An organic compound on analysis gives the following percentage composition of elements present. Carbon 61.85%,hydrogen 10.50% and oxygen 27.65%.determine the empirical formula of the compound.(C=12,H=1,O=16)

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

Step 1: Convert the percentage composition to grams.
The given percentages can be directly converted to grams:
Carbon: 61.85% = 61.85 grams
Hydrogen: 10.50% = 10.50 grams
Oxygen: 27.65% = 27.65 grams

Step 2: Convert the grams to moles.
To convert grams to moles, divide the grams by the molar mass of each element.
Molar mass of carbon = 12 g/mol
Molar mass of hydrogen = 1 g/mol
Molar mass of oxygen = 16 g/mol

Moles of carbon = 61.85 grams / 12 g/mol ≈ 5.1558 moles
Moles of hydrogen = 10.50 grams / 1 g/mol = 10.50 moles
Moles of oxygen = 27.65 grams / 16 g/mol ≈ 1.7281 moles

Step 3: Divide each element's mole value by the smallest mole value.
Dividing each element's mole value by the smallest mole value gives us the ratio of the elements in the compound.

Ratio of carbon = 5.1558 moles / 1.7281 moles ≈ 2.9887 ≈ 3
Ratio of hydrogen = 10.50 moles / 1.7281 moles ≈ 6.0718 ≈ 6
Ratio of oxygen = 1.7281 moles / 1.7281 moles = 1

The empirical formula of the compound is C3H6O.

To determine the empirical formula of the compound, we need to find the simplest, whole number ratio of atoms present in the compound.

1. Convert the percentage of each element into grams. Assume we have 100g of the compound for easy calculation.

Carbon: (61.85g / 100g) x 100g = 61.85g
Hydrogen: (10.50g / 100g) x 100g = 10.50g
Oxygen: (27.65g / 100g) x 100g = 27.65g

2. Convert the grams of each element into moles by dividing the mass by the molar mass of each element.

Carbon: 61.85g / 12g/mol = 5.154 moles
Hydrogen: 10.50g / 1g/mol = 10.50 moles
Oxygen: 27.65g / 16g/mol = 1.728 moles

3. Find the simplest, whole number ratio by dividing each mole value by the smallest number of moles.

Carbon: 5.154 moles / 1.728 moles ≈ 2.98
Hydrogen: 10.50 moles / 1.728 moles ≈ 6.07
Oxygen: 1.728 moles / 1.728 moles = 1

Rounded to the nearest whole number:
Carbon: 3
Hydrogen: 6
Oxygen: 1

Therefore, the empirical formula of the compound is C3H6O.