Phenylbutazone can be synthesised, in a multi-step process, starting from compound A. Elemental microanalysis showed that compound A has a composition, by mass, of 50∙0% C; 5∙60% H; 44∙4% O Calculate the empirical formula of compound A?

1 Answer
May 8, 2018

"C"_3"H"_4"O"_2

Explanation:

An 100color(white)(l)"g" sample of compound A would, by the mass ratio, contain

  • 50.0color(white)(l)"g" of carbon "C"
  • 5.60color(white)(l)"g" of carbon "H"
  • 44.4color(white)(l)"g" of carbon "O"

which corresponds to approximately

  • 50.0/12~~25/6color(white)(l)"mol" of "C"
  • 5.60/1~~28/5color(white)(l)"mol" of "H"
  • 44.4/16~~25/9color(white)(l)"mol" of "O"

where 12, 1, and 16 the respective relative atomic mass of the three elements (rounded to the nearest whole number).

Coefficients in an empirical formula are expected to be whole numbers. Thus it would be necessary eliminate any fraction in these expressions. Multiplying the mass of the sample by 90- the least common multiple of the three denominators- can serve this purpose. That is, there are

  • 375color(white)(l)"mol" of "C",
  • 504color(white)(l)"mol" of "H", and
  • 250color(white)(l)"mol" of "O"

in a 9*10^3color(white)(l)"g" sample of compound A.

Dropping the last digit of 50color(darkblue)(4) gives the approximation 504~~500=4*125, such that it shares the common divisor 125 with the rest two figures.

Factoring 125 out of the ratio gives

n("C"):n("H"):n("O")=375:color(darkblue)(504):250
color(white)(n("C"):n("H"):n("O"))~~(3*color(red)(cancel(color(black)(125)))):(color(darkblue)(4*color(red)(cancel(color(darkblue)(125))))):(2*color(red)(cancel(color(black)(125))))
color(white)(n("C"):n("H"):n("O"))=3:4:2

Hence the empirical formula would be

"C"_3"H"_4"O"_2