O")#

#=# #ul"0.000849 mols O"#

As a result, the subscripts can now be determined. Dividing by the smallest mols normalizes to the most reduced subscripts, and we then scale it up based on the actual molar mass. The result SHOULD be all whole numbers.

#("0.00425 mols C")/("0.000849 mols O") = 5.006 ~~ 5"C" : 1"O"#

#("0.00425 mols H")/("0.000849 mols O") = 5.006 ~~ 5"H" : 1"O"#

Thus, the empirical formula is #color(blue)("C"_5"H"_5"O")#. THIS hypothetical molar mass is calculated to be:

#5 xx "12.011 g/mol" + 5 xx "1.008 g/mol" + 1 xx "15.999 g/mol" = ul"81.094 g/mol"#

And the actual molar mass was #"162 g/mol"#, so the molecular formula is twice this:

#color(blue)(barul(|stackrel(" ")(" ""C"_10"H"_10"O"_2" ")|))#

METHOD 2

You could just use the definition of a percent along with the masses of #"C"# and #"H"#. As before:

#0.187 cancel("g CO"_2) xx cancel("1 mol CO"_2)/(44.01 cancel("g CO"_2)) xx cancel("1 mol C")/cancel("1 mol CO"_2) xx ("12.011 g C")/cancel("1 mol C")#

#=# #"0.0510 g C"#

#"0.0383 g H"_2"O" xx cancel("1 mol H"_2"O")/(18.015 cancel("g H"_2"O")) xx cancel("2 mol H")/cancel("1 mol H"_2"O") xx ("1.008 g H")/cancel("1 mol H")#

#=# #"0.00429 g H"#

And from here, the mass percents are:

#"0.0510 g C"/("0.0689 g C"_x"H"_y"O"_z) xx 100% = ul(74.02% "C")#

#"0.00429 g H"/("0.0689 g C"_x"H"_y"O"_z) xx 100% = ul(6.23% "H")#

#=> 100.00% - 74.02%"C" - 6.23%"H" = ul(19.75% "O")#

And given the molar mass, we can then find the subscripts of the molecular formula right away:

#0.7402 xx "162 g/mol" = "119.91 g"/"x mols C" ~~ bb10 xx "12.011 g C/mol"#

#0.0623 xx "162 g/mol" = "10.09 g"/"y mols H" ~~ bb10 xx "1.008 g H/mol"#

#0.1975 xx "162 g/mol" = "31.995 g/"/"z mols O" ~~ bb2 xx "15.999 g O/mol"#

And we still get the molecular formula of #color(blue)("C"_10"H"_10"O"_2)#.

The empirical formula is then known from reducing to smallest whole numbers, giving #color(blue)("C"_5"H"_5"O")#.

We should be using Average Atomic masses of the constituents. For sake of simplicity whole numbers instead have been used.

#187 " mg of CO"_2# are produced by #68.9" mg of sample."#
We know that #44 " g mole CO"_2# contains #"12 g mole of C"#

Obtained weight of #"CO"_2# will contain
#(187xx10^-3)/44=0.00425# moles of carbon
which also is equal to #12/44xx(187xx10^-3)="0.0510 g of C"#

We also know that #18 "g mole H"_2"O"# contains #"2 g mole of H"#

Obtained weight of #"H"_2"O"# will contain
#(38.3xx10^-3)xx2/18=0.00426# moles of hydrogen
which also is equal to #2/18xx(38.3xx10^-3)="0.00426 g of H"#

From Law of conservation of mass, the mass of #"O"# in the sample is:

#= 0.0689 - 0.0510 " C"- "0.00426 H"#

#=0.01364" g of O"#

In number of mols is

#0.01364/16 =0.0008525" mols"#

Ratio of moles of constituents is

#"C:H:O"="0.00425:0.00426:0.0008525#

Dividing with the lowest mole number we get

#"C:H:O"="4.985:4.997:1#

Ratio of constituents to nearest whole number

#"C:H:O"="5:5:1#

Therefore empirical formula is #"C"_5"H"_5"O"#.

Molar mass based on empirical formula is

#5 xx 12 + 5 xx 1 + 1 xx 16=81" g"#

Given molar mass is #162" g/mol"#.
Hence molecular formula becomes

#162/81xx("C"_5"H"_5"O")#

#"C"_10"H"_10"O"_2"#