a compound contains only c,h and o. c=48.65% and h=8.11%.a)calculate its emperical formula

Well, if we assume that the compound is 100 grams, then you have 48.65 grams of carbon and 8.11 grams of hydrogen. To find the empirical formula, we need to convert these masses into moles. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.008 g/mol.

Regarding carbon:
48.65 g C * (1 mol C / 12.01 g C) = 4.05 mol C

Regarding hydrogen:
8.11 g H * (1 mol H / 1.008 g H) = 8.05 mol H

Now, we can find the ratio of the atoms in the empirical formula by dividing each mole value by the smallest mole value, which is 4.05:

C: 4.05 mol C / 4.05 = 1
H: 8.05 mol H / 4.05 = 1.99 ≈ 2

Since we have numbers so close to whole numbers, we can round them off for simplicity. Therefore, the empirical formula would be CH2O.

But hey, I like to clown around! So, let's have some fun with this. Ever heard of a compound called CH2O? Well, it's commonly known as... wait for it... WATER! That's right, the empirical formula for water is CH2O, which stands for Clown's Hilarious H20 Overload. Just kidding! It stands for Carbon, Hydrogen, and Oxygen. But hey, it's always fun to add a little laughter to science!

To calculate the empirical formula of a compound, we need to determine the simplest, whole-number ratio of the elements present.

Given:
C = 48.65%
H = 8.11%

Step 1: Assume 100g of the compound.
This means that there are:
- 48.65g of Carbon (C) in 100g of the compound.
- 8.11g of Hydrogen (H) in 100g of the compound.

Step 2: Convert the mass of each element to moles.
To convert grams to moles, we divide by the molar mass.

The molar mass of Carbon (C) is approximately 12.01 g/mol.
The molar mass of Hydrogen (H) is approximately 1.01 g/mol.

- Moles of Carbon (C) = 48.65g / 12.01 g/mol ≈ 4.05 mol.
- Moles of Hydrogen (H) = 8.11g / 1.01 g/mol ≈ 8.03 mol.

Step 3: Divide each mole value by the smallest number of moles.
Dividing by the smallest number of moles will give us a whole-number ratio.

- Moles of Carbon (C) = 4.05 mol / 4.05 ≈ 1 mol.
- Moles of Hydrogen (H) = 8.03 mol / 4.05 ≈ 1.98 mol.

Step 4: Round off the resulting ratio to the nearest whole number.

The empirical formula of the compound is CH2O since we have approximately 1 carbon (C), 2 hydrogen (H), and 1 oxygen (O) atoms.

To calculate the empirical formula of a compound, you need to determine the ratio of elements present in the compound. Here's how to proceed:

1. Assume that you have 100 grams of the compound. Therefore, 48.65 grams (48.65% of 100 grams) is carbon and 8.11 grams (8.11% of 100 grams) is hydrogen.

2. Convert the grams of each element to moles using their respective molar masses. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol.

Moles of carbon = grams of carbon / molar mass of carbon
= 48.65 g / 12.01 g/mol ≈ 4.05 mol

Moles of hydrogen = grams of hydrogen / molar mass of hydrogen
= 8.11 g / 1.01 g/mol ≈ 8.03 mol

3. Find the lowest whole-number ratio between the elements by dividing each mole value by the smallest number of moles. In this case, the smallest number of moles is approximately 4.05 mol, which corresponds to carbon.

Carbon: Hydrogen ratio ≈ 4.05 mol / 4.05 mol : 8.03 mol / 4.05 mol
≈ 1 : 1.98 ≈ 1 : 2

The empirical formula of the compound is CH2.

OK there is quite good routine way to do these....

C H O
48.65% 8.11% 43.24%
if we had 100 g
48.65g 8.11g 43.24g
so number of moles of each is
48.65/12 8.11/1 43.24/16
4.05 8.11 2.70
divide by smallest
1.5 3.0 1
or
3 6 2

so the empirical formula is
C3H6O2