analysis of a sample of an organic compound showed it to containd 39.9% of carbon,6.9% of hydrogen and 53.2% of oxygen.A,calculate it empirical formular.8,if its relative molecular mass,what is the molecuar formular of the compound.
If you take a 100 g sample you have 39.9 g C, 6.9 g H, and 53.2 g O.
mols C = 39.9/12 = ?
mols H = 6.9/1 = ?
mols O = 53.2/16 = ?
Now find the ratio of C to H to O in whole numbers with the smallest number being no less than 1. The easy way to do this is to divide the smallest number from above mols by itself. Divide the other numbers by that same small number. Your formula will be CxHyOz. This is the empirical formula. Find the empirical formula mass. Then empirical formula mass x a whole number = molar mass. You have the empirical formula mass and you have the molar mass (but you didn't give it in the post. Solve for the whole number. Post your work if you get stuck.
it's important
1/moles . 1moles .1moles
To find the empirical formula of the compound, we need to determine the simplest whole-number ratio of the elements present in it. Here's how you can do it:
Step 1: Assume we have a 100g sample of the compound, which makes calculations easier.
Step 2: Convert the percentages into grams. In a 100g sample, we would have 39.9g Carbon, 6.9g Hydrogen, and 53.2g Oxygen.
Step 3: Convert the grams of each element to moles using their respective atomic masses. The atomic mass of Carbon (C) is 12.01 g/mol, Hydrogen (H) is 1.01 g/mol, and Oxygen (O) is 16.00 g/mol.
Moles of Carbon (C) = 39.9 g / 12.01 g/mol ≈ 3.32 mol
Moles of Hydrogen (H) = 6.9 g / 1.01 g/mol ≈ 6.83 mol
Moles of Oxygen (O) = 53.2 g / 16.00 g/mol ≈ 3.33 mol
Step 4: Find the simplest whole-number ratio of the moles of each element. Divide each of the mole values by the smallest one, which in this case is approximately 3.32 mol:
Carbon (C): 3.32 mol / 3.32 mol ≈ 1
Hydrogen (H): 6.83 mol / 3.32 mol ≈ 2
Oxygen (O): 3.33 mol / 3.32 mol ≈ 1
Step 5: Write the empirical formula using the whole-number ratio found in the previous step.
The empirical formula for the compound is CH2O.
To calculate the molecular formula, we need to know the relative molecular mass of the compound.
Step 6: Divide the relative molecular mass of the compound by the empirical formula mass to get the repeating unit factor.
Relative Molecular Mass of the compound = Given value (let's say X)
Empirical formula mass of CH2O = (12.01 * 1) + (1.01 * 2) + (16.00 * 1) = 30.03 g/mol
Repeating unit factor = X / 30.03
Step 7: Determine the molecular formula
Multiply the subscripts in the empirical formula by the repeating unit factor.
Molecular Formula = (C1 * repeating unit factor) + (H2 * repeating unit factor) + (O1 * repeating unit factor)
Simplifying if possible, you would have the molecular formula of the compound.
I hope this helps you determine both the empirical and molecular formulas of the compound!