Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of #CO_2# and 1.070 g of #H_2O#. What is the empirical formula of the compound?

Here is the equation with masses:

#"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2#

Use 32.00 g/mol for the molar mass of #"O"_2#
Use 18.03 g/mol for the molar mass of #"H"_2"O"#
Use 44.01 g/mol for the molar mass of #"CO"_2#

We do not know the molar mass of the hydrocarbon so we allow #a, b#, and #c# to be any positive real number.

#"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900 g")/("44.01 g/mol") "CO"_2#

Perform the division:

#"C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO"_2#

Matching coefficients, we get:

#a = 0.04317#
#b = 2 xx 0.05935 = 0.1187#
#c = 0.05935 + 2 xx 0.04317 - 2 xx 0.06210 = 0.02149#

Divide every number by 0.02149:

#a/0.02149 = 2.009#
#b/0.02149 = 5.523#
#c/0.02149 = 1#

Multiply every number by 2:

#2.009 xx 2 = 4.018#
#5.523 xx 2 = 11.05#
#1 xx 2 = 2#

Round off each number to the nearest integer.

#4.018 ≈ color(white)(l)4#
#11.05 ≈ 11#
#color(white)(ll)2color(white)(ml) =color(white)(ll) 2#

The empirical formula is #"C"_4"H"_11"O"_2#.

First, we calculate the masses of #"C"# and #"H"# from the masses of their oxides (#"CO"_2# and #"H"_2"O"#).

#"Mass of C" = 1.900 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.5185 g C"#

#"Mass of H" = 1.070 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1197 g H"#

#"Mass of C + Mass of H" = "0.5185 g + 0.1197 g" = "0.6382 g"#

This is less than the mass of the sample.

The missing mass must be caused by #"O"#.

#"Mass of O = 0.9827 g - 0.6382 g = 0.3445 g"#

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

#bb("Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(ml)×2color(white)(mm)"Integers")#
#color(white)(ml)"C" color(white)(XXXm)0.5185 color(white)(mll)0.04317color(white)(Xmlll)2.005color(white)(m)4.010color(white)(Xmmml)4#
#color(white)(ml)"H" color(white)(XXXm)0.1197 color(white)(mll)0.1188 color(white)(mmml)5.518 color(white)(m)11.04color(white)(Xmmm)11#
#color(white)(ml)"O" color(white)(XXXm)0.3445 color(white)(mll)0.02153 color(white)(mmm)1 color(white)(mmml)2color(white)(Xmmmmm)2#

The empirical formula is #"C"_4"H"_11"O"_2#.

Note: This is an impossible empirical formula.

The molecular formula must have an even number of #"H"# atoms, e.g. #"C"_8"H"_22"O"_4#.

A compound with 8 #"C"# atoms can contain no more than 18 #"H"# atoms.