How do you find the empirical formula?

Question

If 0.4647g of a compound containing C, H, O atoms was burned in oxygen to yield 0.8635g of O2 and 0.1761g of H2O, how do you find the empirical formula?

If the molar mass of the compound is 142.11g/mol, what is the molecular formula?

I did copy the problem correctly, but I think the original problem is wrong, since it doesn't have carbon anywhere in the products.

Also, how do you figure out the moles, when the grams of each element is not given to you?

Also, how do you figure out the moles, when the grams of each element is not given to you?

Answer 1

To find the empirical formula you must change the grams to moles of each element present in the compound.

Once you have the number of moles of each element you find the smallest number of moles. Divide each number of moles by this smallest number and you will get a whole-number ratio. These whole numbers are your subscripts in the empirical formula.

Next use the periodic table to find the total mass of the empirical formula. Divided the molar mass given in the second part of this problem by the empirical mass. This gives a whole-number that you multiply the empirical subscripts by to get the molecular formula.

I would try to work the problem, but I am not sure you have it copied correctly. You do not account for carbon in your products.

Answer 2

well an emperical fromula is the formula with the smallest whole number mole ratios of the elements... for example

C6H12O6 - is the molecular formula
CH2O - IS THE EMPERICAL FORMULA...

EXAMPLE QUESTION:

A compound of Ag(silver) has the following analytical composition 63.50%(Ag). 8.257(N) and 28.257(O). Calculate the empirical formula.

Given: Ag-63.50%=63.5g
N- 8.25%=8.25g
O- 28.25%=28.25g
Multiply the given and find the atomic mass in which you divide the mole by.

Ag= 63.5g* 1mol /108g=0.587 mol
N= 8.25g * 1 mol /14=0.589 mol
O= 28.25g * 1 mol /16g= 1.765 mol


Divide by the lowest value next.

Ag=0.587 /0.587=1
N=0.589 /0.587=1.003
O= 1.765 /0.587 =3.006

Ag:N:O
Ag1N1O3

the final answer would be AgNO3