combustion analysis of 63.8 mg of C,H, and O containing compound produced 145.0 mg of CO2 and 59.38 mg of H2O. What is the empirical formula for the compound?

C3H6O

1. Convert 145.0 mg CO2 to grams C, then divide by 63.8 mg to obtain %C.

2. Convert 59.38 mg H2O to grams H, then divide by 63.8 mg to obtain
%H.
3. %O = 100% - %C - %H
4. Take a 100 g sample which will give you the percentages in #3 as grams.
5. Convert grams to mols. mols = g/atomic mass.
6. Find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (thereby getting 1.000 for that one), then divide the other numbers by the same small number.
This is the empirical formula..
Post your work if you get stuck.

To determine the empirical formula of a compound using combustion analysis, we need to find the moles of each element present in the compound. Here's how you can calculate the empirical formula:

Step 1: Calculate the moles of CO2 produced:
Molar mass of CO2 = 12.01 g/mol (carbon) + (2 * 16.00 g/mol) (oxygen) = 44.01 g/mol

Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of CO2 = 145.0 mg / 44.01 g/mol

Step 2: Calculate the moles of H2O produced:
Molar mass of H2O = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol

Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of H2O = 59.38 mg / 18.02 g/mol

Step 3: Calculate the moles of carbon, hydrogen, and oxygen in the compound:
(Note: The moles of oxygen can be calculated by subtracting the moles of carbon and hydrogen from the moles of CO2 and H2O respectively.)

Moles of C = Moles of CO2
Moles of H = (Moles of H2O * 2) (since there are 2 hydrogen atoms per water molecule)
Moles of O = (Moles of CO2 * 2) + Moles of H2O (since there are 2 oxygen atoms per CO2 molecule and 1 oxygen atom per H2O molecule)

Step 4: Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio:
Divide the above values by the smallest moles value obtained in step 3.

Step 5: Determine the empirical formula:
The empirical formula is the ratio of the elements in the compound using whole numbers. Round the decimal values obtained in step 4 to the nearest whole number.

Now, let's calculate the empirical formula using the given data:

Moles of CO2 = 145.0 mg / 44.01 g/mol = 3.297 mol
Moles of H2O = 59.38 mg / 18.02 g/mol = 3.293 mol

Moles of C = 3.297 mol
Moles of H = 3.293 mol * 2 = 6.586 mol
Moles of O = (3.297 mol * 2) + 3.293 mol = 9.887 mol

Dividing by the smallest value, which is 3.293 mol:

Moles of C = 3.297 mol / 3.293 mol = 1
Moles of H = 6.586 mol / 3.293 mol = 2
Moles of O = 9.887 mol / 3.293 mol = 3

Therefore, the empirical formula for the compound is CH2O.

CHO