i offered their portion masses divided by each element"s family member atomic fixed then separated thru by the the smallest mole ratio.

You are watching: What is the empirical formula of a compound composed of 3.25 hydrogen 19.36

moles that carbon= #19.36/12=1.613# moles of hydrogen= #3.25/1=3.25#moles the oxygen= #77.39/16=4.8369#

divide thru v the smallest ratio

C #1.613/1.613#H #3.25/1.613#O #4.8369/1.613#

=CH2O3 The empirical formula is the SIMPLEST entirety number ratio defining constituent facets in a species...

As v all these problems, the normal rigmarole is to assume a #100*g# mass and also then interrogate the molar quantities of each element.

#"Moles of hydrogen,"# #(3.25*g)/(1.00794*g*mol^-1)=3.220*mol.#

#"Moles that carbon,"# #(19.36*g)/(12.011*g*mol^-1)=1.612*mol.#

#"Moles the oxygen,"# #(77.39*g)/(16.00*g*mol^-1)=4.84*mol.#

And we division thru by the smallest molar amount to obtain the empirical formula...

#"Empirical formula"-=C_((1.612*mol)/(1.612*mol))H_((3.220*mol)/(1.612*mol))O_((4.84*mol)/(1.612*mol))-=CH_2O_3# Meave60
Feb 27, 2022

The empirical formula is #"CH"_2"O"_3"#.

Explanation:

An empirical formula represents the lowest totality number proportion of elements in a compound. There are several measures involved.

1. identify the mole of each aspect by dividing its given mass by its molar mass. As soon as percentages include up come 100%, we can directly convert percentage to mass in grams.

2. determine the mole ratios in between each element and the lowest variety of moles by dividing the mole of each aspect by the lowest number of moles. This action determines the subscripts because that the elements.

3. If all mole ratios are whole numbers, you have the lowest entirety number ratio and also can write them together subscripts for the compound.

4. If one or much more mole proportion is not a totality number, then every ratios need to be multiply by a element that will certainly make all ratios whole numbers.

Moles that elements

Since molar fixed is a fraction (g/mol), I prefer to division by the molar massive by multiply by its mutual (mol/g). I think it provides what happens through the units more clear.

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#"H":# #3.25color(red)cancel(color(black)("g H"))xx(1"mol H")/(1.008color(red)cancel(color(black)("g H")))="3.22 mol H"#

#"C":# #19.36color(red)cancel(color(black)("g C"))xx(1"mol C")/(12.011color(red)cancel(color(black)("g C")))="1.61 mol C"#

#"O":# #77.39color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="4.84 mol O"#

Mole ratios

Because mole cancel and also mole ratios room dimensionless, ns am not going to label the moles.