Question

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?

Answer 1

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

Explanation:

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`.

Answer 2

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

Explanation:

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.