With all these kinds of problems, we assume a massive of #100*g# and also we division thru by the atom mass of each component:

#"Moles of carbon"=(75.69*g)/(12.011*g*mol^-1)=6.30*mol.#

#"Moles that hydrogen"=(8.80*g)/(1.00794*g*mol^-1)=8.73*mol.#

#"Moles the oxygen"=(15.51*g)/(15.999*g*mol^-1)=0.969*mol.#

We division thru through the lowest molar amount (that the oxygen) to acquire a psychological empirical formula the #C_(6.50)H_9O#. But, by definition, the empirical formula is the SIMPLEST totality NUMBER ratio specifying constituent atom in a species. So we double this trial formula come get............

You are watching: What is the empirical formula of ibuprofen

#C_13H_18O_2#

For completeness, we can quote a molecular formula IF WE had actually a molecular mass decision for Ibuprofen.......but us do, that is #206.29* g*mol^-1#

Now the molecular formula is constantly a many of the empirical formula, i.e:

#206.29*g*mol^-1=nxx(13xx12.011+18xx1.00794+2xx16.0)*g*mol^-1.#

Clearly, #n=1#, and also the molecluar formula is the very same as the empirical formula.

Meave60
might 2, 2017

The empirical formula for ibuprofen is #"C"_13"H"_18"O"_2"#. The is also its molecule formula.

Explanation:

The empirical formula of a link represents the lowest totality number ratio of the elements in it. Since the percentages the carbon, hydrogen, and also oxygen include up come 100%, you deserve to directly transform them right into grams.

The moles of each element must be determined. Then the mole ratio of each aspect is calculation by splitting by the smallest number of moles. To calculation the empirical formula, you might need come manipulate the mole ratios, such together multiplying by two.

Determine the moles of every Element

Divide the provided mass by the molar fixed of every element. The molar mass of an facet is the atomic weight (relative atomic mass) ~ above the periodic table in grams/mole, or g/mol.

#"C":##(75.69color(red)cancel(color(black)("g")))/((12.011color(red)cancel(color(black)("g")))/(1"mol"))="6.302 mol"#

#"H":##(8.80color(red)cancel(color(black)("g")))/((1.008color(red)cancel(color(black)("g")))/(1"mol"))="8.73 mol"#

#"O":##(15.51color(red)cancel(color(black)("g")))/((15.999color(red)cancel(color(black)("g")))/(1"mol"))="0.9694 mol"#

Determine the Mole proportion for each Element.

Divide the moles of each aspect by the smallest number of moles.

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#"C":##(6.302color(red)cancel(color(black)("mol")))/(0.9694color(red)cancel(color(black)("mol")))=6.5#

#"H":##(8.73color(red)cancel(color(black)("mol")))/(0.9694color(red)cancel(color(black)("mol")))=9.01#

#"O":##(0.9694color(red)cancel(color(black)("mol")))/(0.9694color(red)cancel(color(black)("mol")))=1#

Empirical Formula

Since #6.5# is not a whole number, multiply all of the ratios by #2# so that they space all entirety numbers.