When 6.853 mg of sex hormone was burned in a combustion analysis,19.73 mg of CO2 and 6.391 mg of H2O were obtained. What is the empirical formula of the compound ?? Please be detailed...help me.

0.01973 g of CO2 has 0.000448 moles of CO2

molar mass of CO2 = 44 g/ mole
there is 1 mole of C in CO2 so moles of C in the compound =
0.000448409 moles
mass of C = 0.00538 g
0.006391 g of H2O has 0.000355 moles of H2O
molar mass of H2O = 18 g/ mole
there are 2 moles of H in H2O so moles of H in the
compound = 0.000710 moles
mass of H = 0.00072 g
mass of H + C = 0.00610 g
mass of sample = 0.006853 g
mass of O 0.000756 g
moles of O = 0.0000473 moles
molar ratio of C : H : O = 0.00045 : 0.00071 : 0.000047
smallest number 0.0000473
divide the ratio by the smallest number we get
molar ratio of C : H : O = 9.49 15.02 1
multiply by 2 to get whole numbers
molar ratio of C : H : O = 18.97099624 : 30.0429128 : 2
empirical formula = C19H30O2
empirical formula mass = 290 g
which is the same as the molecular mass
so the empirical formula is also the molecular formula
C19 H30 O2. That's what I did please help...

I don't agree with all of your numbers but most are close, anyway, and I don't think it will make much difference in the empirical formula. I don't know what else you want. Apparently you didn't type in the entire problem since it says nothing about the molar mass nor how you know the "missing" element is O.

To determine the empirical formula of a compound, we need to calculate the ratio of the elements present in it. In this case, we have the mass of carbon dioxide (CO2) and water (H2O) produced when burning the sex hormone.

1. Start by converting the given masses of CO2 and H2O to moles using their molar masses:
- Molar mass of CO2 = 12.01 g/mol (carbon) + 2(16.00 g/mol) (oxygen) = 44.01 g/mol
- Molar mass of H2O = 2(1.01 g/mol) (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol

Moles of CO2 = mass of CO2 / molar mass of CO2
= 19.73 mg / 44.01 g/mol
= 0.4482 mol

Moles of H2O = mass of H2O / molar mass of H2O
= 6.391 mg / 18.02 g/mol
= 0.3546 mol

2. Now, we need to determine the number of carbon (C) and hydrogen (H) atoms in the empirical formula. To do this, we can use the mole ratio between CO2 and H2O.

The ratio between the carbon atoms in CO2 and the hydrogen atoms in H2O is 1:1. Therefore, the number of carbon atoms is equal to the number of hydrogen atoms.

Number of carbon atoms = Number of hydrogen atoms

Since 1 mole of CO2 contains 1 mole of carbon and 2 moles of oxygen, we have:
Moles of carbon = Moles of CO2 = 0.4482 mol

Since 1 mole of H2O contains 2 moles of hydrogen and 1 mole of oxygen, we have:
Moles of hydrogen = 2 * Moles of H2O = 2 * 0.3546 mol = 0.7092 mol

3. To simplify the ratio, divide the number of moles of each element by the smallest of the two (carbon in this case):
Carbon: 0.4482 mol / 0.4482 mol = 1
Hydrogen: 0.7092 mol / 0.4482 mol = 1.583

Rounding off the ratio to the nearest whole number, we have:
Carbon: 1
Hydrogen: 2

Now we have the empirical formula of the compound, which is CH2.

To determine the empirical formula of the sex hormone compound, we need to analyze the given data.

1. Convert the mass of CO2 and H2O obtained to moles:
- Mass of CO2 = 19.73 mg
- Molar mass of CO2 = 44.01 g/mol
- Therefore, moles of CO2 = (19.73 mg / 44.01 g/mol)

- Mass of H2O = 6.391 mg
- Molar mass of H2O = 18.02 g/mol
- Therefore, moles of H2O = (6.391 mg / 18.02 g/mol)

2. Calculate the moles of carbon and hydrogen present:
- In 1 mole of CO2, there is 1 mole of carbon.
- In 1 mole of H2O, there are 2 moles of hydrogen.
- Therefore, moles of carbon = moles of CO2
- Moles of hydrogen = 2 * moles of H2O

3. Determine the simplest whole number ratio of carbon to hydrogen:
- Divide the moles of carbon and hydrogen by the smallest value obtained from step 2.
- Round the results to the nearest whole number.

4. Write the empirical formula using the whole number ratio obtained in step 3.

Let's now calculate the empirical formula step by step:

1. Calculate the moles of CO2 and H2O:
- Moles of CO2 = (19.73 mg / 44.01 g/mol) = 0.448 mol
- Moles of H2O = (6.391 mg / 18.02 g/mol) = 0.354 mol

2. Calculate the moles of carbon and hydrogen:
- Moles of carbon = 0.448 mol
- Moles of hydrogen = 2 * 0.354 mol = 0.708 mol

3. Determine the simplest whole number ratio:
- Divide the moles of carbon and hydrogen by the smallest value (0.354 mol).
- Carbon's ratio = (0.448 mol / 0.354 mol) = 1.266
- Hydrogen's ratio = (0.708 mol / 0.354 mol) = 2.0

- Rounded to the nearest whole number:
- Carbon's ratio = 1
- Hydrogen's ratio = 2

4. Write the empirical formula:
- The empirical formula of the compound is CH2.

Therefore, the empirical formula of the sex hormone compound is CH2.