Combustion analysis of 0.343 g of a compound containing C, H, and O produces 0.753 g of CO2 and 0.411 g of H2O. Mass spectral analysis shows that the compound has a molar mass around 120 g mol-1. What is the compound's:
Empirical Formula:-----
Molecular Formula:-----
The problem doesn't tell you how much oxygen so you must calculate that.
Convert 0.753 g CO2 to grams C.
Convert 0.411 g H2O to grams H.
Add those two and subtract from 0.343 to find grams oxygen.
Now convert each of those grams to moles.
g C/12 = moles C
g H/1 = moles H
g O/16 = moles O.
Find the ratio in small whole numbers to each other. An easy way to do that is to divide the smallest number by itself which assures you of getting 1.0000 for that one, then divide all the other numbers by that same small number. Round to the nearest whole number to obtain the empirical formula. Determine the mass of the molecule as if the empirical formula is the molecular formula. The problem gives additional information that the molecular formula is approximately 120; therefore, 120/your empirical formula mass = some number which you should round to the nearest whole number (which I will call x). Then the compounds molecular formula will be (empirical formula)x.
To check your values, I obtained C3H8O for the empirical formula, x of 2 to make the molecular formula C6H16O2. Post your work if you get stuck.
To determine the compound's empirical formula, we need to find the ratios of the elements present in the given mass of the compound. We'll calculate the moles of carbon, hydrogen, and oxygen in the compound based on the given data.
1. Moles of CO2:
- Molar mass of CO2 = 12 g/mol (C) + 2*16 g/mol (O) = 44 g/mol
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 0.753 g / 44 g/mol ≈ 0.0171 mol
2. Moles of H2O:
- Molar mass of H2O = 2*1 g/mol (H) + 16 g/mol (O) = 18 g/mol
- Moles of H2O = mass of H2O / molar mass of H2O = 0.411 g / 18 g/mol ≈ 0.0228 mol
3. Moles of carbon in the compound:
- Moles of carbon = Moles of CO2 = 0.0171 mol
4. Moles of hydrogen in the compound:
- Moles of hydrogen = 2 * Moles of H2O = 2 * 0.0228 mol = 0.0456 mol
5. Moles of oxygen in the compound:
- Moles of oxygen = Moles of carbon + Moles of hydrogen - Moles of H2O
= 0.0171 mol + 0.0456 mol - 0.0228 mol = 0.0399 mol
Now, we need to convert these moles into whole-number ratios by dividing each value by the smallest value, which is 0.0171 mol.
Moles of carbon = 0.0171 mol / 0.0171 mol = 1
Moles of hydrogen = 0.0456 mol / 0.0171 mol ≈ 2.67 (rounded to 2.67)
Moles of oxygen = 0.0399 mol / 0.0171 mol ≈ 2.33 (rounded to 2.33)
The empirical formula represents the simplest whole-number ratio of atoms in a compound. Therefore, the empirical formula of the compound is CH2O.
To determine the compound's molecular formula, we need to know its molar mass. Given that the compound has a molar mass around 120 g/mol, we can calculate the ratio between the molar mass of the compound and the empirical formula mass.
1. Empirical formula mass = 12 g/mol (C) + 2 g/mol (H) + 16 g/mol (O) = 30 g/mol
2. Ratio = Molar mass of the compound / Empirical formula mass
= 120 g/mol / 30 g/mol
= 4
Multiply the subscripts of the empirical formula (CH2O) by the ratio to obtain the molecular formula.
Molecular formula = (CH2O)4 = C4H8O4
So, the compound's:
Empirical Formula: CH2O
Molecular Formula: C4H8O4