Upon combustion, 4.30 grams sample of a compound containing only carbon, hydrogen, and oxygen produced 8.59 g of CO2 and 3.52g of H2o. what is empirical formula of the compound ?

Convert 8.59g CO2 to grams C.

8.59 x (atomic mass C/molar mass CO2) =?

Convert 3.52 g H2O to g H (atoms).
3.52 g H2O x (2*atomic mass H/molar mass H2O) =?

Then g O = 4.30-g C - g H = ?

Now convert grams each to mols.
mols C = grams C/atomic mass C
mols H = grams H/atomic mass H
mols O = grams O/atomic mass O

Now find the ratio to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself, then follow suit with the other numbers. Round to whole numbers.
Post your work if you get stuck.

I got a weird answer. C1.9H4O. Do u think this is right?

To find the empirical formula of the compound, we need to determine the ratio of the elements present in it.

1. Start by finding the moles of carbon dioxide (CO2) produced:
- The molar mass of CO2 = 12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/mol.
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 8.59 g / 44.01 g/mol = 0.1954 mol.

2. Next, find the moles of water (H2O) produced:
- The molar mass of H2O = 2(1.01 g/mol) (H) + 16.00 g/mol (O) = 18.02 g/mol.
- Moles of H2O = mass of H2O / molar mass of H2O = 3.52 g / 18.02 g/mol = 0.1953 mol.

3. Now, we need to find the moles of carbon, hydrogen, and oxygen present in the compound.
- Moles of carbon = moles of CO2 = 0.1954 mol.
- Moles of hydrogen = 2 * (moles of H2O) = 2 * 0.1953 mol = 0.3906 mol.
- Moles of oxygen = (moles of CO2) + (moles of H2O) = 0.1954 mol + 0.1953 mol = 0.3907 mol.

4. Divide the moles by the smallest value of moles obtained:
- Divide moles of carbon, hydrogen, and oxygen by 0.1953 mol (which is the smallest value).

Moles of carbon / Smallest value = 0.1954 mol / 0.1953 mol ≈ 1
Moles of hydrogen / Smallest value = 0.3906 mol / 0.1953 mol ≈ 2
Moles of oxygen / Smallest value = 0.3907 mol / 0.1953 mol ≈ 2

5. The ratio of the elements gives us the empirical formula. Thus, the empirical formula of the compound is CH2O.

To determine the empirical formula of the compound, you need to find the ratio of the elements in the compound based on the given mass of each element.

First, let's find the moles of CO2 and H2O produced:

1. Moles of CO2:
- Calculate the molar mass of CO2 (carbon dioxide): 1 atom of C (carbon) + 2 atoms of O (oxygen)
= 12.01 g/mol + (2 × 16.00 g/mol) = 44.01 g/mol
- Find the moles of CO2 by dividing its mass by its molar mass:
moles of CO2 = mass of CO2 / molar mass of CO2
= 8.59 g / 44.01 g/mol ≈ 0.1953 mol

2. Moles of H2O:
- Calculate the molar mass of H2O (water): 2 atoms of H (hydrogen) + 1 atom of O (oxygen)
= (2 × 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
- Find the moles of H2O by dividing its mass by its molar mass:
moles of H2O = mass of H2O / molar mass of H2O
= 3.52 g / 18.02 g/mol ≈ 0.1951 mol

Next, we need to find the mole ratios by dividing the number of moles of each element by the lowest number of moles.

1. Moles of carbon (C):
- In 1 mol of CO2, there is 1 mol of carbon (C)
- Moles of C = moles of CO2 ≈ 0.1953 mol

2. Moles of hydrogen (H):
- In 1 mol of H2O, there are 2 mols of hydrogen (H)
- Moles of H = 2 × moles of H2O ≈ 2 × 0.1951 mol ≈ 0.3902 mol

3. Moles of oxygen (O):
- In 1 mol of CO2, there are 2 mols of oxygen (O)
- Moles of O = 2 × moles of CO2 ≈ 2 × 0.1953 mol ≈ 0.3906 mol

The next step is to convert the mole ratios to whole numbers by dividing each value by the smallest number of moles.

Dividing all of the values by 0.1951 (the smallest number of moles), we get:

C: 0.1953 mol / 0.1951 mol ≈ 1
H: 0.3902 mol / 0.1951 mol ≈ 2
O: 0.3906 mol / 0.1951 mol ≈ 2

The empirical formula of the compound is therefore CH2O.