Acompound contains only carbon, hydrogen, and oxygen. combustion of 9.612 mg of the compound yields 14.41 mg co2 and 3.93 mg h2o. the molar mass of the compound is 176.1 g/mol. what are the empirical and molecular formulas of the compound? (type your answer using the format co2 for co2.)

Formula used for determining the number of moles:

number of moles = -(1)

Mass of = 14.41 mg (given)

Mass of = 3.93 mg (given)

Number of moles of from =

=

Number of moles of from =

=

Mass of carbon  =

=

Mass of hydrogen =

=

Mass of oxygen = total mass of the compound - mass of carbon- mass of hydrogen

=

=

Now, number of moles of oxygen =

=

To identify the empirical formula; divide the number of moles of carbon, hydrogen and oxygen with least number of moles.

Thus,

Now, multiply the numbers with 3 to get the hydrogen number in whole number, we get

The empirical formula =

The empirical weight =

=

Divide the molecular weight with empirical weight to find the factor between empirical formula and molecular formula  , we get

=

=

Now, multiply the numbers of carbon, hydrogen and oxygen with 2 to get molecular formula, we get

Number of carbon = 6

Number of hydrogen = 8

Number of oxygen = 6

Thus, molecular formula is .

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !